I have a particle in an infinite square well (the box is from 0 to $a$), in the state described by the function

$$\psi (x) = \begin{cases} Ax(a-x) & \mathrm{for }\;\;\;\;0<x<a,\qquad \\ 0 \qquad &\text{otherwise}. \end{cases}$$

I have to determine the most likely value of energy and the probability to obtain a value of $E = \frac{9\hbar^2 {\pi}^2}{2ma^2} $.

To solve the second question I thought that $E$ iss the classic solution for energy in a potential well with $n=3$. So I can calculate $\langle3| \psi\rangle$ $-$ in which $3$ is the solution wave function with $n=3$ $-$ and that is it? Right?

But what about first question? Do I have to calculate $\langle H \rangle$ and compare it with a solution of the potential well?

I also have to determine the evolution of the wave function for $t>0$ when at $t=0$ we turn off the potential well, any hints?

  • 2
    $\begingroup$ This is an example of a "good homework" question. $\endgroup$
    – Ali
    Commented Jan 3, 2014 at 12:17

1 Answer 1


First normalize the state to find $A$.

Then you need to express the state as a superposition of the stationary states of the infinite square well: $$ \psi\left(x\right) = A x \left(a-x\right) = \sum_{n=1}^\infty c_n \psi_n\left(x\right), $$ where $\psi_n\left(x\right) = \sqrt{2/a} \sin\left(n \pi x / a\right)$ is the $n$-th stationary state. You can do this using the orthogonality of the stationary states, $$ \int_0^a dx \ \psi^*_m\left(x\right) \psi_n\left(x\right) = \frac{2}{a} \int_0^a dx \ \sin\left(\frac{m \pi x}{ a}\right) \sin\left(\frac{n \pi x}{ a}\right) = \delta_{mn}, $$ by integrating the equation above: $$ \begin{align} \int_0^a dx \ \psi^*_m\left(x\right) \left[A x \left(a-x\right)\right] &= \int_0^a dx \ \psi^*_m\left(x\right) \left[ \sum_{n=1}^\infty c_n \psi_n\left(x\right) \right] \\ &= \sum_{n=1}^\infty c_n \left[ \int_0^a dx \ \psi^*_m\left(x\right)\psi_n\left(x\right) \right]\\ &= \sum_{n=1}^\infty c_n \delta_{m n} \\ &= c_m \end{align} $$ I'll leave the $c_n = A \sqrt{2/a} \int_0^a dx \ \sin\left(n \pi x / a\right) x \left(a-x\right)$ integral for you to work out.

Once you have the $c_n$'s, the most likely value of a measurement of the energy is the energy corresponding to the stationary state with maximum $c_n$.

To find the probability of measuring $9 \hbar^2 \pi^2 / 2 m a^2$ for the energy, determine the stationary state that this energy corresponds to, and compute $\left|c_n\right|^2$.

For the time evolution, since the potential is $0$ everywhere after $t=0$, it is a free particle, and the general solution is: $$ \Psi\left(x,t\right) = \frac{1}{\sqrt{2 \pi}} \int_{-\infty}^\infty dk \ \phi\left(k\right) \exp\left[i\left(k x + \frac{\hbar k^2}{2 m} t\right)\right], $$ where $$ \phi\left(k\right) = \frac{1}{\sqrt{2 \pi}} \int_0^a dx \ \Psi\left(x,0\right) \exp\left(-i k x\right) = \frac{A}{\sqrt{2 \pi}} \int_0^a dx \ x\left(a-x\right) \exp\left(-i k x\right) . $$ So, now you just have to do this integral.

  • 1
    $\begingroup$ This is mathematically cute. $\endgroup$ Commented Jan 3, 2014 at 17:59
  • $\begingroup$ Truly illuminating. So I've got to find the coefficients $b_n$ of the Fourier series. But should I integrate from $-\pi$ to $\pi$ also in this case? And $f(x)sin(nx)$ or $f(x)sin(\frac{\pi n}{a}x)$? $\endgroup$
    – Charlie
    Commented Jan 3, 2014 at 22:40
  • $\begingroup$ I made a few edits which should clear things up. $\endgroup$
    – Eric Angle
    Commented Jan 4, 2014 at 2:37
  • $\begingroup$ Sorry, but I didn't get how to find the $c_n$ from that. I thought that, since $\psi_n$ is a $sin$ function, all I've got to do is to find the coefficients $b_n$ to expand the $f(x)$ with Fourier. $\endgroup$
    – Charlie
    Commented Jan 4, 2014 at 9:14
  • $\begingroup$ I just don't understand. What do I need the orthogonality property for? $\endgroup$
    – Charlie
    Commented Jan 4, 2014 at 13:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.