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Quantum Theory
PH3210
Dr Stephen West and Dr Andrew Ho
Department of Physics,
Royal Holloway, University of London,
Egham, Surrey,
TW20 0EX.
1
Contents
1 Books and Further reading 4
1.1 Books . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4
1.2 Other Resources . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4
2 Introduction to Wavefunctions 5
2.1 Schrodinger formulation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5
2.2 ψ(x, t) and Probability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7
2.3 Energy eigenstates and superposition . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12
2.4 Practice Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15
3 Operators and Observables 16
3.1 Hermitian Operators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17
3.2 Practice Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21
4 Dirac (Bra-Ket) notation 22
4.1 Recovering Wavefunctions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30
4.1.1 Manipulation with the momentum Operator . . . . . . . . . . . . . . . . . . . . 32
4.2 Rules for Amplitudes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33
5 Conserved Quantities 35
5.1 Complete sets of Quantum Numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42
5.2 The Uncertainty Principle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43
6 Simple Harmonic Oscillator (SHO) 48
6.1 Operator Method. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49
7 The Periodic Potential 54
2
7.1 Bloch Waves. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 54
8 Angular Momentum 57
8.1 Angular Momentum Eigenvalues and Eigenfunctions . . . . . . . . . . . . . . . . . . . . 58
8.2 Angular Momentum: A better method. . . . . . . . . . . . . . . . . . . . . . . . . . . . 59
8.3 Meaning of Angular Momentum Operator Results . . . . . . . . . . . . . . . . . . . . . 63
8.4 A close look at Spin . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 65
8.5 Stern-Gerlach experiment . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 68
8.6 Addition of Angular Momentum . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 72
9 Approximate Methods I 76
9.1 Time-independent perturbation theory . . . . . . . . . . . . . . . . . . . . . . . . . . . 76
9.2 Ex. 1: perturbed particle in a box . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 77
9.3 Ex. 2: Quadratic Stark effect . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 78
10 Approximate Methods II 79
10.1 Degenerate perturbation theory (time independent) . . . . . . . . . . . . . . . . . . . . 79
10.2 Linear Stark effect . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 80
11 Approximate Methods III 81
11.1 Effect of external magnetic field on the Hydrogen-like atom . . . . . . . . . . . . . . . . 81
11.1.1 Orbital angular momentum . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 81
11.1.2 Spin angular momentum . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 81
11.1.3 Relativistic effect: spin-orbit coupling . . . . . . . . . . . . . . . . . . . . . . . . 82
11.1.4 Strong field Zeeman effect . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 82
11.1.5 Weak field Zeeman effect . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 83
12 Approximate Methods IV 85
3
12.1 Variational Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 85
12.1.1 Variational principle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 85
12.1.2 Ex.1: 1D simple harmonic oscillator . . . . . . . . . . . . . . . . . . . . . . . . . 86
12.1.3 Ex.2: Ground state energy of Helium . . . . . . . . . . . . . . . . . . . . . . . . 87
13 Approximate Methods V 87
13.1 Time-dependent Perturbation Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . 88
13.2 Application to a special case: sinusoidal or constant in time perturbation . . . . . . . . 90
14 Revision on Hydrogen-like atom 91
14.1 Hydrogen-like atoms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 91
14.2 Angular (θ, φ) equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 92
14.3 Radial (r) equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 92
4
1 Books and Further reading
1.1 Books
• Bransden and Joachain: Quantum Mechanics, Longman, 2nd edition 2000. (530.12.BRA)†
• C Cohen-Tannoudji, B Diu, F Laloe, Quantum Mechanics, vols 1 and 2, John Wiley 1977. (530.12
COH)
• S Gasiorowicz, Quantum Mechanics, John Wiley, 2003. (530.12 Gas)
• F Mandl, Quantum Mechanics, John Wiley, 1992. (530.12 Man)
• Many other good books...
1.2 Other Resources
• Advanced but excellent notes/book can be found at
http://www-thphys.physics.ox.ac.uk/people/JamesBinney/QBhome.htm
†This is the main book for the course.
5
2 Introduction to Wavefunctions
2.1 Schrodinger formulation
• De Broglie postulated that we associate a wave with every particle.
λ =h
por λ =
~p
where ~ = h/2π and λ = λ/2π.
• Schrodinger introduced the wavefunction
“Wavefunction”: ψ(x, t)
and took ψ(x, t) to satisfy partial differential equation
Hψ(x, t) = i~∂ψ(x, t)
∂t“Schrodinger equation”
• H is the “Hamiltonian” - partial differential operator for the total energy
• For a particle moving in 1D
H =p2
2m+ V (x)
where m is the mass of the particle and V is the potential.
• In QM the momentum operator
p = −i~ ∂
∂x
• 1DSE then reads
− ~2
2m
∂2ψ
∂x2+ V (x)ψ = i~
∂ψ
∂t
6
• Solution using separation of variables
ψ(x, t) = φ(x)T (t)
so [− ~2
2m
∂2φ(x)
∂x2+ V (x)φ(x)
]T (t) =
[i~∂T (t)
∂t
]φ(x)
divide both sides by ψ = φT
1
φ
[− ~2
2m
∂2φ(x)
∂x2+ V (x)φ(x)
]=i~T
∂T (t)
∂t
• LHS is f(x) and RHS is a g(t). Only way they can be equal is if both equal to a constant (call it
E).
• We get two equations
− ~2
2m
∂2φ
∂x2+ V (x)φ = Eφ (2.1)
i~∂T (t)
∂t= ET (2.2)
• Eq.(2.2) easy to solve
T (t) = e−iEt/~ × constant.
• Eq.(2.1) is the “Time Independent Schrodinger Equation— or TISE, can be hard to solve,
depends on V .
• Choose easy example - free particle, V (x) = 0. TISE becomes
− ~2
2m
∂2φ
∂x2= Eφ
with solutions
φ ∼ eikx, e−ikx
where
~2k2
2m= E
7
• Putting it all together
ψ(x, t) ∼ ei(kx−Et/~) Wave moving in +ve x direction
ψ(x, t) ∼ e−i(kx+Et/~) Wave moving in −ve x direction
• So we find the angular frequency ω = E/~
E = ~ω = hν
where ν is frequency and we see that the solutions automatically incorporate Einstein relation.
• Also for classical free particle E = p2/2m, so we get
E =~2k2
2m=
p2
2m
→ p = ~k = ~2π
λ=h
λ
automatically incorporates De Broglie “matter wave” relation.
But what does ψ(x, t) mean?
• ψ(x, t) cannot be a physical wave like an oscillating string, or EM wave of Maxwell theory
• TDSE is complex eqn, solutions are inherently complex (real and imag parts of ψ do not separately
solve eqn)
• Something that is complex cannot be directly measured.
• So what is this wavefunction?
2.2 ψ(x, t) and Probability
• To understand ψ(x, t) must introduce
Probablility Density: P (x, t)
• Basic idea is that can no longer be certain of exact position of particle
8
• Only that Probability of measuring it between x and x+ δx is
Prob(in x to x+ δx) = P (x, t)δx
and a postulate of QM is that
P (x, t) = ψ(x, t)∗ ψ(x, t) = |ψ(x, t)|2
• We can use this to find averages or expectation values
〈x〉 ≡∫ ∞−∞
xP (x, t)dx =
∫ ∞−∞
x |ψ(x, t)|2 dx
similarly
⟨x2⟩ ≡ ∫ ∞
−∞x2P (x, t)dx =
∫ ∞−∞
x2 |ψ(x, t)|2 dx
• Total probability of finding particle anywhere must be 1, so
1 =
∫ ∞−∞
P (x, t)dx =
∫ ∞−∞|ψ(x, t)|2 dx
which the normalisation condition on ψ(x, t).
• This must be imposed on ψ(x, t) as SE is linear in ψ.
• If ψ is a solution then so is ψ×constant
• Constant is then fixed by the normalisation condition
• Complication for plane waves
ψ(x, t) = Gei~ (px−Et)
so ∫ ∞−∞|ψ(x, t)|2 dx =
∫ ∞−∞|G|2 dx→∞
⇒ Plane wave is not normalisable!
9
• Reason: Plane waves are not physically realisable.
They exist for all time and are spread over all space
• More physical situation is a “plane wave” confined to some finite region, (e.g. experimental
apparatus), then ∫ L
−L|G|2 dx = 1
⇒ G =1√2L
Look at an old example, the infinite square well
• Solving the Schrodinger equation for a number of examples should be familiar from PH2210
• Quick look at the infinite square well, with potential as depicted in Fig 2.1
V = 0
0 a x
V =∞ V =∞
IIIIII
Figure 2.1: Infinite Square Well
• Since V 6= v(t) solutions of form
ψ(x, t) = φ(x)e−itE/~
where
− ~2
2m
∂2φ
∂x2+ V (x)φ = Eφ
10
• In regions where V =∞, φ must be zero for finite E solution (as K.E ≥ 0)
• Inside well
− ~2
2m
∂2φ
∂x2= Eφ as V=0
so
φ = Aeikx +Be−ikx with~2k2
2m= E
• Now have to match inside to outside solutions via boundary conditions
φ must be continuous
At x = 0, 0 = A+B
At x = a, 0 = Aeika +Be−ika = 2iA sin ka
• Either A = 0 (reject this) or
sin ka = 0, ⇒ k =nπ
a, n = 1, 2, 3...
• Substitute back in to expression for energy
E =~2
2m
π2
a2n2, n = 1, 2, 3...
• Possible energies of particle are quantised
• Typical for particles bound in a potential well (atoms, ...)
• Also have to normalise solution
φn(x) = A sinnπx
a
• so we have
1 =
∫ ∞−∞|φn(x)|2 dx = A2
∫ a
0
sin2 nπx
adx = A2 a
2
⇒ A =
√2
a
11
• Putting it all together the spacial part of the wavefunction reads
φn(x) =
√2
asin
nπx
a
• Note that φn are orthogonal∫ ∞−∞
φ∗n(x)φm(x)dx =2
a
∫ a
0
sinnπx
asin
mπx
adx
=
1 : n = m
0 : n 6= m
n = 1
n = 2
n = 3
n = 4
n = 5
E = 4E1
E = 9E1
E = 16E1
E = 25E1
0 a
∞ ∞
E =2π2
2ma2≡ E1
Figure 2.2: Wavefunctions for Infinite Square Well
• From Fig. 2.2 we see that the ground state, n = 1 has no nodes
12
• First excited state, n = 2 has one node
• It should be clear that the physics of low n states is very different from the classical expectation
• Classically: consider a ball bouncing back and forward in the well
P (x) =1
a
just a uniform distribution. So
〈x〉 =
∫ a
0
xP (x) dx =1
a
∫ a
0
x dx =a
2⟨x2⟩
=
∫ a
0
x2P (x) dx =a2
3
• But quantum
〈x〉1 =a
2same⟨
x2⟩
1=
a2
3− a2
2π2different from classical
where the subscript 1 means we are using the n = 1 state.
• As n→∞ find that 〈x2〉 → classical result The probability distribution becomes more spread out
• This is an example of the general behaviour
2.3 Energy eigenstates and superposition
• Wavefunction of the form
ψn(x, t) = φn(x) e−iEnt/~
is special, it is an Energy Eigenstate or Stationary State (same thing)
• Equivalently:
Hψn(x, t) =
[− ~2
2m
∂2φ(x)
∂x2+ V (x)φ(x)
]ψn(x, t)
= i~∂
∂tψn(x, t) = Enψn(x, t) by TDSE
13
So ψn(x, t) is an Eigenfunction of H with Eigenvalue En
• Physically:
Energy Eigenstate ⇔ Particle is in state of definite energy (here En)
⇔ If energy is measured, guaranteed result En
Superpositions
ψn(x, t) = φn(x)e−iEnt/~
is not the most general solution to TDSE.
• Since TDSE is linear equation, general solution is sum of above
ψ(x, t) =∑n
anφn(x)e−iEnt/~
where an are arbitrary complex coefficients.
Example: Superposition of 2 energy eigenstates
ψ(x, t) = a1φ1(x)e−iE1t/~ + a2φ2(x)e−iE2t/~
act with energy operator
Hψ(x, t) = a1E1φ1(x)e−iE1t/~ + a2E2φ2(x)e−iE2t/~ 6= const× ψ(x, t)
• But ψ is a solution of the TDSE, just not an energy eigenstate as E1 6= E2
• What is the physical significance of this?
• For ψ to be valid wavefunction it must be normalised
1 =
∫|ψ|2 dx =
∫ a
0
dx[a∗1φ1e
iE1t/~ + a∗2φ2eiE2t/~
] [a1φ1e
−iE1t/~ + a2φ2e−iE2t/~
]=
∫ a
0
dx[|a1|2 φ2
1 + |a2|2 φ22 + a1a
∗2φ1φ2e
i(E2−E1)t/~ + a∗1a2φ1φ2ei(E1−E2)t/~]
14
but φ1 and φ2 are orthogonal and are normalised so find
1 = |a1|2 + |a2|2
looks like a sum of probabilities, thus: interpretation of superposition
• If measurement made of energy of particle with this wavefunction will get the following result:
E1 with amplitude a1, probability |a1|2
E2 with amplitude a2, probability |a2|2
• The expectation value of energy should be⟨H⟩
= E1 |a1|2 + E2 |a2|2
we can check this by doing the calculation directly.
⟨H⟩
=
∫ a
0
dx ψ∗ H ψ
which is the definition of⟨H⟩
in state ψ
⟨H⟩
=
∫ a
0
dx[a∗1φ1e
iE1t/~ + a∗2φ2eiE2t/~
] [a1E1φ1e
−iE1t/~ + a2E2φ2e−iE2t/~
]= E1 |a1|2 + E2 |a2|2
using orthogonality of φs. This is the result we expected.
• Note that⟨H⟩
is time independent
• Not all expectation values are time independent for energy superpositions, e.g.
〈p〉 =
∫ a
0
dx ψ∗p ψ =
∫ a
0
dx ψ∗(−i~ ∂
∂xψ
)= −2i~
a
∫ a
0
dx
[a∗1 sin
πx
aeiE1t/~ + a∗2 sin
2πx
aeiE2t/~
] [a1π
acos
πx
ae−iE1t/~ + a2
2π
acos
2πx
ae−iE2t/~
]where we have used the spatial wavefunctions for the infinite square well as our example.
15
• Using the following useful identities∫ a
0
dx sinπx
acos
πx
a= 0 =
∫ a
0
dx sin2πx
acos
2πx
a
and for m 6= n
∫ a
0
dx sinmπx
acos
nπx
a=
am
π(m2 − n2)[1− (−1)m+n] =
m+ n 6= even
2am
π(m2 − n2)
(2.3)
we arrive at the result
〈p〉 = −i~8
3a
[a1a
∗2ei(E2−E1)t/~ − c.c.
]where c.c. is the complex conjugate expression. Taking a simplified case where the coefficients
are both real (a1, a2 ∈ R)
〈p〉 =16~3a
a1a2 sin(E2 − E1)t
~
Momentum oscillates in time!!!!
2.4 Practice Exercises
• (1) Show that in the case of the infinite square well presented in the notes
〈x〉1 =a
2⟨x2⟩
1=
a2
3− a2
2π2.
What are these values for the nth state? That is 〈x〉n and 〈x2〉n. Do these values get closer to the
classical values as we increase n? Is this what we expect?
