Embed Size (px)
Citation preview
Quick Explanations Josh Powell
The purpose of these notes is to capture in writing my take on various topics and ideas of interest tome at the time I feel I’ve gained an understanding of them. This will hopefully allow me to make quickreferences back to them at future points when I need to review these ideas. Perhaps they will someday beof use to others as well.
-Josh PowellDecember 9, 2006
1
1 Classical Mechanics
1.1 Noether’s Theorem
Given a Lagrangian L = L(qi, qi), make a coordiante variation given by qi → qi + δqi. Then,
L(qi, qi) −→ L(qi + δqi, qi + δqi)
= L(qi, qi) +(∂L
∂qiδqi +
∂L
∂qiδqi
)+O(δq2)
= L+ δL
If δL is a total time derivative of some function, call this function f , then we can define a conserved charge
Q =∂L
∂qiδqi − f
The total time derivative of Q can be shown to vanish by using the Euler-Lagrange equations of the motion
Q =(d
dt
∂L
∂qi
)δqi +
∂L
∂qiδqi − df
dt
=∂L
∂qiδqi +
∂L
∂qiδqi − df
dt
= δL− df
dt= 0
by the definition of f .
A simple extension of this idea to conserved currents in field theories can be made. See Section ??
1.2 Generating Functions of Canonical Transformations
Adapted from Landau and Lifshitz, Vol. 1, Third Ed.The principle of least action can yield the canonical equations of motion through independently varying thedegrees of freedom in
δ
∫ (∑pidq
i −Hdt)
= 0
If we change to a different set of coordiantes, Qi, Pi, the canonical equations of motion would instead beobtained by varying
δ
∫ (∑PidQ
i −H ′dt)
= 0
The resulting dynamics as described by these coordinates will match those of the original set if the twointegrands differ by a total differential dF of some F = F (qi, pi, t). That is, if∑
pidqi −Hdt =
∑PidQ
i −H ′dt+ dF
which we rewrite asdF =
∑pidq
i −∑
PidQi + (H ′ −H)dt
Hence, we get ∂F∂qi = pi, ∂F
∂Qi = −Pi and ∂F∂t = H ′−H. Note in particular that if F is time-dependent, the two
Hamiltonians will differ in a way more than just rewriting the old Hamiltonian in terms of the new coordiates.
2
Suppose now that q and P are independent of one another, rather than q and Q. Legendre transform!Add
∑PiQ
i to both sides of the boxed equation above
d(F +
∑PiQ
i︸ ︷︷ ︸≡F2(q,P,t)
)=∑
pidqi +∑
QidPi + (H ′ −H)dt
Whence ∂F2∂qi = pi, ∂F2
∂Pi= Qi and ∂F2
∂t = H ′ −H. A similar procedure is followed if p and Q, or p and P arethe independent variables.
1.3 Inertia Tensor
Two frames, one inertial the other rotating, both with origin at a system’s center of mass. Go from inertialframe to rotating frame: ~v = d~r
dt → ~v + ~ω × ~r. Then, in the rotating frame the kinetic energy is
T = 12
∑m(~v + ~ω × ~r)2
= 12
∑m(v2 + 2~v · ~ω × ~r + (~ω × ~r)2
)= 1
2MV 2CM + 1
2~v × ~ω ·∑
m~r︸ ︷︷ ︸= 0
+ 12
∑m(~ω × ~r)2
= 12MV 2
CM + 12
∑m[ω2r2 − (~ω · ~r)2
]In analogy with the 2-dim. case in which Erot = 1
2Iω2, we define the rotational energy of a rigid body with
angular velocity ~ω to be
Erot = 12
∑m[ω2r2 − (~ω · ~r)2
]= 1
2
∑m [ωiδijωj − ωiriωjrj ]
= 12ωiωj
∑m(r2δij − rirj)
so that
Iij =∑
m(r2δij − rirj) =∫d3~x ρ(~x)
(δijx
2 − xixj
)Steiner’s Theorem gives the inertia tensor about the point separated from the center of mass by a vector ~a:
I ′ij = Iij +M(a2δij − aiaj)
1.4 Action-Angle Variables
Adapted from Landau and Lifshitz, Vol. 1, Third Ed.Suppose the Hamiltonian H depends on a slowly varying parameter λ that imparts a small time dependence.
dE
dt=∂H
∂t=∂H
∂λ
dλ
dt
Furthermore, suppose that if λ were kept at a fixed value, the dynamics of the system would be periodicwith period τ . We say that λ varies slowly if dλ
λ dtτ . Since λ varies slowly in time and changes negligibly
per period, we can pull a factor dλdt out of a time average over one cycle:⟨
∂E
∂t
⟩=
⟨dλ
dt
∂H
∂λ
⟩=
dλ
dt
⟨∂H
∂λ
⟩=
dλ
dt· 1τ
∫ τ
0
∂H
∂λdt
3
Now, we claim that the quantity
J =12π
∮p dq
is an adiabatic invariant. This quantity has the meaning of 1/2π times the area of the ellipse traced outin phase space over a period. J is called the action variable, and takes E, the energy of the system as aparameter.
We now see the adiabatic invariance explicitly:
〈J〉 =12π
∮ (∂p
∂E
⟨∂E
∂t
⟩+∂p
∂λ
dλ
dt
)dq
=12π
∮ (∂p
∂E
dλ
dt
⟨∂H
∂λ
⟩+∂p
∂λ
∂λ
∂t
)dq
=12π
∮ ( ∂p∂E
〈∂H/∂λ〉∂p/∂λ︸ ︷︷ ︸
=−〈∂H/∂p〉
+1) ∂p∂λ
∂λ
∂tdq = 0
The period is easily obtained given J . Note that q = ∂H∂p → dt = dq
∂H/∂p . But
τ =∫ τ
0
dt =∮
dq
∂H/∂p
=∮
∂p
∂Hdq = 2π
∂J
∂E
Therefore,∂E
∂J=
2πτ
= ω = φ
where φ is the associated angle variable.
2 Mathematical Methods
This section will be about various mathematical methods that find uses in a broad variety of settings. Thismeans that it isn’t natural or appropriate to group them in with only, say, quantum mechanics when someother arena lays equal claim to them. I imagine it will be useful mostly as a bunch of topics that othersections reference for more details or motivation of a particular method.
2.1 Complex Variables
2.1.1 A connection between 1/x and δ(x)
In the realm of normal functions of the variable x, the solutions to the differential equation
−x dfdx
= f
take the form f(x) = C/x. Should we extend the allowed set of solutions to distributions, whose propertiesare determined by their behavior within an integral, then we get yet another class of solutions: f(x) = C δ(x).Ronen Plesser has offered up the following as a connection between these two:
Consider the limitlimε→0
( 1x′ − x+ iε
− 1x′ − x− iε
)= 2πiδ(x′ − x)
A typical way of establishing this might be to combine the two fractions in the limit, get the resultingLorentzian, and note that the limiting procedure sends the support of this Lorentzian to a single point while
4
Figure 1: The two propagator terms in the limit.
simultaneously preserving the area for all (positive) values of ε. Hence, one gets a valid representation ofthe δ-function. Instead, let’s view these two terms in the context of a contour integral, such as one used forevaluating a propagator. This is depicted in Fig. (1). Here, the first term in the limit corresponds to thelower of the two contours, which will dip below the axis at x′ = x. The second term will rise above this point,but it’s negative sign can be incorporated as a reversal of its orientation as a contour. Away from the valuex′ = x, the two contours cancel each other exactly. Near this point, the limit sends them to a positivelyoriented loop around the point x, hence justifying the equation of this expression with 2πiδ(x′ − x).
2.2 Transformations
Physics and mathematics use a wide variety of techniques which transform systems from one set of coordinatesor variables to another. I will detail some of these.
2.2.1 Legendre Transform
Adapted from Greiner’s book on thermodynamics.Given a differentiable function f(x), consider its derivative f ′(x). We’re interested in finding a function ofp(x) := f ′(x) that encodes all the same information as f(x) does. Let’s call this function g(p). That is,there is a well-defined mapping f(x)↔ g(p). We posit the following:
g = f − xp with p =df
dx
At first, this looks to be a function g(x, p), rather than simply g(p). We need to show that it is truly only afunction of p = f ′(x).
dg = df − p dx− x dp
=(df − df
dxdx)− x dp
= −x dp
where we used the definition of p to see that the first two differential terms cancel. Hence, g depends onlyon the variable p. We have the conjugate pair
df = p dx
dg = −x dp
which is related to the situation in classical mechanics wherein a coordinate and its conjugate momentumcan be swapped and become, respectively, a “generalized momentum” and “generalized coordinate” undera canonical transformation.
5
To complete the transformation, we need to be able to invert p = f ′(x) to get x = (f ′)−1(p). Thisrequires that f ′ be bijective between coordinates x and slopes p. We now substitute and have
g(p) = f((f ′)−1(p)
)− p · (f ′)−1(p)
Perhaps this will be fully clear with an example. Take f(x) = x2. Then, f ′ = 2x : R −→ R which isbijective. So, p = 2x↔ x = 1
2p. From here we find
g = f − xp = x2 − x · 2x= 1
4p2 − 1
2p2 = − 1
4p2
The inverse transformation will set g′ = −x as indicated in the conjugate pair above.
Legendre transforms find a great deal of use in mechanics – classical mechanics, as well as statisticalmechanics and thermodynamics. See §3.1.1 for a view of their role in thermodynamics.
3 Thermodynamics and Statistical Mechanics
3.1 Basics
3.1.1 Thermodynamics
The following are distilled from both Kittel & Kroemer and Reif, both are clear and thorough introductionsto statistical thermodynamics.The first law of thermodynamics is the only one that truly needs to be independently postulated. The secondand third fall out automatically from a proper treatment of entropy in the context of statistical mechanics.The first law can be boiled down to the statement: “There is thermal energy.” In situations where thethermal energy can change, ie heat flows, one must include it in statements of conservation of energy. Inparticular, if ones transfers a tiny bit of energy dE into a system, it can come in the form of mechanicalwork δW on the system or as heat flow δQ into the system.1
dE = δW + δQ
Why δW and δQ rather than dW and dQ? The work and heat depend on the whole history of the processperformed, and so, aren’t proper differentials. Think of the work required to drag a block on a surfacewith friction around a closed loop. Clearly work is performed though the net displacement is null. Thatequivalent is true with heat flow. All this isn’t pedantic. It tells us that one can run loops in configurationspace such as those used in Carnot cycles. Additionally, these two quantities are not differences; the notionof “difference between works” is meaningless. I’m simply trying to indicate that these quantities are verysmall but not proper differentials. Reif discusses this in some detail.
Take a system at absolute temperature τ = β−1 and put it in thermal contact with another system at aslightly different temperature and do no mechanical work on the system. The flow of thermal energy will besmall compared to the total internal energy already present in the system: δQ E. What is the increasein g, the number of possible microconfigurations?
ln g(E + δQ)− ln g(E) =∂ ln g∂E
δQ ≡ β δQ
So, defining σ = ln g to be the entropy and setting 1/τ = ∂σ/∂E, we have
dσ =δQ
τ
1This will later need one more term when the particle number is allowed to change.
6
If the energy of the system depends on external parameters xi, then the generalized forces associatedwith these parameters are given by
Xi,r = +∂
∂xiεr(x1, x2, . . .)
The strange sign above results from our interest in the work done on and not by the system. The exactvalue of this force depends on the particular microconfiguration, r. Averaging over these gives Xi. Hence,when we change these parameters by dxi respectively, the work done by the system is
work done by the system = −∑
i
Xidxi = −δW
This quantity is the negative of the work done on the system, δW , which was featured in the First Law ofThermodynamics above. Examples include
Type X dxPressure p −dV
Surface Tension S dAGeneric Linear Force F dx
Magnetization M dHPolarization P dD
The strange sign on −dV arises because it takes a decrease in volume to do positive work on the system viaits pressure. This is the only exception I know of.
Among all the thermodynamic variables one might choose to consider, the thermodynamic potentials arethe most important. These are E the internal energy, F the Helmholtz free energy, G the Gibbs free energyand H the enthalpy.2 A differential change in E is given by
dE = τdσ − pdV
In general, the second term on the right hand side will be Xdx, but I have taken the common case X = pand dx = −dV to be used in all the following equations. If we allow the particle number in the system tochange through chemical contact without another system or a reservoir, we must set
dE = τdσ − pdV + µdN
From here, we define
F = E − τσG = E + pV − τσH = E + pV
Given the above, it is easy to compute dF , dH and dG as well. The quantities left as differentials in theseexpressions will the be natural variables or parameters of these quantities. For E, these are σ, V and N asseen above. That is, E = E(σ, V,N). By contrast, the Gibbs free energy is G = G(τ, p,N). The astute readerwill note that these other quantities are Legendre transforms of the internal energy E. See Section 2.2.1 formore about Legendre transformations. The information encoded in each of these is the same. This result iseminently reasonable, as the system is independent of our choice of description. What changes is the ease ofa particular computation. One tries to choose a thermodynamic potential at least one of whose differentialsvanishes for the problem under consideration. For example, an isobaric process, one at a constant pressure,will satisfy dp = 0. As a result, a term vanishes in the differential of H = H(σ, p,N) and G = G(τ, p,N).The decision to use one or the other might be made by further information about a conserved quantity: say,is the process isothermic or isentropic.
2Greiner gives yet another: Φ = E − τσ − µN which he calls the grand potential. Apparently, this potential finds its use indiscussions of isothermal problems with fixed chemical potential.
7
3.1.2 Maxwell Relations
Maxwell relations are derived from evaluating a second derivative of a given energy variable with respect totwo of its parameters. The result must be the same when we switch the order of the derivatives, so we arriveat certain identities. For example, with E = E(σ, V,N) we have
∂2E
∂V ∂σ=
∂2E
∂σ∂V( ∂
∂V
)σ,N
(∂E∂σ
)V,N
=( ∂
∂σ
)V,N
(∂E∂V
)σ,N
∴( ∂τ∂V
)σ,N
= −( ∂p∂σ
)V,N
The others are all obtained from similar considerations.
3.1.3 Computing Thermodynamic Variables from Statistical Quantities
Given the partition functionZ =
∑r
e−εr/τ =∑
r
e−βεr
we want to compute the internal energy of the system. This is the average energy of each of the consituents,which we take to be distributed according to the thermal distribution. That is
E = ε =∑
r
εrP (εr)
=∑
r
εr ·e−βεr∑s e
−βεs
= Z−1∑
r
(− ∂
∂βe−βεr
)= − ∂
∂βlnZ
Likewise, one can compute the generalized force X conjugate to the variable x if given the partitionfunction.
X =∑
r
(∂εr∂x
) e−βεr∑s e
−βεs
= −Z−1∑
r
( 1β
∂
∂xe−βεr
)= − 1
β
∂
∂xlnZ
As an example, we compute the mean pressure of a system with a known partition function.
p =1β
∂
∂VlnZ
8
From here, we compute the entropy given the partition function.
d lnZ =∂ lnZ∂x
dx+∂ lnZ∂β
dβ
= −βX dx− E dβ= −βδW − d
(βE)
+ β dE
d(lnZ + β E
)= β
(dE − δW
)=δQ
τ= dσ
∴ σ = lnZ + βE
Technically, equality of the differentials means that the functions equal only up to an additive constant. Buttaking the τ → 0 limit reveals this constant vanishes.
Note quickly that Pr := P (εr) = e−βεr/Z implies βεr = − ln(PrZ). Then,
σ = lnZ + βE = lnZ + β∑
r
εrPr
= lnZ −∑
r
Pr ln(PrZ)
= lnZ − lnZ∑
r
Pr −∑
r
Pr lnPr = −∑
r
Pr lnPr
This is a pretty little point of contact with the standard definition of entropy from the theory of information.I will try to write more about this connection later.
We can use the expression σ = lnZ + βE to write the Helmholtz free energy
F = E − τσ = E − τ(lnZ + βE
)= E − τ lnZ − E = −τ lnZ
or working backwardsZ = e−F/τ
4 Classical Field Theories
4.1 Noether’s Theorem
Let L(φ, ∂µφ) be the Lagrangian density of a field theory. Make the transformation φ → φ + δφ. If theaction
∫d4xL changes only by a surface term, the dynamics remain the same under this transformation.
The transformation is said then to be a symmetry of the system. Under a transformation L gains at most a4-divergence: L → L+ ∂µf
µ. What is the actual change in L under this transformation?
