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L.J. Sham April 10, 2009
Physics 130A Problem Set 2
- Solutions of the Schr¨odinger equation? (I) (cos kx)e−iωt; (II) cos(kx − ωt); (III) (sin kx)e−iωt; (IV) sin(kx − ωt); (V) e+ikx+iωt
(a) Given that k and ω are both real and positive, decide in each case whether it is a solution of the Schr¨odinger equation for a free particle in one dimension. (b) For those wave functions which satisfy the Schr¨odinger equation, find the correspond- ing probability current density.
- Momentum wave function
(a) Show that if the momentum wave function of a free particle in one dimension at a certain time is given by:
ϕ(p) =
σ ℏ
e−σ|p|ℏ
− 1 , (1)
then its wave function is a Lorentzian:
Ψ(x) =
(2σ^3 /π)^1 /^2 x^2 + σ^2
(b) Find the mean energy and its uncertainty for the particle. (c) If the momentum wave function of a free particle in one dimension at a certain time is a Lorentzian:
ϕ(p) =
(2λ^3 /π)^1 /^2 p^2 + λ^2
find, using the preceding results and without any more integration, its position wave function.
- Application of the uncertainty principle I Before you do this problem and the next, you may wish to study the example below. By using the uncertainty principle in the form ∆x∆p ≈ ℏ 2 , estimate the ground-state energy for a harmonic oscillator in one dimension. The harmonic oscillator is a particle of mass m and subject to a restoring force proportional to its displacement with the force constant mω^2.
- Application of the uncertainty principle II It is possible to confine an electron to move in one dimension in a semiconductor quantum wire. If the electron so confined is attracted by a fixed Coulomb charge in the wire giving it a potential energy
V (x) = −
e^2 |x|
where x is the position of the electron measured from the fixed charge, estimate the ground- state energy.
Applications of the Uncertainty Principle
– A particle under gravity
Consider a familiar problem in mechanics. A particle falls under gravity towards an impenetrable floor. According to classical mechanics, the ground state (the state of least energy) is one in which the particle is at rest on the floor. Let us measure the distance vertically from the floor and call it x. Thus, we know the position of the particle in its ground state (x = 0) and also its momentum (p = 0). This contradicts the uncertainty principle. What is the ground state energy according to quantum mechanics? The potential energy of the particle is
V (x) = mgx if x > 0 , = +∞ if x < 0.
Just plug this into the Schr¨odinger equation and solve it. We shall do this sort of thing later on. Let us try a simple estimate here. The ground state will differ from the classical solution by having an uncertainty in position of ∆x and momentum ∆p where
∆p ∼ ℏ/∆x. (6)
Then the energy is approximately,
E ∼
(∆p)^2 2 m
ℏ^2
2 m(∆x)^2
Minimizing the energy with respect to ∆x, we obtain
∆x ∼
ℏ^2
m^2 g
(me m
× 1. 11 × 10 −^3 meter. (8)
where me is the mass of the electron. From the uncertainty principle, we have deduced that a particle cannot rest on a floor even under the pull of gravity. Even in the lowest energy state, the particle bounces up and down with a range given by (8), which is surprisingly large for a microscopic particle such as an electron (several order of magnitude larger than the Bohr radius). It shows how weak gravitational force is in comparison with the electromagnetic force.
(c) If the momentum wave function of a free particle in one dimension at a certain time is a Lorentzian:
ϕ(p) =
(2λ^3 /π)^1 /^2 p^2 + λ^2
find, using the preceding results and without any more integration, its position wave function.
