Harmonic potentials, raising and lowering operators

Problem:

The orthonormal set of wave functions for the stationary states of the harmonic oscillator with U(x) = ½mωx2 is
n(η) = Nn Hn(η) exp(-½η2)}, with η = (mω/ħ)½x.
The Hermite polynomials Hn(η) satisfy the recurrence relations
ηHn(η)  = nHn-1(η) + ½Hn+1(η)  and dHn(η)/dη = 2nHn-1(η).
The normalization constants Nn are given by
N0 = π¼, Nn+1 = (2(n+1))Nn.

(a)  Show that the matrix elements of X can be expressed as
<m|X|n> = (nħ/(2mω))½δm,n-1 + ((n+1)ħ/(2mω))½δm,n+1.
(b)  Derive a similar expression for the matrix elements of X2.
(c)  The ladder operators for the harmonic oscillator have the properties
a|n> = √(n+1) |n+1>,   a|n> = √(n) |n-1>.
Derive the expressions for a and a in terms of η and d/dη.

Hint: Start by investigating the action of the operator  d/dη on Ψn(η).

Solution:

• Concepts:
The eigenfunctions of the harmonic oscillator, recurrence relations.
• Reasoning:
To evaluate the matrix elements, we have to evaluate integrals.  This is done with the help of recurrence relations and normalization conditions.
• Details of the calculation:
(a)  <m|X|n> =  (ħ/(mω))½<m|η|n>, <m|n> = δm,n.
<m|η|n> = ∫-∞Φm*(η)ηΦn(η)dη = NmNn-∞Hm(η)ηHn(η) exp(-η2) dη
(use recurrence relation)
<m|η|n> = NmNnn∫-∞Hm(η)Hn-1(η) exp(-η2) dη + NmNn½∫-∞Hm(η)Hn+1(η) exp(-η2) dη
= NmNn(NmNn-1)-1 n∫-∞Φm(η)Φn-1(η) dη + NmNn(NmNn+1)-1 ½∫-∞Φm(η)Φn+1(η)dη
= (Nn/Nn-1)nδm,n-1 + ½ Nn/Nn+1m,n+1 = (n/2)½δm,n-1 + ½(2(n+1))½δm,n-1.
<m|X|n> = (nħ/(nmω))½δm,n-1 + ((n+1)ħ/(2mω))½δm,n+1.

(b)  <m|X2|n> =  (ħ/(mω))<m|η2|n>.
<m|η2|n> = ∫-∞Φm*(η)η2Φn(η)dη = NmNn-∞(mHm-1(η) + ½Hm+1(η))(nHn-1(η) + ½Hn+1(η)) exp(-η2) dη
= NmNn[(mn/(Nm-1Nn-1))δm-1,n-1 + ½n/(Nm+1Nn-1))δm+1,n-1 + ½m/(Nm-1Nn+1))δm-1,n+1 + ¼/(Nm+1Nn+1))δm+1,n+1
= ½(mn)½δm,n + ½(n(m+1))½δm+2,n + ½(m(n+1))½δm-2,n + ¼(2(m+1)2(n+1))½δm,n
= (n + ½)δm,n + ½(n(n-1))½δm+2,n + ½(m(m-1))½δm-2,n.
<m|X2|n> = (ħ/(mω))(n + ½)δm,n + ½(ħ/(mω))(n(n-1))½δm+2,n + ½(ħ/(mω))(m(m-1))½δm-2,n.

(c)  dΦn(η)/dη = Nn(dHn(η)/dη)exp(-½η2) - NnHn(η) η exp(-½η2)
= Nn2nHn-1(η) exp(-½η2) - Nn(nHn-1(η) + ½Hn+1(η)) exp(-½η2).
ηΦn(η) = NnHn(η) η exp(-½η2) = Nn(nHn-1(η) + ½Hn+1(η)) exp(-½η2).
n(η)/dη + ηΦn(η) =  Nn2nHn-1(η) exp(-½η2) = 2n(Nn/Nn-1n-1(η) = √(2n)Φn-1(η)
We have
a|n> = √(n) |n-1>,
therefore a = (1/√2)[η + d/dη].
Similarly,
n(η)/dη - ηΦn(η) =  -NnHn+1(η) exp(-½η2) = -(Nn/Nn+1n+1(η) = -√(2(n+1))Φn+1(η).
We have
a|n> = √(n+1) |n+1>.
therefore a = (1/√2)[η - d/dη].