L(φ, ∂µφ) −→L(φ+ δφ, ∂µφ+ ∂µδφ)
= L(φ, ∂µφ) +δLδφδφ+
δLδ(∂µφ)
∂µδφ
= L(φ, ∂µφ) + ∂µ
[δL
δ(∂µφ)δφ
]+[δLδφ− ∂µ
δLδ(∂µφ)︸ ︷︷ ︸
= 0
]δφ
9
where the second term in parentheses vanishes according to the Euler-Lagrange equations of motion. Whatremains gets set equal to our anticipated 4-divergence ∂µf
µ:
∂µ
[δL
δ(∂µφ)δφ
]= ∂µf
µ
∴ ∂µ
[δL
δ(∂µφ)δφ− fµ
]= 0
Hence, we define our conserved 4-current and associated scalar charge as
jµ =δL
δ(∂µφ)δφ− fµ
Q =∫d3x j0(x)
It’s then easy to show that the charge Q satisfies the equation Q = 0.
4.2 Noether Procedure: Making Global Symmetries Local
See Green, Schwarz & Witten for more information on the Noether procedure.The Noether procedure is a means of “gauging” a global symmetry. That is, given a system exhibiting
a global symmetry, it is the process of making the symmetry hold locally. Essentially, this idea boils downto adding terms to the action which make it invariant under a variation whose strength can change withlocation. Let’s make this notion precise. Consider a field φ and an action functional S0[φ, ∂φ] on it. A globalsymmetry is one for which
S0[φ, ∂φ] = S0[φ+ ε δφ, ∂(φ+ ε δφ)]
for an infinitessimal constant ε. Now let ε depend on location: ε = ε(x). In general, φ → φ + ε(x)δφ willnot be a symmetry of the action. However, since this symmetry is good if ε(x) is a constant, it must be thechanges in ε which spoil the invariance of the action. We should expect the integrand of δS to be proportionalto ∂µε. Write the variation as
δS0 =∫d4xJµ∂µε
where Jµ is the resulting current. Now, any variation about a field configuration φ(x) leaves S unchanged ifφ satisfies the equations of motion. In such a case, the above variation must vanish too, even though ε waspromoted to being a “local constant”, which is just jargon for a smooth function of position. We can thenperform partial integration to find
∂µJµ = 0
and so the current Jµ is conserved for on-shell field configurations. This result provides another view of thecurrent obtained in §4.1.
The Noether procedure says the following: posit a gauge potential Aµ which transforms under thevariation φ→ φ+ ε(x)δφ as δAµ = ∂µε. Then, augment the action of the theory to be
S = S0 −∫d4xJµAµ
takign Jµ to be the current obtained above and leaving Aµ unspecified but with its variation as described.δS0 cancels against the variation of Aµ in the second term in the action.
δS = δS0 −∫d4xJµδAµ −
∫d4x (δJµ)Aµ = −
∫d4x(δJµ)Aµ
10
But Jµ also gets varied and leaves another term in δS. If this remaining term does not vanish then we mustcontinue the process. With some luck, the procedure eventually comes to an end and a local symmetry forthe theory has been constructed. The result is an action whose change under the constructed variations – yes,there the variations themselves can receive modifications – will vanish regardless of the field configurationand whether it is on-shell or off-shell. We say then that we have gauged the original symmetry.
4.2.1 Example: Complex Scalar Field
A complex scalar field φ with action
S0 =∫d4x 1
2
(∂µφ∂
µφ∗ −m2φ∗φ)
possesses a global symmetry which rotates the phases of the fields: φ → eiθφ and φ∗ → e−iθφ∗. Take θ tobe infinitessimal: δφ = iθφ and δφ∗ = −iθφ∗. The variation of S0 under this transformation is easily seento vanish (to first order in θ). We now treat θ(x) as a function on spacetime. The variation is then
δS0 =∫d4x 1
2
(∂µ(iθφ)∂µφ∗ + ∂µφ∂
µ(−iθφ∗)−m2(−iθφ∗)φ−m2φ∗(iθφ))
=∫d4x 1
2
(iφ∂µφ∗ − iφ∗∂µφ
)∂µθ
The current is therefore Jµ = 12 i(φ∂
µφ∗ − φ∗∂µφ). Now posit a gauge potential Aµ which transforms underthis variation as δAµ = ∂µθ. We modify the action appropriately.
S = S0 + S1 =∫d4x 1
2
(∂µφ∂
µφ∗ −m2φ∗φ)−∫d4xJµAµ
The variation of the action under δφ = iθφ, δφ∗ = −iθφ∗ and δAµ = ∂µθ takes the form
δS = δS0 −∫d4xJµδAµ −
∫d4x (δJµ)Aµ = −
∫d4xφ∗φAµ∂
µθ
The first two terms in the middle expression cancel, by design. But there is a remaining term which notvanish. Fortunately for us, we do not need to consider another gauge field which couples to this remainingcurrent, since ∂µθ is already the variation of the first gauge field. We make the following manipulations.
AµδJµ = φ∗φAµ∂
µθ = φ∗φAµδAµ
= φ∗φδ( 12AµA
µ) = δ( 12φ
∗φAµAµ)
It was okay to pull φ∗φ into the variation because this combination is gauge invariant – that is, it’s unchangedunder the variation we’re considering. The fact that AµδJ
µ is a total variation allows us to finish the Noetherprocedure. We add one more term to the action, which is the spacetime integral of the quantity being variedin the last line above.
S =∫d4x 1
2
(∂µφ∂
µφ∗ −m2φ∗φ)−∫d4xJµAµ +
∫d4x 1
2φ∗φAµA
µ
=∫d4x 1
2
((∂µφ− iAµφ)(∂µφ∗ + iAµφ∗)−m2φ∗φ
)=∫d4x 1
2
(DµφD
µφ∗ −m2φ∗φ)
where we’ve defined Dµφ = ∂µφ − iAµφ and Dµφ∗ = ∂µφ
∗ + iAµφ∗. These quantities are gauge covariant
derivatives. We call them covariant because Dµφ→ becomes iθDµφ under the variation, and Dµφ∗ becomes
−iθDµφ∗. From here, note that DµφD
µφ∗ is actually gauge invariant, just like φ∗φ. The action S is now
11
left entirely unchanged under the transformation, even for field configurations for which the equations ofmotion do not hold.
As it stands, the field Aµ is non-dynamical, owing to its lack of a kinetic term. At this point, it merelyimposes a constraint which will make the equations of motion for φ and φ∗ obtained from S match thoseobtained from the original action S0. Should we choose to give Aµ a kinetic term this claim is no longertrue. The kinetic term will need to be gauge invariant as well. If it were the standard kinetic term foran electromagnetic field, then we’d obtain a complex (ie, charged) scalar coupled to electromagnetism. Iftreated quantumly, this would be the action for scalar QED.
4.3 Electrodynamics
4.3.1 Some Hints to Remember Basics
For the most part, Maxwell’s Equations are straightforward to recall. The only thing that sometimes getsmixed up for me is which of the two vector-valued equations gets which sign. The easiest way to rememberthis is to note that Lenz’s law involves the most famous minus sign in all of physics: an increasing magneticflux through a loop of wire gives a negatively-oriented current in the loop. That is, the changing magneticflux induces a current, which then produces a magnetic field that opposes the changing external magneticfield. At the differential level, this means that positive n · ∂ ~B/∂t corresponds to a negative n · (∇× ~E) field.This reasoning suggests
∇× ~E +∂ ~B
∂t= 0
The two get opposite signs so that
∇× ~H − ∂ ~D
∂t= ~J
All in all, we have
∇ · ~D = ρ ∇ · ~B = 0
∇× ~E +∂ ~B
∂t= 0 ∇× ~H − ∂ ~D
∂t= ~J
The next situation that requires some remembering is the boundary conditions for electric and magneticfields at the interface of two media. In the absence of surface charges and surface currents, the setup is easy
n ·(~D2 − ~D1
)= 0 n ·
(~B2 − ~B1
)= 0
n×(~E2 − ~E1
)= 0 n×
(~H2 − ~H1
)= 0
In their presence, though, the right hand sides of some of these need to be modified. The surface chargedensity σ is a scalar and clearly belongs in a scalar-valued equation, whereas the surface current ~K is avector and so belongs in a vector-valued equation.
n ·(~D2 − ~D1
)= σ n ·
(~B2 − ~B1
)= 0
n×(~E2 − ~E1
)= 0 n×
(~H2 − ~H1
)= ~K
A quick consideration of the units serves to confirm that this is the right choice.
4.4 General Relativity
4.4.1 Schwarzschild Solution with an Arbitrary Cosmological Constant
We seek the static, spherically symmetric metric that solves the vacuum Einstein Equation with a cosmologi-cal constant Λ which will reduce to the ordinary Schwarzschild solution when Λ→ 0. The Einstein Equation
12
with cosmological constant is
Gµν + Λgµν = Rµν − 12Rgµν + Λgµν = 0
As in the discussion leading up to the regular Schwarzschild solution, the desire for a static, sphericallysymmetric metric will constrain the metric to the following form
ds2 = −f(r)dt2 + h(r)dr2 + r2dθ2 + r2 sin2θ dφ2
The easiest way to solve for the Riemann tensor is to use frame fields ea and impose on them the torsionfree-condition dea =
∑b e
b ∧ ωab. The natural choice for frame fields is
e0 =√f dt e1 =
√h dr
e2 = r dθ e3 = r sin θ dφ
Now use the torsion-free condition to solve for the connection 1-forms.
de0 =f ′
2√fdr ∧ dt = e1 ∧ ω0
1 + e2 ∧ ω02 + e3 ∧ ω0
3
−→ ω02 ∝ dθ and ω0
3 ∝ dφ and ω01 =
f ′
2√fh
dt
de1 = 0 = e0 ∧ ω10 + e2 ∧ ω1
2 + e3 ∧ ω13
−→ ω10 ∝ dt and ω1
2 ∝ dθ and ω13 ∝ dφ
de2 = dr ∧ dθ = e0 ∧ ω20 + e1 ∧ ω2
1 + e3 ∧ ω23
−→ ω20 ∝ dt and ω2
3 ∝ dφ and ω21 =
1√hdθ
de3 = sin θ dr ∧ dt+ r cos θ dθ ∧ dφ = e0 ∧ ω30 + e1 ∧ ω3
1 + e2 ∧ ω32
−→ ω30 ∝ dt and ω3
1 =sin θ√hdφ and ω3
2 = cos θ dφ
If we inspect the proportionalities, we’ll find that they either agree with an explicit expression for a connectionform given elsewhere, or they are in direct conflict with another proportionality which means that theconnection form must vanish. The two that must vanish for this reason are ω0
2 = ω03 = 0. Given ωa
b
we can lower the first index using the metric ηab = diag(−1, 1, 1, 1) on these abstract indices, and then useantisymmetry ωab = −ωba. Just to summarize, I’ll write them all again below.
ω01 = ω1
0 =f ′
2√fh
dt ω02 = ω2
0 = 0
ω03 = ω3
0 = 0 ω12 = −ω2
1 = − 1√hdθ
ω13 = −ω3
1 = − sin θ√hdφ ω2
3 = −ω32 = − cos θ dφ
13
We now compute the curvature 2-forms Rab = dωa
b+∑
c ωa
c∧ωcb. Note that if a term vanishes, I’ll surpress
it here for compactness and clarity.
R01 = dω0
1 =d
dr
( f ′
2√fh
)dr ∧ dt
R02 = ω0
1 ∧ ω12 =
( f ′
2√fh
dt)∧(− 1√
hdθ)
= − f ′
2h√fdt ∧ dθ
R03 = ω0
1 ∧ ω13 =
( f ′
2√fh
dt)∧(− sin θ√
hdφ)
= −f′ sin θ
2h√fdt ∧ dφ
R12 = dω1
2 + ω13 ∧ ω3
2 = d(− 1√
hdθ)
+(− sin θ√
hdφ)∧(cos θ dφ
)=
h′
2h3/2dr ∧ dθ
R13 = dω1
3 + ω12 ∧ ω2
3 = d(− sin θ√
hdφ)
+( 1√
hdθ)∧(cos θ dφ
)=h′ sin θ2h3/2
dr ∧ dφ
R23 = dω2
3 + ω21 ∧ ω1
3 = d(− cos θ dφ
)+( 1√
hdθ)∧(− sin θ√
hdφ)
=(1− 1
h
)sin θ dθ ∧ dφ
We convert the abstract indices back to Lorentz indices and then compute the components of the Ricci tensor.The conversion is Rµ
νρσ =∑
a,b eaµeb
ν(Rab)ρσ. But the matrix ea
µ and its inverse are both diagonal, sothis conversion is fairly trivial and I’ll just lump this in with the computation of the Ricci tensor. Uponinspecting the basis elements of Ra
b we see that every one of them is dxa ∧ dxb, which means that theoff-diagonal components of the Ricci tensor vanish automatically.
R00 = (ea=1µ=1)(eb=0
ν=0)(Ra=1b=0)10 + (e22)(e00)(R2
0)20 + (e33)(e00)(R30)30
=
√f
h
d
dr
( f ′
2√fh
)+
f ′
2rh+
f ′
2rh=
√f
h
d
dr
( f ′
2√fh
)+f ′
rh
R11 = (e00)(e11)(R01)01 + (e22)(e1ν=1)(R2
1)21 + (e33)(e11)(R31)31
= −
√h
f
d
dr
( f ′
2√fh
)+
h′
2rh+
h′
2rh= −
√h
f
d
dr
( f ′
2√fh
)+h′
rh
R22 = (e00)(e22)(R02)02 + (e11)(e22)(R1
2)12 + (e33)(e22)(R32)32
= − rf′
2fh+rh′
2h2+(1− 1
h
)R33 = (e00)(e33)(R0
3)03 + (e11)(e33)(R13)13 + (e22)(e33)(R2
3)23
= −rf′ sin2θ
2fh+rh′ sin2θ
2h2+(1− 1
h
)sin2θ
We then compute the Ricci scalar
R = R00 +R1
1 +R22 +R3
3 = g00R00 + g11R11 + g22R22 + g33R33
= − 1fR00 +
1hR11 +
1r2R22 +
1r2 sin2θ
R33
= 2[ 1r2
(1− 1
h
)+
h′
rh2− f ′
rfh− 1√
fh
d
dr
( f ′
2√fh
)]Since Rµν vanishes off the diagonal and so does gµν , the Einstein Equation is automatically fulfilled forµ 6= ν. On diagonal, it takes the form
Rµµ + gµµ
(Λ− 1
2R)
= 0 (no sum)
14
Substituting in the known expressions for the Ricci tensor and scalar, we get√f
h
d
dr
( f ′
2√fh
)+f ′
rh− f
[Λ− 1
r2
(1− 1
h
)+
h′
rh2− f ′
rfh− 1√
fh
d
dr
( f ′
2√fh
)]= 0
−
√h
f
d
dr
( f ′
2√fh
)+h′
rh+ h[Λ− 1
r2
(1− 1
h
)+
h′
rh2− f ′
rfh− 1√
fh
d
dr
( f ′
2√fh
)]= 0
− rf′
2fh+rh′
2h2+(1− 1
h
)+ r2
[Λ− 1
r2
(1− 1
h
)+
h′
rh2− f ′
rfh− 1√
fh
d
dr
( f ′
2√fh
)]= 0
The fourth equation will actually just be exactly sin2θ times the third one. This gives no new information,and we do not need to consider it if we keep the third one. Divide the first of these equations by f , thesecond by h and the third by r2.
