Exploring Quantum Physics - Week 3 Extra Credit Question 9

Question

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Suppose $\psi(x, \frac{T}{2}) = \alpha\psi(\beta x + \gamma L, 0)$, we are the values of $\alpha$, $\beta$ and $\gamma$?

Solution:

We have $\phi(x, t) = \sum_{n=1}^{\infty} a_n \psi_n(x) \exp \left(-iE_n t/\hbar \right)$. Notice that the value of $t$ only impacts the exponential term, so let's focus there.

Substitute in $E_n = \frac{\pi^2 \hbar^2 n^2}{2mL^2}$ and $T = \frac{8mL^2}{h}$, we have the exponential term at $t = \frac{T}{2}$ equals to $\exp \left(-i\frac{\pi^2 \hbar^2 n^2}{2mL^2} \frac{8mL^2}{h}/\hbar \right) =  \exp \left(-in^2\pi \right) = (-1)^n$.

The last equality comes from nowhere, really, it is just an observation on those values, now we know that by shifting time by $\frac{T}{2}$, we are inverting the odd terms and keeping the even terms untouched.

Next we wanted to derive the values for $\alpha$, $\beta$ and $\gamma$ to achieve the effect of inverting the odd terms and leaving the even terms untouched. The identity $\sin(x + n\pi) = (-1)^n \sin(x)$ comes in really handy. It is really telling us that set $\gamma$ to 1.

Thanks for the discussion forum's reminder, I should make sure the range I passed into $\psi$ lies between $0$ and $L$, so it leaves me no choice but set $\beta$ to -1 to maintain that constraint.

Last but not least, setting $\beta$ to -1 inverted the terms unconditionally by 1, so we need to set $\alpha$ to -1 to flip it back, so the answer is -1,-1,1