• (2) Using Eq. (2.3) show that
〈p〉 =16~3a
a1a2 sin(E2 − E1)t
~
for the wavefunction ψ(x, t) = a1φ1(x)e−iE1t/~ + a2φ2(x)e−iE2t/~ where φn(x) =√
2a
sin nπxa
. (Try
and prove Eq. (2.3))
16
3 Operators and Observables
• Numerical values of classical quantities (e.g. Momentum and energy) become differential operators
in QM
x- momentum : p = −i~ ∂
∂x
energy : H =p2
2m+ V (x) = − ~2
2m
∂2
∂x2+ V (x)
• This is a general principle of QM
Every physical observable corresponds to an operator, Q
• Measured numbers in experiments are related to operators via eigenvalue equations
• In general we have
Q χn(x) = qn χn(x)
Eigenfunction χn(x) labelled by n
Eigenvalue qn(x) labelled by n
• The set of all qn is called the spectrum
• Example of energy
H φn = Enφn
• Eigenvalue En, Eigenfunction φn
• General principle of QM is
Measurements of a physical observable always give a
result which is one of the eigenvalues of corresponding operator Q
17
3.1 Hermitian Operators
• We know that experiments only return real measurements, this means that the class of QM
operators must have real eigenvalues
• This class of operator are called Hermitian operators
• So
Every operator representing a physical observable
must be a Hermitian operator
• Definition of Hermitian operator; for 1d QM an operator Q is Hermitian if∫ ∞−∞
χ∗ Q ψ dx =
∫ ∞−∞
(Q χ)∗ ψ dx
for any functions χ(x), ψ(x) which are normalisable and vanish at x = ±∞
What kinds of operators are Hermitian?
• i) The Position operator x∫ ∞−∞
χ∗ x ψ dx =
∫ ∞−∞
(x∗χ)∗ ψ dx =
∫ ∞−∞
(xχ)∗ ψ dx
so since x is a real number
x† = x
where the dagger is notation for Hermitian conjugate.
• ii) The potential energy V (x)∫ ∞−∞
χ∗ V (x) ψ dx =
∫ ∞−∞
(V (x)∗χ)∗ ψ dx
so as long as coefficients in V (x) are real we have
V (x)† = V (x)
which is what we expect as a complex potential does not make sense.
18
• iii) Momentum operator∫ ∞−∞
χ∗(−i~ ∂
∂x
)ψ dx = −i~
∫ ∞−∞
χ∗∂ψ
∂xdx
we need to get the operator acting on χ, so integrate by parts
RHS = −i~
[χ∗ ψ]∞−∞ −∫ ∞−∞
∂χ
∂x
∗ψ dx
Since x is real, so ∂
∂xis also real and
∂χ
∂x
∗=
(∂χ
∂x
)∗so
RHS =
∫ ∞−∞
(−i~∂χ
∂x
)∗ψ dx X
Thus p is hermitian. Note that the “i” in the definition of p is crucial
• iv) Kinetic energy operator
− ~2
2m
∂2
∂x2
is hermitian. Proof by integration by parts twice.
• v) Hamiltonian operator Since H = T + V and both T and V are Hermitian, H† = H also. Sums
of Hermitian operators are Hermitian.
Proof of the reality of eigenvalues
Let Qχn = qnχn, thus ∫ ∞−∞
χ∗mQχn dx =
∫ ∞−∞
χ∗mqnχn dx
but Q is hermitian so
LHS =
∫ ∞−∞
(Qχm)∗χn dx =
∫ ∞−∞
(qmχm)∗χn dx =
∫ ∞−∞
q∗mχ∗mχn dx
Putting the LHS side and the RHS of Eq. (3.1) together
0 = (qn − q∗m)
∫ ∞−∞
χ∗mχn dx
19
now chose m = n
0 = (qn − q∗n)
∫ ∞−∞|χn|2 dx = (qn − q∗n)
⇒ qn = q∗n qn is real
But there is more...choose n 6= m and use reality of qs
0 = (qn − qm)
∫ ∞−∞
χ∗mχndx
Thus χm and χn are orthogonal functions. This should be familiar from our infinite square well
example where the wavefunctions had the forms
φn =
√2
asin
πnx
a
We know from direct calculation that∫ ∞−∞
φ∗nφm dx = 0
• But yet more..
• For φms that have fourier series any function can be expanded as
f(x) =∞∑n=1
anφn(x)
This is true in general using eigenfunctions of Hermitian operators.
Theorem:
Let χn(x) be the eigenfunction of any hermitian operator
Then any normalisable function f(x) can be written as
f(x) =∞∑n=1
anχn(x)
20
• We say that the χn(x)s form a complete set of functions (or states).
• It is this expansion theorem that enables a connection between operators and probabilities
• Let’s see how...
• Suppose at a given time (choose t = 0) we have a wavefunction ψ(x, 0) and we measure Q
• Can expand ψ(x, 0) in eigenfunctions of Q
ψ(x, 0) =∑n
anχn(x)
finding ans is simple: multiply Eq. (3.1) by χ∗m(x) and integrate
RHS =
∫ ∞−∞
∑n
χ∗m an χn dx =∑n
∫ ∞−∞
an χ∗m χndx =
∑n
an δnm = am
LHS =
∫ ∞−∞
∑n
χ∗mψ(x, 0) dx
so
am =
∫ ∞−∞
∑n
χ∗mψ(x, 0) dx
Now substitute expanded ψ(x, 0) in equation for⟨Q⟩
⟨Q⟩ψ
=
∫(ψ(x, 0))∗ Q ψ(x, 0) dx =
∫ (∑n
anχn(x)
)∗Q
(∑m
amχm(x)
)
=
∫ (∑n
anχn(x)
)∗∑m
am qmχm(x)
=∑n,m
a∗namqm
∫χ∗n χm dx︸ ︷︷ ︸δmn
=∑n
|an|2 qn
Remember that |an|2 is the probability that result of a single measurement of Q gives qn.
qns are the possible results of single measurements of Q.
21
• Example: Back to the infinite square well. We know Hφn = Enφn with
φn =
√2
asin
πnx
aEn =
n2π2~2
2ma2
Note that these are not momentum eigenfunctions
pφn = −i~∂φn∂x
= −i~√
2
a
nπ
acos
πnx
a6= const × φn
• Easy to find the momentum eigenfunctions though
φpn =1√aeipnx/~ satisfies − i~∂φpn
∂x= pn φpn
Now write the energy eigenfunctions in terms of φpn
φn =
√2
a
1
2i
(einπx~/a~ − e−inπx~/a~)
=1
i√
2φpn −
1
i√
2φ−pn
where pn = nπ~a
.
• So if a particle is in energy eigenstate φn and we measure its momentum we find
+pn with probability =∣∣∣ 1i√
2
∣∣∣2 = 12
−pn with probability=∣∣∣ −1i√
2
∣∣∣2 = 12
3.2 Practice Exercises
• (1) Prove that the kinematic operator
− ~2
2m
∂2
∂x2
is Hermitian.
22
4 Dirac (Bra-Ket) notation
• More powerful and more general than wavefunctions
• Wavefunctions can be linearly superposed with complex coefficients
• Mathematical structure is a complex linear vector space
Basic object is “state ket” |ψ〉
which represents the state of the system
• A particular set of states are energy eigenstates
H |n〉 = En |n〉
• We also have complex conjugates of wavefunctions and eigenstates.
• Represent these by “state Bras”
〈ψ| or 〈n|
• The overlap of wavefunctions ∫dx φ(x)∗ ψ(x)
generalises to the unitary inner product for bras and kets
inner product = 〈φ|ψ〉
• It is clear from this generalisation that
〈φ|ψ〉 = 〈ψ|φ〉∗
• A Ket is normalised if
〈ψ|ψ〉 = 1
and |ψ〉 and |φ〉 are orthogonal if
〈ψ|φ〉 = 〈φ|ψ〉 = 0
23
• The complex vector space possessing the structure of an inner product, which describes the possible
states of the system is called the
Hilbert Space
of the system. Sometimes denoted by H.
• Like all linear spaces H has a dimension determined by maximal number of linearly independent
vectors
c1 |u1〉+ c2 |u2〉+ . . .+ cN |uN〉 = 0
satisfied only if c1 = c2 = . . . = cN = 0.
• A difference with vector spaces is that
Dimension of H is typically ∞
E.g. we know particle in 1D infinite square well has ∞ number of energy eigenstates
φn(x)↔ |n〉which are linearly independent.
• However, later we will meet very important physical systems where H is finite dimensional (spin,
...) and cannot be described by wavefunctions
• Existence of inner product 〈 | 〉 allows us to construct complete sets of orthonormal kets (a basis
for H) via Schmidt procedure.
Aside on Schmidt procedure
• Given a set of α linearly independent, normalisable functions (call them ψα) we can always con-
struct a new set of α mutually orthogonal functions (call them φα) via the Schmidt procedure.
• Start with the function ψ1 we write the following series
φ1 = ψ1
φ2 = a21φ1 + ψ2
φ3 = a31φ1 + α32φ2 + ψ3
...
φα = aα1φ1 + aα2φ2 + . . .+ aα,α−1φα−1 + ψα
24
We need to find the forms of the a coefficients. Starting with a21, we multiply φ2 by φ∗1 and
integrate over all space ∫φ∗1φ2dV = a21
∫φ∗1φ1dV +
∫φ∗1ψ2dV
but by definition the φ states are orthogonal and as φ1 = ψ1 and the ψ are normalised we are left
with
0 = a21 +
∫φ∗1ψ2dV
giving
a21 = −∫φ∗1ψ2dV.
This leads to the full expression for φ2 as
φ2 = −[∫
φ∗1ψ2dV
]φ1 + ψ2
which is easily shown to be orthogonal to φ1 as desired. Next, we wish to find the form for φ3.
To do so we first multiply the expression for φ3 above by φ1 and integrate over all space to get
0 = a31
∫φ∗1φ1dV + a32
∫φ∗1φ2dV +
∫φ∗1ψ3dV.
Again the φ are orthogonal and φ1 is normalised (the other φs are not) and we find
a31 = −∫φ∗1ψ3dV.
To find a32 we follow a similar procedure but multiply by φ∗2. In this case we have
a32 = −∫φ∗2ψ3dV∫φ∗2φ2dV
.
We can now write
φ3 = −[∫
φ∗1ψ3dV
]φ1 −
∫φ∗2ψ3dV∫φ∗2φ2dV
φ2 + ψ3
We repeat this procedure until we reach the final state φα which has the form
φα = −α−1∑i=1
∫φ∗iψαdV∫φ∗iφidV
+ ψα.
25
• We can redo this exercise using Dirac notation, write the series out again
|φ1〉 = |ψ1〉|φ2〉 = a21 |φ1〉+ |ψ2〉|φ3〉 = a31 |φ1〉+ a32 |φ2〉+ |ψ3〉
...
|φα〉 = aα1 |φ1〉+ aα2 |φ2〉+ . . .+ |ψα〉
Following the same procedure as above to find a21 we find in Dirac notation
〈φ1|φ2〉 = a21 〈φ1|φ1〉+ 〈φ1|ψ2〉 .
Due to orthogonality 〈φ1|φ2〉 = 0 and φ1 is already normalised we can rearrange and
a21 = −〈φ1|ψ2〉 .
then
|φ2〉 = − |φ1〉 〈φ1|ψ2〉+ |ψ2〉 .
Similarly we find
|φ3〉 = − |φ1〉 〈φ1|ψ3〉 − |φ2〉 〈φ2|ψ3〉〈φ2|φ2〉 + |ψ3〉 .
The final state is then given by
|φα〉 = −α−1∑i=1
|φi〉 〈φi|ψα〉〈φi|φi〉 + |ψα〉
To have an orthonormal set we need to normalise these states. This is simply done and the result
for the normalised states |Um〉 as
|Um〉 =|φm〉√〈φm|ψm〉
Now we have a complete set of orthonormal kets (a basis for H) with
|Um〉 with 〈Un|Um〉 = δnm
• If we have such a basis then any state |ψ〉 of system can be expanded
|ψ〉 =dim H∑n=1
cn |Un〉.
26
The cns determined by taking inner product of this equation with 〈Um|
〈Um|ψ〉 =dim H∑n=1
cn 〈Um|Un〉 = cm.
Thus the expansion for |ψ〉 is
|ψ〉 =dim H∑n=1
|Un〉 〈Un|ψ〉
• The bras and kets are related by taking the adjoint (complex conjugate transpose for usual vectors)
so
|ψ〉 =dim H∑n=1
cn |Un〉
(|ψ〉)† =
dim H∑n=1
cn |Un〉†
〈ψ| =dim H∑n=1
c∗n 〈Un|
The normalisation implies
1 = 〈ψ|ψ〉 =dim H∑n=1
c∗n 〈Un|dim H∑m=1
|Um〉 cm =∑n,m
c∗ncm 〈Un|Um〉 =∑n,m
c∗ncmδmn =∑n
|cn|2
• In QM also have (hermitian) operators which represent observables
• Key principle of QM is that all operators are linear, i.e. if
|ψ〉 = c1 |U1〉+ c2 |U2〉
then
O |ψ〉 = c1(O |U1〉) + c2(O |U2〉)
27
• One exception to this, the “time-reversal” operator T which is anti-linear - this is an advanced
(and interesting) topic...
• Operators also act on vector space of bras linearly (O can act “left” or “right”)
〈ψ| O = (c∗1 〈U1|+ c∗2 〈U2|)O = c∗1(〈U1| O) + c∗2(〈U2| O)
• Since bras and kets are exchanged by taking adjoint, useful to define action of adjoint on operators
Adjoint O† of operator O defined by
〈ψ| O† |φ〉 ≡ 〈φ| O |ψ〉∗
• Hermitian (or “self-adjoint”) operators satisfy
O† = O
• For hermitian ops (only)
〈ψ| O |φ〉 ≡ 〈φ| O |ψ〉∗
The quantity 〈φ| O |ψ〉 is called the
“matrix element” of O between states |φ〉 and |ψ〉
• If |Un〉 is an orthonormal basis then
〈Um| O |Un〉
is a matrix element in the conventional sense.
• A surprisingly useful operator secretly appears in the expansion of general |ψ〉
|ψ〉 =dim H∑n=1
|Un〉 〈Un|ψ〉
• Since this holds for any ket |ψ〉 we find
28
I =dim H∑n=1
|Un〉 〈Un|
where I is the “identity operator”
which leaves all states unchanged
I |φ〉 =dim H∑n=1
|Un〉 〈Un|φ〉 = |φ〉
by expansion theorem for |φ〉.
• The matrix element of I is (in some a basis)
〈Up| I |Uq〉 =dim H∑n=1
〈Up|Un〉 〈Un|Uq〉 =dim H∑n=1
δpnδnq = δpq
• Thus written explicitly as a matrix I is represented as
I ↔
1 0 0 0 . . .
0 1 0 0 . . .
0 0 1 0 . . .
0 0 0 1 . . ....
......
.... . .
︸ ︷︷ ︸dimH
dimH
The unit matrix.
• Another important example is the matrix element of a product of operators
〈Up| AB |Uq〉 = 〈Up| AIB |Uq〉︸ ︷︷ ︸can always insert I
=∑n
〈Up| A |Un〉 〈Un| B |Uq〉 =∑n
ApnBnq
just usual matrix multiplication of matrices representing A and B.