1√fh
d
dr
( f ′
2√fh
)+
f ′
rfh−[Λ− 1
r2
(1− 1
h
)+
h′
rh2− f ′
rfh− 1√
fh
d
dr
( f ′
2√fh
)]= 0
− 1√fh
d
dr
( f ′
2√fh
)+
h′
rh2+[Λ− 1
r2
(1− 1
h
)+
h′
rh2− f ′
rfh− 1√
fh
d
dr
( f ′
2√fh
)]= 0
− f ′
2rfh+
h′
2rh2+
1r2
(1− 1
h
)+[Λ− 1
r2
(1− 1
h
)+
h′
rh2− f ′
rfh− 1√
fh
d
dr
( f ′
2√fh
)]= 0
It’s valid to divide by f and h, since if they were to vanish, we would have a degenerate metric. If we addthe first and second of these equations, we get
f ′
rfh+
h′
rh2= 0
f ′
f+h′
h= 0
d
drlnh = − d
drln f
lnh = ln f−1 + C
∴ h =k
f
for k > 0. We can rescale the time coordinate so that k = 1. With this equality in mind, we can simplifythe first of the equations above to
f ′′
2+f ′
r−[Λ− 1− f
r2+f ′
r+f ′
r+f ′′
2
]= 0
−Λ +1− fr2− f ′
r= 0
f + rf ′ = 1− r2Λd
dr(rf) =
d
dr(r − 1
3r3Λ)
f = 1 +C
r− r2
3Λ
and so
ds2 = −(1 +
C
r− r2
3Λ)dt2 +
(1 +
C
r− r2
3Λ)−1
dr2 + r2 dθ2 + r2 sin2θ dφ2
This solution for f(r) actually solves all of the three equations above, though we only needed to use thefirst and second. We see that if Λ→ 0, we get the standard Schwarzschild solution. Unless Λ is particularly
15
small, it is impossible to do a large r expansion that reveals C = −2GM/c2 or just −2M for G = c = 1.Hence, I’ll leave the constant of integration as C.
Let a test particle’s worldline be parametrized by τ , which is taken to be proper time for a timelikeworldline and some affine parameter for a lightlike one. Define uµ = dxµ
dτ . Note the invariance of the metricunder θ → π − θ. It implies that if our initial position and velocity lie in what we dub the equitorial plane,it will remain there for all time. But by appropriately picking our coordinate system, we can always makethis the case. Doing so, we have
−κ = gabuaub = −
(1 +
C
r− r2
3Λ)t2 +
(1 +
C
r− r2
3Λ)−1
r2 + r2φ2
where κ = 0 for a lightlight geodesic and κ = +1 for a timelike geodesic. Since our system is, by design,static, we have a timelike Killing field ξa = ( ∂
∂t )a with associated constant of the motion
E = −gabξaub =
(1 +
C
r− r2
3Λ)t
and another Killing field ψa = ( ∂∂φ )a by rotational symmetry which gives the conserved quantity
L = gabψaub = r2φ
We can rewrite the geodesic equation above in terms of E and L to get
12E2 =
12r2 +
12
(1 +
C
r− r2
3Λ)(L2
r2+ κ)
which we can interpret at the equation for radial motion of a particle in the effective potential
Veff =12
(1 +
C
r− r2
3Λ)(L2
r2+ κ)
=12κ− 1
6L2Λ− 1
6κr2Λ +
κC
2r+L2
2r2+CL2
2r3
Since a lightlike particle has κ = 0, the cosmological constant only has the effect of shifting the effectivepotential by a constant proportional to L2. This does’t change the dynamics at all, as long as the particledoesn’t couple to some interaction that can change the value of L. For a timelike particle, this same termarises plus another that grows like r2 times the cosmological constant. This means that, depending on thesign of Λ, all particles are bound or all orbits are unstable when viewed on the distance and energy scalesset by the size of the cosmological constant.
4.4.2 Schwarzschild Solution of Extended Line Mass
For the same reasons as the 3+1-dim Schwarzschild solution took a similar form, the 2+1-dim case mustlook like:
ds2 = −f(r)dt2 + h(r)dr2 + r2dθ2
We take the dreibein 1-forms to be
e0 =√f dt
e1 =√h dr
e2 = r dθ
We will determine our connection 1-forms ωab using the Maurer-Cartan structure equations:
dea =∑
b
eb ∧ ωab
16
which we know will specify the connection 1-forms fully.
de0 =f ′
2√fdr ∧ dt = e1 ∧ ω0
1 + e2 ∧ ω02
de1 = 0 = e0 ∧ ω10 + e2 ∧ ω1
2
de2 = dr ∧ dθ = e0 ∧ ω20 + e1 ∧ ω2
1
Since the first of these has no dθ dependence though the form e2 does, it must be true that ω02 ∝ dθ so that
the wedge product e2 ∧ ω02 vanishes. Likewise from the third of these equations, ω2
0 = ω02 ∝ dt. Hence,
ω20 = ω0
2 = 0
Now, the first equation gives
f ′
2√fdr ∧ dt = e1 ∧ ω0
1
=√h dr ∧ ω0
1
∴ ω01 =
f ′
2√fh
dt = ω10
and similarly the third gives
dr ∧ dθ = e1 ∧ ω21
=√h dr ∧ ω2
1
∴ ω21 =
1√hdθ = −ω1
2
Technically, we haven’t ruled out the possibility that ω01 has another component in dr that vanished under
the above wedge product, and that ω21 has a second term in dr as well. But if either of these two possible
extra terms were to exist, the middle Maurer-Cartan equation could not be satisfied. Hence, we’ve derivedthe exact form of each of the connection 1-formsThe remaining three possibilities vanish by anti-symmetry:ω0
0 = ω11 = ω2
2 = 0. We now use the following expression for the curvature 2-form:
Rab = dωa
b +∑
c
ωac ∧ ωc
b
Symmetry and anti-symmetry in the indices will allow us to compute only the following three components,and then to deduce the rest from these as needed later on.
R01 = dω0
1 + >
0
ω00 ∧ ω0
1 + ω01 ∧
>0
ω11 +
>0
ω02 ∧ ω2
1
= d( f ′
2√fh
dt)
=d
dr
( f ′
2√fh
)dr ∧ dt
From now on, for compactness, I’ll omit terms that vanish due to one of the connection forms vanishing.
R02 = ω0
1 ∧ ω12 =
f ′
2√fh
dt ∧ − 1√hdθ
=f ′
2h√fdθ ∧ dt
R12 = dω1
2 = d(− 1√
hdθ)
=h′
2h3/2dr ∧ dθ
17
It’s easy to read the components of the mixed-index tensor Rabρσ = (Ra
b)ρσ from this. We simply that notethat, for example, (dr∧dt)ρσ = δρ1δσ0−δρ0δσ1. We then complete the job by converting our abstract indicesback to Lorentz indices:
Rµνρσ =
∑a,b
eµaeν
b(Rab)ρσ
But rather than doing this for all indices, we should only care about those that are important to the Riccitensor components of interest. We look back at the results for the Riemann tensor, still cast as a curvature2-form and notice that R0
1 has only non-zero components ρ = 0, σ = 1 and vice-versa. Similarly, R02 is
non-vanishing only for ρ = 0, σ = 2 or ρ = 2, σ = 0, and R12 for ρ = 1, σ = 2 and ρ = 2, σ = 1. Therefore,
the only components of the Ricci tensor that don’t vanish automatically are R00, R11 and R22. This is good,since we’ll eventually setting all of the components equal to 0, as the stress energy tensor is non-vanishingonly on a single point. Let’s compute these three Ricci tensor components:
R00 = R0000 +R1
010 +R2020
= 0 +1√h·√f · d
dr
( f ′
2√fh
)+
1r·√f · f ′
2h√f
=
√f
h
d
dr
( f ′
2√fh
)+
f ′
2rh
and similarly
R11 = −
√h
f
d
dr
( f ′
2√fh
)+
h′
2rh
R22 = − f ′
2rh+
h′
2rh
The Einstein Equation tells us that these components should also be set equal to 0, except on the one pointin the support of the stress-energy tensor. The easiest way to solve the resulting differential equations is thedivide the first one, R00 = 0, by f and the second by h, and then add the two equations
f ′
2rfh+
1√fh
d
dr
( f ′
2√fh
)= 0
h′
2rh2− 1√
fh
d
dr
( f ′
2√fh
)= 0
∴f ′
2rfh+
h′
2rh2= 0
Hence, f ′/f + h′/h = 0. The third of the Einstein Equations above says that f ′ = h′. Note that I’mmultiplying and dividing liberally by f(r), h(r) and r, since I’m working away from the origin r = 0 wherethe Ricci flat spacetime is assumed to be non-singular (why would it be?) and so f, h 6= 0. I feel this isjustified considering that expressions like 1/
√h etc. have appeared in physically meaningful quantities like
the Ricci tensor’s components. But,f ′
f+h′
h= 0 −→ f =
C
h
for arbitrary C > 0. Now notice that f ′ = −Ch′/h2. Since R22 = 0 −→ f ′ = h′, we can say h′ = −Ch′/h2 →h′(1 + C/h2) = 0. Hence, either h′ = 0 so that h = C/f is some constant, or we get h2 = −1/C, meaningh is imaginary, which is unphysical. With the appropriate rescaling, we finally derive that our metric takesthe form
ds2 = −Kdt2 +1Kdr2 + r2dθ2
18
5 Quantum Mechanics
5.1 Multiplication of Spherical Harmonics
Adapated from Gottfriend & YanI will state the results from Gottfriend & Yan here now without proof. I will later come back and motivatethis and derive it in my own fashion.
∫dΩY ∗
LM (Ω)Yl1m1(Ω)Yl2m2(Ω) =
√(2l1 + 1)(2l2 + 1)
4π(2L+ 1)〈l1,m2; l2,m2|LM〉〈l1, 0; l2, 0|L, 0〉
If we take 2L = l1 + l2 + l3, there is sort-of compact closed form for the second Clebsch-Gordon coefficienton the right hand side:
〈l1, 0; l20|l3, 0〉 = (−1)l3+L L!√
2l3 + 1(L− l1)!(L− l2)!(L− l3)!
√(l1 + l2 − l3)!(l2 + l3 − l1)!(l3 + l1 − l2)!
(2L+ 1)!for 2L even
= 0 for 2L odd
Note that this is a corrected version of what Gottfried & Yan had misprinted in the Appendix.
5.2 Projection Theorem
Adapted from Cohen-Tannoudji, Diu, Laloe, Vol. 2Imagine one has a spherical tensor operator T of rank L, and a basis on state space that includes the angularmomentum observables J2 and Jz, where all other quantum numbers are labelled collectively by α. Then,the content of the Wigner-Eckart theorem is that T ’s matrix elementshave a simple dependence on the Jz
value m:〈αjm|TLM |α′j′m′〉 = 〈αj||TL||α′j′〉 · Cjm
LMj′m′
where the first factor on the right hand side is known as the reduced matrix element of the operator TLM
between the two subspaces, and CjmLMj′m′ is the appropriate Clebsch-Gordon coefficient.
In the special case that L = 1, so that T is a vector operator, we can derive a further result known as theprojection theorem. The content of this theorem is that the operator T is proportional to the total angularmomentum of the system. Begin by noting that we can define J± = Jx ± i Jy and similarly T± = Tx ± i Ty
and then compute
[J+, T+] = [J−, T−] = 0[J±, T∓] = ±2~Tz
Because J+ and T+ commute, we can write
〈α, j,m+ 2|J+T+|α, j,m〉 = 〈α, j,m+ 2|T+J+|α, j,m〉
We have taken α′ = α and j′ = j. I.e., the two substaces are the same between the bra and the ket in theabove equation. Between the two operators on each side of this equation, insert a complete set of states:
I =∑
α′,j′,m′
|α′, j′,m′〉〈α′, j′,m′|
Only one of these inserted states will produce non-zero matrix element of J+, so we are left with
〈α, j,m+ 2|J+|α, j,m+ 1〉〈α, j,m+ 1|T+|α, j,m〉 = 〈α, j,m+ 2|T+|α, j,m+ 1〉〈α, j,m+ 1|J+|α, j,m〉
19
Since these matrix elements of J+ do not vanish, we can divide by the product of the one on each side tofind
〈α, j,m+ 2|T+|α, j,m+ 1〉〈α, j,m+ 2|J+|α, j,m+ 1〉
=〈α, j,m+ 1|T+|α, j,m〉〈α, j,m+ 1|J+|α, j,m〉
In the above, we can take m to be any value from −j up to j − 2. The result is that all of these ratios takethe same value.
〈α, j,m+ 1|T+|α, j,m〉 = A+α,j〈α, j,m+ 1|J+|α, j,m〉
We can repeat the above analysis with J− and T− since these two operators commute as well. Omittingthe details, one arrives at the statement
〈α, j,m− 1|T−|α, j,m〉 = A−α,j〈α, j,m− 1|J−|α, j,m〉
5.3 Screening in Helium
Adapted from “Group Theory and Quantum Mechanics” by Tinkham, and Gottfried & Yan. Tinkham’s textis particularly thorough while still concise.
Screening is the effect by which electrons in shells whose probability density lies mostly contained insidethe expected radius of another shell serve to reduce the pull towards the nucleus felt by the outer lyingelectrons. I’ll show here the mathematical details of the calculation confirming the suspicion that this effectshould occur. The ground state of helium will occur for both of the electrons in the 1s state, with theirspin states necessarily anti-symmetrized. Under this configuration though, we expect screening to be fairlysubdued, as neither electronic state really lies outside the bulk of the other. So, we look at an excited state.I have picked the 1s2p state to perform this calculation. We consider both the configurations with the spinstates symmetrized and anti-symmetrized, the 2 3P and 2 1P states in the chemist’s terminology. The way wewill feel the screening effect is to add the mutual repulsion of the electrons to the base Hamiltonian whichconsists of two hydrogenic systems. Without this mutual electronic repulsion, the desired wavefunctionswould be
ψ1s(~r) = R1s(r)Y00(r)
with
R1s(r) = 2( Za0
)3/2
e−Zr/a0
andψ2p(~r) = R2p(r)Y1m(r)
with
R2p(r) =1
2√
6
( Za0
)5/2
r e−Zr/2a0
with Z = 2 for helium.
Two coupled electrons in a spin triplet will have their spin states symmetrized, while those in a spinsinglet will be anti-symmetrized. Hence, in order to maintain overall anti-symmetrization for the electrons’quantum states, as the Pauli Exclusion Principle insists, we must anti-symmetrize the spatial wavefunctionsin the spin triplet configuration, and symmetrize the spatial wavefunctions in the spin singlet.
ψ(~r1, ~r2) = 1√2
[ψ1s(~r1)ψ2p(~r2)± ψ2p(~r1)ψ1s(~r2)]
Without the interaction between the electrons, the correct values for Z in the above wavefunctions wouldbe Z = 2 if we want to talk about helium. Instead, we let the two values of Z float, and use the variationaltechnique to minimize the energy within the configuration space of all Z ′ and Z ′′, the two effective nuclearcharges as seen by the 1s and 2p orbitals respectively. To help us remember that we’ve replaced the values
20
Z in the radial wavefunctions R1s(r) and R2p(r), we write then explicitly with the paramter: R1s(r;Z ′) andR2p(r;Z ′′). Now let’s compute the the expectation of the Hamiltonian
H =(~p1
2
2m− 2e2
r1
)+(~p2
2
2m− 2e2
r2
)+
e2
r12
for r12 = |~r1 − ~r2|. In general, R1s(r;Z ′) is not a single electron solution of the helium Hamiltonian, unlessZ ′ = 2. We deal with this by rewriting this Hamiltonian as follows
~p12
2m− 2e2
r1=(~p1
2
2m− Z ′e2
r1
)+
(Z ′ − 2)e2
r1
The expectation of the helium Hamiltonian is now easy to read off from this:
〈ψnl(~r1;Z ′)|(~p1
2
2m− 2e2
r1
)|ψnl(~r1;Z ′)〉 = 〈ψnl(~r1;Z ′)|
(~p12
2m− Z ′e2
r1
)|ψnl(~r1;Z ′)〉+ 〈ψnl(~r1;Z ′)|
( (Z ′ − 2)e2
r1
)|ψnl(~r1;Z ′)〉
= − (Z ′)2
n2(13.6 eV) + (Z ′ − 2)e2
Z ′
a0n2
=(Z ′)2 − 4Z ′
n2(13.6 eV)
We can apply this to both the 1s and the 2p orbitals. Hence, the expectation of the overall Hamiltonianreduces to the sum of these two terms plus the expectation of the interaction
〈ψ(~r1, ~r2;Z ′, Z ′′)|H|ψ(~r1, ~r2;Z ′, Z ′′)〉 =(Z ′)2 − 4Z ′
12(13.6 eV) +
(Z ′′)2 − 4Z ′′
22(13.6 eV)
+ 〈ψ(~r1, ~r2;Z ′, Z ′′)|e2
r12|ψ(~r1, ~r2;Z ′, Z ′′)〉
We are glad to see that without the interaction, the energy expectation is minimized at E = −68 eV byZ ′ = Z ′′ = 2.
