Question:
What are the conditions that is true about $ \psi_1 $.
Solution:
We had the equation $ H_0 \psi_1 + V \psi_0 = E_0 \psi_1 + E_1 \psi_0 $.
With the last solution we can drop the last term.
Now consider the inner product of this equation with the eigenfunctions of the Quantum Harmonic Oscillator.
$ \langle \phi_k | H_0 | \psi_1 \rangle + \langle \phi_k | V | \psi_0 \rangle = E_0 \langle \phi_k | \psi_1 \rangle $
Being eigenfunction of the $ H_0 $, we can simplify the first term, and substitute in the eigen energies.
$ \hbar\omega(k+\frac{1}{2})\langle \phi_k | \psi_1 \rangle + \langle \phi_k | V | \psi_0 \rangle = \hbar\omega(\frac{1}{2}) \langle \phi_k | \psi_1 \rangle $
So it obviously simplify to
$ \hbar\omega k\langle \phi_k | \psi_1 \rangle + \langle \phi_k | V | \psi_0 \rangle = 0 $
From this form, it is clear that we need to evaluate $ \langle \phi_k | V | \psi_0 \rangle $. By symmetry, $ \langle \phi_0 | V | \psi_0 \rangle $ and $ \langle \phi_2 | V | \psi_0 \rangle $ are 0, so the key term to evaluate is $ \langle \phi_1 | V | \psi_0 \rangle $
$ \begin{eqnarray*} & & \langle \phi_1 | V | \psi_0 \rangle \\ &=& \int\limits_{\infty}^{\infty}{dz(\frac{1}{\sqrt{2}}\left(\frac{m\omega}{\pi\hbar}\right)^{1/4}\exp(-\frac{m\omega z^2}{2\hbar})2\left(\frac{m\omega}{\hbar}\right)^{1/2}z)mgz(\left(\frac{m\omega}{\pi\hbar}\right)^{1/4}\exp(-\frac{m\omega z^2}{2\hbar})) } \\ &=& 2mg\frac{1}{\sqrt{2}}\left(\frac{m\omega}{\hbar}\right)^{1/2}\left(\frac{m\omega}{\pi\hbar}\right)^{1/2}\int\limits_{\infty}^{\infty}{dz z^2(\exp(-\frac{m\omega z^2}{\hbar})) } \\ &=& 2mg\left(\frac{m\omega}{\hbar}\right)\frac{1}{\sqrt{2\pi}}\int\limits_{\infty}^{\infty}{dz z^2(\exp(-\frac{m\omega z^2}{\hbar})) } \\ \end{eqnarray*} $
The integral can be seen as the variance of a normal distribution with variance = $ \frac{\hbar}{2m\omega} $, so we will do the integral as follow:
$ \begin{eqnarray*} & & 2mg\left(\frac{m\omega}{\hbar}\right)\frac{1}{\sqrt{2\pi}}\int\limits_{\infty}^{\infty}{dz z^2(\exp(-\frac{m\omega z^2}{\hbar})) } \\ &=& 2mg\left(\frac{m\omega}{\hbar}\right)\sqrt{\frac{\hbar}{2m\omega}}\frac{1}{\sqrt{2\pi}\sqrt{\frac{\hbar}{2m\omega}}}\int\limits_{\infty}^{\infty}{dz z^2(\exp(-\frac{m\omega z^2}{\hbar})) } \\ &=& 2mg\left(\frac{m\omega}{\hbar}\right)\sqrt{\frac{\hbar}{2m\omega}}(\frac{\hbar}{2m\omega}) \\ &=& mg\sqrt{\frac{\hbar}{2m\omega}} \end{eqnarray*} $
To complete this, we substitute it back to get the result we need.
$ \begin{eqnarray*} \hbar\omega k \langle \phi_k | \psi_1 \rangle + \langle \phi_k | V | \psi_0 \rangle &=& 0 \\ \hbar\omega \langle \phi_1 | \psi_1 \rangle + \langle \phi_1 | V | \psi_0 \rangle &=& 0 \\ \hbar\omega \langle \phi_1 | \psi_1 \rangle + mg\sqrt{\frac{\hbar}{2m\omega}} &=& 0 \\ \hbar\omega \langle \phi_1 | \psi_1 \rangle &=& -mg\sqrt{\frac{\hbar}{2m\omega}} \\ \langle \phi_1 | \psi_1 \rangle &=& -\frac{mg}{\hbar\omega}\sqrt{\frac{\hbar}{2m\omega}} \\ \end{eqnarray*} $
Lot of work just to show it is non-zero, but we will need it for the next problem.
