Problem statement: an electron is in its fundamental state in an infinite (1-dimensional) potential well, its walls being located at $x=0$ and $x=a$. Suddenly, the right wall moves from $x=a$ to $x=2a$. ¿What is the probability of the electron being in its fundamental state? And what's the probability of it being in its first excited state?

As far as I got: The first piece of information I have (the particle being in its fundamental state in the first potential well) gives me an initial condition my solution should fulfill. That is to say, if we denote as $\Psi(x,t)$ the electron's wave function after the potential expands, then necessarily $\Psi(x,0)=\phi_0(x)$, being $\phi_0(x)$ the particle's fundamental state in the first potential (just its spatial part, I'm unaware if there's a specific term for it). I also know from my lectures this fundamental state is given by:

$$\phi_0(x) = \sqrt{\frac{2}{a}}\sin(\frac{\pi x}{a}), \quad x\in (0,a)$$

and null otherwise. I also know the particle, once in the second potential, will be in a state of superposition of some or all of its bound states:

$$\Psi(x,t)=\displaystyle \sum_{n=1}^\infty C_n\phi_n'(x)e^{-iE_n't/\hbar}$$

where the prime (') symbol denotes these magnitudes are defined in the second potential. Imposing the initial condition, I get:

$$\phi_0(x) = \displaystyle \sum_{n=1}^\infty C_n\phi_n'(x)$$

$$\sqrt{\frac{2}{a}}\sin(\frac{\pi x}{a}) = \displaystyle \sum_{n=1}^\infty C_n\sqrt{\frac{2}{2a}}\sin\left(n\pi\left(\frac{x}{2a}-\frac{1}{2}\right)\right)$$

Now, performing a Fourier trick, I get:

$$\int_0^{2a}\frac{\sqrt{2}}{a}\sin(\frac{\pi x}{a})\sin\left(m\pi\left(\frac{x}{2a}-\frac{1}{2}\right)\right)dx = \int_0^{2a}\displaystyle \sum_{n=1}^\infty C_n \sin\left(n\pi\left(\frac{x}{2a}-\frac{1}{2}\right)\right)\sin\left(m\pi\left(\frac{x}{2a}-\frac{1}{2}\right)\right)\cdot dx$$

The second term is $0$ if $m\neq n$ and $C_n$ for $m=n$, so I get:

$$C_n = \int_0^{2a}\frac{\sqrt{2}}{a}\sin(\frac{\pi x}{a})\sin\left(m\pi\left(\frac{x}{2a}-\frac{1}{2}\right)\right)dx$$

which yields $0$ for every $n$. How can this be?

  • $\begingroup$ Just a note: you can write the trig function as: \sin $\endgroup$ Mar 10 at 17:04

1 Answer 1


The wave function for the first well is equal to zero outside its boundaries (since the walls are infinitely high), therefore in the last integral the limits of integration are not from 0 to 2a, but from 0 to a, which leads to a non-zero value of the integrals Cn for any n. Visually, they are equal to the (non-zero) area under the graph of the product of wave functions, which means a non-zero probability of being at any level n. For large n, it decreases as n increases.


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.