Question:
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What is the fundamental period of the wavefunctions?
Solution:
Notice that we have $\phi(x, t) = \sum_{n=1}^{\infty} a_n \psi_n(x) \exp \left(-iE_n t/\hbar \right)$, it is just a sum of complex exponentials, so let's take a look at their periods.
The periods are just $ \frac{2\pi}{E_n/\hbar} = \frac{2\pi\hbar}{E_n}$ and the energies are $ \frac{\pi^2\hbar^2n^2}{2mL^2} $, so can compute the periods as $ \frac{2\pi\hbar}{\frac{\pi^2\hbar^2n^2}{2mL^2}} = \frac{4mL^2}{\pi\hbar n^2} = \frac{8mL^2}{h n^2}$
Now we see the largest period happens when $ n = 1 $ and it is an integral multiple of all other periods, so $ \frac{8mL^2}{h} $ is the answer.
... snip ...
What is the fundamental period of the wavefunctions?
Solution:
Notice that we have $\phi(x, t) = \sum_{n=1}^{\infty} a_n \psi_n(x) \exp \left(-iE_n t/\hbar \right)$, it is just a sum of complex exponentials, so let's take a look at their periods.
The periods are just $ \frac{2\pi}{E_n/\hbar} = \frac{2\pi\hbar}{E_n}$ and the energies are $ \frac{\pi^2\hbar^2n^2}{2mL^2} $, so can compute the periods as $ \frac{2\pi\hbar}{\frac{\pi^2\hbar^2n^2}{2mL^2}} = \frac{4mL^2}{\pi\hbar n^2} = \frac{8mL^2}{h n^2}$
Now we see the largest period happens when $ n = 1 $ and it is an integral multiple of all other periods, so $ \frac{8mL^2}{h} $ is the answer.
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