We now treat the interaction term.⟨1r12
⟩=∫d3~r1 d
3~r21√2
[ψ∗1s(~r1)ψ
∗2p(~r2)± ψ∗2p(~r1)ψ
∗1s(~r2)
]× 1r12× 1√
2[ψ1s(~r1)ψ2p(~r2)± ψ2p(~r1)ψ1s(~r2)]
=∫d3~r1 d
3~r212[(ψ∗1s(~r1)ψ
∗2p(~r2)ψ1s(~r1)ψ2p(~r2) + ψ∗1s(~r2)ψ
∗2p(~r1)ψ1s(~r2)ψ∗2p(~r1)
)±(ψ∗1s(~r1)ψ
∗2p(~r2)ψ2p(~r1)ψ1s(~r2) + ψ∗2p(~r1)ψ
∗1s(~r2)ψ2p(~r2)ψ∗1s(~r1)
)]× 1r12
=∫d3~r1 d
3~r2[ψ∗1s(~r1)ψ1s(~r1)ψ∗2p(~r2)ψ2p(~r2)± ψ∗1s(~r2)ψ
∗2p(~r1)ψ2p(~r2)ψ1s(~r1)
]× 1r12
There are two terms here, the direct interaction energy∫d3~r1 d
3~r2ψ∗1s(~r1)ψ1s(~r1)ψ∗2p(~r2)ψ2p(~r2)
r12
and the exchange interaction energy∫d3~r1 d
3~r2ψ∗1s(~r2)ψ
∗2p(~r1)ψ2p(~r2)ψ1s(~r1)
r12
The names refer to in which state a particle leaves the interaction compared to that state in which it began.The direct interaction corresponds to a particle entering and leaving in the same state, while each electron
21
leaves in the state in which the other began in an exchange interaction. The ± sign depends on whetherwe symmetrize or anti-symmetrize the two particles’ wavefunctions, which in turn was decided by whetherwe’re in the spin triplet or singlet. The triplet sees the two spatial wavefunctions anti-symmetrized. Hence,the probability of the two particles sitting on top of one another vanishes, ψ(~r1, ~r2) = 0 for ~r1 = ~r2, meaningthe average distance between the two electrons is larger in this case. The repulsion between them is lower onaverage and the interaction contribution is not as large as in the other case. The spin triplet configurationis known as orthohelium. The other choice, a spin singlet sees larger interaction energy between the twoelectrons and is known as parahelium.
We proceed by expanding 1/r12 in spherical harmonics
1r12
=∑lm
4π2l + 1
rl<
rl+1>
Y ∗lm(r1)Ylm(r2)
with r> = maxr1, r2 and r< = minr1, r2. Unlike in the evaluation of the other expectations, which weremanifestly devoid of angular dependence, the interaction integrals involve angular variables. This happensbecause there is an angular term in
1|~r1 − ~r2|
=1√
r21 + r22 − 2r1r2 cos θ12
Integration over all ~r1 and ~r2 will restore overall rotational invariance of the interaction Hamiltonian, butthe calculation is still complicated. Write the direct term as∫ ∞
0
dr1 r21
∫ ∞
0
dr2 r22 |R1s(r1;Z ′)|2|R2p(r2;Z ′′)|2
×∫dΩ1dΩ2Y
∗00(r1)Y
∗1m(r2)
(∑LM
4π2L+ 1
rL<
rL+1>
Y ∗LM (r1)YLM (r2)
)Y00(r1)Y1m(r2)
The angular part is the best to attack first. Following Section (5.1) we determine which values of L and Mproduce a non-zero angular integral.
∫dΩ2Y
∗1m(r2)YLM (r2)Y1m(r2) =
1
2√
πfor L,M = 0
1√5π
for L = 2 and M = 00 otherwise
and ∫dΩ1Y
∗00(r1)Y
∗LM (r1)Y00(r1) =
12√
πfor L,M = 0
0 otherwise
requires that L = 0 and M = 0 in order not to vanish. Hence, only the term with L = 0 and M = 0 in oursum produces a non-zero integral, leaving behind∫
dΩ1dΩ2Y∗00(r1)Y
∗1m(r2)Y00(r1)Y1m(r2)×
1r12
=( 1
2√π
)2 4πr>
=1r>
Similarly, with the exchange integral, only the term with L = 1 and M = 0 survives, so that∫dΩ1dΩ2Y
∗00(r2)Y
∗1m(r1)Y00(r1)Y1m(r2)×
1r12
=( 1
2√π
)2 4π3r<r2>
=13r<r2>
The direct and exchange energies are now determined, up to a factor of e2, by the integrals∫ ∞
0
dr1 r21
∫ ∞
0
dr2 r22|R1s(r1;Z ′)|2|R2p(r2;Z ′′)|2 ×
1r>
22
and13
∫ ∞
0
dr1 r21
∫ ∞
0
dr2 r22R
∗1s(r2;Z
′)R∗2p(r1;Z′′)R1s(r1;Z ′)R2p(r2;Z ′′)×
r<r2>
Evaluate these integrals by breaking the quarter plane r1 ≥ 0 × r2 ≥ 0 into the two regions r1 ≥ r2(r> = r1, r< = r2) and r1 ≤ r2 (r> = r2, r< = r1) to find⟨
e2
r12
⟩=
12Z ′′[1− (Z ′′)4(6Z ′ + Z ′′)
(2Z ′ + Z ′′)5](13.6 eV)± 224
3(Z ′)3(Z ′′)5
(2Z ′ + Z ′′)7(13.6 eV)
We finish by minimizing the expected value of the total Hamiltonian over Z ′ and Z ′′ to get
Z ′ = 1.99 and Z ′′ = 1.09 −→ E = −58.0 eV
for orthohelium 2 3P , andZ ′ = 2.00 and Z ′′ = 0.96 −→ E = −57.7 eV
for parahelium 2 1P .
5.4 Runge-Lenz Vector
Adapted from a lecture by Prof. Tom Mehen of Duke UniversityNeglecting the all interactions but the Coulomb interaction between the proton and the electron, the non-relativistic hydrogen atom has energy levels given by:
En = −13.6 eVn2
The operators L2 and Lz both generate symmetries of the Hamiltonian, since [L2,H] = [Lz,H] = 0. Wecall these the Casimir operator and root, respectively, of an SO(3) symmetry exhibited by this system. Inother words, the hydrogen atom in free space exhibits rotational invariance. All Hamiltonian eigenstatesrelated to one another by the symmetries generated by L2 and Lz have the same energy eigenvalues. Butthis implies only a (2l + 1)-fold degeneracy among the energy eigenvalues, whereas they are truly n2-folddegenerate. We have an accidental symmetry of the system, probably better dubbed a hidden symmetry.
This extended symmetry is due to the Runge-Lenz vector
~Acl = ~L× ~p+me2
r~r
~Aqu = 12 (~L× ~p− ~p× ~L) +
me2
r~r
which is a constant of the motion. Note the following commutation relations among the Runge-Lenz vectorand the angular momentum vector
[Li, Lj ] = i~ εijkLk
[Li, Aj ] = i~ εijkAk
[Ai, Aj ] = −i~ · 2m|H|εijkLk
This is the set of commutation relations for an so(3, 1) algebra, the algebra of the Lorentz group! If we defineAi =
√2mHki then we obtain the more symmetric commutation relations
[Li, Lj ] = i~ εijkLk
[Li, kj ] = i~ εijkkk
[ki, kj ] = i~ εijkLk
23
Note, H < 0, so owing to the imaginary ratio of ki to Ai, we have turned this into the set of commutationrelations for so(4). Defining J±i = 1
2 (Li ± ki), we get
[J+i , J
+j ] = i~ εijkJ
+k
[J−i , J−j ] = i~ εijkJ
−k
[J+i , J
−j ] = 0
which gives two decoupled copies of su(2) ' so(3). That is, so(4) ∼= su(2)⊕ su(2). Then,(~J±)2 = 1
4
(~L2 + ~k2 ± 2~L · ~k
)but ~L · ~k ∝ ~L · ~A = 0. Thus, ( ~J+)2 = ( ~J−)2. It is important to realize that the algebra does not imposethis condition, it is the physics that led us to see these two su(2)’s have the same magnitude. So, our goodquantum numbers are: ( ~J+)2, J+
z , J−z , and the states are |j,m+,m−〉 with −j ≤ m+,m− ≤ j. This yieldsa (2j + 1)2-fold degeneracy. Note now that
−2mE~k2 = ~A · ~A = 2mE(~L2 + ~2) +m2e4
where the ~2 appeared due to the quantum commutator. Solving for the energy E yields
E = − me4
2(~L2 + ~k2 + ~2)
= − me4
2(4 ~J2 + ~2)
= − me4
2~2(2j + 1)2
Then, n = 2j + 1 with degeneracy (2j + 1)2 = n2. We’ve recovered the energy levels and degeneracies forthe hydrogen atom.
5.5 Isospectral Hamiltonians - Precursor to Supersymmetry
A good source for further information regarding this topic is G. Junker’s text Supersymmetric Methods inQuantum and Statistical Physics.
5.5.1 Introduction
Imagine we are given a Hamiltonian H(1) that can be factored into the form H(1) = A†A for some operatorA. We can define another Hamiltonian H(2) = AA†. Our second Hamiltonian is automatically Hermitian.Now consider an eigenstate |φ〉 of the first Hamiltonian that has energy E:
H(1)|φ〉 = E|φ〉A†A|φ〉 = E|φ〉
Operating on both sides with A yields
A(A†A
)|φ〉 = A
(E|φ〉
)(AA†
)(A|φ〉
)= E
(A|φ〉
)H(2)
(A|φ〉
)= E
(A|φ〉
)H(2)|ψ〉 = E|ψ〉
24
Hence, there is an eigenstate |ψ〉 = A|φ〉 of H(2) with this same energy eigenvalue. We could equally well havebegun with the statement that H(2)|ψ〉 = E|ψ〉 and then found a state of H(1) given by |φ〉 = A†|ψ〉 withthe same energy. Note that a Hamiltonian of the two given forms will have only non-negative eigenvalues:〈ψ|A†A|ψ〉 =
∣∣|Aψ〉∣∣2 ≥ 0. Equality holds only if the state is annihilated by A. Analogous statements holdfor A†. In order to have both |φ〉 and |ψ〉 properly normalized, we need
|φ〉 = E−1/2A†|ψ〉|ψ〉 = E−1/2A|φ〉
This reveals an issue with the procedure: what about the case when E = 0? Suppose that |φ〉 has vanishingenergy. The eigenvalue equation then reads
H(1)|φ〉 = A†A|φ〉 = 0
or A†|ψ〉 = 0 so that |φ〉 was originally null. Hence in the case where there’s a state with vanishing energyE = 0, either |ψ〉 or |φ〉 must be a null state. At most one of these two Hamiltonians may have a state withthis vanishing eigenvalue. Additionally, there is only full, unbroken supersymmetry (SUSY) if one of thetwo groundstates indeed does have a vanishing energy eigenvalue. We’ll see later that failure to satisfy thiscondition is equivalent to a ground state which is not invariant under SUSY transformations: a hallmark ofa broken symmetry (at best).
An important class of operators A with which this analysis is often carried out take the form
A =~√2m
d
dx+W (x)
A† = − ~√2m
d
dx+W (x)
Here, W (x) is known as the superpotential. These yield paired Hamiltonians
H(−) := H(1) = A†A = − ~2
2md2
dx2+W (x)2 − ~√
2mW ′(x)
H(+) := H(2) = AA† = − ~2
2md2
dx2+W (x)2 +
~√2m
W ′(x)
The above is often stated with units ~ = 2m = 1. From here on, I will adopt this convention when convenient.
The condition that A|φ0〉 = 0 for the ground state |φ0〉 of H(−) becomes( ddx
+W (x))φ0(x) = 0
which is known as the Riccati equation.
5.5.2 Simple Application: Simple Harmonic Oscillator
To begin, we quickly note that picking W (x) = (mω2/2)1/2 x will generate two paired Hamiltonians both ofwhich are simple harmonic oscillators, one with potential ~ω higher than the other.
V1,2 =12mω2x2 ± 1
2~ω
We see that one of these two Hamiltonians has a ground state energy of E = 0 and so, SUSY is anunbroken symmetry relating these two potentials. Now we repeatedly use the fact that supersymmetry is
25
Figure 2: The energy levels of the simple harmonic oscillator are evenly spaced.
relating one system to a shifted version of itself. Because the potential is the same up to a constant shift,this means the ground state of the second Hamiltonian has energy E = ~ω. Hence, there is a state of thefirst Hamiltonian with the same energy, its first excited state. The existence of this state in the first poten-tial means the second Hamiltonian has an equivalent first excited state, just shifted. This, then, yields bysupersymmetry a second excited state of the first Hamiltonian. This process can be continued ad infinitumand explains why the energy levels of a simple harmonic oscillator are evenly spaced.
In reality, these results are nothing new; we could have obtain them using the familiar a and a† operatorsfrom the typical algebraic treatment of the harmonic oscillator. In going through this analysis, one sees thatthe operators A and A† are performing the roles of the operators a and a†. Indeed, W (x) = (mω2/2)1/2 ximplies
A =√
~ω a
A† =√
~ω a†
Regardless, it’s pleasing that the forms of the ladder operators are suggested by this more general supersym-metric technique. We’ll see the same in other contexts.
5.5.3 Two Quick Examples
Taking our superpotential to be W (x) = tanhx, we get two paired potentials
V1(x) = 1 V2(x) = 1− 2 sech2x
The second of these is the Rosen-Morse potential, which is relevant to the study of Fermi liquids and tothe Korteweg-de Vries equation. I have no idea about either of these topics. But it does have two nice andreadily observed properties. First, because the constant potential has no bound states, the Rosen-Morsepotential can have at most one, corresponding to a zero-mode. Indeed, we find such a zero energy solutiondoes exist and takes the form
φ0(x) = C sechx
The second observation is that the famous reflectionless behavior of the Rosen-Morse potential follows im-mediately from the fact that it is paired to a constant potential.
Another example, with no interesting corollaries that I know of, is the superpotential W (x) = Θ(x) −Θ(−x). The two paired potentials are then V (x) = 1±2δ(x), an attractive and repulsive δ-function potential.
26
5.5.4 Treatment of the Poschl-Teller Potential
We can use the method of isospectral Hamiltonians to evaluate the spectrum of a rather difficult potentialby relating it to a very simple system. Take W (x) = tanx. This choice gives
A =d
dx+ tanx
A† = − d
dx+ tanx
We get the two paired Hamiltonians:
H(−) = A†A = − d2
dx2− 1
H(+) = AA† = − d2
dx2+ 2 sec2 x− 1
The second of these has been previously studied under the name of the Poschl-Teller potential.
-1.5 -1 -0.5 0.5 1 1.5
5
10
15
20
25
Figure 3: The Poschl-Teller potential V2(x) = 2 sec2 x− 1, which is paired to the free particle in a box.
Since the second potential is naturally limited in spatial extent by the divergence of sec2 x at x = ±π2 ,
we impose the same restriction on the first Hamiltonian as well. The system described by H(−) is thereforerevealed to be a free particle in a box, with eigenstates
φn(x) =
√
2π sinnx for n = 2, 4, 6, . . .√2π cosnx for n = 1, 3, 5, . . .
and corresponding eigenenergies En = n2 − 1. In particular, E1 = 0, so there will not be an equivalenteigenstate of H(+). Furthermore, solving the Riccati equation would yield the ground state of vanishingenergy: ( d
dx+ tanx
)φ0(x) = 0 −→ φ0 = C cosx
Take the eigenspectrum of H(−) over to H(+) to find the energy eigenstates.
ψn(x) =
E−1/2n
(ddx + tanx
)√2π sinnx for n = 2, 4, 6, . . .
E−1/2n
(ddx + tanx
)√2π cosnx for n = 3, 5, 7 . . .
The corresponding energy eigenvalues are of course still En = n2 − 1. Plots of some of the states are shownbelow.
27
-1.5 -1 -0.5 0.5 1 1.5
-1
-0.5
0.5
1
Figure 4: Three lowest energy eigenstates of H(−).
-1.5 -1 -0.5 0.5 1 1.5
-0.75
-0.5
-0.25
0.25
0.5
0.75
Figure 5: Three lowest energy eigenstates ofH(+).
