I have a particle in a 1D infinite potential well. i.e

$$V(x)=\left\{ \begin{array}{c} V(x)=0 \text{ for } |x|<a \\ V(x)=\infty \text{ for } |x| \geq a \\ \end{array}\right. $$

I am given normalised stationary states:

$\phi_n(x) = \alpha_n \sin\left(\dfrac{\pi n(x+a)}{2a}\right)$ where $n=1,2,3,...$

$(1)$ I have to find the $\alpha_n$ and the corresponding energy eigenvalues $E_n$.

For this i just integrated the given $\phi_n(x)$ from $-a$ to $a$ and set it equal to 1 as normalizable. I got $\dfrac{1}{\alpha_n} = \dfrac{2a}{\pi n}(1-(-1)^n)$ from which i have concluded $\alpha_n = 0$ when $n$ is even and $\dfrac{\pi n}{4a}$. Not sure if this is right as the given states were for $n=1,2,3,...$? But once I have these then finding $E_n$ is fine.

$(2)$ Secondly, the bit I'm really stuck on. I am told given state $\psi(x) \propto x $ for $0<x<a$, but $\psi(x)=0$ for $x \leq 0$ or $x \geq a$. Give an explicit expansion of $\psi(x)$ in terms of the $\psi_n(x)$.

I literally have no idea. I've never seen a question like this before where a state is proportional to some function and to write it using the stationary states. I know the stationary states span the space of allowed states so it is possible. It looks like some sort of fourier series question but it is of course not periodic along the real line? A walkthrough of how to do this question would be really appreciated as I've never seen one like this.


1 Answer 1


For the first part, it is not $\phi_n$ that has to be normalized, it is $|\phi_n|^2$, so that's what you have to integrate over and set equal to $1$.

For the second part, this is a pretty standard thing and you are correct that this is "some sort of Fourier series". Here's a hint. The eigenfunctions of the Schroedinger equation not only span the space of allowed states, they are also orthogonal to each other with respect to the scalar product $\langle f, g \ rangle = \int f(x) g(x)$. The integration limits are, technically, $-\infty$ to $\infty$, but since all allowed states must be $0$ outside of $|x| < a$, we can just integrate from $-a$ to $a$.

So now, we know that $\int_{-a}^a \phi_n(x) \phi_m(x) = \delta_{nm}$. For your decomposition, make the following "ansatz": $$\psi(x) = \sum_{n=1}^\infty c_n \phi_n(x)$$ As you said: The eigenfunctions span the allowed states, so such a decomposition must exist. Now multiply both sides with $\phi_m(x)$ and integrate, and that will tell you how to compute $c_m$.

  • $\begingroup$ Thanks! The only thing I'm still unclear is the fact it was $\psi \propto x$ not $\psi = x$ so in the ansatz i can only say $\psi(x) = kx$ for some $k$? Or does this not matter as long as it's normalisable k can be changed? $\endgroup$
    – guest313
    Commented May 19, 2016 at 21:17
  • $\begingroup$ EDIT: Also, it's clear how to proceed if $\psi(x)= x$ over $-a<x<a$ however it says $\psi(x)=0$ for $x \leq0$ so again I don't see how to use orthogonality in these two seperate regions, given the orthogonality relation is an integral from $-a$ to $a$ $\endgroup$
    – guest313
    Commented May 19, 2016 at 21:25
  • $\begingroup$ I have $c_m = \dfrac{2a}{\pi m}(-1)^{m+1}$ ??? $\endgroup$
    – guest313
    Commented May 19, 2016 at 21:52
  • $\begingroup$ Oh yes, use $kx$ where $k$ is some normalization constant that you can compute if you want. Just use the orthogonality as if $\psi(x) = x$ for $-a < x < a$, then restrict the solution to $x > 0$. $\endgroup$
    – Lagerbaer
    Commented May 24, 2016 at 16:49

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