Problem:

(a)  Construct the normalized, linear combination of one-dimensional harmonic-oscillator states of the form |ψ> = c00> + c11>, with c0 and c1 real, such that the expectation value of the position operator X is maximized.  Here |Φ0> and |Φ1> refer to the ground state and the first excited state, respectively.
(b)  For the above state evaluate the expectation values of the momentum and parity operators.
Useful formulas:

a|n> = √(n)|n-1>,  a|n> = √(n+1)|n+1>.
X = (ħ/(2mω))½(a + a),  P = i(mħω/2)½(a - a).
<x|0> = (mω/(πħ))¼exp(-½mωx2/ħ).
-∞exp(-a2x2)dx = √π/(2a)  (a > 0).

Solution:

• Concepts:
Postulates of quantum mechanics, the 1D harmonic oscillator.
• Reasoning:
The mean value of any observable A is given by <ψ|A|ψ>.  We are asked to maximize the mean value of the observable x, if the particle is in a linear superposition of two eigenstates of the 1D harmonic oscillator.
• Details of the calculation:
(a)  |ψ> = c00> + c11>.  Normalization: |c0|2 + |c1|2 = 1.
Since c0 and c1 are real, we have c02 + c12 = 1.
<X>  = <ψ|X|ψ> = (ħ/(2mω))½<ψ|(a + a)|ψ>
= (ħ/(2mω))½[c0c10|a|Φ1> + c0c11|a0>]
= 2(ħ/(2mω))½c0c1 = 2(ħ/(2mω))½c0(1 – c02)½.
d<X>/dc0 = 0 yields c0 = 1/√2.  We then need c1 = 1/√2 to have a maximum.
|ψ> = (1/√2)|Φ0> + (1/√2)|Φ1>.
(We need c0 and c1 to have the same sign, obviously choosing both coefficients negative would also be ok.)
(b)  <P> = ½i(mħω/2)½[<Φ0|a|Φ1> - <Φ1|a0>] = 0.
Denote the parity operator by Π.
Π|Φ0> = |Φ0>, Π|Φ1> = -|Φ1>.
<Π> = ½[<Φ00> - <Φ11>] = 0.

Problem:

Consider a particle in a harmonic oscillator potential with Hamiltonian H = p2/(2m) + ½mωx2.
Its state vector at  t = 0 is |ψ(0)> = exp(-|α|2/2)∑0n/(√n!))|n>,
where the |n> are the orthonormal eigenstates of H.
(a)  Show that |ψ(0)> is an eigenstate of the lowering operator a and find the eigenvalue.
(b)  If the energy of the particle is measured at t = 0, what values of E can be found and with what probability?
(c)  H = ħω(aa + 1).  Find <H> and ΔH at t = 0.
(d)  Find |ψ(t)> and show that it is still an eigenstate of a.  Find the eigenvalue.
(e)  Find <x>(t).  Why is the state |ψ(0)> called a quasi classical state?

Solution:

• Concepts:
The eigenstates of the harmonic oscillator, raising and lowering operators.
• Reasoning?
The system is in a given linear superposition of eigenstates of the harmonic oscillator.
We can do the problem by letting a and a operate on those eigenstates.
(a)  a|n> = n½|n-1>,  a|n> = (n+1)½|n+1>.
• Details of the calculations:
(a) a|ψ(0)> = exp(-|α|2/2)∑0n/(√n!))a|n> = α exp(-|α|2/2)∑1n(√n)/(√n!))|n-1>
= α exp(-|α|2/2)∑1n-1/√((n-1)!))|n-1> = α exp(-|α|2/2)∑0n/(√n!))|n> = α|ψ(0)>.
The eigenvalue of a is α.
(b)  The possible values are En = (n + ½)ħω.  En is found with probability  exp(-|α|2)(α2n/n!).
(c)  <H> = ħω<ψ(0)|(aa + 1)|ψ(0)> = ħω(<aψ(0)|aψ(0)> + ½) = ħω(|α|2 + ½).
<H2> = ħ2ω2<ψ(0)|(aa + 1)2|ψ(0)> = ħ2ω2(|α|4 + 2|α|2 + ¼).
ΔH = (<H2> - <H>)½ = ħω|α|.
ΔH/<H> = |α|/(|α|2 + ½) ≈ 1/|α| << 1 if |α|  >> 1.
(d)  U(t, 0)|ψ(0)> = exp(-iHt/ħ)|ψ(0)> = exp(-|α|2/2)∑0n/(√n!))exp(-i(n + ½)ωt)|n>
= exp(-iωt/2) exp(-|α|2/2)∑0n/(√n!))exp(-inωt)|n> = exp(-iωt/2) exp(-|α'|2/2)∑0(α'n/(√n!))|n>,
with α' = α exp(-iωt), |α'| = |α|.
|ψ(t)> is an eigenstate of a as shown in part (a) with eigenvalue α'.
(e)  <x(t)> = (ħ/(2mω))½ <a + a>(t) = (ħ/(2mω))½(<aψ(t)|ψ(t)> + <ψ(t)|aψ(t)>)
= (ħ/(2mω))½(α'* + α') = (2ħ/(mω))½Re(α exp(-iωt)).
<x>(t) oscillates with frequency ω like the classical coordinate x.  For large |α| the uncertainty in the energy is small compared to the value of the energy.

