Exploring Quantum Physics - Week 7 Question 6

Question:

What is the exact energy?

Solution:

To solve the equation, we use a substitution $z = \zeta - z_0$.

$\begin{eqnarray*} -\frac{\hbar^2}{2m}\frac{d^2\Psi}{dz^2} + \frac{m\omega^2}{2}z^2\Psi + mgz\Psi = E_{exact} \Psi \\ -\frac{\hbar^2}{2m}\frac{d^2\Psi}{d\zeta^2} + \frac{m\omega^2}{2}(\zeta - z_0)^2\Psi + mg(\zeta - z_0)\Psi = E_{exact} \Psi \\ -\frac{\hbar^2}{2m}\frac{d^2\Psi}{d\zeta^2} + \frac{m\omega^2}{2}(\zeta^2 - 2\zeta z_0 + z_0^2)\Psi + mg(\zeta - z_0)\Psi = E_{exact} \Psi \\ \end{eqnarray*}$

Now we see a convenient choice for $z_0$ would be $\frac{g}{\omega^2}$, because that would give

$\begin{eqnarray*} -\frac{\hbar^2}{2m}\frac{d^2\Psi}{d\zeta^2} + \frac{m\omega^2}{2}(\zeta^2 - 2\zeta z_0 + z_0^2)\Psi + mg(\zeta - z_0)\Psi = E_{exact} \Psi \\ -\frac{\hbar^2}{2m}\frac{d^2\Psi}{d\zeta^2} + \frac{m\omega^2}{2}(\zeta^2 - 2\zeta \frac{g}{\omega^2} + z_0^2)\Psi + mg(\zeta - z_0)\Psi = E_{exact} \Psi \\ -\frac{\hbar^2}{2m}\frac{d^2\Psi}{d\zeta^2} + \frac{m\omega^2}{2}(\zeta^2 + z_0^2)\Psi - mgz_0\Psi = E_{exact} \Psi \\ -\frac{\hbar^2}{2m}\frac{d^2\Psi}{d\zeta^2} + \frac{m\omega^2}{2}\zeta^2\Psi = (E_{exact} + mgz_0 - \frac{m\omega^2}{2}z_0^2) \Psi \\ \end{eqnarray*}$

Therefore, for ground state, we have $,$.

$\begin{eqnarray*} \frac{h\omega}{2} &=& E_{exact} + mgz_0 - \frac{m\omega^2}{2}z_0^2 \\ E_{exact} &=& \frac{h\omega}{2} - mgz_0 + \frac{m\omega^2}{2}z_0^2 \\ &=& \frac{h\omega}{2} - mg\frac{g}{\omega^2} + \frac{m\omega^2}{2}(\frac{g}{\omega^2})^2 \\ &=& \frac{h\omega}{2} - \frac{mg^2}{\omega^2} + \frac{m\omega^2g^2}{2\omega^4} \\ &=& \frac{h\omega}{2} - \frac{mg^2}{\omega^2} + \frac{mg^2}{2\omega^2} \\ &=& \frac{h\omega}{2} - \frac{mg^2}{2\omega^2} \end{eqnarray*}$