Solution –
(a) Let k = p/ℏ. We have found it easier to use k which may be regarded either as a wave vector or momentum in units of ℏ. We note first that the momentum wave function is normalized, ∫ (^) ∞
−∞
dp |ϕ(p)|^2 = 2ℏ
0
dk
σ ℏ
e−^2 σk^ = 1. (5)
The Fourier transform of the wave function defined in Lecture 4 is consistent with the normalized wave function in k, ∫ (^) ∞
−∞
dk 2 π
| ψ˜(k)|^2 = 1. (6)
Note that the metric of 1 / 2 π associated with any integration over k is the price for avoiding 1 /
2 π in the definition of the Fourier integrals. Thus, the expression in k space is,
ψ^ ˜(k) =
2 πℏ ϕ(ℏk) =
2 πσ e−σ|k|. (7)
Then, by the inverse Fourier transform theorem,
Ψ(x) =
dk 2 π
eikx^ ψ˜(k) =
σ 2 π
−∞
dk eikxe−σ|k|
σ 2 π
[∫ ∞
0
dk ek(ix−σ)
]
2 σ π
[
ix − σ
]
2 σ π
σ x^2 + σ^2
(b) The mean energy for the free particle is,
〈 Hˆ〉 =
dk 2 π
ψ˜∗(k)ℏ
(^2) k 2 2 m
ψ˜(k) = 2
0
dk σ e−^2 σk^
ℏ^2 k^2 2 m
=
ℏ^2
8 σ^2 m
0
dt t^2 e−t^ substituting t = 2σk
ℏ^2
4 σ^2 m
Similarly,
〈 Hˆ^2 〉 =
dk 2 π
ψ˜∗(k)
[
ℏ^2 k^2 2 m
] 2
ψ^ ˜(k) = 2
0
dk σ e−^2 σk
[
ℏ^2 k^2 2 m
] 2
[
ℏ^2
8 σ^2 m
] 2 ∫ ∞
0
dt t^4 e−t^ substituting t = 2σk
[
ℏ^2
σ^2 m
] 2
[
ℏ^2
σ^2 m
] 2
The energy uncertainty ∆E is given by,
∆E =
〈 Hˆ^2 〉 − 〈 Hˆ〉^2 =
ℏ^2
σ^2 m
ℏ^2
σ^2 m
(c) It is just the inverse of the Fourier transform evaluated in part (a). We just need to sort out the parameters and the normalization constant. The new Fourier transform of the wave function is given by, from Eq. (7),
ψ^ ˜(k) =
2 πℏ ϕ(ℏk) =
2 πℏ
(2λ^3 /π)^1 /^2 ℏ^2 k^2 + λ^2
2 σ^3 /^2 k^2 + σ^2
where we have put λ/ℏ = σ. Then, the inverse Fourier transform is,
ψ(x) =
dk 2 π
eikx^
2 σ^3 /^2 k^2 + σ^2
σe−σ|x|^ =
λ ℏ
e−λ|x|/ℏ, (13)
by contour integral. However, since we are not allowed to integrate any more, we may equate the inverse of Fourier transform of Eq. (8) in part (a) with the last expression in Eq.(7),
√ 2 πσ e−σ|k|^ =
dx e−ikx
2 σ π
σ x^2 + σ^2
By first changing the sign of x, then switching x and k and adjusting the pre-factors so that the right-hand side integrand looks like the last expression of Eq. (12), the last equation becomes,
√ σ e−σ|x|^ =
dk 2 π
eikx^
2 σ^3 /^2 k^2 + σ^2
, QED. (15)
- Application of the uncertainty principle I Before you do this problem and the next, you may wish to study the example below. By using the uncertainty principle in the form ∆x∆p ≈ ℏ 2 , estimate the ground-state energy for a harmonic oscillator in one dimension. The harmonic oscillator is a particle of mass m and subject to a restoring force proportional to its displacement with the force constant mω^2.
The difference is that the potential energy estimate is taken roughly at |x| = ∆x. The ground state expectation value of the energy is then,
E =
(∆p)^2 2 m
e^2 ∆x
By the approximation for ∆p ≈ ℏ/∆x, we have,
E(∆x) ≈
ℏ^2
2 m(∆x)^2
e^2 ∆x
To find the minimum, put the first derivative with respect to ∆x to zero,
E′(∆x) ≈ −
ℏ^2
m(∆x)^3
e^2 (∆x)^2
The results are,
∆x ≈
ℏ^2
me^2
E 0 = E(∆x) ≈
ℏ^2
2 m
[
me^2 ℏ^2
] 2
me^4 ℏ^2
me^4 2 ℏ^2
