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Hoping this is not a silly and stupid question let me ask for help in this problem.

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

$\psi (x) = Ax(a-x) \qquad \mathrm{for }\qquad 0<x<a$, $\qquad 0 \qquad otherwise$.

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 $E$ it's the classic solution for energy in a potential well with $n=3$. So I calculate $\langle3| \psi\rangle$ $-$ in which $3$ is the solution wave function with $n=3$ $-$ and that's it? Right?

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

After I also have to determine the evolution of the wave function for $t>0$ when at $t=0$ we turn off the potential well, and I have no idea how to solve it.

Please forgive me for my English. Hope you'll give me some hints.

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    $\begingroup$ This is an example of a "good homework" question. $\endgroup$ – Ali Jan 3 '14 at 12:17
  • $\begingroup$ Any hints to give me? $\endgroup$ – Charlie Jan 3 '14 at 15:51
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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.

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  • $\begingroup$ This is mathematically cute. $\endgroup$ – dmckee Jan 3 '14 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 Jan 3 '14 at 22:40
  • $\begingroup$ I made a few edits which should clear things up. $\endgroup$ – Eric Angle Jan 4 '14 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 Jan 4 '14 at 9:14
  • $\begingroup$ I just don't understand. What do I need the orthogonality property for? $\endgroup$ – Charlie Jan 4 '14 at 13:01

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