5.5.5 Free Particle in 3 Dimensions
This section is mostly useful to us as a precursor to the next, where we add a Coulombic potential to theHamiltonian. It also finds use in the treatment of scattering off a hard sphere.3 Spherical symmetry of theHamiltonian allows us to write ψklm(r, θ, φ) = Rkl(r)Ylm(θ, φ) with Rkl(r) satisfying[
− ~2
2md2
dr2+
~2l(l + 1)2mr2
]Rkl(r) =
~2k2
2mRkl(r)
I will supress all physical values here and instead just focus on the mathematical technique: set ~ = 2m =k = 1. Switching to the function u = rR(r), we get[
− d2
dr2+l(l + 1)r2
]ul(r) = ul(r)
Now define
Al :=d
dr− l + 1
r
H(−)l := A†lAl = − d2
dr2+l(l + 1)r2
Direct computation reveals the paired Hamiltonian H(+)l = AlA
†l actually equals H(−)
l+1 ! So, given an eigen-
function ul of H(−)l with non-zero energy, we get an eigenfunction Alul of H(−)
l+1 for free: first note that Alul
is an eigenstate of H(+)l and then that H(+)
l = H(−)l+1 . Since there is no attractive potential, all states have
positive energy.
Let’s construct some of the solutions. The base case (l = 0) has the solutions
−u′′0(r) = 0 −→ u0(r) = sin r or cos r
From here, we can repeatedly apply the appropriate Al to generate all the remaining solutions. The firstfew are
l Solution 1 Solution 20 sin r cos r1 sin r
r − cos r cos rr + sin r
2(
3r2 − 1
)sin r − 3
r cos r(
3r2 − 1
)cos r + 3
r sin r
These are known as the Riccati-Bessel functions. Divide through each of these by r to get R(r) and werecognize them as the familiar spherical Bessel functions that we knew all along were the radial wavefunctions.
3If the quantum particles being scattered are photons, this is known as Mie scattering and is needed in understanding theappearance of a great variety of substances, like milk.
28
5.5.6 Hydrogen Atom
Adapted from a lecture by Prof. Robert Jaffe of MITWe have already seen in Section 5.4 that the hydrogen spectrum displays a degeneracy of energy levels –En is n2-fold degenerate – is larger than that which would result purely from rotational invariance of thepotential. The Runge-Lenz vector expands this symmetry to one that accounts fully for these degeneracies.Read about this in Appendix A. Superysymmetry is yet another way to account for this larger than expecteddegeneracy. In fact, we’ll see later that supersymmetry suggests the exact form of the Runge-Lenz vector inthe same way it suggested the form of the harmonic oscillator’s ladder operators. The effective Hamiltonianfor the radial wavefunction of an electron in the hydrogen atom is
Hcoull = − d2
dr2+l(l + 1)r2
− 1r
The label l denotes that this Hamiltonian has had the effect of angular dependence distilled into an effectiveangular momentum barrier (a “kinetic potential”) that depends on l, the quantum number labelling thetotal orbital angular momentum of the electron. Now, define two operators, also taking labels l, by
Al =d
dr− l + 1
r+
12(l + 1)
A†l = − d
dr− l + 1
r+
12(l + 1)
These operators yield partner Hamiltonians
H(−)l = A†lAl = − d2
dr2+l(l + 1)r2
− 1r
+1
4(l + 1)2
= Hcoull +
14(l + 1)2
H(+)l = AlA
†l = − d2
dr2+
(l + 1)(l + 2)r2
− 1r
+1
4(l + 1)2
= Hcoull+1 +
14(l + 1)2
One can solve both Al|φl〉 = 0 and A†l |φl〉 = 0 to find that only the former gives a sensible, normalizableresult. Thus, H(−)
l that has a zero energy groundstate. This means that SUSY is a good symmetry.
Alφl(r) =[ ddr− l + 1
r+
12(l + 1)
]φl(r) = 0
dφl
φl=[ l + 1
r− 1
2(l + 1)
]dr
∴ φl(r) = Crl+1e−r/2(l+1)
We can also compute
H(−)l φl =
[Hcoul
l +1
4(l + 1)2]φl = 0 −→ Hcoul
l φl = − 14(l + 1)2
φl
We now have the functional form and energy in both the shift Hamiltonian H(−)l and unshifted Hamiltonian
Hcoull of this state. The rest of the treatment comes from observing that H(−)
l = Hcoull + 1/4(l + 1)2 and
H(+)l = Hcoul
l+1 + 1/4(l + 1)2 together yield
H(+)l = H
(−)l+1 +
14(l + 1)2
− 14(l + 2)2
29
Figure 6: Combining two adjacent pairs of isospectral Hamiltonians
We can now relate the isospectral pair H(−)l and H(+)
l to the isospectral pair H(−)l+1 and H(+)
l+1 . By extension,
we can combine all pairs of isospectral Hamiltonians, expressing every H(+)l as a shifted copy of H(−)
l+1 . Fromabove, we have computed the functional form of each of the states along the diagonal bounding the set ofstates from below. Repeated application of the appropriate A†l operator will send us leftward across a row,which will end after the l = 0 state is reached. The energy remains the same as we move to the through thel values. Since it was the original, maximal value of l that appears in the energy eigenvalue of |φl〉 underHcoul
l , we define n = lmax + 1 and say
En = − 14n2
It is now important to make two notes about the interpretation of this solution, both stemming fromconventions set when we defined Hcoul
l . First, by choosing ~ = 2m = e = 1, we implicitly set the Bohrradius to a0 = 2. With units restored, the exponential in the above radial wavefunction would actually readexp[−r/a0(l + 1)]. Restoring units in the expression for En gives the standard result as well. Second, wealso solved for u(r) = r R(r), so one must divide by r to get the radial wavefunction.
Figure 7: Scheme for obtaining functional form of states
We should now perform an explicit calculation for a simple case, say l = 2. The lowest energy state inthe l = 2 tower can be seen in Figure 7 to be n = 3 according to the labelling scheme chosen here. Then,
30
the state has radial distribution given by
u3,2(r) = Ar3e−r/6
Move to the l = 1 tower by applying A†1.
u3,1(r) = A†1u3,2(r)
=(− d
dr− 2r
+14
)Ar3e−r/6
= Br2(1− r
12)e−r/6
Finish the row by finding the l = 0 state
u3,0(r) = A†0u3,1(r)
=(− d
dr− 1r
+12
)Br2
(1− r
12)e−r/6
= Cr( r218− r + 3
)e−r/6
With units restored, and converting to R(r), these three become
R3,2(r) = r2e−r/3a0
R3,1(r) = r(1− r
6a0
)e−r/3a0
R3,0(r) =[29( ra0
)2 − 2r
a0+ 3]e−r/3a0
up to an omitted normalization on each of these.
5.5.7 Connecting SUSY to Runge-Lenz Vector
Adapted from Lyman & Aravind J. Phys. A: Math. Gen. 26 (1993) 3307-3311
5.6 Supersymmetry in the Context of Quantum Mechanics
For more information on the following topics, see Supersymmetry: Basics and Concepts by Soni and Singh.One has a supersymmetric quantum mechanical system when the following conditions are met: there is aHamiltonian H acting on the Hilbert space H, along with N self-adjoint operators Qi = Q†
i satisfying
Qi, Qj = Hδij for i, j = 1, . . . , N
We call the Qi’s the supercharges of the theory. Note that by the above H = 2Q21 = . . . = 2Q2
N , so itautomatically follows that
[Qi,H] = 0 for all i
Sometimes the above definition does not involve the self-adjoint condition, instead folding two superchargesinto one complex supercharge4
Qi := 1√2
(Q2i + iQ2i+1
)4The tildes over Q aren’t standard notation. I include them only to distinguish from the self-adjoint case. Since one should
stick to one convention or another, there’s never any ambiguity that requires this extra decoration.
31
Working it out, we see that these will satisfy a slightly different algebra
Qi, Qj = 0
Q†i , Q
†j = 0
Qi, Q†j = Hδij
for all i,j. The two pictures are equivalent.
5.6.1 Famous N = 1 SUSY System: The Pauli Hamiltonian
Given a vector potential ~A(~x), define a self-adjoint supercharge
Q = Q† :=1
2√m
(~p− e
c~A)· ~σ
By design, the corresponding Hamiltonian is supersymmetric.
H = 2Q2 =1
2m(~p− e
c~A)2 − e~
2mc~σ · ~B
We recognize this as nothing other than a spin- 12 particle in a magnetic field. The interaction term has a
Lande g factor of 2. Therefore, classically an electron in a magnetic field possesses a supersymmetry asdescribed here. Quantum effects break this.
5.6.2 N = 2 SUSY QM: The Witten Model
At this point, we can make contact with the discussion of the operators A and A† from the section onisospectral Hamiltonians by identifying one complex supercharge Q with the matrices
Q =(
0 A0 0
)Q† =
(0 0A† 0
)from which it easily follows that Q,Q = Q†, Q† = 0 and
Q,Q† =(H(+) 0
0 H(−)
)where H(−) = A†A and H(+) = AA†. We’ve now folded the two separate systems with identical energyspectra (up to zero modes) into one system living in a Hilbert space H = H(+) ⊕H(−) with a HamiltonianH = H(+) ⊗ I + I⊗H(−).
We look for an operator that will distinguish the two subspaces from one another. Clearly H cannot.Recall that the whole reason we used an anticommutator in H = Q,Q† we to put the same sign in fromof H(+) and H(−) in their respective subspaces. Had we not done so, we would’ve gotten
[Q,Q†] =(AA† 0
0 −A†A
)=(H(+) 0
0 −H(−)
)This is exactly what we want. The eigenvalues of this operator are non-negative on the H(+) subspace andnon-positive on the H(−) subspace. Let’s “divide by” the Hamiltonian to scale out any sensitivity to theexact energy levels and call the result
W :=[Q,Q†]Q,Q†
=(
+I 00 −I
)
32
This is the Witten parity operator. It’s eigenstates are the eigenstates of H, but the eigenvalues now differbetween the two subspaces of H. As it stands, it’s only well-defined on the orthogonal complement tokerH but a variety of regularization procedures resolve this issue. The Witten parity operator implementsa Z2 grading of the Hilbert space. Energy eigenstates are even or odd if they have eigenvalues +1 or −1,respectively, of W . We can also talk about the parity of operators
[W,E] = 0 ←→ E is evenW,O = 0 ←→ O is odd
We’ve noted that whether or not SUSY is a good symmetry is determined by the ground state energy.Define the Witten index by
∆ := Tr (−W ) = dim kerH(−) − dim kerH(+)
All states with energy E > 0 come in pairs: one eigenstate of H(+) and one of H(−), so such states contributenothing to ∆. Cases when ∆ 6= 0 have at least one zero mode and hence correspond to a system with goodSUSY. ∆ = 0, however, does not mean SUSY is necessarily broken, at least in general.
5.6.3 Hints of Superspace
Consider a quantum mechanical system with a complex superchargeQ. SinceH = i ∂∂t is the generator of time
translations and H = Q,Q† gives a relationship between the supercharge and this generator of translation,we ask ifQ is a generator of translations in some coordinate too such that the given anticommutator reduces toa time translation. The fact that this is an anti -commutator suggests that Grassmann variables are natural.We posit a Grassmann valued coordinate θ, and another θ for Q†. We want Q to go like i ∂
∂θ but also need tobring time translations into the mix too so that we’ll get the right behavior under anticommutation. Hence,we’ll need to add terms of the form θ ∂
∂t and θ ∂∂t . After tooling around a bit, we find that
Q = i∂
∂θ− aθ ∂
∂t
Q† = −i ∂∂θ
+ aθ∂
∂t
are the most general linear combinations that satisfy Q,Q = Q†, Q† = 0 and still mix derivatives in θand t. We find
Q,Q† = ∂
∂θ,∂
∂θ
− aaθ, θ ∂
2
∂t2+(ia ∂
∂θ, θ
+ ia ∂
∂θ, θ) ∂
∂t
= i(a+ a)∂
∂t= (a+ a)H
We can pick a and a to suit our convenience.5
We can now make a (unitary) SUSY transformation of our state: |ψ〉 −→ exp( i~G) where
G = iξ∗Q− iQ†ξ = G†
is the Hermitian generator of this transformation. In order for the state to be SUSY invariant, we need
Q|ψ〉 = Q†|ψ〉 = 0
This is a property that we require of the ground state if we want SUSY to be a good (ie, unbroken) symmetry.As we’ve seen, this requires the ground state energy to vanish.
5Some authors seem to pick a = a = 1 and then have Q, Q† = 2H instead.
33
From here, we can define covariant derivatives
D :=∂
∂θ− iaθ ∂
∂t
D† :=∂
∂θ+ iaθ
∂
∂t
that satisfyD,Q† = 0 and D†, Q = 0
This notion will be useful later in the context of a field theory and won’t be developed further here.
6 Quantum Field Theory
Quantum field theory is such a broad topic that it’s difficult to know what to include here in a quickexplanation. I imagine the contents of this section will depend largely on what strikes my fancy, and also onwhat time allows.
6.1 Relativistic Wave Equations
We’ll develop here the relativistic wave equations for the classical scalar, spinor and vector fields with andwithout mass.
6.1.1 Klein-Gordon Equation
We’re familiar from relativistic mechanics with the result that a particle with mass m and momentum p = |~p|has a total energy E given by
E2 = p2c2 +m2c4
Canonical prescription to get a dynamical wave equation is to make the substitutions E → i~ ∂t and pi →−i~ ∂i in the statement of energy conservation. Doing so for the above equation while labelling our systemφ, we get
i~∂
∂tφ(~x, t) = ±
√−~2c2∇2 +m2c4φ(~x, t)
This has one critical problem: the operator on the right hand side is not local! Tweaking the field φ in asmall neighborhood around a given spacetime point (~x0, t0) changes the value this operator takes outside thisneighborhood. Since we want local theories – as much as possible, at least – we seek a different dynamicalequation. Fortunately, we have it in front of us already: we simply use the same equation without takingthe square root first: (
i~∂
∂t
)2
φ(~x, t) =[c2(−i~∇
)2
+m2c4]φ(~x, t)
which becomes
0 =[ ∂2
∂t2−∇2 +m2
]φ(~x, t)
after suppressing the physical constants. So what’s the difference? Firstly, the differential operator is nowperfectly local, which is what we wanted. But the side effect of making this change is that we have nowfolded both positive and negative energy solutions into the same dynamical equation. This will turn out tohave many interesting implications.
Let’s define the symbol
:= ∂µ∂µ = ∂2 =
∂2
∂t2−∇2
34
known as the D’Alembertian operator. The above is now written
( +m2)φ = 0
This is known as the Klein-Gordon equation. It’s merely a statement of conservation of energy-momentum,and therefore holds for all (non-interacting) fields, regardless of the spin this field possesses.
To solve the general dynamical equation problem in classical field theory, one constructs the Green’sfunction G(x, y), which satisfies
( +m2)G(x, y) = −iδ(4)(x− y)
The factor −i is included as a convention. The formal solution to this is found using Fourier transforms
G(x, y) =∫
d4k
(2π)4i
k2 −m2e−ik·(x−y)
The standard treatment is then to push the poles in G(k) = i/(k2 −m2) off the real axis:
G(k) =i
k2 −m2 + iε
and later on take ε→ 0. The three choices of which poles go inside the contour define the retarded, Feynmanand advanced Green’s functions.
6.1.2 Dirac EquationAdapted from Ryder’s text on Quantum Field Theory
Whereas spin-0 fields have trivial behavior under changes of frame, the others do not. Next simplestis the spin- 1
2 case, known as spinors. We begin by looking at the Lorentz group, which consists of spatialrotations as well as Lorentz boosts. The group of orthogonal transformations preserving inner products underthe flat spacetime metric h = diag(1,−1,−1,−1) is O(1, 3). This group has four connected components.We specialize to SO(1, 3), the subgroup of O(1, 3) with positive determinants. There are two connectedcomponents, one containing the identity and the other which has unit determinant because it reverses timeand inverts space. The former of these, denoted SO+(1, 3) is the Lorentz group and consists of the proper,orthochronous isometries of this metric.6 The Lorentz group is six dimensional, corresponding to the threeparameters of a boost and the three of a spatial rotation. Hence, its Lie algebra so(1, 3) has six generators.Call the generators of boosts Ki and those of rotations Ji. They obey the following alegbra
[Ji, Jj ] = iεijkJk
[Ji,Kj ] = iεijkKk
[Ki,Kj ] = −iεijkJk
We can find many specific representations of this algebra, most notably Ji = 12σi and Ki = ± i
2σi. But first,let’s make a change of basis and consider the six generators
Ai = 12 (Ji + iKi)
Bi = 12 (Ji − iKi)
Written in this basis, the algebra becomes
[Ai, Aj ] = iεijkAk
[Bi, Bj ] = iεijkBk
[Ai, Bj ] = 0
6For those interested, O(1, 3)/SO+(1, 3) ∼= Z2 × Z2, the Klein four group.
