In previous post I mentioned that the First Commutation Relation and Second Commutation Relation are equivalent. Here is a proof from Classical Perspective.

We assume one of them to be true. For example, we assume First Commutation Relation to be true. i.e. $[\hat{x}, \hat{p}] = i\hbar\rightarrow \hat{p} = -i\hbar\frac{\partial{}}{\partial{x}}$

Classical Hamiltonian is given by $\hat{H} = \frac{\hat{p}^2}{2m} + \hat{V} = -\frac{\hbar^2}{2m}\frac{\partial{}^2}{\partial{x^2}} + V$

We next compute the commutation between Hamiltonian and Position $[\hat{H},\hat{x}] = [\frac{\hat{p}^2}{2m} + \hat{V},\hat{x}]=\frac{1}{2m}[\hat{p}^2, x] =i\hbar\frac{\hat{p}}{m}$

We see the operator in the last term is precisely the velocity operator $\hat{v}$ $\hat{v}\psi = v\psi$

Therefore from definition of velocity $\hat{v}\psi = \frac{\partial{x}}{\partial{t}}\psi = [\frac{\partial{}}{\partial{t}}, \hat{x}]\psi$

So we conclude that $[\hat{H},\hat{x}]\psi = i\hbar[\frac{\partial{}}{\partial{t}}, \hat{x}]\psi$ $\hat{H} = i\hbar\frac{\partial{}}{\partial{t}}$

It then follows $[\hat{H},\hat{t}] = [i\hbar\frac{\partial{}}{\partial{t}}, \hat{t}] = i\hbar$

Equivalence proved.

(Note it’s only proved for non-relativistic hamiltonian, i still haven’t worked out the most general case at the moment)

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