Answer to Question #126200 in Quantum Mechanics for Kaleem

Question

A nuclear scientist claimed that he has accurately measured the energy and momentum of a
radiation (e.g. Alpha, Beta, Gamma-ray, Neutron etc.) using his most sensitive Geiger Muller
counter. Can you believe his claim is correct? If yes, give a proper justification for his claim and
if no, then also prove that he is wrong

Answer 1

In Quantum Mechanics two quantities can be measured simultaneously, if their operators commute.

The operator for the radiation energy (kinetic) is the following:

$$\hat{T} = -\dfrac{\hbar}{2m}\Delta$$

where $\Delta = \dfrac{\partial^2}{\partial x^2} + \dfrac{\partial^2}{\partial y^2} + \dfrac{\partial^2}{\partial z^2}$ is the Laplacian.

The operator for the momentum is the following:

$$\hat{\mathbf{p}} = -i\hbar\nabla$$

where $\nabla = \left( \dfrac{\partial}{\partial x}, \dfrac{\partial}{\partial y}, \dfrac{\partial}{\partial z}\right)$ is the nabla operator.

These operators commute if the following equation is true:

$$\hat{T}\hat{\mathbf{p}} =\hat{\mathbf{p}}\hat{T}$$

As far as both operators consist of a partial derivatives, the question is whether or not we can change the order of the derivatives. According to the Clairaut's theorem, we can do it. Thus:

$$\hat{T}\hat{\mathbf{p}} = -\dfrac{\hbar}{2m}\Delta(-i\hbar\nabla) = \dfrac{i\hbar^2}{2m}\Delta(\nabla) = \dfrac{i\hbar^2}{2m}\nabla(\Delta) = \hat{\mathbf{p}}\hat{T}$$

Thus, operators comute and we can measure the energy and momentum of a radiation at the same time.

Answer. Yes, his claim is correct.