• Check that our earlier definition of O† agrees with adjoint of matrix
(O†)mn ≡ 〈Um| O† |Un〉 = 〈Un| O |Um〉∗ = (O)∗nm = (Omn)∗T
29
Summary so far...Fundamental Postulates of QM
• I) States of a system are represented by normalised kets |ψ〉 or bras 〈φ| in a Hilbert space (which
varies from system to system)
• II) Observables are represented by linear hermitian operators acting on kets and bras
• III) Such hermitian operators A are assumed to possess a complete set of orthonormal eigenstates,
e.g.
|1〉 with eigenvalue a1
|2〉 with eigenvalue a2
|3〉 with eigenvalue a3
......
...
“Complete set” means that |1〉, |2〉,... forms a basis for the space, so any |ψ〉 can be expanded in
eigenstates of A (and for any A)
|ψ〉 =∑n
|n〉 〈n|ψ〉
• IV) The fundamental probability postulate for measurement is
1) Possible results of measurement of A are eigenvalues of A only.
2) Prob(A = an) = |〈n|ψ〉|23) After measurement of A with result an the state |ψ〉 is reduced to (“collapsed to”)∣∣ψafter
⟩= |n〉
coefficient in front of |n〉 is changed to 1 where |n〉 is an eigenket of a with A |n〉 = an |n〉.
Subsequent measurements of same A on∣∣ψafter
⟩= |n〉 return an with probability =1. As long as
no measurements of other operators B are made at intermediate times...more later.
(Note: if an is a degenerate eigenvalue (i.e. if more than one linearly independent ket has same
an eigenvalues, the procedure of reduction is slightly more involved - advanced topic)
• V) In the absence of a measurement the time evolution of the |ψ(t)〉 describing the state of the
system at time t changes smoothly in time according to the TDSE
i~∂|ψ(t)〉∂t
= H |ψ(t)〉
30
Notes
1) The TDSE is a linear eqn, which is also deterministic, i.e. given the state at t = 0, |ψ(0)〉. The
state at a later time t is uniquely determined as long as no measurements are performed.
2) Thus the probabilistic, non-deterministic aspects of QM are purely due to the collapse of the
state upon measurement.
3) H is the Hamiltonian - the operator corresponding to the energy of the system
4) We can formally integrate the TDSE (H |ψ〉 = i~∂|ψ〉∂t
) from time = t0 to tf
|ψ(tf )〉 = e−iH(tf−t0)/~ |ψ(t0)〉
where the exponential is defined by its power series, e.g.
eO = 1 + O +O2
2!+ . . .
since H = H† (hermitian) the operator
U ≡ e−iH(tf−t0)/~
is unitary,
U †U = ei~H†(tf−t0)/~e−i~H(tf−t0)/~ = 1
so the time evolution is “unitary evolution”.
4.1 Recovering Wavefunctions
• How do we recover wavefunctions?
• Consider position operator x. This has a continuous spectrum of eigenvalues
x |x〉 = x |x〉
• The eigenkets |x〉 are normalised as
〈x|x′〉 = δ(x− x′)
δ(x − x′) is the analogue of 〈n|m〉 = δnm in the discrete case. Some useful identities with delta
functions are ∫ ∞−∞
eikxdk = 2πδ(x)
31
This should be used inside integral as∫ ∞−∞
f(x)δ(x− a)dx = f(a)
In 3-d this becomes ∫ ∞−∞
eik.xd3k = (2π)3δ3(x)
• The expansion of a normalised state |ψ〉 of the particle in terms of position eigenkets reads
|ψ〉 =
∫dx |x〉 〈x|ψ〉
which is the analogue of |ψ〉 =∑
n |n〉 〈n|ψ〉
• General rules of the Dirac formalism tell us the interpretation of 〈x|ψ〉
〈x|ψ〉 = probability amplitude that a particle
in state |ψ〉 is located at x
• I.e. 〈x|ψ〉 is precisely what we previously called the wavefunction
• The description of sates by wavefunctions is called the “x-representation” (or coordinate repre-
sentation)
• Schrodinger’s wave mechanics is the form QM takes if the coordinates of a particle are all one
cares about. (E.g. if no spin, no antiparticle creation, ...)
• All aspects of wave mechanics can be derived from Dirac description, e.g. overlap
〈ψ|φ〉 = 〈ψ|∫dx |x〉 〈x|︸ ︷︷ ︸
inserting I in x-rep
|φ〉
=
∫dx 〈ψ|x〉 〈x|φ〉
=
∫dx 〈x|ψ〉∗ 〈x|φ〉 =
∫dx ψ∗(x) φ(x) X
32
4.1.1 Manipulation with the momentum Operator
• Momentum operator p acting on some eigenstate of momentum (look at one dimension example
p |p〉 = p |p〉
We also have that
〈r|p〉 =1√2π~
eipr/~
Can check the normalisation by using the fact that 〈p|p′〉 = δ(p− p′) and
〈p|p′〉 =
∫dr 〈p|r〉 〈r|p′〉 =
∫dr
1
2π~eipr/~e−ip
′r/~ = δ(p− p′)
Also we can rewrite the following
〈r| p |ψ〉 =
∫dp 〈r|p〉 〈p| p |ψ〉 =
∫dp 〈r|p〉 p 〈p|ψ〉 =
1√2π~
∫dreipr/~p 〈p|ψ〉
= −i~ ∂
∂r
∫dr 〈r|p〉 〈p|ψ〉 = −i~ ∂
∂r〈r|ψ〉
Therefore
〈r| p |ψ〉 = −i~ ∂
∂rψ(r)
Similarly
〈r| p2
2m|ψ〉 = − ~2
2m
∂2ψ(r)
∂r2
33
4.2 Rules for Amplitudes
• In the x-representation (i.e. usual wavefunctions) the amplitudes are found by solving the TDSE
with appropriate boundary conditions
• Now we come to very general rules for probability amplitudes that can be derived from the Dirac
formulation
• Define “an event” in an experiment to be a situation in which all the initial and final conditions
of the experiment are completely specified. I.e. all positions, angular momenta,... if all particles
specified.
Rules
• Rule 1: When an event can occur several alternative ways the amplitude is the sum of the
amplitudes for each way considered separately (so we get interference)
• Rule 2: The amplitude for each separate way an event can occur can be written as the product
of the amplitude for part of the event occurring that way with the amplitude of the remaining
part e.g. Amp(particle x→ y)
y
z
x
Amp(x→ y) = Amp(x→ z).Amp(z → y)
or
〈y|x〉 = 〈y|z〉 〈z|x〉
• Rule 3: If an experiment is performed which is capable in principle of determining which of the
alternative ways is actually taken (so in fact not all final conditions are the same) Then total
probability (not amplitude) is sum of probabilities for each alternative
Ptot = P1 + P2 + ...
34
⇒ interference is lost. NOTE: “capable in principle” doesn’t mean that its necessary for a human
(or other scientific being) to check that all final conditions are the same. (its enough for the state
of one atom to be different whether we are aware or not)
• Useful to manipulate amplitudes even when we don’t know (yet) exactly their value E.g. two slit
interference
Source, SSlit 1
Slit 2
ObservingScreen
x
〈1|S〉 = Amplitude for particle to go from S to slit 1
〈x|1〉 = Amplitude for particle to go from 1 to slit x
similar for 〈2|S〉 and 〈x|2〉. So the total amplitude from source S to x
Amp1+2 = 〈x|1〉 〈1|S〉+ 〈x|2〉 〈2|S〉
where we multiply amplitudes along the route and add the different routes. This leads to
Prob(S → x) =∣∣Amp1+2
∣∣2and because of the interference term this is not simply equal to |〈x|1〉 〈1|S〉|2 + |〈x|1〉 〈1|S〉|2.
35
5 Conserved Quantities
• In classical physics many conserved quantities (e.g. Energy, Momentum, Ang momentum)
What about in QM?
• Consider expectation value of some operator Q which does not have any explicit time-dependence
(e.g. x or −i~∂/∂x)
〈Q〉ψ =
∫ ∞−∞
ψ∗(x, t) Q ψ(x, t)
• Due to the fact that ψ is a function of t we will find that, in general, 〈Q〉ψ is also a function of t...
d〈Q〉ψdt
=
∫ ∞−∞
∂ψ∗
∂tQ ψ + ψ∗ Q
∂ψ
∂t
dx
Use TDSE Hψ = i~∂ψ∂t
d〈Q〉ψdt
=
∫ ∞−∞
(Hψ
i~
)∗Q ψ + ψ∗ Q
(Hψ
i~
)dx
=1
i~
∫ ∞−∞−ψ∗HQ ψ + ψ∗QHψ dx
using the fact that H is hermitian.
d〈Q〉ψdt
=i
~
∫ ∞−∞
ψ∗ (HQ−QH)ψ dx =i
~〈(HQ−QH)〉ψ
• q is a conserved quantity if
d〈Q〉ψdt
= 0
no matter what state ψ the particle is in.
• Only can happen if
(HQ−QH)ψ ≡ [H,Q]ψ = 0
for any normalisable function, ψ. The object [H,Q] is the commutator of H and Q.
Example
36
Simplest case is Q = 1.
d
dt〈ψ|ψ〉 =
i
~〈ψ| [H, 1] |ψ〉 = 0
as [H, 1] = 0 for any operator A.
• Namely, the probability is conserved independent of time, i.e. 〈ψ|ψ〉 = 1
Aside on Probability Current
• ρ = ψ∗ψ is the probability density
∂ρ
∂t=∂ψ∗
∂tψ + ψ∗
∂ψ
∂t(5.1)
but TDSE says for 1D QM
i~∂ψ
∂t= − ~2
2m∇2ψ + V ψ
Using this in Eq. (5.1) we have
∂ρ
∂t=
i
~
(− ~2
2m∇2ψ∗ + V ψ∗
)ψ − ψ∗
(~2
2m∇2ψ + V ψ
)= − i~
2m∇. (ψ∇ψ∗ − ψ∗∇ψ)
Defining
j =i~2m
(ψ∇ψ∗ − ψ∗∇ψ)
and we finally have the conservation equation
∂ρ
∂t+ ∇.j = 0
37
Return to basic conservation equation
d
dt〈ψ|Q |ψ〉 =
i
~〈ψ| [H,Q] |ψ〉
• So far looked at case of Q = 1
• Now explore more generally, with H = p2
2m+ V (x)
• i) Total energy is conserved because
[H,H] = HH −HH = 0
so
d
dt〈ψ|H |ψ〉 = 0 ∀ |ψ〉
• ii) Now consider momentum p
[H, p] =
[p2
2m+ V (x), p
]=
[p2
2m, p
]+ [V (x), p]
=1
2m(p2p− pp2)︸ ︷︷ ︸
=0
+[V (x), p] = [V (x), p]
but [V (x), p] = [V (x),−i~ ∂∂x
]. To work this out we must put arbitrary function on RHS and we
find
[V (x), p] f(x) = i~∂V (x)
∂xf(x)
So...
[V (x), p] = i~∂V (x)
∂x
Not Zero!
Therefore [H, p] = i~ ∂V (x)∂x
and
∂〈p〉∂t
= −⟨∂V (x)
∂x
⟩
Note the similarity of this result to classical equation of motion.
38
• This shows that d 〈p〉 /dt cannot be zero unless ∂V (x)/∂x = 0. But this means that V =Const
and we find linear momentum conservation only whenH is independent of position (system is
independent of position)
• iii) Now take Q = X
[H, x] =
[p2
2m+ V (x), x
]=
[p2
2m,x
]+ [V (x), x]︸ ︷︷ ︸
=0
A useful identity is
[A2, B] = A[A,B] + [A,B]A
Applying this we have
[H, x] =1
2mp[p, x] + [p, x]p
Finally we need [p, x]
[p, x]f(x) =
(−i~ ∂
∂xx+ xi~
∂
∂x
)f(x) = −i~
∂x
∂xf(x) + x
∂f
∂x− x∂f
∂x
= −i~f(x)
This is a fundamental property of x and p operators.
Hence,
[H, x] =1
2mp(−i~) + (−i~)p = −i~ p
m
Thus
∂〈x〉∂t
=〈p〉m
• And position is not concerved unless in special case with 〈p〉 = 0.
• Note: the above equation is analogue to classical x = p/m
• From both the above equation and ∂〈p〉∂t
= −⟨∂V (x)∂x
⟩we see that the classical equations of motion
are obeyed on the average in QM - - EHRENFEST THEOREM
More on commutators
• We have seen that [H,Q] = 0 implies d〈Q〉dt
= 0
• So far we have been working with expectation values
39
• We now would like to ask, does [H,Q] = 0 have special consequences for individual measure-
ments...Answer is yes...
• We know eigenstates of h satisfy
Hφn = Enφn
Recall that this means that if
ψ(x, t) = e−iEnt/~φn(x)︸ ︷︷ ︸not a superposition
then when we measure energy we will get En (with probability =1).
• Now suppose the same functions φn are are also eigenstates of another operator Q
Qφn = qnφn
• This means that if the particle is described by φn then it also has a definite value of Q given byqn
Prob(Q = qn) = 1
• What is the condition that this is possible?
• Act on Hφn = Enφn with Q
QHφn = EnQφn = Enqnφn
• Alternatively act on Qφn = qnφn with H
HQφn = qnHφn = qnEnφn
• This is the same result. Subtracting these two equations
(HQ−QH)φn = 0
Since, by assumption, this is true for all φn
[H,Q] = 0
Thus
40
If [H,Q] = 0 it is possible for the particle to be in state of definite energy
a “stationary state” in which the value of Q is also definite
• What happens if ψ(x, t) is not an eigenstate? E.g. a superposition of 2 eigenstates
ψ(x, t) = a1φ1(x)e−iE1t/~ + a2φ2(x)e−iE2t/~
with
1 = |a1|2 + |a2|2
so
Prob(E = E1) = |a1|2
Prob(E = E2) = |a2|2
Let us compute 〈Q〉
〈Q〉 =
∫ψ∗Qψ dx
=
∫ (a∗1φ1e
iE1t/~ + a∗2φ2eiE1=2t/~) (a1φ1e
−iE1t/~ + a2φ2e−iE1=2t/~)
= q1 |a1|2 + q2 |a2|2
This is just
〈Q〉 = q1 × (Prob in state 1) + q2 × (Prob in state 2)
So if we make a measurement of both H and Q we find
(E1, q1) with prob = |a1|2(E2, q2) with prob = |a2|2
Fundamental point is the following:
If [H,Q] = 0 then can measure both quantities and
simultaneously find precise values of both H and Q
41
V = 0
0 a x
V =∞ V =∞
IIIIII
Figure 5.1: Infinite Square Well
What happens if [H,Q] 6= 0?
We have already studied this when we looked at Q = P in the case of the infinite square well.
[H,P ] = i~∂V
∂x6= 0
Since
at x = 0∂V
∂x→ −∞
andat x = a∂V
∂x→∞
• In this case we saw that If particle has definite energy
ψ(x, t) = e−iEnt/~φn(x)
then
ψ is a superposition of states with different momenta
• Equivalent:
We cannot know both E and P for sure since [H,P ] 6= 0
• General result in QM
42
If A and B are physical (hermitian) ops. with [A,B] 6= 0
then cannot in general simultaneously know values A and B.
