Exploring Quantum Physics - Final Exam Part 1 Question 8

Question:

What are the eigen energies for this system?

$\hat H = -\frac{\hbar^2}{2m} \frac{d^2}{dx^2} + \frac{1}{2} m \omega^2 x^2 - q\mathcal{E}x$

Solution:

Using the given substitution $z = x - \frac{q\mathcal{E}}{\omega^2m}$, we have

$\begin{eqnarray*} \left(-\frac{\hbar^2}{2m} \frac{d^2}{dx^2} + \frac{1}{2} m \omega^2 x^2 - q\mathcal{E}x\right)\psi(x) &=& E\psi(x) \\ \left(-\frac{\hbar^2}{2m} \frac{d^2}{dz^2} + \frac{1}{2} m \omega^2 (z + \frac{q\mathcal{E}}{\omega^2m})^2 - q\mathcal{E}(z + \frac{q\mathcal{E}}{\omega^2m})\right)\psi(z + \frac{q\mathcal{E}}{\omega^2m}) &=& E\psi(z + \frac{q\mathcal{E}}{\omega^2m}) \\ \left(-\frac{\hbar^2}{2m} \frac{d^2}{dz^2} + \frac{1}{2} m \omega^2 (z + \frac{q\mathcal{E}}{\omega^2m})^2 - q\mathcal{E}z - \frac{q^2\mathcal{E}^2}{\omega^2m}\right)\psi(z + \frac{q\mathcal{E}}{\omega^2m}) &=& E\psi(z + \frac{q\mathcal{E}}{\omega^2m}) \\ \left(-\frac{\hbar^2}{2m} \frac{d^2}{dz^2} + \frac{1}{2} m \omega^2 (z^2 + 2z\frac{q\mathcal{E}}{\omega^2m}+(\frac{q\mathcal{E}}{\omega^2m})^2) - q\mathcal{E}z - \frac{q^2\mathcal{E}^2}{\omega^2m}\right)\psi(z + \frac{q\mathcal{E}}{\omega^2m}) &=& E\psi(z + \frac{q\mathcal{E}}{\omega^2m}) \\ \left(-\frac{\hbar^2}{2m} \frac{d^2}{dz^2} + \frac{1}{2} m \omega^2 z^2 + q\mathcal{E}z + \frac{q^2\mathcal{E}^2}{2\omega^2m} - q\mathcal{E}z - \frac{q^2\mathcal{E}^2}{\omega^2m}\right)\psi(z + \frac{q\mathcal{E}}{\omega^2m}) &=& E\psi(z + \frac{q\mathcal{E}}{\omega^2m}) \\ \left(-\frac{\hbar^2}{2m} \frac{d^2}{dz^2} + \frac{1}{2} m \omega^2 z^2 + \frac{q^2\mathcal{E}^2}{2\omega^2m} - \frac{q^2\mathcal{E}^2}{\omega^2m}\right)\psi(z + \frac{q\mathcal{E}}{\omega^2m}) &=& E\psi(z + \frac{q\mathcal{E}}{\omega^2m}) \\ \left(-\frac{\hbar^2}{2m} \frac{d^2}{dz^2} + \frac{1}{2} m \omega^2 z^2 - \frac{q^2\mathcal{E}^2}{2\omega^2m}\right)\psi(z + \frac{q\mathcal{E}}{\omega^2m}) &=& E\psi(z + \frac{q\mathcal{E}}{\omega^2m}) \\ \left(-\frac{\hbar^2}{2m} \frac{d^2}{dz^2} + \frac{1}{2} m \omega^2 z^2\right)\psi(z + \frac{q\mathcal{E}}{\omega^2m}) &=& (E + \frac{q^2\mathcal{E}^2}{2\omega^2m})\psi(z + \frac{q\mathcal{E}}{\omega^2m}) \\ \end{eqnarray*}$

Now we know $E + \frac{q^2\mathcal{E}^2}{2\omega^2m} = \hbar\omega\left(n + \frac{1}{2}\right)$, in other words, $E = \hbar\omega\left(n + \frac{1}{2}\right) - \frac{q^2\mathcal{E}^2}{2\omega^2m}$