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

Expert's answer

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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