5.1 Complete sets of Quantum Numbers
• Suppose we have two operators Q1, Q2 satisfying
[H,Q1] = 0, and [H,Q2] = 0
then
d〈Q1〉dt
andd〈Q2〉dt
and both observables are conserved.
• However if [Q1, Q2] 6= 0 then will not be able to make a definite measurement of both
• If [Q1, Q2] = 0 then always possible to find simultaneous eigenstates of Q1 and Q2 and H. In this
case the wavefunction
ψ(x, t) = e−iEnt/~φn(x)
which has
Definite energy En
Definite Q1 q1
Definite Q2 q2
These are the quantum numbers of the state ψ. In this case [Q1, Q2] = 0, Q1 and Q2 are
“compatible”.
• In the case [Q1, Q2] 6= 0, Q1 and Q2 are “incompatible” and it is possible for us to have state
ψ(x, t) = e−iEt/~φ(x)
with either a
Definite energy En
Definite Q1 q1
43
or
Definite energy En
Definite Q2 q1
But not both
• E.g. Since [p, x] = i~ we can never have a state in which both p and x have definite values
• We say that E, ... form a complete set of Quantum numbers if maximal number of simultaneous
eigenvalues are specified
5.2 The Uncertainty Principle
• We now want to know what happens when we make measurements on some general superposition
• Consider, e.g. energy superposition
ψ(x, t) = a1φ1e−iE1t/~ + a2φ2e
−iE2t/~
• Suppose measurement of energy at t = t0 gives result E1
→ Means we know for sure that energy is E1
→ But this implies that now (for t ≥ t0) system must be in energy eigenstate
ψ(x, t > t0) = φ1e−iE1t/~
Thus
(1) Part of original ψ which has different energy to the one measured disappears
(2) Remaining part has its coeff changed so that new wavefunction is correctly normalised
• Procedure is the so-called “collapse of the wavefunction”
• Applies when any quantity (not just energy) is measured
• “Collapse of the wavefunction” also know as “Projection Postulate”
• Radical proposals such as “many worlds” which tries to explain why we experience collapse.
• In any case important to stress
44
Every experiment ever performed leads to results consistent
with the simple “collapse” postulate and so we don’t need anything more fancy
• Let’s see what happens in an example of a series of measurements
• (1) Start with infinite square well
ψ(x, t) = φn(x)e−iEnt/~, En =~2π2n2
2ma2, φn =
√2
asin
nπx
a
• (2) measure energy: Result: E1, so wavefunction collapses to
ψ(x, t)a = φ1(x)e−iE1t/~,
• (3) Now measure momentum. To work out possible results must rewrite ψa as a sum of momentum
eigenstates
ψa = e−iE1t/~
√2
a
1
2i
(eiπx/a − e−iπx/a) .
This is a superposition of two momentum states one with π~a
and −π~a
.
• Suppose we now measure momentum π~/a ⇒ Wavefunction collapses to
ψb = e−iE1t/~ eiπx/a√a
• (4) Now measure energy again. To work out possible results must rewrite ψa as a sum of energy
eigenstates
eiπx/a√a
=1√a
(cos
πx
a+ i sin
πx
a
)sin πx
ais already an energy eigenstate but we must expand cos πx
ain terms of a Fourier series
1√a
cosπx
a=∞∑n=1
an
(√2
asin
nπx
a
)︸ ︷︷ ︸
remember this is an
energy eigenfunction
45
where
an =1√a
∫ a
0
√2
asin
nπx
acos
nπx
adx =
0 if n = odd
2√
2nπ(n2−1)
if n = even
Thus
eiπx/a√a
=i√a
(φ1(x)) +∑
n even
2√
2n
π(n2 − 1)(φn(x))
So when we measure energy we get
E1 with prob. P (E1) =1
2
E2 with prob. P (E2) =32
9π2
E4 with prob. P (E4) =128
225π2
......
...
En with prob. P (En) =8n2
(n2 − 1)2π2
• Thus even though we started with an energy eigenstate, by making a measurement of momentum
(an operator incompatible with H) we have left a particle in a state of very indefinite energy!
• These considerations lead Heisenberg to his famous uncertainty principle
• Uncertainty principle not separate axiom of QM it is a consequence of rules already stated.
• When there is uncertainty in the outcome of measurement useful to quantify: Define Uncertainty
as
(∆q)2 ≡ ⟨q2⟩− 〈q〉2
∆q = 0 if q takes a single value.
Heisenberg’s Uncertainty principle considers x and p (remember [x, p] = i~) and states
∆x∆p ≥ ~/2
But what does this mean?
• It says that is we prepare a particle in a state whereby its location x is know to within ∆x, then
the uncertainty in its momentum is at least ~/2∆x
Example:
46
• Suppose we have a Gaussian wavefunction
φ(x) =1
a1/2π1/4e−x
2/2a2
clearly
〈x〉 =
∫ ∞−∞
x |φ(x)|2 dx = 0
also ⟨x2⟩
=1
aπ1/2
∫ ∞−∞
x2e−x2/a2
dx =a2
2
so that
⇒ ∆x =[⟨x2⟩− 〈x〉2]1/2 =
a√2
Now for the momentum p
〈p〉 = − i~a√π
∫ ∞−∞
e−x2/2a2 ∂
∂xe−x
2/2a2
dx = 0
and⟨p2⟩
= − ~2
a√π
∫ ∞−∞
e−x2/2a2 ∂2
∂x2e−x
2/2a2
dx = − ~2
a√π
∫ ∞−∞
e−x2/a2
(x2
a4− 1
a2
)dx =
~2a2
using the useful integral ∫ ∞−∞
e−x2/a2
= a√π
Finally we get
∆p =~
a√
2
Putting this all together
∆p ∆x =~
a√
2
a√2
=~2
Exactly the saturation of Heisenbergs’s uncertainty principle bound. In fact, a Gaussian wave-
function gives the least possible value for ∆p ∆x.
• Since ∆p 6= 0 interesting to ask what the momentum distribution is.
47
• We know x-space wavefunction
φ(x) =1
a1/2π1/4e−x
2/2a2
In Dirac notation this is 〈x|φ〉 - “Amplitude for particle in state φ to be found at x.
• We want 〈p|φ〉 - “Amplitude for particle in state φ to be measured with momentum p.
This is just the Fourier transform of φ(x)
φ(p) =1√2π~
∫ ∞−∞
eipx/~ φ(x)
Factor in front of the integral is there so∫∞−∞ dp |φ|2 = 1
Or in Dirac notation
〈p|φ〉 =∑
all values of x
〈p|x〉 〈x|φ〉
where
〈p|x〉 = 〈x|p〉∗ =
(e−ipx/~√
2π~
)∗=eipx/~√
2π~
• Integral for φ(p) can be done by completing the square
φ(p) =1√2π~
∫ ∞−∞
dx e−1
2a2(x−ipa2/~)2+p2a4/~2 =
√a
~π1/2e−p
2a2/2~2
Another Gaussian!
48
6 Simple Harmonic Oscillator (SHO)
• Many systems to leading approximation are the SHO (or many weakly coupled SHOs), so this is
an important case
• Potential energy is
V (x) =1
2kx2
so TISE
− ~2
2m
d2φ
dx2+
1
2kx2 = Eφ
convenient to rescale the x variable by x = αy
− ~2
2mα2
d2φ
dy2+
1
2α2ky2 = Eφ
and insist that coefficients of K.E. and P.E. are the same
~2
2mα2=
1
2kα2
⇒ α2 =~√mk
Then we have
~√k
m
1
2
(−d
2φ
dy2+ y2φ
)= Eφ
Now set√k/m = ω the classical angular frequency, so let
ε =E(~ω2
)i.e., ε is the energy in units of ~ω/2. Thus our TISE becomes
−d2φ
dy2+ y2φ = εφ
It is easy to check that
φ0 = e−y2/2
49
is a solution (and is normalisable), let plug it in...
φ′0 = −ye−y2/2φ′′0 = −y2e−y
2/2 − e−y2/2
So TISE reads
−(y2e−y2/2 − e−y2/2) + y2e−y
2/2 = εe−y2/2
⇒ Solution if ε = 1 ⇒ The energy eigenvalue.
We will soon see that this is the ground state
E0 =~ω2> 0
• What about the other energy eigenstates? - There are two ways of getting rest...
• (I) Try
φ = H(y)e−y2/2
Putting this into the TISE we find
H ′′ − 2yH ′ +H(ε− 1) = 0
This is called Hermite’s equation, and can be solved by Frobeinius series method. See second year
notes for details.
• There are normalisable solutions for ε = 2n+ 1; Hn(y) is a Hermite polynomial
6.1 Operator Method.
• (II): A much more instructive method which generalises to many other problems In y coords our
Hamiltonian operator is
H = − ∂2
∂y2+ y2
Let’s try to “factorise” this Define Operators
a+ = − ∂
∂y+ y a− =
∂
∂y+ y
50
Acting on arbitrary f(y)
a+a−f(y) =
(− ∂
∂y+ y
)(∂
∂y+ y
)f(y)
= −d2f
dy2+ y2f − f
Similarly
a−a+f(y) = −d2f
dy2+ y2f + f
So we learn
• i)
[a+, a−] = −2(6.1)
• ii) We can write TISE as
(a+a− + 1)φ = εφ(6.2)
• iii) Apply a− to ground state wavefunction:
a−φ0 =
(∂
∂y+ y
)e−y
2/2 = 0
Thus
(a+a− + 1)φ0 = 1.φ0
and again we have found ε0 = 1.
• iv) Now act on TISE (the energy eigenvalue eqn, 6.2) with a+ and use commutator [a+, a−]
a+(a+a− + 1)φ = εa+φ
a+(a−a+ − 2 + 1)φ = εa+φ
[a+a−a+ + (−2 + 1)a+]φ = εa+φ
⇒ (a+a− + 1)(a+φ) = (ε+ 2)(a+φ)
So we have learnt
If φ has eigenvalue ε
a+φ has eigenvalue ε+ 2
51
• v) Similarly easy to show that
If φ has eigenvalue ε
a−φ has eigenvalue ε− 2
• Since a−φ0 = 0, we say that
“a− annihilates φ0”
• There is no state with lower energy than φ0 ⇒ φ0 is indeed ground state, with ε = 1
• vi) By repeated application of a+ on φ0 we generate entire spectrum
a+ and a− as “raising and lowering” or “ladder” or “creation and annihilation” operators.
a−
a+
a+
a+φ0
...
...
• Thus spectrum of SHO is
ε0 = 1 E0 = ~ω2
φ0 = e−y2/2
ε1 = 3 E1 = (1 + 12)~ω φ1 =
(− ∂
∂y+ y
)φ0 = 2ye−y
2/2
ε2 = 5 E2 = (2 + 12)~ω φ2 =
(− ∂
∂y+ y
)φ1 = 2(2y2 − 1)e−y
2/2
......
...
leading to, for general n
εn = 2n+ 1 En = (n+1
2)~ω φn =
(− ∂
∂y+ y
)nφ0 = Hn(y)e−y
2/2
This is a complete solution of the problem!
52
• So far we have not normalised the wavefunctions, e.g. φ0(x) = ce−x2/2α2
(note: must convert back
to x)
1 =
∫ ∞−∞
dx |φ0|2 = c2∫ ∞−∞
dx e−x2/α2
= c2π1/2α
⇒ C =1
α1/2π1/4recall α2 =
~√mk
• Similarly can normalise higher φn(x)s...
• Just as important our wavefunctions are also orthogonal∫ ∞−∞
φn(x)φm(x)dx = 0 n 6= m
• This must be true: they are eigenfunctions of hermitian operator H with different eigenvalues εn
and εm.
• Note that the φn(x) go even-odd-even and just like the infinite square well, the nth excited state
has n nodes.
• Features of the ground-state
φ(x)0 =1
α1/2π1/4e−
x2
2α2 , E0 =~ω2, α2 =
~√mk
• A classical particle of total energy ~ω/2 would be confined to a region where
V (x) ≤ E0,1
2kx2 ≤ ~ω
2⇒ x2 ≤ ~ω
k=
~k
√k
m=
k√mk
= α2
• i.e. −α ≤ x ≤ α
• If we go back to TISE
− ~2
2m
d2φ
dx2+
1
2kx2φ = Eφ
at the point where E = V (x) (The limit of the classical motion) we see that
d2φ
dx2= 0
53
this is a point of inflexion for φ. We can calculate the probability of being outside of the classical
region. Quantum effects most significant for low E states.
Prob outside classical region n
0.157 0
0.116 1
0.095 2
Below is a plot of the n = 0 case.
!3 !2 !1 1 2 3
0.2
0.4
0.6
0.8
1.0φ0/√
απ1/4
ααα α α αx
Classically forbidden regionClassically forbidden region
54
7 The Periodic Potential
• Final example of 1-d problems, consider the motion of a particle in a periodic potential of period,
l, so that
V (x+ l) = V (x) (7.1)
An example of such a potential is shown in Figure 7.1. This type of potential can be used to
model the interactions to which an electron is subjected in a crystal lattice consisting of a regular
array of single atoms separated by the distance l.
V (x)
x
l−V0
0 bb− l
Figure 7.1: Periodic potential with rectangle sections, called the Konig-Penny potential.
7.1 Bloch Waves.
• Although a real crystal is of finite length, assume here that Eq. (7.1) is true for all x.
• Thus, if ψ(x) is a solution of the Schrodinger equation corresponding to energy E, then so is
ψ(x+ l).
55
• As SE is a linear equation, any solution ψ(x) can be represented as a linear combination of two
linearly independent solutions ψ1(x) and ψ2(x)
ψ(x) = c1ψ1(x) + c2ψ2(x).
• Now ψ1(x + l) and ψ2(x + l) are also solutions and can be represented as linear combinations of
ψ1(x) and ψ2(x)
ψ1(x+ l) = a11ψ1(x) + a12ψ2(x)
ψ2(x+ l) = a21ψ1(x) + a22ψ2(x)
Thus we can write ψ(x+ l) as
ψ1(x+ l) = (c1a11 + c2a21)ψ1(x) + (c1a12 + c2a22)ψ2(x) = d1ψ1(x) + d2ψ2(x)
• The relationship between the coefficients (c1, c2) and (d1, d2) clearly involves the matrix multipli-
cation (d1
d2
)=
(a11 a21
a12 a22
)(c1
c2
)
• Let us see what happens if we diagonalise the 2× 2 matrix in this equation. We must then solve
the equation ∣∣∣∣∣ a11 − λ a21
a12 a22 − λ
∣∣∣∣∣ = 0
• This generates a quadratic equation for λ, with two solutions λ1 and λ2. If (c1, c2) is a eigenvector
corresponding to one of the eigenvalues λ, we have d1 = λc1 and d2 = λc2.
• Thus among the solutions there are two having the property
ψ(x+ l) = λψ(x)
where λ is a constant factor. This result is known as “Floquet’s Theorem”.
• We see immediately that
ψ(x+ nl) = λnψ(x), n = 0,±1,±2, ...
56
Now let ψλ1 and ψλ2 be two solutions of the SE corresponding to the energy E, which satisfy
ψ(x+ l) = λψ(x), and correspond respectively to the eigenvalues λ1 and λ2 of Eq. (7.2). Let ψ′λ1
and ψ′λ2are their derivatives
W =
∣∣∣∣∣ ψλ1 ψλ2
ψ′λ1ψ′λ2
∣∣∣∣∣denotes something called the Wronskian determinant of ψλ1 and ψλ2 , we have, using ψ(x + l) =
λψ(x),
W (x+ l) = λ1λ2W (x).
