Quantum Well Centred at the Origin Doubled in Size at time t=0. Symmetry seems to render the question incorrect

[kodel420]

Homework Statement

I am currently studying an Introductory Quantum Mechanics course, and one past example question talks about the following. An infinite potential well given by $$V(x) = \begin{cases} 0 \; -\frac{L}{2} \le x \le \frac{L}{2} \\ \infty \; \text{otherwise} \end{cases}$$

At time $t=0$, the particle is in the ground state of the well and the well is doubled in width and now runs from $-L$ to $L$. What is the probability that the particle is found in the new n=2 state [answer : $\frac{1}{\sqrt{2}}$].

Relevant Equations

$$V_i(x) = \begin{cases} 0 \; -\frac{L}{2} \le x \le \frac{L}{2} \\ \infty \; \text{otherwise} \end{cases}$$
$$V_f(x) = \begin{cases} 0 \; -L \le x \le L \\ \infty \; \text{otherwise} \end{cases}$$

Firstly I have found the eigenstates for both the original well and the new well as the following

$$\psi_{n,\frac{L}{2}} = \begin{cases} \sqrt{\frac{2}{L}} \cos{\frac{n \pi x}{L}} \; \; \; \; \; n \text{ odd} \\ \sqrt{\frac{2}{L}} \sin{\frac{n \pi x}{L}} \; \; \; \; \; n \text{ even} \end{cases}$$

$$\psi_{n,L} = \begin{cases} \frac{1}{\sqrt{L}} \cos{\frac{n \pi x}{2L}} \; \; \; \; \; n \text{ odd} \\ \frac{1}{\sqrt{L}} \sin{\frac{n \pi x}{2L}} \; \; \; \; \; n \text{ even} \end{cases}$$

Then using the principle of superposition the original ground state can be written as a superposition of the new eigenstates

$$\psi_{1,\frac{L}{2}} = \sum_{n = 1}^{\infty} {c_n}\psi_{n,L} \; \; \; \; \; c_n = \int_{-L/2}^{L/2} \psi_{1,\frac{L}{2}} \; \psi_{n,L} \;dx$$

and by the Born Rule the probability is just $|c_n|^2$ so we are looking to find $c_2$. My problem is that the integral $$\int_{-L/2}^{L/2} \psi_{1,\frac{L}{2}} \; \psi_{2,L} = 0$$ obviously not giving the desired result. So my question is, am I doing something wrong, or does the symmetry of the well cause an issue, because the same issue does not arise for a well from $0$ to $L$ being extended to $0$ to $2L$

 

[kuruman]

My problem is that the integral $$\int_{-L/2}^{L/2} \psi_{1,\frac{L}{2}} \; \psi_{2,L} = 0$$ obviously not giving the desired result.

That's not obvious to me. Using this integral, I found the first eight non-zero coefficients and plotted the summation to get the graph shown below ($L=1$). It does what it's supposed to do, no?

 

[kodel420]

That's not obvious to me. Using this integral, I found the first eight non-zero coefficients and plotted the summation to get the graph shown below ($L=1$). It does what it's supposed to do, no?

View attachment 344563

I believe this is correct, however my issue is that the question specifically asks for the probability of the second state, which would be the second coefficient squared. Perhaps I have it wrong, but if you omit the even coefficients in that graph I think the same graph should be produced. Thank you for graphing them though, I didn't think to do that.

I also should have probably written out the integral in full for the second coefficient, that's my bad :

$$c_n = \int_{-L/2}^{L/2} \psi_{1,\frac{L}{2}} \; \psi_{2,L} = \int_{-L/2}^{L/2} \sqrt{\frac{2}{L}} \cos{\frac{\pi x}{L}} \frac{1}{\sqrt{L}} \sin{\frac{2 \pi x}{2L}} = 0$$

 

[TSny]

I also should have probably written out the integral in full for the second coefficient, that's my bad :

$$c_n = \int_{-L/2}^{L/2} \psi_{1,\frac{L}{2}} \; \psi_{2,L} = \int_{-L/2}^{L/2} \sqrt{\frac{2}{L}} \cos{\frac{\pi x}{L}} \frac{1}{\sqrt{L}} \sin{\frac{2 \pi x}{2L}} = 0$$

I agree with this result.

I noticed the following. Suppose the well initially extends from x = 0 to x = L and the particle is in the ground state of this well. The right wall is suddenly moved to x = 2L. Then I find that the probability amplitude for finding the particle in the first excited state of the new well is $1/\sqrt 2$.

[Edited to make the initial width of the well equal to L and the final width equal to 2L.]

 

[kuruman]

I believe this is correct, however my issue is that the question specifically asks for the probability of the second state, which would be the second coefficient squared. Perhaps I have it wrong, but if you omit the even coefficients in that graph I think the same graph should be produced. Thank you for graphing them though, I didn't think to do that.

I also should have probably written out the integral in full for the second coefficient, that's my bad :

$$c_n = \int_{-L/2}^{L/2} \psi_{1,\frac{L}{2}} \; \psi_{2,L} = \int_{-L/2}^{L/2} \sqrt{\frac{2}{L}} \cos{\frac{\pi x}{L}} \frac{1}{\sqrt{L}} \sin{\frac{2 \pi x}{2L}} = 0$$

It cannot be otherwise by symmetry. One is integrating an odd function, $\cos{\frac{\pi x}{L}} \frac{1}{\sqrt{L}} \sin{\frac{2 \pi x}{2L}}$ over symmetric limits which means that the integral vanishes. All the $c_n$ terms with $n=~$even must vanish for the same reason. That is as it ought to be because you are expanding an even function and you cannot have odd terms in the expansion.

The energy level diagram consists of all the odd $n$ coefficients. The first excited state has $n=3$, the second excited state $n=5$ and so on.