Question:
What is the first order energy for the Quantum Harmonic Oscillator under gravity?
Solution:
We will use perturbation theory.
First, as usual, we let $ E = \sum\limits_{k = 0}^{\infty}{E_k\lambda^k} $, $ \psi = \sum\limits_{k = 0}^{\infty}{\psi_k \lambda^k} $.
Substitute these back to the Schrodinger equation. We get this by picking the terms with $ \lambda $ to the first order.
$ \lambda(H_0\psi_1 + V\psi_0) = \lambda(E_0\psi_1 + E_1\psi_0) $
That allow us to show, $ E_1 = \langle \psi_0 | V | \psi_0 \rangle $.
We know from the Quantum Harmonic Oscillator that $ \psi_0 = \left(\frac{m\omega}{\pi\hbar}\right)^{1/4} \exp(-\frac{m\omega z^2}{2\hbar}) $.
So the rest is just an integration.
$ \begin{eqnarray*} & & \int\limits_{-\infty}^{\infty}{dx (\left(\frac{m\omega}{\pi\hbar}\right)^{1/4} \exp(-\frac{m\omega z^2}{2\hbar})) mgz (\left(\frac{m\omega}{\pi\hbar}\right)^{1/4} \exp(-\frac{m\omega z^2}{2\hbar}))} \\ &=& \left(\frac{m\omega}{\pi\hbar}\right)^{1/2} mg \int\limits_{-\infty}^{\infty}{dx z (\exp(-\frac{m\omega z^2}{\hbar}))} \\ \end{eqnarray*} $
Lot of typesetting, but it is now obvious the first order energy is 0 because of the symmetry.
What is the first order energy for the Quantum Harmonic Oscillator under gravity?
Solution:
We will use perturbation theory.
First, as usual, we let $ E = \sum\limits_{k = 0}^{\infty}{E_k\lambda^k} $, $ \psi = \sum\limits_{k = 0}^{\infty}{\psi_k \lambda^k} $.
Substitute these back to the Schrodinger equation. We get this by picking the terms with $ \lambda $ to the first order.
$ \lambda(H_0\psi_1 + V\psi_0) = \lambda(E_0\psi_1 + E_1\psi_0) $
That allow us to show, $ E_1 = \langle \psi_0 | V | \psi_0 \rangle $.
We know from the Quantum Harmonic Oscillator that $ \psi_0 = \left(\frac{m\omega}{\pi\hbar}\right)^{1/4} \exp(-\frac{m\omega z^2}{2\hbar}) $.
So the rest is just an integration.
$ \begin{eqnarray*} & & \int\limits_{-\infty}^{\infty}{dx (\left(\frac{m\omega}{\pi\hbar}\right)^{1/4} \exp(-\frac{m\omega z^2}{2\hbar})) mgz (\left(\frac{m\omega}{\pi\hbar}\right)^{1/4} \exp(-\frac{m\omega z^2}{2\hbar}))} \\ &=& \left(\frac{m\omega}{\pi\hbar}\right)^{1/2} mg \int\limits_{-\infty}^{\infty}{dx z (\exp(-\frac{m\omega z^2}{\hbar}))} \\ \end{eqnarray*} $
Lot of typesetting, but it is now obvious the first order energy is 0 because of the symmetry.
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