Turns out that Wronskian determinant of two solutions of the SE equation corresponding to the
same energy eigenvalue E is a constant, from this we deduce
λ1λ2 = 1.
• Returning to Eq. (7.2). If |λ| > 1, it is clear that for x → ∞ ψ grows and grows, if λ < 1 for
x→ −∞ ψ also grows and grows. Therefore, must have |λ| = 1. Given this,
λ1 = eiKl, λ2 = e−iKl.
where K is a real number. We get a full range of values for the λs by restricting the value of K
to the interval
−πl≤ K ≤ π
l.
Therefore, all physically admissible solutions must satisfy the relation
ψ(x+ nl) = einklψ(x), n = 0,±1,±2...
which is the Bloch Condition. If we rewrite this by letting
ψ(x+ nl) = eiKxuK(x)
then it follows that
uK(x+ l) = uK(x)
which is called Bloch’s Theorem. Then we see that the Bloch wave function, ψ(x+nl) = eiKxuK(x)
represents a travelling wave of wavelength 2π/K, whose amplitude uK(x) is periodic with the same
period l as the crystal lattice.
57
8 Angular Momentum
Throughout this section we will work in 3-D.
• Classically angular momentum is given by L = r × p so in QM we have the operator
L = −i~ r ×∇
where ∇ = i ∂∂x
+ j ∂∂y
+ k ∂∂z
. From this we find the individual components as
Lx = ypz − zpyLy = zpx − xpzLz = xpy − ypx
and using pi = −i~ ∂∂xi
Lx = −i~[y∂
∂z− z ∂
∂y
]Ly = −i~
[z∂
∂x− x ∂
∂z
]Lz = −i~
[x∂
∂y− y ∂
∂x
]with total angular momentum
L2 = L2x + L2
y + L2z
How many angular momentum quantum numbers are there?
• In classical case all Li and L2 are all defined, and specifying any 3 completely determines the
angular momentum
• This is not the case in QM, indeed Lx, Ly, Lz and L2 do not all commute with each other.
• In fact we have
[Lx, Ly] = i~Lz[Ly, Lz] = i~Lx[Lz, Lx] = i~Ly
58
so no 2 operators out of Lx, Ly, Lz commute. We can right this compactly as
[Li, Lj] = i~ εijk Lk
where
εijk =
+1 if (i, j, k) is (1, 2, 3), (3, 1, 2) or (2, 3 ,1)
−1 if (i, j, k) is (1, 3, 2), (3, 2, 1) or (2, 1 ,3)
0 otherwise: i = j or j = k or k = i
• So out of Lx, Ly, Lz we can at most have 1 operator which leads to a quantum number.
• Conventional to choose Lz, however, this is not yet the maximal set.
• Easy to show that
[Lx, L2] = 0
Thus, maximal set of commuting ops is L2 and Lz
Can easily show that both Lx and Ly commute with L2 and so we could use an alternative set of
commuting operators L2, Lx or L2, Ly.
8.1 Angular Momentum Eigenvalues and Eigenfunctions
• We will not derive the eigenfunctions and eigenvalues of angular momentum here as they were
covered in detail in 2nd year course but you should be familiar with the derivation - see second
year notes
• The eigenfunctions and eigenvalues are
L2Yl,m = ~2`(`+ 1)Yl,m, LxYl,m = ~mYl,m
where ` = 0, 1, 2, 3, ... and for a fixed `, m = −`,−` + 1, ..., ` − 1, `. The eigenfunctions Yl,m are
called the spherical harmonics and are functions of (θ, φ) and normalised via∫Y ∗l,m(θ, φ)Yl′,m′(θ, φ)dΩ = δ``′δmm′ .
where θ and φ are defined in terms of spherical polar coordinates as
x = r sin θ cosφ, y = r sin θ sinφ, z = r cos θ.
59
• The first few spherical harmonics are
Y0,0(θ, φ) =
√1
4π
Y1,0(θ, φ) =
√3
4πcos θ
Y1,±1(θ, φ) =
√3
8πsin θ e±iφ
8.2 Angular Momentum: A better method.
• Recall for the SHO we had two ways of getting eigenvalues/eigenvectors
– Differential Equations
– “Algebraic” using a+, a− ops
• We can do similar thing for angular momentum.
• We can introduce “step” or “ladder” operators and solve for eigenvalues of L2 and Lz algebraically
• Big difference though: algebraic method shows there are more possible eigenvalues of L2 and Lz
than what is found using differential ops.
• How can this happen? Start from the beginning and propose eigenvalue equations
L2ψ = ~2K2ψ, Lzψ = ~kψ
• To find out what K2 and k are consider
L+ = Lx + iLy, L− = Lx − iLy
these are the angular momentum “step” operators. Note that
[L±, Lz] = [Lx ± iLy, Lz] = ∓~L±
• Now consider a ψk which satisfies
Lzψk = ~kψk
and define a new function ψ′ = L+ψk (if it exists - it might be zero - come back to this).
60
• What Lz value does ψ′ have, i.e. Lz(L+ψk) =?
• Use commutation relation
[Lz, L+] = ~L+ ⇒ LzL+ = ~L+ + L+Lz
• Thus
Lz(L+ψk) = (~L+ + L+Lz)ψk = ~ψ′ + L+~kψk = ~(k + 1)ψ′
• So the state ψ′ must be proportional to a state with Lz eigenvalue ~(k + 1)
L+ψk = c+(k)︸ ︷︷ ︸Normalisation
ψk+1
similarly easy to show that L−ψk satisfies
Lz(L−ψk) = ~(k − 1)(L−ψk)
so similarly we have
L−ψk = c−(k)ψk−1
L+ and L− step up and down tower of Lz eigenvalues by one ~ unit
• What do L+, L− do to L2 eigenvalues
• Since, L2 commutes with Lx and Ly and L± = Lx ± Ly obvious that
[L2, L±] = 0
• This means that if ψk satisfies L2ψk = ~2K2ψk (so label as ψk,K2) then
L2(L+ψk,K2) = L+L2ψk,K2 = L+~2K2ψk,K2 = ~2K2(L+ψk,K2)
and similarly for L−ψk,K2 .
Thus L+, L− do not change L2 eigenvalue
61
• But this implies that we have a tower of Lz eigenstates with eigenvalues
...~(k − 2), ~(k − 1), ~k, ~(k + 1)...
generated by L+ or L− repeatedly acting on ψk. This must be terminated at both ends
• To see this, note that
L2 − L2z = L2
x + L2y + L2
z − L2z = L2
x + L2y
• Now consider an arbitrary member of the tower (say ψk′,K2) and take expectation values of both
sides of L2 − L2z = L2
x + L2y
LHS =⟨k′, K2
∣∣ (L2 − L2z)∣∣k′, K2
⟩= ((~K)2 − (~k′)2)
⟨k′, K2|k′, K2
⟩= ~2(K2 − k′2)
BUT
RHS =⟨k′, K2
∣∣ (L2x + L2
y)∣∣k′, K2
⟩ ≥ 0
since the sum of expectation values of squares of Hermitian operators is always positive.
⇒ ~2(K2 − k′2) ≥ 0
⇒ k′2 ≤ K2
I.e. Lz eigenvalue is bounded.
• Since k′2
is bounded by K2 there must exist a state with highest Lz, call this
ψkmax,K2
label this as |kmax, K2〉 in Dirac notation and a state with lowest Lz
ψkmin,K2
label this as |kmin, K2〉 in Dirac notation.
• The only way this can be consistent with action of L+ and L− is if
L+
∣∣kmax, K2⟩
= 0 i.e. c+(kmax) = 0
and
L−∣∣kmin, K
2⟩
= 0 i.e. c−(kmin) = 0
62
• In fact there is a connection between values kmax and K2 (orkmin and K2)
• To see this need to manipulate operators
L−L+, L+L−
L−L+ = (Lx − iLy)(Lx + iLy) = L2x + L2
y + i[Lx, Ly]
L−L+ = L2 − L2z − ~Lz
Similarly
L+L− = L2 − L2z + ~Lz
Now act on L+ |kmax, K2〉 = 0 with L− and use L−L+ = L2 − L2
z − ~Lz to show
(L2 − L2z − ~Lz)
∣∣kmax, K2⟩
= 0
⇒ ~2(K2 − k2max − kmax)
∣∣kmax, K2⟩
= 0
⇒ K2 = kmax(kmax + 1)
• Imagine starting from the top of the tower |kmax, K2〉 and acting q times with L− until we reach
|kmin, K2〉
• But, |kmin, K2〉 satisfies L− |kmin, K
2〉 = 0 and acting with L+ on this and using L+L− = L2 −L2z + ~Lz gives
(L2 − L2z + ~Lz)
∣∣kmin, K2⟩
= 0
⇒ ~(K2 − k2min + kmin)
∣∣kmin, K2⟩
= 0
then
⇒ K2 = kmin(kmin − 1)
Comparing this with K2 = kmax(kmax + 1) we find that
−kmax = kmin
Moreover, as we get from kmax to kmin in q steps (using L−) must have
kmax − kmin = 2kmax = q
⇒ kmax =q
2q = 0, 1, 2, 3, ...
63
• Conventional to refer to q/2 = j where j = 0, 1/2, 1, 3/2, ... as the angular momentum of state
(j = kmax)
• Finally K2 = kmax(kmax + 1) = j(j + 1)
• Thus we have learnt
J2 |j,m〉 = ~2j(j + 1) |j,m〉 Jz |j,m〉 = ~m |j,m〉j = 0, 1/2, 1, 3/2, ...
m = −j,−j + 1,−j + 2, ..., j − 1, j
• It is also easy to see that there are 2j + 1 possible states of the same j but with different m (that
is there are 2j + 1 different values for m.
8.3 Meaning of Angular Momentum Operator Results
• We have shown that we get both integer and half integer results for angular momentum j.
• But in the second year course it was shown that the solutions to the associated Legendre equation
are only normalisable when K2 = j(j + 1) where j is integer only.
What is going in?
• The extra half-integer eigenvalues arose because we have in fact solved a more general problem
than that of Lx, Ly, Lz, L2 (although we did not know we were!
– Nowhere did we use explicit differential representation for Lz, L2, etc
– We just used the commutation relations
[Li, Lj] = i~εijkLk
which reflects the law of the combination of rotations in 3-dimensions and which must be
satisfied whatever be the nature of the wave functions they rotate
– What we have discovered is “spin” which can take on both integral and odd-half-integral
values
E.g. photon: j = 1, pion: j = 0, Higgs: j = 0, Many Nuclei: j = 0, 1..., Electron: j = 1/2,
Proton: j = 1/2, Many Nuclei: j = 1/2, 3/2, ...
64
• Spin Wavefunctions are not functions of angular coords θ, φ but must instead be represented as
2− state vectors
(.
.
)for j = 1/2
3− state vectors
.
.
.
for j = 1
4− state vectors
.
.
.
.
for j = 3/2...
then the Lx, Ly, Lz.L2 act as matrices on these vectors and “re-shuffle” their components.
• Spin has forced us to to use “Heisenberg’s formulation of QM (“matrix Mechanics”) which is more
general, though more abstract that Schrodinger’s. Heisenberg though of all operators as matrices
• On the other hand, normal “orbital” angular momentum is always integral and therefore can
always be represented by Ylm(θ, φ) and differential operators.
• It is conventional to have a notation which distinguishes “orbital” from “spin” (or intrinsic)
angular momentum. We write
Ji = Li + Si
where i = x, y, z. Ji is the total angular momentum (j = 0, 1/2, 1, 3/2, ...), Li is orbital angular
momentum (l = 0, 1, 2...only) and Si is spin (s = 0, 1/2, 1, 3/2, ...)
• The operators Si obey the same commutation relations
[Sx, Sy] = i~Sz etc
Thus so do the Js
[Jx, Jy] = i~Jz etc
In summary, the eigenvalue equations are
Total
J2 |j, jz〉 = ~2j(j + 1) |j, jz〉 j = 0, 1
2, 1, ...
Jz |j, jz〉 = ~jz |j, jz〉 jz = −j,−j + 1...j − 1, j
65
Orbital
L2 |l,m〉 = ~2l(l + 1) |l, lz〉 l = 0, 1, 2, ...
Lz |l, lz〉 = ~lz |l, lz〉 m = −l,−l + 1...l − 1, l
Spin
S2 |s, sz〉 = ~2s(s+ 1) |s, sz〉 s = 0, 1
2, 1, ...
Sz |s, sz〉 = ~lz |s, sz〉 sz = −s,−s+ 1...s− 1, s
and as Ji = Li + Si we have Jz = Lz + Sz so
jz = m+ sz.
8.4 A close look at Spin
• Let us look at spin 1/2 example. We showed that:
J2 |j, jz〉 = J2
∣∣∣∣12 ,±1
2
⟩= ~2j(j + 1)
∣∣∣∣12 ,±1
2
⟩=
3~2
2
∣∣∣∣12 ,±1
2
⟩Jz
∣∣∣∣12 ,±1
2
⟩= ±~
1
2
∣∣∣∣12 ,±1
2
⟩
• As we have seen, these do not turn up when we solve the Schrodinger, but requires the use of
matrix formulation of Q.M.
• The spin 1/2 case is particularly important as electrons, protons, neutrons, quarks,... all have
spin 1/2
• Since for j = 1/2 it is not possible to represent angular momentum operators as differential
operators
• Consider the matrices:
Sx =~2
(0 1
1 0
)︸ ︷︷ ︸σx
Sy =~2
(0 −ii 0
)︸ ︷︷ ︸
σy
Sz =~2
(1 0
0 −1
)︸ ︷︷ ︸
σz
where σx, σy, σz are the Pauli matrices. These are Hermitian as
S†x = Sx, S†y = Sy, S†z = Sz
So these have real eigenvalues.
66
• Now calculate commutator of Sx and Sy:
[Sx, Sy] =
(~2
)2((
0 1
1 0
)(0 −ii 0
)−(
0 −ii 0
)(0 1
1 0
))= i~Sz
Exactly the angular momentum commutation relation. In general therefore we have [Si, Sj] =
i~εijkSk.
• We can also calculate the matrix for S2
S2 = S2x + S2
y + S2z = ~2 1
2
(1
2+ 1
)(1 0
0 1
)= ~2 1
2
(1
2+ 1
)I
• Exactly what we want for S2 so that we can act on a two component vector which is an eigenvector
of S2 with eigenvalue ~2 12
(12
+ 1).
• Let’s also look at Sz = ~/2
(1 0
0 −1
), what are its eigenvalues?
• Since Sz is diagonal, obvious that
Sz
(1
0
)=
~2
(1 0
0 −1
)(1
0
)=
~2
(1
0
)
⇒(
1
0
)is state with Sz = ~
2
Similarly
(0
1
)is state with sz = −~
2
• We can also find eigenvalues and eigenvectors of Sx as well
det
(−λ ~/2~/2 −λ
)= 0, ⇒ λ2 − (~/2)2 = 0
⇒ λ = ±~/2
This time the eigenvectors are
1√2
(1
1
)for λ = +~/2
1√2
(1
−1
)for λ = −~/2
Sx
67
Similarly for Sy
1√2
(1
i
)for λ = +~/2
1√2
(1
−i
)for λ = −~/2
Sy
68
8.5 Stern-Gerlach experiment
• How do we know that the electron has spin 1/2?