Problem:

A one-dimensional harmonic oscillator has mass m and angular frequency ω.  Denoting the momentum by p and the coordinate by x, we can define the operators
a = αx + iβp,  a = αx – iβp,  where α =√(mω/(2ħ)),  β =1/√(2mωħ).
(a)  Find [a, a].
(b)  Find the Hamiltonian in terms of a and a.
(c)  Let |{|n>} denote the eigenstates of the Hamiltonian with eigenvalues ħω(n + ½). Given a|n> = (n + 1)½|n + 1>,  a|n> = n½|n - 1>, find the expectation value of x4 for the system in the state |n>.

Solution:

• Concepts:
The eigenstates of the harmonic oscillator Hamiltonian, the raising and lowering operators, the expectation value of an operator
• Reasoning:
We are asked to express the harmonic oscillator Hamiltonian in terms of a and a.
• Details of the calculation:
(a)  [p,p] = [x,x] = 0.  [x,p] = iħ.
[a, a] = [(αx + iβp),(αx – iβp)] = iβα[p,x] - iβα [x,p] = 2iħβα = 1.
(b)  H = p2/(2m)  + mω2x2/2.
x = (a + a)/(2α),  p = -i(a - a)/(2β).
x2 = (1/(4α2))(aa + aa + aa + aa), p2 = (1/(4β2))(-aa + aa + aa - aa).
H = (ħω/4)(-aa + aa + aa - aa) + (ħω/4)(aa + aa + aa + aa)
= (ħω/2)(aa + aa) = ħω(½ + aa).
(c)  x2 = (1/(4α2))(aa + aa + aa + aa).
<n|x4|n> = (1/(16α4)) <n|(a + a)4|n> has 16 terms, but only those with equal numbers of a and a can be non-zero.  Only 6 terms survive.
<n|aaaa|n> = (n + 1)(n + 2),  <n|aaaa|n> = n(n - 1),  <n|aaaa|n> = (n + 1)2
<n|aaaa|n> = n2, <n|aaaa|n> = n(n + 1), <n|aaaa|n> = n(n + 1).
<n|x4|n> = (6n2 + 6n + 3).

Problem:

For a one-dimensional simple harmonic oscillator we may define raising and lowering operators
a = (mω/(2ħ))½(X + iP/(mω)),  a = (mω/(2ħ))½(X - iP/(mω)),
with properties
a|n> = √(n) |n - 1>,  n ≠ 0, a|n> = 0, b = 0, and
a|n> = √(n + 1) |n + 1>.
(a)  Show by direct calculation that the ground state of the simple harmonic oscillator satisfies
(ΔX)2(ΔP)2 = ¼|<[x,p]>|2,
and hence is a minimum uncertainty state.
(b)  Consider a coherent state of a one-dimensional oscillator, |b>, defined as |b> = exp(-|c|2/2)∑0(cn/(√n!))|n>.
Show that |b> is an eigenstate of the lowering operator a and find the eigenvalue.
(c)  Show that |Ψ(t)> = U(t,0)|b> is an eigenstate of a and find the eigenvalue,
(d)  Show that |Ψ(t)> = U(t,0)|b> is also a minimum uncertainty state.

Solution:

• Concepts:
Postulates of quantum mechanics, the 1D harmonic oscillator.
• Reasoning:
The mean value of any observable A is <ψ|A|ψ>.
The root-mean -quare deviation is ΔA = (<(A - <A>)2>)½ = (<A2> - <A>2)½.
• Details of the calculation:
(a)  Let  Xs = (mω/ħ)½X,  Ps =  (mωħ)P.
a = (2)(Xs + iPs), a = (2)(Xs - iPs).
Xs = (2)(a + a),  PS = i(2)(a - a),  {Xs,Ps] = i.
(ΔX)2 = <X2> - <X>2,  (ΔP)2 = <P2> - <P>2.
For the harmonic oscillator in its ground state <X> = <P> = 0.
<Xs2> = ½<0|aa + aa + aa + aa|0> = ½<0|aa|0> = ½.
<Ps2> = -½<0|aa - aa - aa + aa|0> = ½<0|aa|0> = ½.
<Xs2><Ps2> = ¼.  <X2><P2> = ¼ħ2.  [X,P] = iħ.  |<[X,P]>|2 = ħ2.
Therefore (ΔX)2(ΔP)2 = ¼|<[X,P]>|2,
the ground state is a minimum uncertainty state.

