Question:
Which of these is the correct expression for the nth eigenenergy $ E_n $ that appears in the figure above, where n=1,2,3,... is an index that labels the energies in increasing order?
Solution
I struggled on this quite a bit because the wave-function is not dependent on time. This is strange to me at first and I finally figured out that we are really abusing the term wavefunction here. Sometimes the position component of the full wavefunction is also called a wavefunction and leading to confusion. The full wave function also contains a term that with exponential dependence to time. But for that part, the energy is in the exponential term and eventually you will only get $ E = E $ as a trivial and useless result. The key to this is solving the problem based on the momentum operator instead.
$ \begin{eqnarray*} \hat{E} &=& \frac{\hat{p}^2}{2m} \\ &=& \frac{(-i\hbar\frac{\partial}{\partial x})^2}{2m} \\ &=& \frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2} \\ \hat{E} \psi(x) &=& \frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2}(\sqrt{\frac{2}{L}})\sin(\frac{2 n \pi x}{L}) \\ &=& \frac{\hbar^2}{2m}(\frac{2n\pi}{L})^2(\sqrt{\frac{2}{L}})(-\sin(\frac{2 n \pi x}{L})) \\ \end{eqnarray*} $
So the energies are given by: $ \frac{\hbar^2}{2m}(\frac{2n\pi}{L})^2 = \frac{\pi^2\hbar^2n^2}{2mL^2} $.
Which of these is the correct expression for the nth eigenenergy $ E_n $ that appears in the figure above, where n=1,2,3,... is an index that labels the energies in increasing order?
Solution
I struggled on this quite a bit because the wave-function is not dependent on time. This is strange to me at first and I finally figured out that we are really abusing the term wavefunction here. Sometimes the position component of the full wavefunction is also called a wavefunction and leading to confusion. The full wave function also contains a term that with exponential dependence to time. But for that part, the energy is in the exponential term and eventually you will only get $ E = E $ as a trivial and useless result. The key to this is solving the problem based on the momentum operator instead.
$ \begin{eqnarray*} \hat{E} &=& \frac{\hat{p}^2}{2m} \\ &=& \frac{(-i\hbar\frac{\partial}{\partial x})^2}{2m} \\ &=& \frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2} \\ \hat{E} \psi(x) &=& \frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2}(\sqrt{\frac{2}{L}})\sin(\frac{2 n \pi x}{L}) \\ &=& \frac{\hbar^2}{2m}(\frac{2n\pi}{L})^2(\sqrt{\frac{2}{L}})(-\sin(\frac{2 n \pi x}{L})) \\ \end{eqnarray*} $
So the energies are given by: $ \frac{\hbar^2}{2m}(\frac{2n\pi}{L})^2 = \frac{\pi^2\hbar^2n^2}{2mL^2} $.
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