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Theme 3 — Postulates and Applications

Parabolic initial state in the infinite well

Keywords: infinite well · expansion in an eigenbasis · energy probabilities · mean energy · time evolution · revival period

Exercise 1 : Parabolic initial state in the infinite well

A particle of mass $m$ is confined in an infinite well between $x=0$ and $x=L$ , with eigenstates and energies

\phi_n(x)=\sqrt{\frac2L}\sin\Bigl(\frac{n\pi x}{L}\Bigr),\qquad E_n=n^2E_1,\qquad E_1=\frac{\pi^2\hbar^2}{2mL^2}.

At time $t=0$ , the particle is prepared in the state

\psi(x,0)=\sqrt{\frac{30}{L^5}}\;x\,(L-x)\quad\text{for }0\leq x\leq L .

You are given $\int_0^Lx(L-x)\sin\bigl(\frac{n\pi x}{L}\bigr)dx=\frac{2L^3}{n^3\pi^3}\bigl(1-(-1)^n\bigr)$ , as well as the sums over odd integers $\sum_{n\,\mathrm{odd}}\frac{1}{n^4}=\frac{\pi^4}{96}$ and $\sum_{n\,\mathrm{odd}}\frac{1}{n^6}=\frac{\pi^6}{960}$ .

Question 1

Verify that

\psi(x,0)

is normalised and satisfies the boundary conditions. Calculate the mean energy

\langle\hat H\rangle

directly from

\psi(x,0)

Solution

The function vanishes at

x=0

and

x=L

: it satisfies the Dirichlet boundary conditions. Its norm is

\frac{30}{L^5}\int_0^Lx^2(L-x)^2dx=\frac{30}{L^5}\int_0^L\bigl(L^2x^2-2Lx^3+x^4\bigr)dx=\frac{30}{L^5}\Bigl(\frac{L^5}{3}-\frac{L^5}{2}+\frac{L^5}{5}\Bigr)=\frac{30}{L^5}\cdot\frac{L^5}{30}=1 .

For the mean energy, we use $\hat H=\hat P^2/(2m)$ inside the well and integration by parts, whose boundary term vanishes because of the boundary conditions:

\langle\hat H\rangle=\frac{\hbar^2}{2m}\int_0^L|\psi'(x)|^2dx=\frac{\hbar^2}{2m}\cdot\frac{30}{L^5}\int_0^L(L-2x)^2dx=\frac{\hbar^2}{2m}\cdot\frac{30}{L^5}\cdot\frac{L^3}{3}=\frac{5\hbar^2}{mL^2}.

We used $\int_0^L(L-2x)^2dx=\frac{L^3}{3}$ , obtained by the change of variable $u=L-2x$ .

Question 2

Expand

\psi(x,0)

in the eigenstate basis, that is, calculate the coefficients

c_n

such that

\psi(x,0)=\sum_nc_n\phi_n(x)

, given by

c_n=\int_0^L\phi_n(x)\,\psi(x,0)\,dx .

Why are the even-index coefficients zero?

Solution

From the given integral,

c_n=\sqrt{\frac2L}\sqrt{\frac{30}{L^5}}\cdot\frac{2L^3}{n^3\pi^3}\bigl(1-(-1)^n\bigr)=\frac{2\sqrt{60}}{n^3\pi^3}\bigl(1-(-1)^n\bigr).

For even $n$ , $1-(-1)^n=0$ and $c_n=0$ . For odd $n$ , $1-(-1)^n=2$ , and

c_n=\frac{4\sqrt{60}}{n^3\pi^3}=\frac{8\sqrt{15}}{n^3\pi^3}.

The even coefficients vanish for reasons of symmetry: $\psi(x,0)$ is symmetric about the centre of the well, $\psi(L-x)=\psi(x)$ , whereas the even-index $\phi_n$ are antisymmetric about this point. Their inner product is therefore zero.

Question 3

What is the probability of measuring the energy

E_1

? Verify that the probabilities sum to one, then recover

\langle\hat H\rangle

from the

c_n

. Compare

\langle\hat H\rangle

with

E_1

Solution

The probability of measuring

E_n

|c_n|^2=\frac{960}{n^6\pi^6}

for odd

n

. For

n=1

P(E_1)=\frac{960}{\pi^6}\simeq\frac{960}{961{,}4}\simeq0{,}9986 .

The initial state is therefore very close to the ground state, which is not surprising since the parabola $x(L-x)$ closely resembles the sinusoidal arch $\sin(\pi x/L)$ . The sum of the probabilities is

\sum_{n\,\mathrm{odd}}\frac{960}{\pi^6n^6}=\frac{960}{\pi^6}\cdot\frac{\pi^6}{960}=1,

as required by Parseval's identity. The mean energy is

\langle\hat H\rangle=\sum_n|c_n|^2E_n=\frac{960}{\pi^6}E_1\sum_{n\,\mathrm{odd}}\frac{n^2}{n^6}=\frac{960}{\pi^6}\cdot\frac{\pi^2\hbar^2}{2mL^2}\cdot\frac{\pi^4}{96}=\frac{5\hbar^2}{mL^2},

in agreement with the first question. Since $E_1=\frac{\pi^2}{2}\frac{\hbar^2}{mL^2}\simeq4{,}93\frac{\hbar^2}{mL^2}$ , we have $\langle\hat H\rangle\simeq1{,}013\,E_1$ , slightly greater than $E_1$ as it should be, since the ground state minimises the mean energy. This is the principle behind the variational method: a well-chosen trial function gives an upper bound, in this case a very good one, for the ground-state energy.

Question 4

Write down

\psi(x,t)

. Do the energy probabilities depend on time? Show that the probability density

|\psi(x,t)|^2

is periodic, and give its period

T

Solution

Each component evolves with its own phase:

\psi(x,t)=\sum_{n\,\mathrm{odd}}c_n\,e^{-iE_nt/\hbar}\,\phi_n(x)=\sum_{n\,\mathrm{odd}}\frac{8\sqrt{15}}{n^3\pi^3}\,e^{-in^2E_1t/\hbar}\,\phi_n(x).

The energy probabilities $|c_ne^{-iE_nt/\hbar}|^2=|c_n|^2$ are constant. By contrast, the position density contains the cross terms $c_mc_n\phi_m\phi_n\cos\bigl((E_n-E_m)t/\hbar\bigr)$ , which depend on time. Since all the energies are integer multiples of $E_1$ , $E_n=n^2E_1$ , all the phases $e^{-in^2E_1t/\hbar}$ simultaneously return to their initial values when $E_1t/\hbar$ is a multiple of $2\pi$ . The wave function is then exactly $\psi(x,0)$ , and the evolution is periodic with period

T=\frac{2\pi\hbar}{E_1}=\frac{4mL^2}{\pi\hbar}.

In fact, since only odd $n$ occur and $n^2-1$ is then divisible by $8$ , the density is periodic even with period $T/8$ : we see that $e^{-in^2E_1t/\hbar}=e^{-iE_1t/\hbar}\,e^{-i(n^2-1)E_1t/\hbar}$ , and the second factor equals $1$ for $t=T/8$ , so that the state differs from the initial state only by a global phase. In the present case, the state is so close to the ground state that the variations in the density remain very small in any event.