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L.J. Sham April 24, 2009
Physics 130A Problem Set 4
- Positive energy eigenvalues
(a) Show that the expectation value of kinetic energy is always non-negative by proving,
〈ψ| Kˆ|ψ〉 =
ℏ^2
2 m
dx
∂ψ∗ ∂x
∂ψ ∂x
[Hint: take one differential operator and integrate by parts. Specify conditions needed.] (b) If the potential energy of a particle V (x) ≥ 0 , show that its energy eigenvalues are always non-negative. [Hint: examine the expectation value of the Hamiltonian in an energy eigenstate as the sum of kinetic and potential energies.]
- First excited state of SHO
(a) Find the functional dependence on position (in dimensionless form) of the first excited state from ψ 1 = c†ψ 0 , starting with the normalized Gaussian form of the ground state. (b) Without evaluating the integral, prove that the wave function for the excited state you have obtained is normalized. (c) Again without differentiation nor integration, verify the eigenenergy of the excited state.
- Griffiths, p. 50, Problem 2.12.
- Deduction of Properties of the State. A quantum one-dimensional harmonic oscillator has mass m, classical angular frequency ω, and the bottom of the potential at the origin of the coordinate axis. At time t = 0, it is in a state with (I) equal probability in the eigenstates with energies (n + 12 )ℏω and (n − 12 )ℏω; (II) zero mean position; and (III) the mean momentum in the positive axis direction.
(a) Find a wave function of the oscillator at time t = 0 satisfying all three conditons. (b) Find the mean values of its position and momentum at a subsequent time t. (c) Consider a classical oscillator with its total energy equal to the mean energy of the above quantum oscillator. Does the classical amplitude of oscillation agree with the largest mean value of the position of the quantum oscillator in (b)?
L.J. Sham April 24, 2009
Physics 130A Problem Set 4
- Positive energy eigenvalues
(a) Show that the expectation value of kinetic energy is always non-negative by proving,
〈ψ| Kˆ|ψ〉 =
ℏ^2
2 m
dx
∂ψ∗ ∂x
∂ψ ∂x
[Hint: take one differential operator and integrate by parts. Specify conditions needed.] (b) If the potential energy of a particle V (x) ≥ 0 , show that its energy eigenvalues are always non-negative. [Hint: examine the expectation value of the Hamiltonian in an energy eigenstate as the sum of kinetic and potential energies.]
Solution –
(a)
〈ψ| Kˆ|ψ〉 = −
ℏ^2
2 m
−∞
dx ψ∗^
∂^2 ψ ∂x^2
= −
[
ψ∗^
∂ψ ∂x
]∞
x=−∞
ℏ^2
2 m
−∞
dx
∂ψ∗ ∂x
∂ψ ∂x
ℏ^2
2 m
−∞
dx
∂ψ∗ ∂x
∂ψ ∂x
provided that ψ(x) and its derivative tend to zero as x → ±∞. For later use, I would add an alternate (periodic) condition, ψ(x = −L) = ψ(x = L) and ψ′(x = −L) = ψ′(x = L) as L → ∞. (b)
〈ψ| Hˆ|ψ〉 =
ℏ^2
2 m
dx
∂ψ∗ ∂x
∂ψ ∂x
dx |ψ(x)|^2 V (x) ≥ 0. (3)
If ψ is an energy eigenstate with eigenvalue E and normalized,
〈ψ| Hˆ|ψ〉 = 〈ψ|E|ψ〉 = E ≥ 0. (4)
- First excited state of SHO
(a) Find the functional dependence on position (in dimensionless form) of the first excited state from ψ 1 = c†ψ 0 , starting with the normalized Gaussian form of the ground state.
Work on just the dimensionless part,
(c + c†)^2 = c^2 + (c†)^2 + cc†^ + c†c = c^2 + (c†)^2 + 2c†c + 1, (10)
using [c, c†] on the third term, cc†, of the middle expression. The expectation value of c^2 is zero because the operator lowers ψn to ψn− 2. Similarly, 〈(c†)2〉 = 0. The state ψn is the eigenstate of the number operator c†c,
c†cψn = nψn. (11)
Hence,
〈ψn|(c + c†)^2 |ψn〉 = 2n + 1. (12)
Finally,
〈x^2 〉 =
mω
n +
(c) We could work out 〈p^2 〉 in the same way as the previous part, using (c − c†)^2. Just for fun, try this. The total energy for state ψn is (n + 12 )ℏω is made up of the kinetic energy part and the potential energy part,
〈H〉 =
2 m
〈p^2 〉 +
mω^2 〈x^2 〉 (14)
(n + 12 )ℏω =
2 m
〈p^2 〉 +
n +
ℏω. (15)
Hence,
〈p^2 〉 = mℏω
n +
(d) Uncertainty relation,
∆x∆p =
〈x^2 〉〈p^2 〉 = ℏ
n +
- Deduction of Properties of the State. A quantum one-dimensional harmonic oscillator has mass m, classical angular frequency ω, and the bottom of the potential at the origin of the coordinate axis. At time t = 0, it is in a state with (I) equal probability in the eigenstates with energies (n + 12 )ℏω and (n − 12 )ℏω; (II) zero mean position; and (III) the mean momentum in the positive axis direction.
(a) Find a wave function of the oscillator at time t = 0 satisfying all three conditons. (b) Find the mean values of its position and momentum at a subsequent time t.
(c) Consider a classical oscillator with its total energy equal to the mean energy of the above quantum oscillator. Does the classical amplitude of oscillation agree with the largest mean value of the position of the quantum oscillator in (b)?
Solution –
(a) The state at t = 0 involves only ψn and ψn− 1 states. Condition (I) tells us that both coefficients have equal moduli in the formula,
Ψ(t = 0) = √^12 (ψn + ψn− 1 eiϕ). (18)
The phase of one of them is arbitrary and dispensable but the relative phase of the other one to the former is essential. Since we need it later,
Ψ(t) = √^12 (ψne−iEnt/ℏ^ + ψn− 1 eiϕe−iEn−^1 t/ℏ) (19)
= e−i
“ n+^12
” ωt (^) √ 1 2
[
ψn + ψn− 1 ei(ϕ+ωt)
]
Then, the position expectation value at time t is,
〈x(t)〉 ≡ 〈Ψ(t)|x|Ψ(t)〉
=
[
〈ψn| + e−i(ϕ+ωt)〈ψn− 1 |
]
x
[
|ψn〉 + |ψn− 1 〉ei(ϕ+ωt)
]
[
〈ψn|x|ψn− 1 〉ei(ϕ+ωt)^ + e−i(ϕ+ωt)〈ψn− 1 |x|ψn〉
]
, since 〈ψn|x|ψn〉 = 0,
= ℜ
[
〈ψn|x|ψn− 1 〉ei(ϕ+ωt)
]
using as a shorthand, = 〈ψn|x|ψn− 1 〉 cos(ϕ + ωt). (21)
For the last line, we have used 〈ψn− 1 |x|ψn〉 as a real integral, = 〈ψn|x|ψn− 1 〉. The amplitude,
〈ψn|x|ψn− 1 〉 =
2 mω
〈ψn|(c + c†)|ψn− 1 〉
2 mω
〈ψn|c†|ψn− 1 〉
nℏ 2 mω
Thus, by condition (II), at t = 0,
〈x〉 =
nℏ 2 mω
cos ϕ = 0. (23)
