# Infinite potential well suddenly expanding

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?

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