Simple one-dimensional potentialstheory.tifr.res.in/~sgupta/courses/qm2008/lec9.pdf · Quantizing the Fourier modes gives a simple ﬁrst quantized ﬁeld theory. If one wants to

Simple one-dimensional potentials

Sourendu Gupta

TIFR Graduate School

Quantum Mechanics 1August 25, 2008

Sourendu Gupta (TIFR Graduate School) Simple one-dimensional potentials QM I 1 / 24

Outline

1 The harmonic oscillator

2 A charged particle in a magnetic field

3 The isotropic two-dimensional harmonic oscillator

4 References


The harmonic oscillator

Outline




4 References



The Hamiltonian for a harmonic oscillator

The harmonic oscillator is a system which obeys, Hooke’s law, i.e., theforce on the oscillator is proportional to the displacement from itsequilibrium position and points towards that position. Hence the potentialis V (x) = mω2x2/2. The harmonic oscillator Hamiltonian is

H =1

2mp2 +

mω2

2x2.

The Hamiltonian above describes ellipses in phase space: this is theclassical motion of harmonic oscillators. Both the position and themomentum vary with time harmonically, hence the name.The quantum Hamiltonian is obtained as usual by using the operatorforms for p and x . In terms of the dimensionless quantities

X =

√

mω

~x and P =

1√mω~

p, one has H =~ω

2

(

P2 + X 2)

.

Note that [X ,P] = i ; a result which follows from the canonicalcommutator.



Examples

1 Near the minimum of any smooth potential, one can approximate thepotential by its Taylor expansion:V (x) = V (x0) + V ′′(x0)(x − x0)

2/2 + · · · , where x0 is position of theminimum, i.e., the point where the first derivative vanishes. Low-lyingvibrational modes of molecules show almost harmonic spectrum as aresult of this general fact.

2 Complicated many-body interactions in a nucleus can be expanded ina similar Taylor series to give a central potential which isapproximately harmonic. The addition of simple extra terms in theHamiltonian then explain the spectra of many complex nuclei.

3 The motion of a charged particle in a magnetic field is a disguisedversion of a harmonic oscillator. Since the particle travels in a helicalpath around the field lines, the momentum components transverse tothe field lines play the roles of X and P.



Example: normal modes and field theories

The vibrations of rigid solids gives a theory of many independent harmonicoscillator. A linear chain of N atoms bound to each other has the lowestorder potential

V = −1

2

N∑

i=1

(xi − xi+1)2 + · · ·

Assume that the chain closes on itself, so that xN+1 = x1. Fouriertransforming the coordinates xi then decomposes the coupled Hamiltonianinto a sum proportional to

∑

k x2k . Thus, the coupled problem decomposes

into a sum of N independent harmonically oscillating Fourier models. Thisis an example of a classical field theory.Quantizing the Fourier modes gives a simple first quantized field theory. Ifone wants to take into account the cubic, quartic etc., terms in thepotential then these higher order terms can be treated as couplingsbetween the oscillator modes.



Raising and lowering operators

Construct the operators

a =1√2(X + iP), and a† =

1√2(X − iP).

Both a and a† are non-Hermitean, and are conjugates of each other. Thecommutator, [a, a†] = 1, follows using that for X and P. Now consider theHermitean operator N = a†a. One sees that [N, a] = −a and [N, a†] = a†.If |z〉 is an eigenvector of N with eigenvalue z , then [N, a]|z〉 = −a|z〉, andhence Na|z〉 = (z − 1)a|z〉, i.e., a lowers the eigenvalue of N by one unit.Using the other commutator, one can show that a† raises the eigenvalue ofN by one unit.For any |ψ〉, let |φ〉 = a|ψ〉. Since 〈φ|φ〉 ≥ 0 and 〈φ|φ〉 = 〈ψ|N|ψ〉, onefinds that every eigenvalue of N must be greater than or equal to zero. Ifthe eigenvalues of N are not integers, then there cannot be a lower boundto the eigenvalues, since a will always lower the eigenvalue by one unit.So, there is a vector |0〉 such that a|0〉 = 0, and N|0〉 = 0.



The eigenvalues of the Hamiltonian

One finds that

N =1

2(X−iP)(X+iP) =

1

2(X 2+P2−1), hence H = ~ω

(

N +1

2

)

.