35
which reveals that so(1, 3) ∼= su(2)⊕su(2)! Since we know all about constructing linear irreps of su(2), whichare labelled by j = 0, 1
2 , 1,32 , . . ., we automatically get that irreps of so(1, 3) are labelled by (j, j′) and act
on a corresponding vector space of dimension (2j + 1)(2j′ + 1).
In this new language, a scalar field transforms under the (0, 0) irrep. We now consider the two simplestnon-trivial irreps, ( 1
2 , 0) and (0, 12 ). In the former case, we get representations for Ai and Bi
Ai = 12 (Ji + iKi) = 1
2σi
Bi = 12 (Ji − iKi) = 0
which means Ji = iKi = 12σi. Let ξ transform under the corresponding representation of SO+(1, 3).
Labelling the three parameters for rotations and three for boosts as ~θ and ~φ respectively, we find ξ transformslike
ξ −→ exp[i~θ · ~J + i~φ · ~K
]ξ
=exp[i(~θ − i~φ ) · ~A+ i(~θ + i~φ ) · ~B
]ξ
=exp[i(~θ − i~φ ) · 1
2~σ]ξ
The second irrep. (0, 12 ) then has Ji = −iKi = 1
2σi. Calling the corresponding spinor η, we have thetransformation properties
η −→ exp[i~θ · ~J + i~φ · ~K
]η
=exp[i(~θ − i~φ ) · ~A+ i(~θ + i~φ ) · ~B
]η
=exp[i(~θ + i~φ ) · 1
2~σ]η
In older language, ξ and η were known as “dotted” and “undotted” spinors, respectively. The modernterms are “right-” and “left-handed” spinors. Collectively, ξ and η are known as the chiral spinors or Weylspinors. The matrices exp[i(θ± i~φ ) · 12~σ] are both elements of SL(2; C). But they’re not independent. If weview exp[i(~θ − i~φ ) · 1
2~σ] as an element of SL(2,C), we implicity are saying that we can adjust both the realand the imaginary terms in the exponential, which means we’ve specified both ~θ and ~φ. Therefore, the matrixexp[i(~θ + i~φ ) · 1
2~σ] is fully determined. It seems tempting to say that the statement so(1, 3) ∼= su(2)⊕ su(2)made about the Lie algebras corresponds to SO+(1, 3) ∼= SL(2; C) made about the Lie groups. But this isn’tquite right. If we specialize to the standard basis for the Pauli matrices in which it holds that σ2σ
∗i σ2 = −σi,
we can make the following statement
σ2 exp[i(~θ + i~φ ) · 1
2~σ]∗σ2 = exp
[i(~θ − i~φ ) · 1
2~σ]
Therefore, the transformation matrices are related by a change of basis to their complex conjugates. Thismeans that we really need to write SO+(1, 3) ∼= SL(2; C)/Z2. Notice also that in the case that ~φ = 0,corresponding to a pure rotation, the two representations transform identically. This why the notion of “twotypes of spin- 1
2 objects” never arises in non-relativistic situations.What if we also require our states to be invariant under parity – ie, spatial inversion? We see that
spatial inversion sends the irrep (j, j′) to (j′, j). To have invariance under this inversion, we need our irrepsto be (j, j) or (j, j′) ⊕ (j′, j). The second case is no longer regarded as reducible. Applied to the spinorsabove, we see that we want the representation ( 1
2 , 0)⊕ (0, 12 ), which is called a Dirac spinor. It consists of a
right-handed and a left-handed spinor put together and transforms as
ψ :=(ξη
)−→
(D(Λ) 0
0 D(Λ)
)(ξη
)where we define D(Λ) = exp[i(θ−i~φ ) · 12~σ] and D(Λ) = σ2D(Λ)∗σ2. This latter definition is basis dependent.
36
What about the equations of motion? First let’s consider the case of massless spinors. Like with allmassless particles, these spinors have no rest frame. Hence, the attempt to mathematically boost our twotypes of spinors into a rest frame using their respectively appropriate transformation laws gives a vanishingresult:
(p0 − ~σ · ~p)ξ(p) = 0
(p0 + ~σ · ~p)η(p) = 0
Since the two are massless, p0 = |~p|, and we can rewrite these two equations as
(~σ · p)ξ(~p) = ξ(p)(~σ · p)η(~p) = −η(p)
These are the Weyl equations for a massless spinor. The operator ~σ ·p corresponds to twice the helicity, whichis ~S · p = 1
2~σ · p for a spin- 12 particle. The two spinors ξ and η are eigenspinors of helicity and their respective
eigenvalues justify us naming them “right-handed” and “left-handed.” The 4 × 4 matrix representing thehelicity operator on Dirac spinors is
γ5 := iγ0γ1γ2γ3
which equals diag(1, 1,−1,−1) in this basis.
In the case of massive spinors, we can indeed boost into a rest frame. There, the helicity operator hasno meaning and the two spinors lose their distinction. We’ve already noted that they behave identicallyunder pure spatial rotations, to which we are limited if we want to stay in the rest frame. Hence, we seeξ(0) = η(0). The parameter ~φ corresponds to how the boost will change the particle’s rapidity. Its magnitudeis φ = 1
2 ln[(1 + β)/(1− β)], where β = v/c is the velocity relative to our frame. We boost our Dirac spinorso that it has 3-momentum ~p.
ξ(p) = exp[ 12~σ · ~φ ]ξ(0) =[√
(γ + 1)/2 + (~σ · p)√
(γ − 1)/2]ξ(0)
η(p) = exp[ 12~σ · ~φ ]η(0) =[√
(γ + 1)/2− (~σ · p)√
(γ − 1)/2]η(0)
Using the relations p0 = γm and ξ(0) = η(0), we get
ξ(p) =p0 + ~σ · ~p
mη(p)
η(p) =p0 − ~σ · ~p
mη(p)
This is essentially the Dirac equation. To make contact with its usual appearance, let’s rearrange it and castit in the form of a single matrix equation(
−m p0 + ~σ · ~pp0 − ~σ · ~p −m
)(ξ(p)η(p)
)= 0
After taking
γ0 =(
0 11 0
)and γi =
(0 −σi
σi 0
)we can write the above in a much more compact form:
(γµpµ −m)ψ(p) = 0
The equivalent equation in position space is obtained by replacing pµ → i∂µ.
37
Since we know how ξ and η transform, we know how their Hermitian adjoints transform as well.
ξ† −→ ξ† exp[−i(~θ + i~φ ) · 1
2~σ]
η† −→ η† exp[−i(~θ − i~φ ) · 1
2~σ]
This suggests that the scalar quantity invariant under the Lorentz group is not
ψ†ψ = ξ†ξ + η†η
but is insteadψ†γ0ψ = ξ†η + η†ξ
We define ψ := ψ†γ0 and write the above Lorentz scalar as ψψ. It is also preserved under spatial inversion.Related is the quantity ψγ5γ which is invariant under orientation-preserving Lorentz transformations butnegates under spatial inversion. That is, it’s a pseudoscalar quantity. In our current basis
ψγ5ψ = η†ξ − ξ†η
One can additionally construct a vector ψγµψ, a pseudovector ψγµγ5ψ and anti-symmetry rank-2 tensorsψ[γµ, γν ]ψ.
The entirety of the above is done in the chiral basis. That is, we used the basis that diagonalized thehelicity matrix γ5 and wrote things out in the chiral spinors ξ and η which both had a definite handedness.Let’s boost now to the the restframe of a massive Dirac spinor, in which the Dirac equation takes the form
(γ0p0 −m)ψ = 0 −→ p0ψ = mγ0ψ
Hence, the matrix to diagonalize when trying to find the energy eigenspinors is γ0! We find it has eigenvalues+1 and −1, and denote the corresponding eigenspinors u and v. It’s not hard to see that u = 1√
2(ξ+ η) and
v = 1√2(ξ − η). In this new basis
γ0 =(
1 00 −1
), γi =
(0 σi
−σi 0
), and γ5 =
(0 11 0
)Dirac spinors transform under the Lorentz group as
ψ =(uv
)−→
(exp[i~θ · 1
2~σ]cosh(φ/2) (~σ · φ) sinh(φ/2)
(~σ · φ) sinh(φ/2) exp[i~θ · 1
2~σ]
cosh(φ/2)
)(uv
)which tells us, in particular, how the eigenspinors u and v transform. That is, we know the forms of u(p)and v(p). The Dirac equation now separates into two equations, one for u and one for v:
(γµpµ −m)u(p) = 0(γµpµ +m)v(p) = 0
These equations of motion will come up again and again when computing cross sections of processes involvingon-shell fermions.
6.1.3 Vector Fields: Maxwell and Proca Equations
We’re now interested in looking at vector fields. Lorentz 4-vectors, we will see shortly, correspond to the( 12 ,
12 ) representation of SO+(1, 3). We’re led to guess this since the vector space this irrep acts on is of
dimension four, and the behavior under purely spatial rotations is 2 ⊗ 2 = 1 ⊕ 3. That is, it separates
38
into the trivial and three-dimensional irreps of SU(2), corresponding to the time and spatial coordinates,respectively. Since ( 1
2 ,12 ) = ( 1
2 , 0) ⊗ (0, 12 ), we take our 4-vector to be ξ ⊗ η, which then transforms under
the Lorentz group asexp[i(~θ − i~φ ) · 1
2~σ]⊗ exp
[i(~θ + i~φ ) · 1
2~σ]
If one computes the 4 × 4 matrices above, they will be in the chiral basis ξ(1) ⊗ η(1), ξ(1) ⊗ η(2), ξ(2) ⊗η(1), ξ(2) ⊗ η(2), and not in the usual basis. We want to figure out the change of basis that recasts thesematrices into more frequently used coordinates. We do a Clebsch-Gordon decomposition:
|0, 0〉 = 1√2
(|↑↓〉 − |↓↑〉
)−→ 1√
2
(ξ(1) ⊗ η(2) − ξ(2) ⊗ η(1)
)|1, 1〉 = |↑↑〉 −→ ξ(1) ⊗ η(1)
|1, 0〉 = 1√2
(|↑↓〉+ |↓↑〉
)−→ 1√
2
(ξ(1) ⊗ η(2) + ξ(2) ⊗ η(1)
)|1,−1〉 = |↓↓〉 −→ ξ(2) ⊗ η(2)
The time coordinate is the one that remains invariant under spatial rotations, which means it’s the anti-symmetric singlet. The rest form the spatial vector in the spherical basis. We conjugate the above Lorentzgroup elements by the unitary transformation described by
S =1√2
0 1 −1 0√2 0 0 0
0 1 1 00 0 0
√2
and get the spherical representation of all the boosts and rotations. That is, we get Jz, J± and the corre-sponding boost operators. We can change to the Cartesian basis using
U =1√2
0 1 −1 0−1 0 0 1−i 0 0 −i0 1 1 0
instead of S. This gives the familiar expressions for rotating and boosting a Lorentz 4-vector in the (t, x, y, z)basis.
6.1.4 Relating Spinors to Tensors
Define the matrix-valued 4-vector σµ = (I, ~σ). From this, we can write down the soldering quantities
σµAB := 1√
2(σµ)AB
σµ
AB:= 1√
2(σ∗µ)AB
We are using here the old dotted and undotted notation for spinors, sometimes called van der Waerdennotation. To review, the dotted spinors are right-handed, and undotted are left-handed. Spinorial indicesare denoted with latin letters and are raised and lowered using the ε-symbol.
ΦA = εABΦB
We now note a few identities following from the above definitions.
σµ
ABσAB
ν = δµν
σµ
AXσBY
µ = δBAδ
YX
σµ
AXσν
BX + σν
AXσµ
BX = εABη
µν
σAXµ σνBX = 1
2ηµνδAC − i
2εµνρτσAX
ρ στBX
39
We now have a way to “translate” between tensors and spinors. One might convert a tensor index into twospinor indices by contracting the tensor index with that of the appropriate soldering term, or go the otherway by with the two spinorial indices.
XAB = xµσABµ ←→ xµ = XABσµ
AB
In this language, the Weyl equations become
∂ABηA = 0
∂ABξB = 0
where we have defined ∂AB := σµ
AB∂µ. Note then that ∂AB∂
CB = 12δ
CA.
6.1.5 Amplitudes and Cross-Sections
6.1.6 The Optical Theorem
6.2 The Standard Model - Updated 7/1/07
Adapted from M. Srednicki, Quantum Field Theory
6.2.1 Field Content
The Standard Model can be described as three identical copies (generations, that is, flavors) of left-handedWeyl spinor fields charged under su(3)⊕ su(2)⊕ u(1) (and thus, up to the factors in the center of the group,has gauge group SU(3)× SU(2)× U(1)) as follows:
Field charge Description Electric charge (Q = T3 + Y )
(1, 2,−1/2) Neutrino-electron electroweak doublet ±1/2− 1/2 = 0−1
(1, 1,+1) Antiparticles to right-handed electron singlet 0 + 1 = 1(3, 2,+1/6) Quark electroweak doublet ±1/2 + 1/6 = 2/3
−1/3
(3, 1,−2/3) Antiparticles to right-handed up-quarks 0− 2/3 = −2/3(3, 1,+1/3) Antiparticles to right-handed down-quarks 0 + 1/3 = 1/3
along with a complex scalar field transforming as (1, 2,−1/2) - the Higgs doublet. (The antiparticles ofright-handed particles are used so that all particles mentioned are left-handed. Ultimately we must completethe theory to include the antiparticles of all of the particles mentioned.)
Note that remembering the relationship between electric charge Q, hypercharge Y and z-component ofsu(2) is enough to remember the hypercharges given the well-known quark electric charges (2/3,−1/3 forthe left-handed up, down quarks respectively).
We will include all possible interactions that are renormalizable (operators with mass dimension ≤ 4).
6.2.2 Free Parameters
There are 19 (real) free parameters in the Standard Model:
Parameter Variables in Lagrangian NumberCharged lepton masses λa 3
Quark masses mI ,m′I 6
Coupling constants gi 3CKM Matrix 4Theta angle θ 1
Of course, this doesn’t account for neutrino masses, which are likely going to require an addition in thefuture . . .
40
6.2.3 Higgs sector
The Higgs sector of the Standard Model is then like scalar QED:
LHiggs = −|Dµφ|2 − V (φ)
for Dµφ = ∂µφ − i(g2Aµ + g1B)φ and V (φ) = 14λ(φ†φ − 1
2v2)2. We can choose τa = 1
2σa, Y = − 1
2I asgenerators for the su(2)⊕u(1). The SU(2)×U(1) is broken to the U(1) of EM at low energies by a non-zeroVEV for φ, which by global SU(2) rotation can be taken as
φ =1√2
[v0
]The part of the Higgs Lagrangian quadratic in gauge fields can be taken to find the new particle masses(with σ± = 1√
2(σ1 ± iσ2),W± = 1√
2(A1 ∓ iA2)):
−LHiggs →⟨(g2(W+τ+ +W−τ− +A3τ3) + g1Y B)2
⟩+ . . .
=(vg2
2
)2
W+ ·W− +v2
8[W 3 B
] [ g22 −g1g2
−g1g2 g21
] [W 3
B
]as 〈σ+σ−〉 = 〈1〉 =
⟨σ3⟩
= 12v
2 and all other relevant expectation values are zero. There is one zeroeigenvalue of the matrix at right, which corresponds to the gauge boson for QED, and the other eigenvalueis 1
2m2Z = 1
8v2(g2
1 + g22). We can find an eigenvector of unit magnitude parameterized in terms of an angle;
then the zero-mass eigenvector is [g22 −g1g2
−g1g2 g21
] [sin(θW )cos(θW )
]= 0
⇒ tan(θW ) =g1g2
Define [ZA
]=[cos(θW ) − sin(θW )sin(θW ) cos(θW )
] [W 3
B
]Then Aµ is the massless eigenstate. θW is called the Weinberg angle, which determines the ratio of U(1)and SU(2) coupling constants as well as the ratio of W and Z boson masses:
MW =vg22
MZ =v
2
√g21 + g2
2 =v
2
√1 + tan2(θW )g2
MW
MZ= cos(θW )
The Higgs v.e.v. can be used to generate mass terms for the leptons through Yukawa couplings, whichmust be adjusted to give the proper masses to the leptons.