• Classically a charged object with angular momentum has a magnetic Dipole Moment
• Quantum Mechanics also leads to this conclusion (although to be precise we need Quantum Field
theory to give a truly consistent understanding of e− magnetic dipole moment.
µ = −gs e
2me
S
where gs is a correction due to quantum effects, the factor e2me
is the classical proportionality
between µ and S, the negative sign is there because the electron’s charge is -1.
• Now we introduce a magnetic field in the z-direction Bz.
• There are two possible values for the interaction energy −B.µ due to the quantisation of Sz.
If sz = ~/2 −B.µ = + gse2me
~2Bz
If sz = −~/2 −B.µ = − gse2me
~2Bz
which means that we have two discrete possible values of magnetic dipole moment.
• The combination of factors
e~2me
≡ µB “Bohr Magneton”
• Of course typically potential energy of a system will be a combination of a spin-independent V (r),
and a spin-dependent term (arising, say, from a B.µ interaction
• Thus typically the form of the energy (Hamiltonian) for a spin 1/2 system (e.g. electron in an
atom) will be
H = H0I +gse
2me
B.S
where H0 could be Coulomb potential + kinetic energy, I is the identity matrix and so the H0
term is spin-independent and B.S = BxSx +BySy +BzSz.
• Thus we can consider a modified form of the TDSE.
HΨ = i~∂Ψ
∂t
69
H is now a 2× 2 matrix as well as a differential operator and Ψ is a two component vector, with
each component being a function of (t, r, θ, φ) i.e.
Ψ =
(ψ1(t, r, θ, φ)
ψ1(t, r, θ, φ)
)
• If B-field is a constant then this equation (modified TDSE) is simple to analyse, write
Ψ = Φ(t, r, θ, φ)u
where u is a two component vector.
• This gives
(H0Φ)u+ Φgse
2me
(B.S)u = i~∂Φ
∂tu (8.1)
this can only be satisfied if
(B.S)u = λu
• If the rest of the problem is spherically symmetric, then we can always choose direction of x, y, z
axes such that z− axis is in the B direction.
• With this choice
B = (0, 0, Bz)
then
BSzu = λu.
Since Sz = ~2
(1 0
0 −1
)we learn that
λ = ±~B2, for u =
(1
0
),
(0
1
)Putting this back into Eq. 8.1, we get two equations(
H0 +gse
2me
~B2
)φ1(t, r, θ, φ) = i~
∂φ1
∂t(H0 − gse
2me
~B2
)φ2(t, r, θ, φ) = i~
∂φ2
∂t
where φ1 and φ2 are top and bottom components of Φ =
(φ1
φ2
).
70
• Now separate variables as usual for both equations
φ1(t, r, θ, φ) = e−iE1t/~φ1(r, θ, φ)
φ2(t, r, θ, φ) = e−iE2t/~φ2(r, θ, φ)
Giving TISE for φ1 and φ2.
• Interpretation is simply that the electrons can either have spin aligned with B or anti-aligned and
energies are shifted by ±gsµBB/2.
• Now consider dipole in non-uniform B-field.
• Figure 8.5 shows a dipole in a non-uniform E. The net force acting on the dipole
Ez
z
z + a
a
+
−
Figure 8.1: EDM in a non-uniform E field.
F = qE(z + a)− qE(z) ' q(E(z) + a∂E
∂Z− E(z)) ' qa
∂E
∂Z= p
∂E
∂Z
Similarly if we have a magnetic dipole in a non-uniform field
F = µ∂B
∂z.
• Now suppose we have a beam of electron going through a region of non-uniform Bz field
71
e− beam
Bz
∂Bz/∂z > 0
observing screen
• The e−s will experience a force wither up or down depending on which way their µ (i.e. Sz) is
pointing
• This is the essence of the Stern-Gerlach experiment - see page 33 of Bransden and Joachain.
• Important points to note
– (1) Classically one would expect to see a continuous distribution of e− deflections on observing
screen (as classically Sz, and thus µz, can take on any value - not quantized)
– (2) The experiment instead shows two spots (for spin 1/ system) on screen→ Sz is quantised
with 2 values
– (3) In detail when one calculates position of spots one finds they are twice as far apart as
expected based upon
µ = − e
2me
S
This is one of the (many) reasons why we are forced to introduce
gs ' 2 = “Gyromagnetic Ratio”
So
µ = − gse
2me
S
72
8.6 Addition of Angular Momentum
• In many situations of physical interest we deal with systems of more than one particle. We
therefore need to know how to deal with the system as a whole in terms of the distinct sub-
systems (i.e. the individual particles).
• Look at the simplest case, the addition of two commuting angular momentum. Thus we have
J = J1 + J2
where J1 and J2 are any two angular momentum corresponding to the independent sub-sytems
(i.e. the two particles) 1 and 2. Similarly Jz = J1z + J2z.
• Let |j1,m1〉 be a normalised simultaneous eigenvector of J21 and J1z so that
J21 |j1,m1〉 = j1(j1 + 1)~2 |j1,m1〉
and
J1z |j1,m1〉 = m1~ |j1,m1〉 .
• Similarly let |j2,m2〉 be a normalised simultaneous eigenvector of J22 and J2z so that
J22 |j2,m2〉 = j2(j2 + 1)~2 |j2,m2〉
and
J2z |j2,m2〉 = m2~ |j2,m2〉 .
• We can construct a normalised simultaneous eigenvector of J21,J
22, J1z, J2z belonging respectively
to the eigenvalues j1(j1 + 1)~2, j2(j2 + 1)~2,m1~ and m2~ is given by the “direct product”
|j1, j2,m1,m2〉 = |j1,m1〉 |j2,m2〉 .
• For a fixed value of j1, m1 can take one of 2j1 + 1 values (−j1,−j1 + 1, ...j1) and for a fixed value
of j2, m2 can take 2j2 + 1 values (−j2,−j2 + 1, ...j2).
• Hence for given values of j1, j2 there are (2j1 +1)(2j2 +1) “direct products” which form a complete
set of orthonormal states.
• Apply Jz to the direct product, (remembering that Jz = J1z + J2z)
Jz |j1, j2,m1,m2〉 = (J1z + J2z) |j1,m1〉 |j2,m2〉 = (J1z |j1,m1〉) |j2,m2〉+ |j1,m1〉 (J2z |j2,m2〉)(m1 +m2)~ |j1,m1〉 |j2,m2〉 = (m1 +m2)~Jz |j1, j2,m1,m2〉
which means that |j1, j2,m1,m2〉 is an eigenfunction of the total Jz with eigenvalue (m1 +m2)~.
73
• This is confirmed by the fact that J21,J
22, J1z, J2z operators commute with Jz = J1z + J2z.
• Now consider the operator J2 (where J = J1 + J2).
J2 = (J1 + J2)2 = J12 + J2
2 + 2J1.J2
• Because all components of J1 commute with all of those of J2 and since [J12,J1] = 0 = [J2
2,J2]
it follows that [J2,J12] = 0 = [J2,J2
2].
• However, as
J1.J2 = J1xJ2x + J1yJ2y + J1zJ2z
and since J1z does not commute with J1x or J1y, we find [J2, J1z] 6= 0 similarly we find [J2, J2z] 6= 0.
• Consequently, simultaneous eigenfunctions of J2 and Jz are eigenfunctions of J12 and J2
2 but not
in general of J1z and J2z
• The bottom line is that the system can be formed in terms of two distinct descriptions.
– In terms of eigenfunctions of J12,J2
2, J1z and J2z or
– in terms of eigenfunctions of J12,J2
2, J2 and Jz.
• We can describe the latter with eigenkets labelled |j2, j1, j,m〉 Thus we have for example
J2 |j2, j1, j,m〉 = j(j + 1)~2 |j2, j1, j,m〉
and
Jz |j2, j1, j,m〉 = m~ |j2, j1, j,m〉
• Like the eigenkets |j1, j2,m1,m2〉 the eigenkets |j2, j1, j,m〉 form a complete orthonormal set.
• We can relate these two sets of states via a unitary transformation by inserting a complete set of
states
|j1, j2, j,m〉 =∑m1,m2
|j1, j2,m1,m2〉 〈j1, j2,m1,m2|j1, j2, j,m〉
where 〈j1, j2,m1,m2|j1, j2, j,m〉 is the Clebsch-Gordon coefficient (sometimes written just as
〈j1, j2,m1,m2|j,m〉.
74
• In order to find out what values of j we can have for given j1 and j2 we notice that since
|j1, j2,m1,m2〉 is an eigenfunction of Jz with eigenvalue (m1 +m2)~ we must have that
m = m1 +m2
so the double sum reduces to a single sum.
In other words the Clebsch-Gordon coefficients must vanish unless m = m1 +m2:
〈j1, j2,m1,m2|j1, j2, j,m〉 if m 6= m1 +m2
• We also know that the maximum values of m1 and m2 are j1 and j2 respectively, so the maximum
size of m is j1 + j2
• In turn this means the maximum size of j is j1 + j2.
• Consider some special cases:
• 1) For j = j1 + j2 and m = j1 + j2 there is only one term on the RHS of
|j1, j2, j,m〉 =∑
m1,m2|j1, j2,m1,m2〉 〈j1, j2,m1,m2|j1, j2, j,m〉 corresponding to m1 = j1 and
m2 = j2. Thus
|j1, j2, (j1 + j2), (j1 + j2)〉 = 〈j1, j2, j1, j2|j1, j2, , (j1 + j2), (j1 + j2)〉 |j1, j2, j1, j2〉
and since the eigenkets are normalised
〈j1, j2, j1, j2|j1, j2, , (j1 + j2), (j1 + j2)〉 = 1
• 2) Now consider the state got which m = j1 + j2 − 1.
• In this case there are two possibilities - m1 = j1,m2 = j2 − 1 or m1 = j1 − 1,m2 = j2
• Thus, a state |j1, j2, j, (j1 + j2 − 1)〉 must be a linear combination of the two linearly independent
eigenkets |j1, j2, j1, (j2 − 1)〉 and |j1, j2, (j1 − 1), j2〉
• There are two such combinations, one of them with j = j1 + j2 and the other (which is orthogonal
to the first) with maximum value of m = j1 + j2 − 1 and therefore must have j = j1 + j2 − 1.
• 3) Move now to a state with m = j1 + j2 − 2, we see that there are 3 linearly independent states
of this find with j = j1 + j2, j = j1 + j2 − 1, j = j1 + j2 − 2.
• If we keep repeating this we find that the minimum value of j is |j1 − j2|, as all combinations have
been exhausted
75
• For each value of j there are 2j + 1 values of m so that the total number of eigenfunctions
|j1, j2, j,m〉 is given byj1+j2∑
j=|j1−j2|
(2j + 1) = (2j1 + 1)(2j2 + 1)
• Let’s look at an example of adding j1 = 1 and j2 = 3/2
Table 1: Allowed values of (m1,m2) and (j,m) for j1 = 1 and j2 = 3/2.
m1 m2 m j
1 3/2 5/2 5/2
0 3/2 3/2 5/2, 3/2
1 1/2
-1 3/2
0 1/2 1/2 5/2, 3/2,1/2
1 -1/2
-1 1/2
0 -1/2 -1/2 5/2, 3/2,1/2
1 -3/2
-1 -1/2
0 -3/2 -3/2 5/2, 3/2
-1 -3/2 -5/2 5/2
Total: (2j1 + 1)(2j2 + 1)=12 Total:∑j1+j2
j=|j1−j2|(2j + 1) = 12
• What we have learned is that for given values of j1 and j2 the allowed values of j are
j = |j1 − j2| , |j1 − j2|+ 1, ..., j1 + j2
so that the three angular momentum quantum j1, j2 and j must satisfy the triangle condition
|j1 − j2| ≤ j ≤ j1 + j2
moreover for each value of j there are 2j + 1 eigenfunctions such that m = −j,−j + 1, ..., 1.
• The Clebsch Gordon coefficients can be determined by applying raising and lowering operators
to |j1, j2, j,m〉 =∑
m1,m2|j1, j2,m1,m2〉 〈j1, j2,m1,m2|j1, j2, j,m〉. See P319 of Bransden and
Joachain.
76
9 Approximate Methods I
Rarely are there exact solutions to physics problems, where the energies and wave functions are known.
This half of the course presents a number of approximate methods to deal with some of these problems.
9.1 Time-independent perturbation theory
• Often, Hamiltonian we want to solve is close to one we know how to solve:
Hλ = H0 + λV (1)
eg. λV describes some external field (electric, magnetic) that we can dial up, with λ the amplitude.
• Suppose at λ = 0, the system is in an exact eigenstate of H0: H0|φa〉 = Ea|φa〉.
• Idea: we want to find for the perturbed system Hλ, the eigenstate |ψ〉
Hλ|ψ〉 = (H0 + λV )|ψ〉 = Eλ|ψ〉 (2).
Key: the perturbed eigenstate |ψ〉 is going to look a lot like the unperturbed one |φa〉, when λ is
small. So try a power series in λ:
Eψ = Ea + λEb + λ2Ec ; |ψ〉 = |φa〉+ λ|b〉+ λ2|c〉 (3)
where Eb, Ec and the states |b〉, |c〉 are to be determined. We have only written terms up to order
λ2. Substitute Eqs. (3) into (2), and equate terms with the same powers of λ on left hand side
and right hand side:
λ0eqn : H0|φa〉 = E0a|φa〉 (nothing new)
λ1eqn : H0|b〉+ V |φa〉 = Ea|b〉+ Eb|φa〉λ2eqn : H0|c〉+ V |b〉 = Ea|c〉+ Eb|b〉+ Ec|φa〉
• λ1 equation: multiply on left by the known 〈φa|, and use thatH0|φa〉 = Ea|φa〉 ⇔ 〈φa|H0 = Ea〈φa|:
Eb = 〈φa|V |φa〉 (4) first order energy change
77
• λ2 equation: again multiply on left by 〈φa|, and using the normalisation 〈φa|φa〉 = 1,
Ec = 〈φa|V |b〉 − Eb〈φa|b〉 (4)
Now need to know what |b〉 is, in terms of states we know. Now for the unperturbed problem,
we have the complete set of eigenstates |k〉 where H0|k〉 = Ek|k〉. (The state |φa〉 belongs to this
set.) This complete set is:
orthogonal : 〈j|k〉 = 0 for j 6= k,
normalized : 〈j|j〉 = 1
So expand |b〉 in terms of these known eigenstates with numerical coefficients bk:
|b〉 =∑k
bk|k〉 (5)
Substitute Eq. (5) into λ1 eqn, and multiply from left by 〈j| which we choose such that 〈j| 6= 〈φa|:
bj =〈j|V |φa〉Ea − Ej =⇒ |b〉 =
∑j
〈j|V |φa〉Ea − Ej |j〉 (6)
Finally, substituting Eq. (6) into Eq. (4), using that |j〉 is an ortho-normal set of H0, we get,
Ec =∑j 6=φa
〈φa|V |j〉〈j|V |φa〉Ea − Ej =
∑j 6=φa
|〈j|V |φa〉|2Ea − Ej (7) second order energy change
• Collecting all, to order λ2, perturbation to the state |φa〉 and its energy Ea:
Eψ = Ea + λEb + λ2Ec = Ea + 〈φa|λV |φa〉+∑j 6=φa
|〈j|λV |φa〉|2Ea − Ej (8)
|ψ〉 = |φa〉+ λ|b〉+ . . . = |φa〉+∑j
〈j|λV |φa〉Ea − Ej |j〉+ . . . (9)
9.2 Ex. 1: perturbed particle in a box
Here we do a simple example to illustrate the formalism.
