2024-08-17
Solutions to Principles of Quantum Mechanics
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目录

Chapter 5 Simple Problems in One Dimension
5.1 The Free Particle
5.2 The Particle in a Box
5.3 The Continuity Equation for Probability
5.4 The Single-Step Potential: A Problem in Scattering
5.5 The Double-Slit Experiment
5.6 Some Theorems

Chapter 5 Simple Problems in One Dimension

5.1 The Free Particle

Exercise 5.1.1 Show that Eq. (5.1.9) may be rewritten as an integral over EE and a sum over the ±\pm index as

U(t)=α=±0[m(2mE)1/2]E,αE,αeiEt/dEU(t)=\sum_{\alpha= \pm} \int_0^{\infty}\left[\frac{m}{(2 m E)^{1 / 2}}\right]|E, \alpha\rangle\langle E, \alpha| \mathrm{e}^{-\mathrm{i} E t / \hbar} \mathrm{d} E

Solution.

E=p22mp=α2mEdp=αm2mEdE\begin{aligned} E=\dfrac{p^{2}}{2m} \rightarrow p&=\alpha\sqrt{2mE}\\ \mathrm{d}p&=\dfrac{\alpha m}{\sqrt{2mE}}\mathrm{d}E \end{aligned}

where α=±1\alpha=\pm 1. Hence

U(t)=+dpppeiEt/=0dpppeiEt/+0+dpppeiEt/=0dEm2mEE,E,eiEt/+0+dEm2mEE,+E,+eiEt/=0+dEm2mEE,E,eiEt/+0+dEm2mEE,+E,+eiEt/=α=±0[m(2mE)1/2]E,αE,αeiEt/dE\begin{aligned} U(t)&=\int_{-\infty}^{+\infty} \mathrm{d}p\,|p\rangle\langle p| \mathrm{e}^{-\mathrm{i}Et/\hbar}\\&=\int_{-\infty}^{0}\mathrm{d}p\,|p\rangle\langle p| \mathrm{e}^{-\mathrm{i}Et/\hbar}+\int_{0}^{+\infty}\mathrm{d}p\,|p\rangle\langle p| \mathrm{e}^{-\mathrm{i}Et/\hbar}\\ &=\int_{-\infty}^{0}\mathrm{d}E \dfrac{-m}{\sqrt{2mE}}\,|E, -\rangle\langle E, -| \mathrm{e}^{-\mathrm{i}Et/\hbar}+\int_{0}^{+\infty}\mathrm{d}E\dfrac{m}{\sqrt{2mE}}\,|E, +\rangle\langle E, +| \mathrm{e}^{-\mathrm{i}Et/\hbar}\\ &=\int_{0}^{+\infty}\mathrm{d}E \dfrac{m}{\sqrt{2mE}}\,|E, -\rangle\langle E, -| \mathrm{e}^{-\mathrm{i}Et/\hbar}+\int_{0}^{+\infty}\mathrm{d}E\dfrac{m}{\sqrt{2mE}}\,|E, +\rangle\langle E, +| \mathrm{e}^{-\mathrm{i}Et/\hbar}\\ &=\sum_{\alpha= \pm} \int_0^{\infty}\left[\frac{m}{(2 m E)^{1 / 2}}\right]|E, \alpha\rangle\langle E, \alpha| \mathrm{e}^{-\mathrm{i} E t / \hbar} \mathrm{d} E \end{aligned}
 ~\tag*{$\blacksquare$}

Exercise 5.1.2 By solving the eigenvalue equation (5.1.3) in the XX basis, regain Eq. (5.1.8), i.e., show that the general solution of energy EE is

ψE(x)=βexp[i(2mE)1/2x/](2π)1/2+γexp[i(2mE)1/2x/](2π)1/2\psi_E(x)=\beta \frac{\exp \left[i(2 m E)^{1 / 2} x / \hbar\right]}{(2 \pi \hbar)^{1 / 2}}+\gamma \frac{\exp \left[-i(2 m E)^{1 / 2} x / \hbar\right]}{(2 \pi \hbar)^{1 / 2}}

[The factor (2π)1/2(2 \pi \hbar)^{-1 / 2} is arbitrary and may be absorbed into β\beta and γ\gamma.] Though ψE(x)\psi_E(x) will satisfy the equation even if E<0E<0, are these functions in the Hilbert space?

Solution.

