Discussion
Which of the following operator commutes with Hamiltonian of one-dimensional oscillator. x and Px are position and momentum operators, a and a+ are annihilation and creation operators :
Answer: D
The operator that commutes with the Hamiltonian is \( a^{+}a \), the number operator.
For the oscillator \( H = \hbar\omega\left(a^{+}a + \frac{1}{2}\right) \), so \( H \) is a function of \( a^{+}a \) alone and \( [H, a^{+}a] = 0 \). The energy eigenstates are the number states \( |n\rangle \).
The ladder operators do not commute with it: \( [H, a] = -\hbar\omega a \) and \( [H, a^{+}] = \hbar\omega a^{+} \), which is why they lower and raise the energy. The momentum does not either: \( [H, P_{x}] = i\hbar m\omega^{2}x \), since the potential depends on \( x \).
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