Originally Posted by

**albi** b)

Let $\displaystyle D_i$ denote differentiation over i-th coordinate. We have one particle with mass m. The energy operator should be like that:

$\displaystyle E = -\frac{\hbar^2}{2m} \nabla^2 = -\frac{\hbar^2}{2m} \sum_i D^2_i $

Momentum operator (coordinate k):

$\displaystyle p_k = -i\hbar D_k $

I will focus on the commutator: $\displaystyle [p_k, E]$

$\displaystyle [p_k, E] = \frac{i\hbar^3}{2m}\left[D_k, \sum_i D^2_i \right] = \frac{i\hbar^3}{2m}\sum_i [D_k, D^2_i] $

I shall assume $\displaystyle i \neq k$:

$\displaystyle [D_k, D^2_i] = D_kD^2_i - D^2_iD_k$

Let's assume the wavefunction belongs to $\displaystyle C^\infty$ class (mathematiclly very restrictive assumption but I consider it as "physical")

As I remember there was some theorem which states that we can change sequence of differentiation (I guess the assumption was that the second derivative is continuous). It was important theorem but I don't remember it's name :( .

So under this assumptions for wavefunction we should get:

$\displaystyle D_kD^2_i \psi = D_kD_iD_i \psi = D_iD_kD_i \psi = D_iD_iD_k \psi = D^2_iD_k \psi$

So the commutator $\displaystyle [D_k, D^2_i]$ is zero

I guess it is obvious that: $\displaystyle [D_k, D^2_k] = 0$

So $\displaystyle [p_k, E] = 0$

Which means that operators $\displaystyle p_k$, $\displaystyle E$ commute.

Thus we can simulataneously measure momentum and energy with infinite precision.

**It seems to me a little strange because **the question "Give the conditions under which this physical quantities are simultaneously measurable. Give an example of quantum system having such property." suggests that they should not commute (otherwise giving the example is trivial - any system has such proprety). **Could you check if my reasoning is correct?**