# Energy and time evolution of a particle in a potential well

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

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?

• This is an example of a "good homework" question. – Ali Jan 3 '14 at 12:17

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.

• This is mathematically cute. – dmckee --- ex-moderator kitten Jan 3 '14 at 17:59
• 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 $f(x)sin(\frac{\pi n}{a}x)$? – Charlie Jan 3 '14 at 22:40
• I made a few edits which should clear things up. – Eric Angle Jan 4 '14 at 2:37
• 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. – Charlie Jan 4 '14 at 9:14
• I just don't understand. What do I need the orthogonality property for? – Charlie Jan 4 '14 at 13:01