78
• Particle in a box, length a, perturbed by a small potential
H = H0 + λV (x) , H0 =p2x
2m+ Vbox(x)
Vbox(x) =
0 : 0 < x < d,
∞ : otherwise.
λV (x) =
ε : 0 < x < b < d,
0 : otherwise.
• For H0, the complete set (n = 1, 2, 3, . . .) of energy eigenstates are:
φn(x) = 〈x|n〉 =
√2
dsin(nπx
d
), En =
1
2m
(~nπd
)2
• first order energy change for a given state |φa〉 = |n〉:
Eb = 〈n|λV |n〉 = ε
∫ b
0
dx φ∗n(x)φn(x) = ε
(b
d− 1
2nπsin
2nπb
d
)• first order change to the state |n〉: using a notation similar to that of the previous section, we
need to find the coefficient bk in the series |b〉 =∑
k |k〉, where now, the label k = 1, 2, 3, . . . refers
to the complete set of eigenstates for H0 above. This is worked out in Problem sheet 4.
9.3 Ex. 2: Quadratic Stark effect
• H atom: turn on electric field E = −∇Φ
• energy change due to potential Φ is just adding up the effect of the potential at the electron at
position re and at the nucleus at rp:
e(Φ(rp)− Φ(re)) ≈ −e r ·∇Φ(r) = e r ·E,
where r = re − rp .
• Choose E parallel to z-direction: E = E z (z is the unit vector in the z-direction), and let the
perturbation be:
λV = e r ·E = e E z
• The unperturbed Hamiltonian is just that of the hydrogen atom, see last lecture. Then, the
complete set of eigenstates for the unperturbed H0 is just the set of ψnlm(r, θ, φ) = Rnl(r)Ylm(θ, φ),
or in a brief ket notation: |n l m〉.
79
• first order energy change to the ground state |1 0 0〉:
λEb = eE〈1 0 0|z|1 0 0〉 = 0
again by parity argument: here in 3 dimensions, r → −r, then z → −z also, but since the ground
state is spherically symmetric, |1 0 0〉 → |1 0 0〉.General principle:
If a perturbation takes the state |a〉 to some state that is completely orthogonal to state |a〉,then there is no first order energy correction to this state |a〉. In equation: if V |a〉 =
∑k 6=a ck|k〉,
then Eb =∑
k 6=a ck〈a|k〉 = 0.
• second order energy change to the ground state |1 0 0〉:
λ2Ec = (eE)2
(nlm)6=(100)∑n,l,m
|〈n l m| z |1 0 0〉|2E100 − Enlm
• Now, [Lz, z] = 0, implying that both |n l m〉 and z |n l m〉 are eigenstates of Lz. Then
〈n l m| z |1 0 0〉 6= 0 iff m = 0, to have the same Lz for the bra and ket.
• Parity argument: again, under z → −z, |1 0 0〉 → |1 0 0〉, which means we require that under
parity change, |n l m〉 → −|n l m〉. This restricts l to be odd integers. In fact, actually doing the
integrals, only l = 1 is allowed.
• We finally get:
λ2Ec = (eE)2∑n≥2
|〈n 1 0| z |1 0 0〉|2E100 − En10
10 Approximate Methods II
10.1 Degenerate perturbation theory (time independent)
In derivation of second order perturbation theory, if somehow, there is a state |k〉 6= |φa〉, but yet,
they have the same energy, Ek = Ea, then the denominator in Eq. (7) blows up, ie, the second order
energy correction is infinite! Key point: if there exists distinct states with the same energy, then the
perturbation may connect these states, and the perturbation may no longer be small.
Instead of developing the general theory of degenerate perturbation, we shall just look at an example
in the Hydrogen atom in an electric field E = E z (as in previous section).
80
10.2 Linear Stark effect
• Look now at the unperturbed eigenstate |2 0 0〉 with unperturbed energy E200, affected by the
same perturbation as before: λV = eEz.
• Key: in Hydrogen atom, energy does not depend on angular momentum l, but just on n from
the radial part. Hence, the state |2 0 0〉 has unperturbed energy E210 = E200. Furthermore,
A ≡ 〈2 1 0| z |2 0 0〉 6= 0 : perturbation mixes up states of same energy.
• Since it takes no energy to connect between these states, simplify by just keeping only these states
in computing the perturbed energy. In matrix notation:
H = H0 + eEz =
(E200 0
0 E200
)+ eE
(0 A∗
A 0
)=
(E200 eEA∗eEA E200
)(10.1)
• Finding the perturbed energies means solving the time independent Schrodinger Equation in
matrix form:∑
bHab vb = Eva for a 2 component eigenvector v with components va. From linear
algebra, this involves solving:
0 = det (H − E1) = det
(E200 − E eEA∗eEA E200 − E
)(10.2)
• Solution: let ± label the two possible solutions. Then the energies and their corresponding
eigenvectors are:
E± = E200 ± eE√A∗A (10.3)
v± =1√2
(|2 0 0〉 ± |2 1 0〉) =1√2
(1
±1
)(10.4)
• Physics: from electromagnetism, total dipole moment: p = p0 +αE, where p0 is the permanent
dipole moment, the second term comes from induced dipole moment from the applied electric field
E, with polarizability of the atom α. Energy change is then δE = −p ·E.
– Quadratic Stark effect: no linear in E part, because for the ground state |1 0 0〉, p0 = 0.
Hence δE = −α|E|2.– Linear Stark effect: there is a linear in E part, which means the states |v±〉 must have a
permanent dipole moment such that δE = −p0 ·E. (The state |2 0 0〉 itself does not have
a permanent dipole.) Quiz: show that p0 ∝ 〈v+|ez|v+〉 = e(A + A∗)/2 6= 0. Try doing the
actual integrals to get the value of A itself.
81
11 Approximate Methods III
11.1 Effect of external magnetic field on the Hydrogen-like atom
When the hydrogen like atom is placed in a magnetic field, there are three perturbations to consider: i)
effect on orbital angular momentum, ii) effect on intrinsic angular momentum (spin), and iii) relativistic
effect— spin-orbit coupling. Reference: see eg. Bransden and Jochain Ch.12.3.
11.1.1 Orbital angular momentum
For a particle of charge Q, moving in external magnetic field B: replace momentum by p→ p−QA,
where the vector potential B = ∇×A. Hence the kinetic energy of the atom becomes:
HKE =1
2m(p+ eA)2
• First simplification: drop A2 part because its energy will be small compared to the p · eA term,
for experimentally relevant B strength (even at 103 Tesla).
• In experiment, B(x) is essentially a constant (no x-dependence) over the size of the atom. Thus,
take A = 12B × r, and rearranging the terms to the form B · r × p = B ·L, we get:
δHL(B) =e
2m(p ·A+A · p) =
e~2m
B ·L (i)
This is the energy change due to the orbital angular momentum of the charged electron. Classi-
cally, the circulating electron has an orbital angular momentum e~2m
.
11.1.2 Spin angular momentum
Electron also has an intrinsic angular momentum (spin) S which contributes further:
δHS(B) =e~mB · S (ii)
Note carefully, compared to (i), there is no factor of 1/2 here. Ultimately, this is confirmed by experi-
ments.
82
11.1.3 Relativistic effect: spin-orbit coupling
Going beyond the Schrodinger equation to include relativistic effects (Dirac equation), there is an extra
term that couples the orbital and spin angular momentum:
δHLS = ξ(r) L · S =1
2m2c2Ze2
4πε0
1
r3L · S (iii)
where as before, Z is the nuclear charge (but we still have only 1 electron).
Combining all ( i + ii + iii ), and taking the magnetic field to be along z direction,
H = H0 +µB~
(Lz + 2Sz)B +1
2m2c2Ze2
4πε0
1
r3L · S (1)
where we take the unperturbed Hamiltonian to be that for the Hydrogen-like atom (previous lectures):
H0 = − ~2
2m∇2 − Ze2
4πε0
1
r(2)
Now, we study different regimes of B strength.
11.1.4 Strong field Zeeman effect
• When B is large enough to ignore spin orbit coupling L · S term (iii).
H ≈ H0 +µB~
(Lz + 2Sz)B
– quantum numbers of the one electron atom:
H0 |n l ml〉 = En |n l ml〉 ,
n = 1, 2, 3, . . . l = 0, 1, 2, . . . , n− 1 ml = 0,±1,±2, . . . ,±l
– new: spin angular momentum quantum numbers: for a spin S = 1/2 electron,
S2 → s(s+ 1) ~2 =3
4~2, Sz → ms~ = ±1
2~
Without magnetic field, the energy of the hydrogen-like atom does not depend on these
spin quantum numbers. So we can write the eigenstates of H0 with all these good quantum
numbers: H0 |n l ml s ms〉 = En |n l ml s ms〉
83
– Now the perturbation δH(B) = µB~ (Lz + 2Sz)B commutes with H0, so we can simultane-
ously diagonalise H0 and δH(B), using the eigenstates of H0. Thus, the perturbed energies
are:
En ml ms = En + µB (ml + 2ms)B (3)
B field splits the degeneracies of spin ↑ vs. spin ↓ and for the different Lz.
• Add back spin-orbit term (iii) to the strong field Zeeman case: treat δHLS as a perturbation.
First order perturbation theory then gives:
En l ml ms = En ml ms + 〈n l ml s ms| ξ(r) L · S |n l ml s ms〉Remembering that eigenstates of H0 factor into radial (Rnl(r)) and angular (Ylml) and spin parts,
– angular and spin part:
〈l ml s ms| L · S |l ml s ms〉 = 〈l ml s ms| Lz · Sz |l ml s ms〉 = ml ms ~2
because 〈l ml s ms| Lx |l ml s ms〉 = 〈l ml s ms| Ly |l ml s ms〉 = 0
– radial part: (for l > 0, otherwise, this term is zero)
λnl = ~2〈n l| ξ(r) |n l〉 = ~2
∫ ∞0
dr r2 Rnl(r) ξ(r) Rnl(r) = −(
e2
4πε0~c
)2z2
n
Enl(l + 1/2)(l + 1)
– Collecting all, (this is known as the Paschen-Back effect):
En l ml ms = En ml ms +ml msλnl = En ml ms −(
e2
4πε0~c
)2Z2
n
ml ms Enl(l + 1/2)(l + 1)
(4)
In addition to the ml, ms splitting of energies in the strong field Zeeman effect,
there is now also the n, l splitting in the Paschen-Beck effect.
11.1.5 Weak field Zeeman effect
• Now, treat B-dependent terms (i) and (ii) as perturbations, and group the spin-orbit (L ·S) term
(iii) with the unperturbed H0.
• Key: From previous lectures on addition of the spin and orbital angular momentum, define
J = L+S. Now these operators J2, Jz,L2,S2 all commute with L ·S, hence the good quantum
numbers that label the eigenstates of H0 + ξ(r) L · S are now the eigenvalues of these operators:
j,mj, l, s. For a single electron, s can only be 1/2 and we drop this label.
84
• So instead of the previous states |n l ml s ms〉, we label the states by |n l j mj〉. Thus,
J2|n l j mj〉 = j(j + 1)~2 |n l j mj〉 j = l ± s = l ± 1/2
J |n l j mj〉 = mj ~ |n l j mj〉 mj = −j,−j + 1, . . . , j
This is another example of expanding one set of states in another basis. In particular, inserting a
completeness relation∑
ml ms|n l ml ms〉〈n l ml ms| = 1,
|n l j mj〉 =
(∑ml ms
|n l ml ms〉〈n l ml ms|)|n l j mj〉 =
∑ml ms
cn lmlms ; jmj|n l j mj〉
where cn lmlms ; jmj= 〈n l ml ms|n l j mj〉 are the so-called Clebsch-Gordon coefficients.
• In this new basis, we can compute the energy change for H0 due to the spin-orbit term. Again,
we can separate the L · S from the ξ(r) part.
– For the angular part, useful trick: rewrite L · S = (J2 − L2 − S2)/2. Then this operator
acting on the state |n l j mj〉 just gives the eigenvalue ~2[j(j + 1) − l(l + 1) − s(s + 1)]/2
with s = 1/2.
– For the radial part, for l > 0, (just as in the previous section):
〈n l|ξ(r)|n l〉 =1
2m2c2Ze2
4πε0
Z3
a30n
3l(l + 1/2)(l + 1)
Combining:
δEnjl = 〈n l j mj|ξ(r) L · S|n l j mj〉=
1
2~2
[j(j + 1)− l(l + 1)− 3
4
]1
2m2c2Ze2
4πε0
Z3
a30n
3l(l + 1/2)(l + 1)
• It can be shown that the perturbation due to the weak magnetic field is then:
δEmj = 〈n l j mj| µB~ (Lz + 2Sz)B |n l j mj〉 = gL µB mj B
gL = 1 +j(j + 1) + s(s+ 1)− l(l + 1)
2j(j + 1)(Lande g − factor)
• Finally, collecting all, the perturbed energy in the weak B-field regime is:
Enjmj l = En + δEnjl + δEmj
85
12 Approximate Methods IV
12.1 Variational Method
This approximation method does not require a small perturbation to a known problem, unlike for
perturbation theory. So this can be useful, but the method does rely on having good guesses for the
trial wavefunction, and gives a bound for the ground state energy only.
12.1.1 Variational principle
We want to solve the problem:
H|ψi〉 = Ei|ψi〉
This defines the energy eigenstates and eigenvalues (energies) for the HamiltonianH. AsH is Hermitian,
we can always find (or construct) a complete orthonormal basis such that 〈ψi|ψj〉 = δij.
• Now we do not know what these states or energies are. We can guess a state Φvar which can be
written generally as:
|Φvar〉 =∑i
ci|ψi〉
• Then the expectation of the energy is:
Evar(Φvar) =〈Φvar|H|Φvar〉〈Φvar|Φvar〉 =
∑i |ci|2 Ei∑i |ci|2
(12.1)
• Now subtract the ground state energy E0 on both sides of Eq. 12.1:
Evar(Φvar)− E0 =
∑i |ci|2 (Ei − E0)∑
i |ci|2(12.2)
By definition, Ei ≥ E0, (equality if and only if state i = 0 is the ground state). Thus every term
on the right hand side of Eq. 12.2 is positive, which implies that the left side is also positive. This
leads to:
Variational Principle: Any approximate trial wavefunction gives an energy that is always above (or
at best equal to) the true ground state energy E0.
86
Evar(Φvar) =〈Φvar|H|Φvar〉〈Φvar|Φvar〉 ≥ E0
Strategy: Suppose the trial wavefunction depends on a parameter a. Then Evar(Φ(a)) can be mini-
mized with respect to a to get the best approximation to the ground state energy.
Best approximation when:dEvar(Φvar(a))
da= 0
• Special case: suppose our trial wavefucntion is in fact the ground state |ψ0〉 with energy E0, plus
a little bit of some excited state |ψe〉 at energy Ee (> E0):
|Φvar〉 =1√
1 + |ce|2(|ψ0〉+ ce|ψe〉)
(This state is already normalised.) We assume |ce| 1.
