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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.

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Why do you expect they wouldn't be subject to interference? You're right, $x$ and $H$ generally don't commute, meaning they can't be simultaneously measured. – Michael Brown Jan 23 '13 at 2:15
What do you exactly mean by interference? – Jorge Jan 23 '13 at 9:28
@Burzum It means that between $E$ and $x$ there is no interference of measurements, i.e. $[E,x]=0$ (or?) $[H,x]=0$. That's what I understand. – Ana Jan 23 '13 at 13:09
@MichaelBrown Actually I thought that as the momentum $p_y$ and $x$ are not subject to interference, i.e. $p_y$ and $x$ are not complementary operators ($[p_y,x]=0$), then in analogy $E$ and $x$ "should" not be complementary operators. But, in fact I have no reasons to expect that. Actually this is an exercise of the book "A course in Statistical Thermodynamics" by Kestin & Dorfman. – Ana Jan 23 '13 at 13:14
up vote 2 down vote accepted

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. $\langle \hat{p} \rangle = 0$.

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I have a doubt. Is it correct to use $H$ instead of $E$? Because I'm looking for a statement between energy and position. – Ana Jan 23 '13 at 14:01
In Quantum mechanics each measurable quantity has a Hermitian operator associated with it. The operator associated with the energy $E$, is the Hamiltonian $H$. This means any question you have about measurements of the energy, will be answered by performing calculations on the operator $H$. In this case you're asking about the uncertainty of $E$, so you need to use $H$ in the above formula of the uncertainty principle. – Joe Jan 23 '13 at 14:16
You are right. So, the answer is there is no interference between the simultaneous measurements of position and energy of a system only if the potential $V(x)$ is symmetric. Because in that case $\psi$ is symmetric or asymmetric, then $\langle p\rangle=\frac{\hbar}{i}\int\psi^{*}\partial_{x}\psi dx=0$. Right? – Ana Jan 23 '13 at 21:15
That's almost right. First of all, even if the potential is symmetric, you still need the system to be in an energy eigenstate. Only the eigenstates are guaranteed to be symmetric or antisymmetric. Secondly, it's not true that simultaneous measurement is possible only if the potential is symmetric. There might be other cases in which you get $\langle [H,x] \rangle=0$. All you can say is that this is one case where it is possible. – Joe Jan 24 '13 at 6:58
OK, I got it. Thanks! – Ana Jan 24 '13 at 15:32

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