In XX basis, equation (5.1.3) is

xHE=ExE\langle x\mid H\mid E\rangle=E\langle x\mid E\rangle

which becomes

22md2dx2ψE(x)=EψE(x)-\dfrac{\hbar^{2}}{2m}\dfrac{\mathrm{d}^{2}}{\mathrm{d}x^{2}}\psi_{E}(x)=E\psi_{E}(x)

The most general solution is

ψE(x)=A+e+i2mEx+Aei2mEx\psi_{E}(x)=A_{+}\mathrm{e}^{+\frac{\mathrm{i}}{\hbar}\sqrt{2mE}x}+A_{-}\mathrm{e}^{-\frac{\mathrm{i}}{\hbar}\sqrt{2mE}x}

where

xE=βxE,++γxE,\langle x\mid E\rangle=\beta\langle x\mid E,+\rangle+\gamma\langle x\mid E,-\rangle
β=A+2πγ=A2π\begin{aligned} \beta&=A_{+}\sqrt{2\pi\hbar}\\ \gamma&=A_{-}\sqrt{2\pi\hbar} \end{aligned}

If E<0E<0, these solutions are not in the Hilbert space, since then the two terms grow exponentially as x±x\to\pm\infty.

 ~\tag*{$\blacksquare$}

Exercise 5.1.3 We have seen that there exists another formula for U(t)U(t), namely, U(t)=eiHt/U(t)=e^{-i H t / \hbar}. For a free particle this becomes

U(t)=exp[i(2t2md2dx2)]=n=01n!(it2m)nd2ndx2n(5.1.18)U(t)=\exp \left[\frac{i}{\hbar}\left(\frac{\hbar^2 t}{2 m} \frac{d^2}{d x^2}\right)\right]=\sum_{n=0}^{\infty} \frac{1}{n!}\left(\frac{i \hbar t}{2 m}\right)^n \frac{d^{2 n}}{d x^{2 n}}\tag{5.1.18}

Consider the initial state in Eq. (5.1.14) with p0=0p_0=0, and set Δ=1,t=0\Delta=1, t^{\prime}=0 :

ψ(x,0)=ex2/2(π)1/4\psi(x, 0)=\frac{e^{-x^2 / 2}}{(\pi)^{1 / 4}}

Find ψ(x,t)\psi(x, t) using Eq. (5.1.18) above and compare with Eq. (5.1.15).

Hints : (1) Write ψ(x,0)\psi(x, 0) as a power series:

ψ(x,0)=(π)1/4n=0(1)nx2nn!(2)n\psi(x, 0)=(\pi)^{-1 / 4} \sum_{n=0}^{\infty} \frac{(-1)^n x^{2 n}}{n!(2)^n}

(2) Find the action of a few terms

1,(it2m)d2dx2,12!(it2md2dx2)21,\quad \left(\frac{i \hbar t}{2 m}\right) \frac{d^2}{d x^2}, \quad \frac{1}{2!}\left(\frac{i \hbar t}{2 m} \frac{d^2}{d x^2}\right)^2

etc., on this power series.

(3) Collect terms with the same power of xx.

(4) Look for the following series expansion in the coefficient of x2nx^{2 n} :

(1+itm)n12=1(n+1/2)(itm)+(n+1/2)(n+3/2)2!(itm)2+\left(1+\frac{i t \hbar}{m}\right)^{-n-12}=1-(n+1 / 2)\left(\frac{i \hbar t}{m}\right)+\frac{(n+1 / 2)(n+3 / 2)}{2!}\left(\frac{i t \hbar}{m}\right)^2+\cdots

(5) Juggle around till you get the answer.

Exercise 5.1.4 Consider the wave function

ψ(x,0)=sin(πxL),xL/2=0,x>L/2\begin{aligned} \psi(x, 0) & =\sin \left(\frac{\pi x}{L}\right), & & |x| \leqslant L / 2 \\ & =0, & & |x|>L / 2 \end{aligned}

It is clear that when this function is differentiated any number of times we get another function confined to the interval xL/2|x| \leqslant L / 2. Consequently the action of

U(t)=exp[i(2t2m)d2dx2]U(t)=\exp \left[\frac{i}{\hbar}\left(\frac{\hbar^2 t}{2 m}\right) \frac{d^2}{d x^2}\right]

on this function is to give a function confined to xL/2|x| \leqslant L / 2. What about the spreading of the wave packet?