6.2.4 Lepton Sector
We will deal with one generation for now; the others work in exactly the same way.Other than the masses acquired via the Higgs mechanism, the other main aspect of the lepton sector we
have left to express is the interactions that arise via the covariant derivatives of lepton fields. First rewrite
41
the gauge fields A3, B in terms of A,Z:
g2A3µT
3 + g1BµY = g2(cWZ + sWA)T 3 + g1(−sWZ + cWA)Y
= g2sW (T 3 +g1g2
cot(θW )Y )A− g1sW (Y − g2g1
cot(θW )T 3)Z
= (g2sW )QA− (g2sW ) tan(θW )(Q− (1 + cot(θW )2)T 3)
= eQA+e
sW cW(T 3 − s2WQ)Z
for e ≡ g2sW the EM coupling constant (here defined as a positive number, that is, the electron has charge−e) and sW , cW sine and cosine of the Weinberg angle. Then lepton interactions are easy to write; theleft-handed kinetic term provides currents for W±, Z, and A, while the right-handed kinetic term providesanother part of the Z,A currents.
L = ψLiD/ψL + ψRiD/ψR
→ g2√2
(W+
µ J−,µ +W−
µ J+,µ)
+e
sW cWZµJ
µZ + eAµJ
µEM
for
J−,µ = ψLτ−ψL = νγµeL
J+,µ = ψLτ+ψL = eLγ
µν
JµZ = ψ(τ3PL − s2WQ)ψ =
12(eLγ
µeL − νγµν)− s2WJµEM
JµEM = −ψγµψ
For low energies relative to the W , Z boson masses we can integrate out their effects to produce
Zµ →e
sW cWM2Z
JµZ
W± → g2√2M2
W
J±
−Leff(including mass part)→ g22
2M2W
J+µ J
−,µ +g22
2M2Zs
2W c2W
J2Z
=e2
2s2WM2W
(J+
µ J−,µ + J2
Z
)≡ 2√
2GF
(J+
µ J−,µ + J2
Z
)for Fermi constant GF defined as e2
4√
2s2W M2
W
.
6.2.5 QCD Sector
The form of the interactions in the QCD sector are much like those of the Lepton sector - covariant derivativeslead to trilinear interactions with gluons, W , Z bosons, and the photon. The main unique feature here is thatthe analog of the charged currents in the quark sector provide an extra flavor mixing which results from thepotential difference between the gauge and mass eigenstates between the u and d mass terms (suppressingelectroweak indices):
Lquark mass = −yIJuRIφ qLJ − y′IJdRIφ qLJ + h.c.
−−−−−−→Higgs vev
−mIJuRIuLJ −m′IJdRIuLJ + h.c.
→∑
I
−mIuRIuLI −m′IdRIdLI + h.c.
42
wherein the last step - the diagonalization of the mass matrices - was accomplished by sending uI → UIJuJ
and dI → DIJdJ for unitary matrices U,D in flavor space. In neutral currents the effect cancels, but incharged currents there is an effect. The quark electroweak interactions are analogous to the lepton ones,except now
J−,µ = ψLτ−ψL = dLIγ
µuLI
→ dLIγµD†
IKUKJuLJ
≡ dLIγµV †
IJuLJ
J+,µ → uLIγµVIJdLJ
with neutral currents of the same form as the lepton currents except involving quark spinors. The matrixV is the quark mixing matrix, also known as the Cabibo-Kobayashi-Maskawa (CKM) matrix. This unitary3×3 matrix can be described by 9 real parameters. After taking into account the fact that the overall phaseof any of the 6 quarks is meaningless, though a common rephasing amounst all the quarks does not effectV , we see that 5 of the parameters can be removed, leaving V determined by 4 parameters - three angleswhich control quark mixing and one phase which controls CP violation.
6.3 Anomalies
6.4 Spinors
6.5 Conformal Field Theory (?)
7 String Theory
Adapted from M. B. Green, J. H. Schwarz, and E. Witten. Superstring Theory, Volumes I and II
7.1 Conventions
7.2 Classical Strings
7.2.1 Polyakov Action - Motivation
String theory is (or originally was) a theory of extended one-dimensional objects. We want a (at first,globally) Lorentz-invariant way of embedding the world sheets of strings into spacetime. For inspiration, welook at how this works in 1 +D dimensional special relativity. A covariant Lagrangian that reproduces theequations of special relativity for massive particles is
S = −m∫ds
= −m∫dτ√−X2
= −m∫dτ
√X2 − (X0)2
where s is the proper length of the particle worldline (and thus the action is manifestly Lorentz-invariant),X(τ) is any parameterization of the particle worldline. Variation of this action gives equation of motion
d
dτ
(X√−X2
)= 0
⇒ X√−X2
= C
43
The Lagrangian was written in a way in the first line that is manifestly independent of the parameter chosen,and the form in latter lines can be seen to be independent of parameter by explicit substitution. If we selectτ as the proper time, then the factor inside the square root is 1, and X becomes the proper-velocity of theparticle. Then the equation of motion tells us that the four-velocity is constant, as expected.
We want to generalize this to massless particles and furthermore to put this action in a form wherequantization might be easier (as the square-root part will be difficult). Thus we introduce an einbein e(τ):
S′ =∫dτ(e−1X2 + em2
)(1)
Here e is non-dynamical, and has equation of motion
−X2 + e2m2 = 0 (2)
which can then be used to recover the original Lagrangian by substitution. However, if we use this substitutedform, we still have (2) as a condition on our system. This is actually a gauge choice parameterized by e.Selecting e = 1
m is equivalent to selecting τ as the proper time. (1) is then in some sense equivalent to theoriginal Lagrangian. In fact, what we have done is introduce a gauge symmetry - a local reparameterizationsymmetry of the theory. If we give e transformation properties under reparameterization to τ ′(τ) such that
e(τ)dτ = e(τ ′)dτ ′
⇒ e(τ) = e(τ ′)dτ ′
dτ
then the latter Lagrangian is also invariant under reparameterization. Then we can actually use the latterLagrangian and maintain all of the symmetries of the original Lagrangian as long as we remember to imposethe condition (2).
This is the motivation to keep in mind as we develop the Polyakov action: We start with an actionmotivated by special relativity. Then we introduce a gauge invariance that we can fix to return to theoriginal Lagrangian. This new form of the Lagrangian is easier to quantize. However, as we will see lateron, we need to verify that the gauge invariance we have introduced is maintained at the quantum level orelse our theory ceases to make sense - that is, is inequivalent to the original theory we wanted to quantize!This is what ultimately introduces certain constraints in string theory on the dimensionality of spacetime(D = 26 in bosonic string theory and D = 10 in superstring theory) and on the conformal dimension ofphysical states.
7.2.2 Polyakov Action - Description
Now let’s do the same thing for an object with one spatially-extended direction. Instead of one coordinateto parameterize a worldline, we need two coordinates to parameterize a worldsheet. These two coordinatesare traditionally τ, σ, and there is a tendency to thing of τ as “time”, but it is important to remember thatthis is not the case unless one gauge-fixes X0 = τ .
The generalization proper time from special relativity is proper area. That is,The Polyakov action is
SP = −T2
∫d2σ√−hhαβ∂αX
µ∂βXµ
where T is the string tension and h has Lorentzian signature. This is also equal to
T =1
2πα′=
1πl2
for α′ the Regge slope and l = 2α′ the string length. α′ is the unique dimensionful parameter of string theory(it is determined from G, ~, c) and has units of (length)2. Many formulas simplify with α′ = 1
2 , and we willuse this convention unless otherwise specified.
44
Varying this with respect to hαβ we get
δSP = − 12π
∫d2σδhαβ
√−h(∂αX
µ∂βXµ − 1
2hαβh
γδ∂γXµ∂δXµ
)which, in fact, is how one calculates the stress-energy tensor for such a theory:
Tαβ ≡ −2T
1√−h
∂S
∂hαβ
= ∂αXµ∂βX
µ − 12hαβh
γδ∂γXµ∂δXµ (3)
(This is one definition of the stress-energy tensor that is equivalent to other more conventional methods ofdefining it in a wide-variety of cases. This is the one most relevant to general relativity. The constant outfront is conventional for string theory.)
Thus, we need the condition Tαβ = 0 in order for our theory to be sensible. This gives us (by taking thequare root of the determinant of both sides of (3))
√−G =
12hγδGγδ
√−h
which implies that the Polyakov action with the constraint Tαβ = 0 is equivalent to the Nambu-Goto action.
7.2.3 Symmetries of the Lagrangian
The symmetries present in a choice of h are the most important to state as we need eventually to gauge fixaway all of the freedom introduced in h.
The Lagrangian is invariant under an infinitesimal reparameterization σ′ = σ + ξ:
δXµ = ξα∂αXµ
δhαβ = ξγ∂γhαβ − ∂(αξγhβ)γ
and also under a Weyl scaling of the worldsheet metric
δhαβ = Λhαβ
We can use these two transformations to bring the worldsheet metric to conformal gauge, but this does notfix some of the gauge invariance as a combination of these two can result in no change to the metric. For aflat worldsheet this would be a combination such that
Ληαβ = ∂(αξβ)
In complex coordinates, the important constraints are
∂ξ = ∂ξ = 0
The reparameterization symmetry algebra is that of the conformal group in two dimensions, which is infinite-dimensional. We characterize these transformations by
z → z + ξ = z + anzn
z → z + ξ = z + anzn
⇒ Generators are ln ≡ zn+1∂, ln ≡ zn+1∂
[lm, ln] = [zm+1∂, zn+1∂]
= (m− n)zm+n+1∂ = (m− n)lm+n
[lm, ln] = 0, [lm, ln] = (m− n)lm+n
45
The ln, ln form the Virasoro algebra, and l−1, l0, l1 and their barred counterparts generate the group ofglobally defined conformal transformations (through exponentiation), while the remaining part of the algebragenerates local transformations.
It’s useful to mention a few other symmetries and their conserved currents:
Symmetry Conserved CurrentWorldsheet translations T0β from beforeSpacetime translations Pµ
α = 1π∂αX
µ
Spacetime Lorentz transformations Mµνα = 1
π (Xµ∂αXν −Xν∂αX
µ)
7.2.4 Classical equations of Motion
One solution to the Tαβ = 0 constraint is to choose hαβ = eφηαβ . This is called conformal gauge, and is oneof the most useful forms for the worldsheet metric. In fact, any worldsheet metric can be brought to thisform for appropriate φ so that the Gauss-Bonnet formula is satisfied. In this gauge the Lagrangian reducesto
LP,CG = − 12π
(∂αXµ∂αXµ) = − 1
2π
(−X2 +X ′2
)⇒ Xµ =
(− ∂2
∂τ2+
∂2
∂σ2
)Xµ = 0
with subsidiary conditions
T00, T11 = 0→ X2 +X ′2 = 0
T01 = 0→ X ·X ′ = 0
⇔(X ±X ′)2 = 0
7.3 Three Roads to Quantization
Since a number of our methods in this section and later will involve mode expansions of the wave equation,we do this here. There are different mode expansions depending on whether one has open or closed strings,and for open strings, depending on the boundary conditions for the string. However, since we’re not goingto be talking about strings connecting to objects (D-branes), the most important expansions for us are forclosed strings and Neumann (X ′(σ = 0) = X ′(σ = π) = 0) boundary conditions on open strings.
For closed strings (Xµ(σ = 0) = Xµ(σ = π)), we have left-moving and right-moving standing waves. Infact the most general solution of Xµ = 0 is a linear combination of left and right-moving solutions:
Xµ = xµ + 2α′pµτ + i√
2α′∑n 6=0
1n
(αµ
ne−2in(τ−σ) + αµ
ne−2in(τ+σ)
)Xµ
R,L =12xµ + α′pµ(τ ∓ σ) + i
√α′
2
∑n 6=0
1n
([αµ
n, αµn] e−2in(τ∓σ)
)=
12xµ − i
2α′pµ log [z, z] + i
√α′
2
∑n 6=0
1n
([αµ
n, αµn] [z, z]−n
)=
12xµ − i
4pµ log [z, z] +
i
2
∑n 6=0
1n
([αµ
n, αµn] [z, z]−n
)[∂XR, ∂XL] = − i
2
∑n
[αµn, α
µn][z−n−1, z−n−1]
where the bracketed notation or the ± notation is used to indicate things that change for the right-movingversus left-moving expansion. Here we used z = e2i(τ−σ), z = z∗, which are exponentiated versions of complex
46
coordinates we have previously used (the exponentiation maps the string worldsheets to the). The modecoefficients, which will become operators, are xµ, αµ
0 = αµ0 ≡ 1
2pµ, αµ
n, αµn.
For the open string we have only one copy of oscillators
Xµ = xµ + 2α′pµτ + i√
2α′∑n 6=0
1nαµ
ne−inτ cos(nσ)
and hence α0 ≡ pµ.Mode-based quantization involves deriving commutation relations for the mode operators through pos-
tulating equal τ commutators:
[X(τ, σ), X(τ, σ′)] = −iπδ(σ − σ′), [X(τ, σ), X(τ, σ′)] = [X(τ, σ), X(τ, σ′)] = 0
This gives[αµ
m, ανn] = mδm+nη
µν
Then, using the form of the stress energy tensor we have
Lm =12
∑n
α−nαm+n
up to potential normal ordering ambiguities for m = 0. This ambiguity will be captured in the coefficient a,which will later be determined.
7.3.1 Old Covariant Quantization
In this somewhat ad-hoc method of quantization, we simply impose the Virasoro conditions that on physicalstates Li − δi,0a = 0 for i ≥ 0.
7.3.2 Light-cone Quantization
7.3.3 Modern Covariant Quantization (with BRST)
L0 =12α2
0 +∞∑
n=1
αµ−nαnµ
≡ 12α2
0 +N
and for the closed string there is an equivalent result and definition for L0, N . This gives that for physicalstates of open and closed strings we have
Open : a =12p2 +N ⇒ m2 = −2a+ 2N
Closed : a =12
(12p
)2
+N ⇒ m2 = −8a+ 8N
7.4 The Two Superstring Formalisms
8 Differential Geometry
8.1 Pushforwards and Pullbacks
For more information, see Nakahara or Gockeler and SchuckerGiven a smooth map f : M → N from a manifold M to another N , we seek an induced linear mapping
47
Figure 8: A depiction of a pushforward of a vector.
f∗ : TpM → Tf(p)N that takes a vector V ∈ TpM and sends it to a vector f∗v in Tf(p)N . This is known asthe pushforward7 of the vector v under the map f . A vector is defined as an element in the tangent spaceat a point p ∈M equal to the derivative of a curve c : R→M with respect to its argument. That is,
v =dc
dt
∣∣∣∣p
for some c(t). Now, the curve c(t) gets mapped under f into the manifold N to give another curve f c(t),which can then be differentiated in t at the point f(p). The resulting vector is the pushforward f∗v as it isdefined. This procedure is illustrated in Fig. (8).
This geometric picture is fine conceptually, but to do computations one might want to work this definitioninto something more practical. Let’s take a coordinate patch U ⊂M that contains the point p and anotherpatch V ⊂ N that contains f(p). These coordinate patches are locally equivalent to Rm and Rn, respectively.
φ : U → Rm
ψ : V → Rn
are injective and smooth. We can think of these two maps as being an ordered set of their coordiantes. Thatis, φ = (xj) = (x1, . . . , xm) and ψ = (yi) = (y1, . . . , yn). We get well-defined mappings ψf φ−1 : Rm → Rn
that connect the coordinate systems on the two patches U and V . In particular, the ith component of thismap is
yi = yi f φ−1(x1, . . . , xm)
which will appear in our computation shortly.In the patch U , the components of the vector v are8
vj =d
dt
(xj c
)∣∣∣p
while the components in V of the vector f∗v are
(f∗v)i =d
dt
(ψ f c(t)
)∣∣∣f(p)
=d
dt
(yi f φ−1︸ ︷︷ ︸
yi(x)
φ c(t)︸ ︷︷ ︸x(t)
)∣∣∣f(p)
=∑
j
∂
∂xj
(yi(x)
) ddt
(xj c(t)
)∣∣∣f(p)
=∑
j
J ij v
j
7Other names for f∗ include the induced mapping, as already given, and the differential mapping of f .8What does it mean to say that the coordinate patch U on the manifold M gives us a basis on the tangent space at p?