Thus, the eigenvalues of H are E = ~ω(n + 1/2), for integer n ≥ 0. Thelowest state satisfies

(

mω

~x +

d

dx

)

ψ0(x) = 0, i .e., φ0(x) ∝ exp

(

−mωx2

2~

)

.

Since this is a first order differential equation, there is an unique solution.Thus, the lowest eigenvalue of N (and hence, of H) is unique. Then, byinduction with a† we can show that none of the states are degenerate.If |n − 1〉 = cna|n〉, then 〈n − 1|n − 1〉 = |cn|2n〈n|n〉. If |n − 1〉 isnormalized, then normalization of |n〉 requires cn = 1/

√n. Then

|n〉 = a†|n − 1〉/√n = (a†)n|0〉/√

n!.



All wavefunctions at one go

The generating function for the wavefunctions is—

G (z ; x) =

∞∑

n=0

zn

√n!ψn(x) =

∞∑

n=0

zn

√n!〈x |(a

†)n√n!

|ψ0〉 = 〈x |eza† |ψ0〉.

Now, from the Baker-Campbell-Hausdorff formula, we find thatexp(za†) = exp(zX/

√2) exp(−izP/

√2) exp(−z2/4). Hence

G (z ; x) = exp

(

−z2

4+ zx

√

mω

2~

)

〈x |e−izP/√

2|ψ0〉.

Then using the action of the exponential on the bra, and the normalizedground state wavefunction

ψ0(x) =(mω

π~

)1/4e−mωx2/2~,

we find that

G (z ; x) =(mω

π~

)1/4exp

(

−mωx2

2~+

√

2mω

~zx − z2

2

)

.



The Hermite polynomials

Define the function f (x) = exp(−x2) and its n-th derivative,f (n)(x) = (−1)nHn(x) exp(−x2). The Hn(x) are called Hermitepolynomials (prove that they are polynomials). By directdifferentiation one can obtain the recurrence relation

Hn(x) =

[

2x − d

dx

]

Hn−1(x).

From the definition of the Hermite polynomials it is clear that

e−z2+2zx = e

x2f (x − z)2 =

∞∑

n=0

zn

n!Hn(x).

From the recurrence relation we can write down the explicit forms—

H0(x) = 1

H1(x) = 2x

H2(x) = 4x2 − 2.

Note that the even numbered polynomials have even parity.Sourendu Gupta (TIFR Graduate School) Simple one-dimensional potentials QM I 10 / 24


The harmonic oscillator wave functions

Comparing the recurrence relation for harmonic oscillator wave functions

G (z ; x) =(mω

π~

)1/4exp

(

−mωx2

2~+

√

2mω

~zx − z2

2

)

with that for Hermite polynomials,

e−z2+2zx = e

x2f (x − z)2 =

∞∑

n=0

zn

n!Hn(x),

we find that

ψn(x) =(mω

π~

)1/4 1√2nn!

Hn(X )e−X 2/2.

Note that the parity flips for every successive n. The wavefunction ψn

(and hence the Hermite polynomial, Hn) has exactly n zeroes. The zeroesof successive polynomials are interleaved. The Hermite polynomials areorthogonal under the measure dx exp(−x2).



Classical-quantum correspondence

Since X = (a + a†)/√

2 and P = −i(a − a†)/√

2, and the operators a anda† have vanishing diagonal elements in the eigenbasis of N, it is clear that〈n|X |n〉 = 〈n|P|n〉 = 0. Squaring each of these operators, we find that〈n|X 2|n〉 = 〈n|P2|n〉 = n + 1/2. Clearly, then one has 〈V 〉 = 〈T 〉 = E/2.These relations are the same as for a classical harmonic oscillator.The commutators [H,X ] = −iP and [H,P] = iX follow from thecommutators of N with a and a†. Then

d〈X 〉dt

=d

dt〈ψ|eiHt/~X e

−iHt/~|ψ〉 = i〈[H,X ]〉= ~ω〈P〉

d〈P〉dt

= −~ω〈X 〉.

These equations are the same as the classical Hamilton’s equations.



