Question #126200
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
1
Expert's answer
2020-07-14T08:51:49-0400

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

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


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

where Δ=2x2+2y2+2z2\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:


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

where =(x,y,z)\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:

T^p^=p^T^\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:


T^p^=2mΔ(i)=i22mΔ()=i22m(Δ)=p^T^\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.


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