What are the conditions that is true about $ \psi_1 $.
Solution:
We had the equation $ H_0 \psi_1 + V \psi_0 = E_0 \psi_1 + E_1 \psi_0 $.
With the last solution we can drop the last term.
Now consider the inner product of this equation with the eigenfunctions of the Quantum Harmonic Oscillator.
$ \langle \phi_k | H_0 | \psi_1 \rangle + \langle \phi_k | V | \psi_0 \rangle = E_0 \langle \phi_k | \psi_1 \rangle $
Being eigenfunction of the $ H_0 $, we can simplify the first term, and substitute in the eigen energies.
$ \hbar\omega(k+\frac{1}{2})\langle \phi_k | \psi_1 \rangle + \langle \phi_k | V | \psi_0 \rangle = \hbar\omega(\frac{1}{2}) \langle \phi_k | \psi_1 \rangle $
So it obviously simplify to
$ \hbar\omega k\langle \phi_k | \psi_1 \rangle + \langle \phi_k | V | \psi_0 \rangle = 0 $
From this form, it is clear that we need to evaluate $ \langle \phi_k | V | \psi_0 \rangle $. By symmetry, $ \langle \phi_0 | V | \psi_0 \rangle $ and $ \langle \phi_2 | V | \psi_0 \rangle $ are 0, so the key term to evaluate is $ \langle \phi_1 | V | \psi_0 \rangle $
$ \begin{eqnarray*} & & \langle \phi_1 | V | \psi_0 \rangle \\ &=& \int\limits_{\infty}^{\infty}{dz(\frac{1}{\sqrt{2}}\left(\frac{m\omega}{\pi\hbar}\right)^{1/4}\exp(-\frac{m\omega z^2}{2\hbar})2\left(\frac{m\omega}{\hbar}\right)^{1/2}z)mgz(\left(\frac{m\omega}{\pi\hbar}\right)^{1/4}\exp(-\frac{m\omega z^2}{2\hbar})) } \\ &=& 2mg\frac{1}{\sqrt{2}}\left(\frac{m\omega}{\hbar}\right)^{1/2}\left(\frac{m\omega}{\pi\hbar}\right)^{1/2}\int\limits_{\infty}^{\infty}{dz z^2(\exp(-\frac{m\omega z^2}{\hbar})) } \\ &=& 2mg\left(\frac{m\omega}{\hbar}\right)\frac{1}{\sqrt{2\pi}}\int\limits_{\infty}^{\infty}{dz z^2(\exp(-\frac{m\omega z^2}{\hbar})) } \\ \end{eqnarray*} $
The integral can be seen as the variance of a normal distribution with variance = $ \frac{\hbar}{2m\omega} $, so we will do the integral as follow:
$ \begin{eqnarray*} & & 2mg\left(\frac{m\omega}{\hbar}\right)\frac{1}{\sqrt{2\pi}}\int\limits_{\infty}^{\infty}{dz z^2(\exp(-\frac{m\omega z^2}{\hbar})) } \\ &=& 2mg\left(\frac{m\omega}{\hbar}\right)\sqrt{\frac{\hbar}{2m\omega}}\frac{1}{\sqrt{2\pi}\sqrt{\frac{\hbar}{2m\omega}}}\int\limits_{\infty}^{\infty}{dz z^2(\exp(-\frac{m\omega z^2}{\hbar})) } \\ &=& 2mg\left(\frac{m\omega}{\hbar}\right)\sqrt{\frac{\hbar}{2m\omega}}(\frac{\hbar}{2m\omega}) \\ &=& mg\sqrt{\frac{\hbar}{2m\omega}} \end{eqnarray*} $
To complete this, we substitute it back to get the result we need.
$ \begin{eqnarray*} \hbar\omega k \langle \phi_k | \psi_1 \rangle + \langle \phi_k | V | \psi_0 \rangle &=& 0 \\ \hbar\omega \langle \phi_1 | \psi_1 \rangle + \langle \phi_1 | V | \psi_0 \rangle &=& 0 \\ \hbar\omega \langle \phi_1 | \psi_1 \rangle + mg\sqrt{\frac{\hbar}{2m\omega}} &=& 0 \\ \hbar\omega \langle \phi_1 | \psi_1 \rangle &=& -mg\sqrt{\frac{\hbar}{2m\omega}} \\ \langle \phi_1 | \psi_1 \rangle &=& -\frac{mg}{\hbar\omega}\sqrt{\frac{\hbar}{2m\omega}} \\ \end{eqnarray*} $
Lot of work just to show it is non-zero, but we will need it for the next problem.
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