How would you prove that the simultaneous measurements of position and energy are not subject to interference?
I was thinking in calculate the commutation relation between $x$ and $H$ (Because $\Delta E=\Delta H$), but I realized that $[H,x]\neq0$, so I tried to use a more general expression for the Uncertainty Principle that says that if $H_1$ and $H_2$ are Hermitian operators then $\Delta H_1 \Delta H_2\geq\frac{1}{2}|\langle [H_1,H_2]\rangle|$, but again, $[H,x]\neq0$. Can you suggest me a way to do this? Thanks.
Quantum Mechanics – Uncertainty Relation and Energy-Position Interference Explained
heisenberg-uncertainty-principlequantum mechanics
Best Answer
You're right to use the general form of the Uncertainty Principle, namely: $$ \Delta H_1 \Delta H_2\geq\frac{1}{2}|\langle [H_1,H_2]\rangle|. $$ However, note that in the right hand side you have the expectation value of the commutator, so even if $[H,x] \neq 0$ it can still be that $\langle [H,x] \rangle = 0$. If this is the case then you can simultaneously measure position and energy.
For example, if you have a simple one-dimensional Hamiltonian with a potential: $$ H = \frac{\hat{p}^2}{2 m} + V(x), $$ then you can easily show that $$ [H,x] = -\frac{i \hbar}{m} \hat{p}.$$ Now you just have to check whether your system happens to be in a quantum state for which the expectation value of the momentum vanishes, i.e. $\langle \hat{p} \rangle = 0$.