5.2 The Particle in a Box

Exercise 5.2.1 A particle is in the ground state of a box of length LL. Suddenly the box expands (symmetrically) to twice its size, leaving the wave function undisturbed. Show that the probability of finding the particle in the ground state of the new box is (8/3π)2(8 / 3 \pi)^2.

Exercise 5.2.2 (a) Show that for any normalized ψ,ψHψE0|\psi\rangle,\langle\psi|H| \psi\rangle \geqslant E_0, where E0E_0 is the lowest-energy eigenvalue. (Hint : Expand ψ|\psi\rangle in the eigenbasis of HH.)

(b) Prove the following theorem: Every attractive potential in one dimension has at least one bound state. Hint: Since VV is attractive, if we define V()=0V(\infty)=0, it follows that V(x)=V(x)= V(x)-|V(x)| for all xx. To show that there exists a bound state with E<0E<0, consider

ψα(x)=(απ)1/4eαx2/2\psi_\alpha(x)=\left(\frac{\alpha}{\pi}\right)^{1 / 4} e^{-\alpha x^2 / 2}

and calculate

E(α)=ψαHψα,H=22md2dx2V(x)E(\alpha)=\left\langle\psi_\alpha|H| \psi_\alpha\right\rangle, \quad H=-\frac{\hbar^2}{2 m} \frac{d^2}{d x^2}-|V(x)|

Show that E(α)E(\alpha) can be made negative by a suitable choice of α\alpha. The desired result follows from the application of the theorem proved above.

Exercise 5.2.3 Consider V(x)=aV0δ(x)V(x)=-a V_0 \delta(x). Show that it admits a bound state of energy E=ma2V02/22E=-m a^2 V_0^2 / 2 \hbar^2. Are there any other bound states? Hint: Solve Schrödinger's equation outside the potential for E<0E<0, and keep only the solution that has the right behavior at infinity and is continuous at x=0x=0. Draw the wave function and see how there is a cusp, or a discontinuous change of slope at x=0x=0. Calculate the change in slope and equate it to

ε+ε(d2ψdx2)dx\int_{-\varepsilon}^{+\varepsilon}\left(\frac{d^2 \psi}{d x^2}\right) d x

(where ε\varepsilon is infinitesimal) determined from Schrödinger's equation.

Exercise 5.2.4 Consider a particle of mass mm in the state n|n\rangle of a box of length LL. Find the force F=E/LF=-\partial E / \partial L encountered when the walls are slowly pushed in, assuming the particle remains in the nnth state of the box as its size changes. Consider a classical particle of energy EnE_n in this box. Find its velocity, the frequency of collision on a given wall, the momentum transfer per collision, and hence the average force. Compare it to E/L-\partial E / \partial L computed above.

Exercise 5.2.5 If the box extends from x=0x=0 to LL (instead of L/2-L / 2 to L/2L / 2) show that ψn(x)=(2/L)1/2sin(nπx/L),n=1,2,,\psi_n(x)=(2 / L)^{1 / 2} \sin (n \pi x / L), n=1,2, \ldots, \infty and En=2π2n2/2mL2E_n=\hbar^2 \pi^2 n^2 / 2 m L^2.

Exercise 5.2.6 Square Well Potential. Consider a particle in a square well potential:

V(x)={0,xaV0,xaV(x)= \begin{cases}0, & |x| \leqslant a \\ V_0, & |x| \geqslant a\end{cases}

Since when V0V_0 \rightarrow \infty, we have a box, let us guess what the lowering of the walls does to the states. First of all, all the bound states (which alone we are interested in), will have EV0E \leq V_0. Second, the wave functions of the low-lying levels will look like those of the particle in a box, with the obvious difference that ψ\psi will not vanish at the walls but instead spill out with an exponential tail. The eigenfunctions will still be even, odd, even, etc.

(1) Show that the even solutions have energies that satisfy the transcendental equation

ktanka=κ(5.2.23)k \tan k a=\kappa\tag{5.2.23}

while the odd ones will have energies that satisfy

kcotka=κ(5.2.24)k \cot k a=-\kappa\tag{5.2.24}

where kk and iκi \kappa are the real and complex wave numbers inside and outside the well, respectively. Note that kk and κ\kappa are related by

k2+κ2=2mV0/2(5.2.25)k^2+\kappa^2=2 m V_0 / \hbar^2\tag{5.2.25}

Verify that as V0V_0 tends to \infty, we regain the levels in the box.

