Question:
What is the value of $ \langle j | x | k \rangle $ where $ | j \rangle $ and $ | k \rangle $ are the j and k eigen function of the Quantum Harmonic Oscillator?
Solution:
As a disclaimer, I didn't quite solve the problem completely in the exam. But I get the correct answer.
The key annoying piece is the $ x $ inside the sandwich. We just break it down into ladder operators.
$ \hat{x} = \sqrt{\frac{2\hbar}{2m\omega}}(\hat{a}^{\dagger} + \hat{a}) $.
Substitute this back into $ \langle j | x | k \rangle $, we have got
$ \begin{eqnarray*} & & \langle j | x | k \rangle \\ &=& \langle j | \sqrt{\frac{2\hbar}{2m\omega}}(\hat{a}^{\dagger} + \hat{a}) | k \rangle \\ &=& \sqrt{\frac{2\hbar}{2m\omega}} \langle j | (\hat{a}^{\dagger} + \hat{a}) | k \rangle \\ &=& \sqrt{\frac{2\hbar}{2m\omega}} (\langle j | \hat{a}^{\dagger} | k \rangle + \langle j | \hat{a} | k \rangle)\\ &=& \sqrt{\frac{2\hbar}{2m\omega}} (\sqrt{k+1}\langle j | k + 1 \rangle + \sqrt{k} \langle j | k - 1 \rangle)\\ &=& \sqrt{\frac{2\hbar}{2m\omega}} (\sqrt{k+1}\delta_{j, k + 1} + \sqrt{k} \delta_{j, k - 1})\\ \end{eqnarray*} $
What is the value of $ \langle j | x | k \rangle $ where $ | j \rangle $ and $ | k \rangle $ are the j and k eigen function of the Quantum Harmonic Oscillator?
Solution:
As a disclaimer, I didn't quite solve the problem completely in the exam. But I get the correct answer.
The key annoying piece is the $ x $ inside the sandwich. We just break it down into ladder operators.
$ \hat{x} = \sqrt{\frac{2\hbar}{2m\omega}}(\hat{a}^{\dagger} + \hat{a}) $.
Substitute this back into $ \langle j | x | k \rangle $, we have got
$ \begin{eqnarray*} & & \langle j | x | k \rangle \\ &=& \langle j | \sqrt{\frac{2\hbar}{2m\omega}}(\hat{a}^{\dagger} + \hat{a}) | k \rangle \\ &=& \sqrt{\frac{2\hbar}{2m\omega}} \langle j | (\hat{a}^{\dagger} + \hat{a}) | k \rangle \\ &=& \sqrt{\frac{2\hbar}{2m\omega}} (\langle j | \hat{a}^{\dagger} | k \rangle + \langle j | \hat{a} | k \rangle)\\ &=& \sqrt{\frac{2\hbar}{2m\omega}} (\sqrt{k+1}\langle j | k + 1 \rangle + \sqrt{k} \langle j | k - 1 \rangle)\\ &=& \sqrt{\frac{2\hbar}{2m\omega}} (\sqrt{k+1}\delta_{j, k + 1} + \sqrt{k} \delta_{j, k - 1})\\ \end{eqnarray*} $
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