• Then,
Evar(Φvar) =〈Φvar|H|Φvar〉〈Φvar|Φvar〉 =
E0 + |ce|2Ee1 + |ce|2 ≈ E0 + |ce|2(Ee − E0) > E0
Important property: error in wavefunction ∼ |ce|, yet error in ground state energy is only ∼ |ce|2.
12.1.2 Ex.1: 1D simple harmonic oscillator
H =p2x
2m+
1
2mω2x2 H|n〉 = En|n〉
Of course in this example, we know the exact energies En = (n + 1/2) ~ω, and exact eigenstates |n〉,eg. the ground state φn(x) ∝ exp−[(x/α)2/2]. But suppose we do not. Now the ground state is even
in x, so lets try a normalised wavefunction that is also even in x:
Φvar(x, a) =
√2a3
π
1
x2 + a2.
Then,
Evar(Φvar(a)) = 〈Φvar(a)|H|Φvar(a)〉 =~2
4ma2+
1
2mω2a2
By the variational principle, to find the best approximate energy, minimize Evar with respect to the
parameter a:
dEvar(Φvar(a))
da= 0 =⇒ a2
min =~√
2mω=⇒ Evar(Φvar(amin)) =
1√2~ω .
This is indeed bigger than the true ground state energy of ~ω/2.
87
12.1.3 Ex.2: Ground state energy of Helium
Simplest model of He (which has 2 electrons and a nuclear charge Z = +2e):
HHe = − ~2
2m
(∇2r1
+∇2r2
)− 2e2
4πε0
(1
r1+
1
r2
)+
e2
(4πε0) |r1 − r2|The last term describes the Coulomb interaction between the two electrons. The first two terms are just
like two copies of Hydrogen-like atom, with kinetic energy and Coulomb energies between the nucleus
(charge +2e) and one electron at r1 and another at r2.
• If we drop the interaction between the two electrons, the ground state wavefunction would be:
φ0var(r1, r2) = ψ100(r1) ψ100(r2) =
8
πa30
e−2(r1+r2)/a0 =⇒ Evar(φ0) = 2EH(Z=2)0
where EH(Z=2)0 is just the ground state energy for one electron in a hydrogen-like atom of nuclear
charge Z = 2. That the variational He energy is just twice the hydrogen-like atom energy is
because when there is no interaction between the two electrons, the He Hamiltonian just factors
out to precisely two copies of Hamiltonian of the hydrogen-like atom.
• We can do better: Key physics: each electron sees the nuclear charge partially screened by the
other electron. So we can try an effective nuclear charge Z < 2, i.e.,
Φ0var(r1, r2, Z) =
1
π
(Z
a0
)3
e−(r1+r2) Z/a0
• Calculation gives that (E1 is the Rydberg energy):
〈Φ0var(Z)|HHe|Φ0
var(Z)〉 =
(−2Z2 +
27
4Z
)E1
Minimizing this with respect to Z gives:
Zmin =27
16≈ 1.688 < 2
Indeed there is some screening.
13 Approximate Methods V
In experiment, we sometimes apply a perturbation like an external force like electric field etc, starting
at some particular time. Also, it is common to apply a perturbation that is inherently time-depedent,
like a sinusoidally varying electric field. Thus, we need to be able to solve perturbation theory which
has time dependence: this in fact reveals very rich quantum dynamics.
88
13.1 Time-dependent Perturbation Theory
We start with the unperturbed Hamiltonian with its associated unperturbed energies and unperturbed
eigenstates: H0|n〉 = En|n〉.
• Now at time t = 0, we switch on a perturbation λV (t):
H(t < 0) = H0 , H(t ≥ 0) = H0 + λV (t)
Then we have the time-depedent Schrodinger equation (TDSE):
i~d
dt|ψ(t)〉 = [H0 + λV (t)] |ψ(t)〉 (13.1)
which describes the time evolution of the state |ψ(t)〉.
• Let us assume that the system is prepared initially at t ≤ 0 to be in the unperturbed eigenstate
|i〉 where H0|i〉 = Ei|i〉:
|ψ(t = 0)〉 = |i〉
• We want to find the probability that due to the perturbation, there is a transition from |i〉 to
some other state |f〉, where |f〉 is also an unperturbed eigenstate of H0:
Pfi(t) = |〈f |ψ(t)〉|2
(The subscript i is to remind ourselves that the intial state is |i〉.)
• Since both initial and final states are from the set of unperturbed eigenstates, which form a
complete, orthonormal set, we expand the unknown state |ψ(t)〉 in these:
|ψ(t)〉 =∑n
cn(t)|n〉
Substitute this into TDSE, and multiply by a bra 〈k| to the left:
i~d
dtck(t) = ck(t)Ek +
∑n
λ Vkn(t) cn(t) (13.2)
where we have defined the matrix element Vkn(t) = 〈k|V (t)|n〉. We have as usual, used the
orthonormality property 〈k|n〉 = δk n .
– For λ = 0 (no perturbation), clearly, the solution to Eq.13.2 is just ck(t) = bk exp(−iEkt/~),
with bk a constant that can be determined from the initial condition.
89
– Key: we guess that when λ 6= 0 (but still small), ck(t) still looks quite like bk, at least shortly
after the pertubration is switched on, so we guess:
ck(t) = bk(t) exp(−iEkt/~) (13.3)
The idea is essentially that exp(−iEkt/~) deals with the time evolution of the H0 bit, and the
new coefficient bk(t) deals with any leftover time dependence coming from the perturbation.
Substituting Eq.13.3 into Eq.13.2, cancelling common terms and multiplying by eiEkt/~ on both
sides gives:
i~d
dtbk(t) = λ
∑n
exp(i ωknt) Vkn(t) bn(t) (13.4)
where ωkn = (Ek − En)/~ is the energy difference between states |k〉 and |n〉 . So far, this is still
an exact equation derived from the TDSE.
• Now, just as for time-independent perturbation theory, we expand the unknown coefficient bk(t)
in a series in the small parameter λ:
bk(t) = b(0)k (t) + λ b
(1)k (t) + λ2 b
(2)k (t) + · · · (13.5)
where the superscripts (0), (1) etc identify the order of the coefficients.
Again, just as for time-independent perturbation theory, substitute Eq.13.5 into Eq.13.4 and
equate term by term the bits multiplying a particular order in λ:
λ0eqn : i~ ddtb(0)k (t) = 0
λ1eqn : i~ ddtb(1)k (t) =
∑n
exp(i ωknt) Vkn(t) b(0)n (t)
(One can see that generally, the left hand side i~ ddtb(q)k (t) is proportional to on the right had side
one order lower b(q−1)n (t), but we do not need this here.)
• For the λ0 equation, we can immediately solve it, and use the initial condition that |ψ(t = 0)〉 = |i〉,which means the only non-zero coefficient is for k = i:
λ0 eqn =⇒ b(0)k (t) = b
(0)k (t = 0) = δk i (13.6)
We can also integrate the λ0 equation directly, and use the initial condition of Eq.13.6 to get rid
of the sums (since only one term n = i is non zero and contributes to the sum):
λ1 eqn =⇒ b(1)k (t) =
1
i~∑n
∫ t
0
dt exp(i ωknt) Vkn(t) b(0)n (t) =
1
i~
∫ t
0
dt exp(i ωki t) Vki(t) (13.7)
This is the key result we need for calculating transition probabilities:
90
• The transition probability from |i〉 to some other state |f〉 at t is:
Pfi(t) = |〈f |ψ(t)〉|2 =
∣∣∣∣∣∑n
bn(t)〈f |n〉∣∣∣∣∣2
= |bf (t)|2 ≈∣∣∣b(0)f (0) + λ b
(1)f (t)
∣∣∣2For the state |f〉 6= |i〉, we then get:
Pfi(t) ≈∣∣∣λ b(1)
f (t)∣∣∣2 =
(λ
~
)2 ∣∣∣∣∫ t
0
dt exp(i ωfi t) Vfi(t)
∣∣∣∣2 (8)
13.2 Application to a special case: sinusoidal or constant in time pertur-
bation
Let V (t) = V0 cos(Ω t), with Ω the frequency of the perturbation. Then
Vfi(t) = Vfi cos(Ω t) , Vfi = 〈f |V0|i〉
Doing the time integral in Eq.8 gives:
Pfi(t,Ω) ≈(λ|Vfi|
2~
)2 ∣∣∣∣exp [it(ωfi + Ω)]− 1
ωfi + Ω+
exp [it(ωfi − Ω)]− 1
ωfi − Ω
∣∣∣∣2 (9)
A special case of Eq.(9) is when the frequency of the perturbation Ω = 0, which is a constant pertur-
bation (except that it is switched on at t > 0):
Pfi(t,Ω = 0) ≈(λ|Vfi|
~
)2 ∣∣∣∣exp [it ωfi]− 1
ωfi
∣∣∣∣2 =
(λ|Vfi|
~
)2 [sin (t ωfi/2)
ωfi/2
]2
(9a)
• Selection rules: For either Ω = 0 or Ω 6= 0 cases, if the perturbation matrix element Vfi =
〈f |V0|i〉 = 0, there can be no transition at all from |i〉 to this state |f〉 at any t. This is the basis
for selection rules in atomic and molecular physics.
• Resonant transition: From Eq. 9, if Ω ≈ ±ωfi, the probability very quickly grows to one
(and then the perturbation theory breaks down). This means the system very quickly makes the
transition into this specific state |f〉 when the resonance condition Ω = ±ωfi is satisfied. Lets take
Ω > 0. Then, if Ef > Ei, the system absorbs a quanta of energy of ~Ω from the perturbation
to go from |i〉 to state |f〉. Conversely, if Ef < Ei, the system has an induced emission when
going from |i〉 to state |f〉.
91
Eg. absorption case: dropping the ωfi + Ω term being much smaller than the resonant term
ωfi + Ω, we get:
Pfi(t,Ω > 0) ≈(λ|Vfi|
2~
)2 ∣∣∣∣exp [it (ωfi − Ω)]− 1
(ωfi − Ω)
∣∣∣∣2 =
(λ|Vfi|
2~
)2 [sin (t (ωfi − Ω)/2)
(ωfi − Ω)/2
]2
This has a maximum at ωfi = Ω, with height(λ|Vfi|
2~
)2
t2 and hence grow quickly with time. The
peak at that position has width 4π/t which gets sharper in time.
14 Revision on Hydrogen-like atom
These notes are essentially a revision of Y2 quantum mechanics: it also serves as a list of learning
outcomes.
14.1 Hydrogen-like atoms
single e− in a central (Coulomb) potential of a charge Ze nucleus
• Central potential: only depends on radial distance r between e− and nucleus; Coulomb means 1/r
dependence:
V (r) = − Ze2
(4πε0)r
Hamiltonian is then:
H = − ~2
2m∇2 − Ze2
(4πε0)r(1)
and this is to be used in the time-independent Schrodinger equation Hψ(r) = Eψ(r).
• Solution using separation of variables
ψ(r)→ ψ(r)nlm = Rnl(r)Ylm(θ, φ) (2)
Remember: in spherical polar, (x = r sin θ cosφ, y = r sin θ sinφ, z = r cos θ):
∇2 → 1
r2
∂
∂r
(r2 ∂
∂r
)− L2
~2r2L2 = −~2
[1
sin θ
∂
∂θ
(sin θ
∂
∂θ
)+
1
sin2 θ
∂2
∂φ2
]where L is the angular momentum operator. Substituting (2) into (1) leads to a separation into
two equations: radial and angular.
92
14.2 Angular (θ, φ) equation
Thanks to central potential: [L2, V (r)] = 0. But also, [L2,− ~2
2m∇2] = 0. Thus [L2, H] = 0 and in fact
[L, H] = 0. This means:
• Wave function of H atom is also eigen-fucntion of L operator. Indeed,
L2 Ylm(θ, φ) = l(l + 1) ~2 Ylm(θ, φ) Ylm(θ, φ) = Pml (θ, φ) eimφ (3)
where the orbital angular momentum quantum number l and azimuthal quantum number m
satisfies:
l = integers ≥ 0, −l ≤ m ≤ l
and the associated Laguerre polynomials:
l m Pml (θ, φ)
0 0 1
1 0 cos θ
1 ± 1 sin θ
2 0 3 cos2 θ − 1
2 ± 1 sin θ cos θ
2 ± 2 sin2 θ
• Ylm(θ, φ) are spherical harmonics, they satisfiy orthonormality:∫ 2π
0
dφ
∫ π
0
dθ sin θ Y ∗l′m′(θ, φ) Ylm(θ, φ) = δl l′ δm m′
Or in bra-ket notation:
〈l′,m′|l,m〉 = δl l′ δm m′
where δl l′ = 1 if and only if l = l′, otherwise when l 6= l′, δl l′ = 0.
14.3 Radial (r) equation
• Substituting (3), (2) into (1) gives the radial equation:[− ~2
2m
(1
r2
∂
∂r
(r2 ∂
∂r
)− L2
~2r2
)+l(l + 1)~2
2mr2− Ze2
(4πε0)r
]Rnl(r) = EnRnl(r) (4)
93
Remember that by definition of polar coordinates, r > 0.
• Eq. (4): no m-dependence =⇒ energies En has no m-dependence, i.e. (2l + 1) fold degen-
eracy (same energy for −l ≤ m ≤ l). This comes from the central potential which is spher-
ically symmetric =⇒ H that is rotationally invariant, and so energy cannot depend on m.
(Remember, [L, H] = 0, and m is the projection of the angular momentum on the z-axis:
Lz = −i~ ∂∂φ, LzYlm(θ, φ) = m~ Ylm(θ, φ).)
• Behaviour of Rnl(r): define unl(r) = rRnl(r). Then Eq. (4) becomes:
− ~2
2m
d2unl(r)
dr2+ Veff (r)unl(r) = Enunl(r) , Veff (r) = − Ze2
(4πε0)r+l(l + 1)~2
2mr2(5)
This is a 1-dimensional Schrodinger equation, with the first term in Veff for the Coulomb potential,
the second part is for the centrifugal potential.
• General solution of Eq. (5) involves the Laguerre polynomials with a new quantum number n, in
addition to the angular momentum l, with the restriction:
n = 1, 2, 3, . . . , l = 0, 1, 2, . . . , n− 1
n l Rnl(r)
1 0 2(za
)3/2exp− (Zr
a
)2 0 2
(za
)3/2 (1− zr
2a
)exp− (Zr
2a
)2 1 1√
3
(z2a
)3/2 ( zra
)exp− (Zr
2a
) a = 4πε0~2
me2≡ a0 is the Bohr radius.
The corresponding energies are of the Bohr form:
En = − m
2~2
(Ze2
(4πε0)r
)21
n2n = 1, 2, 3, . . .
Note: En has no angular momentum l dependence: this is special to Coulomb interaction.
• These are bound state energies: En < 0: ie it costs energy to take an electron far away from
the nucleus. Indeed, the exp− (Zra
)bit ensures the electron to be close to the nucleus. The
Schrodinger equation (5) also has E > 0 solutions: these are scattering states where a far away
energetic electron can just ”bounce off” the nucleus and run off to very far away again. We do
not study these here.
94
• Spectroscopic notation: the different l wavefunctions (also called orbitals in atomic physics) are
given names:l = 0 1 2 3 · · ·
label s p d f · · ·
95
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