Through the point p we can construct curves locally non-constant in only one of the basis directions at a time. We can do thisonce for each of the dim M directions at p. This spans the vector space TpM and is the basis I’m referring to.
48
Figure 9: We put coordinate patches around the relevant points.
where J is the Jacobian matrix of f : M → N . Note the middle step just involves the chain rule fromelementary calculus. It is obvious now how one would push forward a type-(p, 0) tensor.
Note that we do not get to map a covector in, say, T ∗pM , into a form in T ∗f(p)N . This is clear from thefact that the Jacobian matrix has an index i that pertains to our coordinates in N and one j that pertainsto our coordiantes in M . As a result, the geometrical object v ∈ TpM , which can exist independent of anymapping we may someday perform on it, can only have its index j contracted with the lower index of J .A covector, by contrast, would need to be contracted with the raised index. But this index deals with themanifold N . Instead, we go the other way. That is, for ω ∈ T ∗f(p)N , we have some f∗ω ∈ T ∗pM called thepullback of ω under f . The components of f∗ω are
(f∗ω)j =∑
i
J ijωi
A more geometrical motivation for, and the actual definition of, a pullback is that it is the mapping f∗ suchthat for any ω ∈ T ∗f(p)N and any v ∈ TpM , we have
〈f∗ω, v〉 = 〈ω, f∗v〉
where 〈 , 〉 is the inner product of a covector and a vector in the appropriate manifold. It is also straightforward to generalize to a pullback on a type-(0, q) tensor. Pullbacks of forms are relevant to many areas ofmath and physics.
8.2 Levi-Civita Done Right
One first encounters the Levi-Civita symbol εijk···m as that object in an n dimensional space for whichε123···n = +1 and each exchange of two indices causes a flip in sign, eg ε2134···n = −ε1234···n. That is, it’s thecompletely antisymmetric “tensor” with as many indices as our space has dimensions. We’ll see shortly thatit’s not truly a tensor. Let’s observe that on a Riemannian manifold (N, g) coordinatized by (x1, . . . , xn),
49
we can view ε as
ε :=1n!εabc···mdx
a ∧ dxb ∧ · · · ∧ dxm
= dx1 ∧ · · · ∧ dxn
How does ε transform under a change of coordinates φ : N → N : (x′i) 7→ (xi)?
ε′ := φ∗ε =∣∣∣∂x′∂x
∣∣∣εwhere |∂x′/∂x| is the determinant of the Jacobian matrix
J ij =
∂
∂x′jyφ∗dxi
This is not a tensorial transformation! We need to find a means to counter that determinant prefactor.Our Riemannian metric now comes to the rescue. In coordinates (xi), let the components of g be gij . TheRiemannian metric is invariant under coordinate changes, g′ := φ∗g = g. This means it’s components musttransform under coordinate changes as
g′ij =∂xm
∂x′i∂xn
∂x′jgmn
so the determinant of the metric transforms as
det g′ =∣∣∣ ∂x∂x′
∣∣∣2 det g
Definine ε := |det g|1/2ε. Then, ε′ := φ∗ε = ε, so our object is a true tensor. This is the volume form.
8.3 Frame Fields
8.3.1 Motivation and Definition
8.3.2 Trick for Quick Transformations
A technique that may be useful in determining the frame field in one coordinate system in terms of that ofanother given the transformation between the two coordinate systems. It’s so trivial I’m not sure I’ll keep ithere. A demonstration is more effective than a description. Consider the transformation between Cartesianand spherical polar coordinates: z = r cos θ, etc.
z = r cos θ~∇z = z
= r∂
∂r(r cos θ) +
θ
r
∂
∂θ(r cos θ)
= cos θ r − sin θ θ
and so on. In a simple case like this transformation, a picture and some simple geometric reasoning inone’s head might be faster, but this method is not too difficult to apply even in general situations, such astranforming to/from ellipsoidal coordinates, etc.
8.4 The Cartan Calculus
There are three fundamental operations that can be performed on form fields on a manifold: the exteriorderivative d, the contraction operator with respect to vector field X, iX , and the Lie derivative with respectto vector field X, LX . The goal of this subsection is to specify these operations and give formulas linkingthem. The Lie derivative can be further extended to general tensors.
50
8.4.1 The Exterior Derivative
This operation should be familiar, but its axiomization is useful: The exterior derivative is the unique linearmapping that increases form degree by one and satisfies:
1. d2 = 0
2. iXdf ≡ df(X) = X(f) for function (0-form) f .
3. d(α ∧ β) = dα ∧ β + (−1)kα ∧ dβ for k-form α.
8.4.2 The Contraction
We can define the contraction as an operation from n+ 1-forms to n forms via
(iXω)(V1, . . . , Vn) = ω(X,V1, . . . , Vn)
or axiomize via assuming linearity and that
1. iXω = ω(X) for one forms ω.
2. iX(α ∧ β) = (iXα) ∧ β + (−1)klα ∧ (iXβ) for k-,l-forms α, β.
8.4.3 The Lie Derivative
The Lie derivative is a way of examining the change in a form that is “more primitive” than the covariantderivative in the sense that it does not require a metric. It is useful to motivate the Lie derivative in termsof flows: For any vector field V in a neighborhood of point p we can define an a family of diffeomorphismsφV,λ for an interval around zero and a neighborhood around p by
φV,λ(x) = exp(λV )x
which is uniquely characterized by the differential equation
d
dλφV,λ(x) = V (φV,λ(x))
The Lie derivative on forms is then defined by:
LV ω =d
dλ(φV,λ)∗(ω(φV,λ(x)))
∣∣∣λ=0
The Lie derivative on forms can be axiomized by:
1. LXf = Xf for function f
2. (LXω)(V ) = LX(ω(V )) + ω(LXV ) = X(ω(V )) + ω([X,V ]) for one-form ω
(Though we haven’t defined LXV yet, but take we can take [X,V ] as the definition for now and willsee that this is the definition motivated by the flow method later.)
3. LX(α ∧ β) = (LXα) ∧ β + α ∧ (LXβ)
(This part will also be motivated by the general tensor extension of the Lie derivative.)
51
8.4.4 Formulas in Index Notation
Take ω to be a p-form.We will indicate forms in index notation via the notation:
ω = ωα1...αpdxα1 ⊗ · · · ⊗ dxαp
= ω[α1...αp]dxα1 ∧ · · · ∧ dxαp
where the antisymmetrization indicated by brackets includes a 1p! . (We can always choose to define the ω’s
such that ωα1...αp = ω[α1...αp].)The formula for d is provable from the axiomization above as
(dω)α1...αp+1 = ∂[α1ωα2...αp+1]
For the contraction we get(iXω)α1...αp−1 = p Xµωµα1...αp−1
For the Lie derivative, first examine the one-form case:
LV ω =d
dλ(φV,λ)∗(ω(φV,λ(x)))
∣∣∣λ=0
=d
dλ(φV,λ)∗(ωµdx
µ∣∣∣φV,λ(x)
)∣∣∣λ=0
=d
dλωµ(φV,λ(x))[d(φV,λ(x))]µ
∣∣∣λ=0
=1λ
(ωµ(xα + λV α)(dxµ + λ∂νV
µdxν − ωµdxµ +O(λ2)
) ∣∣∣λ=0
= [V ν(∂νωµ) + ων(∂µVν)]dxµ
⇒ (LV ω)µ = V ν(∂νωµ) + ων(∂µVν)
Then, similarly, for the Lie derivative for higher forms we get
(LV ω)α1...αp = V α∂αωα1...αp + pωα[α2...αp∂α1]V
α
Using these formulas we can prove Cartan’s Magic Formula:
(iXdω)α1...αp = (p+ 1)Xα∂αω[α1...αp]
(diXω)α1...αp = p ∂[α1|(Xαωα|α2...αp])
= −pXα∂αω[α1...αp] + pωα[α2...αp∂α1]X
α
= −pXα∂αωα1...αp + pωα[α2...αp∂α1]X
α
⇒ (diX + iXd)ω = LXω
8.4.5 Summary of Invariant Formulas
1. d2(= d, d) = 0
2. d, iX = LX (Cartan’s Magic Formula)
3. [d, LX ] = 0
4. [LX , LY ] = [LX , LY ] = L[X,Y ]
5. [LX , iY ] = i[X,Y ]
52
6. iX , iY = 0
This is clear from the first of the two definitions, but can be seen from the second definition by inductionand linearity: The property is clearly true on two-forms of the form α∧ β for α, β one-forms (which issufficient by linearity for all two-forms) and thus is true for all n-forms by linearity on p-,1-forms α, β
iX iY (α ∧ β) = (iX iY α) ∧ β + (−1)p−1(iY α)β(X) + (−1)p(iXα)β(Y )
iX , iY (α ∧ β) = iX , iY (α) ∧ β + (−1)p−1 ((iY α)β(X)− (iY α)β(X)) + (−1)p ((iXα)β(Y )− (iXα)β(Y ))= 0
8.4.6 Extending the Lie Derivative
The Lie derivative can be extended to operate on vectors, and eventually general (m,n) tensors. Its extensionto vectors in terms of flows is
LV W =d
dλ(φV,−λ)∗(W (φV,λ))
∣∣∣λ=0
which leads to a formula in index notation:
LV W =d
dλ(φV,−λ)∗(Wµ∂µ
∣∣∣φV,λ(x)
)∣∣∣λ=0
=d
dλW ν(φV,λ(x))(∂µ(φV,−λ)ν)∂ν
∣∣∣λ=0
=1λ
(W ν(xµ + λV µ)(δµ
ν − λ∂νVµ)∂µ −Wµ∂µ +O(λ2)
) ∣∣∣λ=0
= (V ν∂νWµ −W ν∂νV
µ)∂µ
⇒ (LV W )µ = V ν(∂νWµ)−W ν(∂νV
µ) = [V,W ]µ
8.5 Killing vectors
A Killing vector is one for whichLV g = 0
We can formula this in index notation:
LV gµν = V ρ∂ρgµν + gρν(∂µVρ) + gµρ(∂νV
ρ)= ∂µ(gρνV
ρ) + ∂ν(gµρVρ)− V ρ(∂µgρν + ∂νgρµ − ∂ρgµν)
= ∂µ(gρνVρ) + ∂ν(gµρV
ρ)− 2V ρgρα C αµν
= ∇(µVν) = 0
8.6 The Gauss-Bonnet Theorem
8.6.1 Viewed as an Index Theorem
8.6.2 Example: a 1-Torus
The metric
[gij ] =[(a+ b cosφ)2 0
0 b2
]can be written in terms of frames as gij = ea
iebjδab if we take the frames to be
eθ = (a+ b cosφ)dθ
eφ = b dφ
53
We can determine the desired connection 1-forms by imposing the torsion free condition dea =∑
b eb ∧ ωa
b
to find:
deθ = −b sinφdφ ∧ dθ = eφ ∧ ωθφ
deφ = 0 = eθ ∧ ωφθ −→ ωφ
θ ∝ dθ
Combining these two equations givesωφ
θ = −ωθφ = sinφdθ
Of course ωθθ and ωφ
φ vanish by antisymmetry. We now compute the Riemann curvature Rab = dωa
b +∑c ω
ac ∧ ωc
b.Rθ
φ = dωθφ = − cosφdφ ∧ dθ = cosφdθ ∧ dφ
This is a form and the indices are still abstract indices, so we must convert to coordinate indices.
Rijkl =
∑a,b
eaieb
j(Rab)kl
which gives
Rθφθφ =
1a+ b cosφ
· b · cosφ =b cosφ
a+ b cosφ
This is enough, we’ve shown in class that the Riemann curvature tensor of a two dimensional space has onlyone independent component. To make the antisymmetry results valid, we lower the raised index
Rθφθφ = gθθRθφθφ = b cosφ (a+ b cosφ) = Rφθφθ , etc.
Hence,
Rφθφθ = gφφRφθφθ =
cosφb
(a+ b cosφ)
which allows us to compute the elements of the Ricci tensor, which is diagonal.
Rφφ = Rφφφφ +Rθ
φθφ =b cosφ
a+ b cosφ
Rθθ = Rφθφθ +Rθ
θθθ =cosφb
(a+ b cosφ)
Then, the Ricci scalar is
R = Rφφ +Rθ
θ = gφφRφφ + gθθRθθ =1b2
b cosφa+ b cosφ
+1
(a+ b cosφ)2cosφb
(a+ b cosφ)
=2 cosφb
1a+ b cosφ
Noting that√g = b (a+ b cosφ) we get
R√g = 2 cosφ
From here, it’s obvious that the given integral vanishes, since the∫ 2π
0dφ cosφ = 0. This is the specific
instance of Gauss-Bonnet on a particular metric on the torus, which has Euler characteristic χ = 0.
9 Algebraic Topology
The topics discussed in the Algebraic Topology subject will eventually cover the basics of homological algebra,its application to solving problems of interest, and also a discussion of the geometric (ie, simplicial, cellular,etc) view of algebraic topology as well.
54
9.1 Connected Sum of Genus-g Surface and a Crosscap
Given a closed, orientable genus-g Riemann surface, one can remove the interior of a disk. Call this resultingsurface A. Then, ∂A ∼= S1. Let B denote a Mobius strip, whose boundary is also homeomorphic to S1.Then, let f : ∂B → ∂A be a homeomorphism. We’re interested in the homology of the resulting surfaceA ∪f B, or just A ∪B for simplicity.The planar representation of the genus-g surface makes it clear that if we poke a single hole in the surface’s
Figure 10: The decomposition of our space into two subspaces A and B.
2-cell we can deformation retract the resulting surface A onto the wedge sum of 2g circles. But we havepoked such a hole in the set up of the problem. We poked it and opened it up slightly so that we could sewthe Mobius strip in its place. Additionally, it’s obvious that the Mobius strip deformation retracts onto acircle. The overlap A ∩ B of the two spaces is shown in the diagram as the shaded region. By design, it ishomeomorphic to S1 times an interval. We can now write down the Mayer-Vietoris sequence that will let uscompute the homology over any ring R. We’ll compute for both Z and Z2.
Figure 11: Standard planar representation of a genus-g closed, orientable Riemann surface.
· · · ∂ // Hk(A ∩B;R)i∗ // Hk(A;R)⊕Hk(B;R)
f∗ // Hk(A ∪B;R) ∂ // Hk−1(A ∩B;R)i∗ // · · ·
Since the homology groups of A are isomorphic to those of2g∨S1 and those of B to S1, we can compute most
of this sequence readily. We know that Hk(A;R) ⊕Hk(B;R) ∼= 0 and that Hk−1(A ∩ B;R) ∼= 0 for k ≥ 3,so we immediately get that Hk(A ∪B) is trivial for these k values. Now let’s first find H2(A ∩B;R).
· · · // H2(A;R)⊕H2(B;R)f∗ // H2(A ∪B;R) ∂ // H1(A ∩B;R)
i∗ // H1(A;R)⊕H1(B;R) // · · ·
where i∗ : x 7→ (2x,−x). Why is this? We use the First Isomorphism Theorem for groups on ∂ : H2(A∪B)→H1(A ∩B) in conjunction with exactness.
H2(A ∪B)/ ker ∂ ∼= im ∂ = ker i∗ = 0
55
Exactness gives us that ker ∂ = im f∗ = 0 is the trivial group, so that
H2(A ∪B;R) ∼= 0
Now on to H1(A ∪B;R).
· · · // H1(A;R)⊕H1(B;R)f∗ // H1(A ∪B;R) ∂ // H0(A ∩B;R)
i∗ // H0(A;R)⊕H0(B;R) // · · ·
with f∗ : (x, y) 7→ x+ 2y and i∗ : x 7→ (x,−x). The First Isomorphism Theorem on f∗ gives us
R⊕2g+1/ ker f∗ ∼= im f∗ = ker ∂
We can see that
ker f∗ =
2Z R = ZZ2 R = Z2
which lets us compute
ker ∂ =
Z2g ⊕ Z2 R = ZZ2g
2 R = Z2
56
