Exploring Quantum Physics - Week 4 Question 3

Question:

Calculate the expectation value of the kinetic, $\langle \hat T \rangle = \langle \frac{\hat p^2}{2m}\rangle$, and potential, $\langle \hat V\rangle = \langle \frac{1}{2}m\omega^2 \hat x^2\rangle$, energy operators in the $n$-th excited state of the harmonic oscillator, $|n\rangle$

Solution:

The problem hint on not required to do integration, which is great. In fact, there is a easy way to just pick the right answer, the sum of the energies must be $\hbar \omega\left(n + \frac{1}{2}\right)$ and there is only one answer that matches this, which is good for sanity check.

Now let's actually computes the kinetic energy, in the previous problem, we have the momentum operator, so we just have to apply that twice.

$\begin{eqnarray*} \hat{p} \psi_n &=& i\sqrt{\frac{m\hbar\omega}{2}}(\hat{a}^\dagger - \hat{a})\psi_n \\ &=& i\sqrt{\frac{m\hbar\omega}{2}}(\hat{a}^\dagger\psi_n - \hat{a}\psi_n) \\ \hat{p}^2 \psi_n &=& i\sqrt{\frac{m\hbar\omega}{2}}(\hat{a}^\dagger - \hat{a})(i\sqrt{\frac{m\hbar\omega}{2}}(\hat{a}^\dagger\psi_n - \hat{a}\psi_n)) \\ &=& \frac{m\hbar\omega}{2}(\hat{a} - \hat{a}^\dagger)(\hat{a}^\dagger\psi_n - \hat{a}\psi_n) \\ &=& \frac{m\hbar\omega}{2}(\hat{a}\hat{a}^\dagger\psi_n - \hat{a}\hat{a}\psi_n - \hat{a}^\dagger\hat{a}^\dagger\psi_n + \hat{a}^\dagger\hat{a}\psi_n) \\ &=& \frac{m\hbar\omega}{2}((1+\hat{a}^\dagger\hat{a})\psi_n - \hat{a}\hat{a}\psi_n - \hat{a}^\dagger\hat{a}^\dagger\psi_n + \hat{a}^\dagger\hat{a}\psi_n) \\ &=& \frac{m\hbar\omega}{2}(\psi_n - \hat{a}\hat{a}\psi_n - \hat{a}^\dagger\hat{a}^\dagger\psi_n + 2\hat{a}^\dagger\hat{a}\psi_n) \\ &=& \frac{m\hbar\omega}{2}(\psi_n - \hat{a}\hat{a}\psi_n - \hat{a}^\dagger\hat{a}^\dagger\psi_n + 2n\psi_n) \\ &=& \frac{m\hbar\omega}{2}((2n + 1)\psi_n - \hat{a}\hat{a}\psi_n - \hat{a}^\dagger\hat{a}^\dagger\psi_n) \\ &=& \frac{m\hbar\omega}{2}((2n + 1)\psi_n - C_1\psi_{n-2} - C_2\psi_{n+2}) \\ \frac{\hat{p}^2}{2m} \psi_n &=& i\sqrt{\frac{\hbar\omega}{4}}(\hat{a}^\dagger - \hat{a})(i\sqrt{\frac{m\hbar\omega}{2}}(\hat{a}^\dagger\psi_n - \hat{a}\psi_n)) \\ \end{eqnarray*}$

$C_1$ and $C_2$ are simply constant that I don't bother computing. Now we know the wavefunctions are orthonormal, so the final answer for the kinetic energy is $\frac{\hbar\omega}{4}(2n+1)$.