Schrodinger and Heisenberg pictures

We have chosen a representation of quantum evolution in which theoperators corresponding to time-independent classical variables remaintime independent, and the states evolve by the action of the unitaryevolution operator. This is called the Schrodinger picture of quantumdynamics.Observables 〈ψ′(t)|O|ψ(t)〉 are time dependent through unitary evolutionof states 〈ψ′(0)|U†(t)OU(t)|ψ(0)〉. Therefore they remain unchanged ifwe use the Heisenberg picture, in which states are time independent andoperators evolve with time through the adjoint action of the unitaryevolution operator. The time evolution of expectation values can becomputed exactly as for the harmonic oscillator. Since the analogousequations hold in any state, one can interpret this as the operatorevolution equation in the Heisenberg picture—

dO

dt= i [H,O].



The thermal density matrix

For a single harmonic oscillator placed inside a heat bath, one finds thepartition function

Z (β) = Tr exp(−Hβ) = e−~ωβ/2

∞∑

n=0

e−~ωβn =

exp(−~ωβ/2)

1 − exp(−~ωβ),

where β = 1/kBT . Since ρ(T ) = exp(−Hβ)/Z , the expectation value ofthe energy is

〈H〉 =1

ZTr H exp(−βH) = − 1

Z

dZ

dβ= −d log Z

dβ.

Using the expression for Z above, we get

〈H〉 =1

2~ω +

~ω

exp(~ωβ) − 1.

The Planck spectrum begins to emerge.Sourendu Gupta (TIFR Graduate School) Simple one-dimensional potentials QM I 14 / 24

A charged particle in a magnetic field

Outline




4 References



The classical theory

The Hamiltonian of a particle of charge e and mass m in a magnetic field is

H =1

2m(P − eA)2, B = ∇× A.

The classical equations of motion which follow from this are

dr

dt=

p

m,

dp

dt=

e

mp × B.

Taking the z-direction to be the direction of B, we obtain, pz(t) = pz(0), and

d

dt

(

px

py

)

= iωσ2

(

px

py

)

,

where the cyclotron frequency is given by ω = eB/m. On scaling the time by a

factor ω, it is clear that these equations of motion can be obtained from a

fictitious Hamiltonian H ′ = (p2x + p2

y )/2. The harmonic solutions are obtained

from the above equations by an unitary transformation to p± = (px ± ipy )/√

2.

The solutions are p±(t) = p±(0) exp(∓iωt). Since the phase space is 6d and the

motion is integrable, there are three conserved quantities: E , pz and |p±|.Sourendu Gupta (TIFR Graduate School) Simple one-dimensional potentials QM I 16 / 24


Quantization

Introduce P = p − eA. Then [Pj ,Pk ] = iǫjklBl . With the choice made ofthe coordinate axes, one sees that [Pj ,Pz ] = 0 for all j . Since H = P2/2m,this implies that [H,Pz ] = 0. The remainder of the Hamiltonian is

H ′ = (P2x + P2

y )/2m, where [Px ,Py ] = ieB.

Since [H,H ′] = 0, N is a conserved quantum number. Therefore theHamiltonian is exactly that of the harmonic oscillator. The eigenvalues arecalled Landau levels, and are given by

E (N, kz) =~

2

2mk2z + ~ω

(

N +1

2

)

.

The third operator which commutes with H has a continuous spectrum,hence the Landau levels are infinitely degenerate. (Find the thirdcommuting operator)


The isotropic two-dimensional harmonic oscillator

Outline




4 References



The energy eigenvalues and eigenvectors

The isotropic harmonic oscillator in two dimensions is specified by the two positionvariables x1 and x2 and the two conjugate momenta p1 and p2. Isotropy impliesthat the angular frequency, ω is the same in all directions. Then introducing thescaled quantities Xi = xi

√

mω/~ and Pi = p/√

mω~, one has the Hamiltonian

H =~ω

2(P2

1 + P22 + X 2

1 + X 22 ).

Introducing the shift operators aj = (Xj + iPj)/√

2 and their Hermitean

conjugates, a†j , as before, one can show that H can be written in terms of two

number operators Nj = a†j aj in the form H = ~ω[(N1 + 1/2) + (N2 + 1/2)]. As a

result, the energy eigenstates can be specified in the form |n1, n2〉 where ni are

the eigenfunctions of Ni . The energies of these states are E = ~ω(n1 + n2 + 1).