(b)   a|b> = exp(-|c|2/2)∑0(cn/(√n!))a|n> = c exp(-|c|2/2)∑1(cn(√n)/(√n!))|n-1>
= c exp(-|c|2/2)∑1(cn-1/√((n-1)!))|n-1> = c exp(-|c|2/2)∑0(cn/(√n!))|n> = c|b>.
The eigenvalue of a is c.

(c)  |Ψ(t)> = U(t,0)|b> = exp(-iHt/ħ)|b>
= exp(-|c|2/2)∑0(cn/(√n!))exp(-i(n + ½)ωt)|n>
= exp(-iωt/2) exp(-|c|2/2)∑0(cn/(√n!))exp(-inωt)|n>
=  exp(-iωt/2) exp(-|c'|2/2)∑0(c'n/(√n!))|n>,
with c' = c exp(-iωt), |c'| = |c|.
|ψ(t)> is an eigenstate of a  with eigenvalue c’.

(d)  |ψ(t)> is an eigenstate of the lowering operator.
<X(t)> = (ħ/(2mω))½<a + a>(t)
= (ħ/(2mω))½(<aψ(t)|ψ(t)> + <ψ(t)|aψ(t)>)
= (ħ/(2mω))½(c'* + c').
<X(t)>2 = (ħ/(2mω))(c'* + c')2.
<X2(t)> = (ħ/(2mω))<a†2 + a2 + 2aa + 1>(t)
= (ħ/(2mω))((c'* + c')2 + 1).
(ΔX)2 = (ħ/(2mω)).
<P(t)> = i(mωħ/2)½<a - a>(t) = i(mωħ/2)½(c'* - c').
<P(t)>2 = -(mωħ/2)(c'* - c')2.
<P2(t)> = -(mωħ/2)<a†2 + a2 - 2aa - 1>(t)
= -(mωħ/2)((c'* - c')2 - 1).
(ΔP)2 = (mωħ/2).
ΔXΔP = ħ/2.
The coherent state is also a minimum uncertainty state.

Problem:

No potential is harmonic for arbitrarily large displacements from the origin.  Eventually nonlinearities set in.  But, as long as the displacement is small, so that the lowest order quadratic term dominates, we can treat the potential as harmonic.  However, there is zero point motion of the quantum state.  If the extent of the ground state is large compared to the range over which the potential is quadratic, the spectrum in no way looks like that of a SHO.

Consider the potentials V(x) = V0sin2kx.  Find the characteristic oscillation frequency near an equilibrium point.  Determine the extent of the ground state for the corresponding SHO.  By comparing this to the characteristic scale over which the potential is harmonic, estimate the number of levels for which the energy spectrum looks harmonic (i.e. equally spaced).

Solution:

• Concepts:
Taylor series expansion about a local potential minimum
• Reasoning:
For an equilibrium point x0 we have dV/dx|x0 = 0, d2V/dx2|x0 > 0.
V(Δx) = ½d2V/dx2|x0*(Δx)2
Denote d2V/dx2|x0 = V0''.  Then the oscillation frequency is ω = (V0''/m)½.
• Details of the calculation:
We have a periodic function with multiple equilibrium points.  We can pick any of them.
sin kx ~ kx + (1/6)(kx)3,  V0sin2kx = V0(k2x2 + ⅓(kx)4) = ½V0''x2 + ...  .
ω = (2V0k2/m)½.
The value of x when the next term in the expansion of the potential becomes as important as k2x2 is x1 = (3)½/k.
Then k2x2 = ⅓(kx)4.
We need Δx << x1.  For the harmonic oscillator Δx = <x2>½
x = (ħ/(2mω))½(a + a),  a|n> = (n)½|n-1>,  a|n> = (n+1)½|n+1>.
x2 = (ħ/(2mω))(a + a)(a + a), <x2> = (ħ/(mω))(n + ½)
We therefore need (ħ/(mω))n  << 3/k2, or n << (18mV0)½/(ħk).