(2) Equations (5.2.23) and (5.2.24) must be solved graphically. In the (α=ka,β=κa)(\alpha=k a, \beta=\kappa a) plane, imagine a circle that obeys Eq. (5.2.25). The bound states are then given by the intersection of the curve αtanα=β\alpha \tan \alpha=\beta or αcotα=β\alpha \cot \alpha=-\beta with the circle. (Remember α\alpha and β\beta are positive.)

(3) Show that there is always one even solution and that there is no odd solution unless V02π2/8ma2V_0 \geqslant \hbar^2 \pi^2 / 8 m a^2. What is EE when V0V_0 just meets this requirement? Note that the general result from Exercise 5.2.2b holds.

5.3 The Continuity Equation for Probability

Exercise 5.3.1 Consider the case where V=VriViV=V_r-i V_i, where the imaginary part ViV_i is a constant. Is the Hamiltonian Hermitian? Go through the derivation of the continuity equation and show that the total probability for finding the particle decreases exponentially as e2Vit/e^{-2 V_i t / \hbar}. Such complex potentials are used to describe processes in which particles are absorbed by a sink.

Exercise 5.3.2 Convince yourself that if ψ=cψ~\psi=c \tilde{\psi}, where cc is constant (real or complex) and ψ~\tilde{\psi} is real, the corresponding j\mathbf{j} vanishes.

Exercise 5.3.3 Consider

ψp=(12π)3/2ei(pr)/\psi_{\mathbf{p}}=\left(\frac{1}{2 \pi \hbar}\right)^{3 / 2} e^{i(\mathbf{p}\cdot\mathbf{r}) / \hbar}

Find j\mathbf{j} and PP and compare the relation between them to the electromagnetic equation j=ρv\mathbf{j}=\rho \mathbf{v}, v\mathbf{v} being the velocity. Since ρ\rho and j\mathbf{j} are constant, note that the continuity Eq. (5.3.7) is trivially satisfied.

Exercise 5.3.4 Consider ψ=Aeipx/+Beipx/\psi=A e^{i p x / \hbar}+B e^{-i p x / \hbar} in one dimension. Show that j=(A2B2)p/mj=\left(|A|^2-|B|^2\right) p / m. The absence of cross terms between the right- and left-moving pieces in ψ\psi allows us to associate the two parts of jj with corresponding parts of ψ\psi.

5.4 The Single-Step Potential: A Problem in Scattering

Exercise 5.4.1 Evaluate the third piece in Eq. (5.416) and compare the resulting TT with Eq. (5.4.21). [Hint: Expand the factor (k122mV0/2)1/2\left(k_1^2-2 m V_0 / \hbar^2\right)^{1 / 2} near k1=k0k_1=k_0, keeping just the first derivative in the Taylor series.]

Exercise 5.4.2 (a) Calculate RR and TT for scattering of a potential V(x)=V0aδ(x)V(x)=V_0 a \delta(x). (b) Do the same for the case V=0V=0 for x>a|x|>a and V=V0V=V_0 for x<a|x|<a. Assume that the energy is positive but less than V0V_0.

Exercise 5.4.3 Consider a particle subject to a constant force ff in one dimension. Solve for the propagator in momentum space and get

U(p,t;p,0)=δ(ppft)ei(p3p3)/6mfU\left(p, t ; p^{\prime}, 0\right)=\delta\left(p-p^{\prime}-f t\right) e^{i\left(p^{\prime 3}-p^3\right) / 6 m\hbar f}

Transform back to coordinate space and obtain

U(x,t;x,0)=(m2πit)1/2exp{i[m(xx)22t+12ft(x+x)f2t324m]}U\left(x, t ; x^{\prime}, 0\right)=\left(\frac{m}{2 \pi \hbar i t}\right)^{1 / 2} \exp \left\{\frac{i}{\hbar}\left[\frac{m\left(x-x^{\prime}\right)^2}{2 t}+\frac{1}{2} f t\left(x+x^{\prime}\right)-\frac{f^2 t^3}{24 m}\right]\right\}

[Hint: Normalize ψE(p)\psi_E(p) such that EE=δ(EE)\left\langle E \mid E^{\prime}\right\rangle=\delta\left(E-E^{\prime}\right). Note that EE is not restricted to be positive.]

5.5 The Double-Slit Experiment

5.6 Some Theorems