Equivalently, one could write the eigenstates as |N, n1〉 and energies as

E = ~ω(N + 1). Since n1 does not enter the expression for the energy, and it has

the range 0 ≤ n1 ≤ N, the eigenvalue is (N + 1)-fold degenerate. Therefore there

is a larger symmetry in the problem.



Extended symmetry

The degeneracy of the states, |N, n1〉 due to the multiplicity of n1 meansthat if one changes n1 and n2 simultaneously, while keeping N = n1 + n2

fixed, then the energy does not change. An operator of the form a†1a2 does

precisely this. The two Hermitean operators s1 = a†1a2 + a

†2a1 and

s2 = ia†1a2 − ia

†2a1 acting on a level |N, n1〉 produce linear combinations of

|N, n1 − 1〉 and |N, n1 + 1〉. Check this logic by evaluating [H, s1] and[H, s2]. Construct the full algebra using [s1, s2], and commutatorswith any new operators formed in this process)This process generates the combinations s0 = N1 + N2, s1, s2 ands3 = N1 − N2. The commutation relations between these operators is thesame as that between Pauli matrices. (Check that in the subspace ofE = 2~ω these operators are exactly the Pauli matrices). However,when acting in the eigenspace of larger values of E , the opertors arerepresented by larger matrices. Compute the operator S2 = s2

1 + s22 + s2

3 inthe degenerate subspace of any E . Also compute the matrix representationof s3 in this subspace.



The symmetry group SU(2)

An arbitrary (new) linear combination of the degenerate eigenstates of theisotropic two-dimensional harmonic oscillator is generated by the unitary

matrix U = exp(

i∑

j θjsj

)

. Note that Det U = 1 (because the trace of its

logarithm is zero). Since these linear combinations all have the sameenergy, all these U must commute with the Hamiltonian.In particular, this is true of the two-dimensional subspace with N = 1. All2 × 2 unitary matrices with unit determinant form a group. This is calledthe group SU(2). Since all these matrices commute with H, thesymmetry group of this problem is SU(2). The higher dimensionalmatrices generated by the above prescription do not exhaust all possibleunitary matrices of that size, but a subgroup which is isomorphic to SU(2).These matrices of different sizes are called different representations ofSU(2). The trace of S2 in each representation is characteristic of thatrepresentation. The Hermitean operators s1, s2 and s3 are called thegenerators of SU(2), or elements of the algebra su(2).



A problem

Consider the isotropic harmonic oscillator in three dimensions. In analogywith the construction we have presented here, find the complete group ofsymmetries of this problem: it is called SU(3).

1 Construct the complete algebra of operators from Hermiteancombinations of the bilinears of the shift operators which leave theenergy unchanged.

2 Find the commutators of these operators, and construct thecompletion of this algebra. How many operators are there in thealgebra?

3 Find a complete set of commuting operators among these.

4 In the degenerate space of eigenstates corresponding to the energyeigenvalue E = 5~ω/2, construct the representations of the elementsof the algebra.

5 Construct the representation of the algebra in the space of energyeigenstates with eigenvalue E = 7~ω/2.


References

Outline




4 References


References

References

1 Quantum Mechanics (Non-relativistic theory), by L. D. Landau and E.M. Lifschitz. The material in this lecture are scattered throughchapters 3 and 15 of this book. The later chapter deals with the topicof Landau levels.

2 Quantum Mechanics (Vol 1), C. Cohen-Tannoudji, B. Diu and F.Laloe. Chapter 5 of this book discusses the harmonic oscillator. Read,in particular, the many applications of this model potential.

3 Solid State Physics, by N. W. Ashcroft and N. D. Mermin. Chapter14 of this book deals with the many applications of Landau levels tosolid state physics phenomena.

4 The article by M. Harvey in the book Advances in Nuclear Physics,vol 1 (Plenum Press, New York) discussed the Elliott Model, which isan application of the SU(3) symmetry of the three dimensionalharmonic oscillator to problems in nuclear physics.

5 Classical groups for Physicists, by B. G. Wybourne. This book ishighly recommended for a good exposition on Lie groups.


Documents

Simple one-dimensional potentialstheory.tifr.res.in/~sgupta/courses/qm2008/lec9.pdf · Quantizing the Fourier modes gives a simple ﬁrst quantized ﬁeld theory. If one wants to