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$\frac{d}{dt}$$\hat{H}$ = $\frac{i}{\hbar}$$[\hat{H},\hat{H}]$ +$\frac{\partial{\hat{H}}}{\partial{t}}$

That's as far as I've got. I do not know much about the Heisenberg equation or even what it represents. Could someone give me a beginners intro to it ?

I do have one idea : $\hat{H}$ = $i\hbar$$\frac{d}{dt}$

I've been told that if there is no time dependence then $\frac{\partial{\hat{H}}}{\partial{t}}$ in the Heisenberg equation goes to 0.

I am not sure if the Hamiltonian has no time dependence because of that derivative wrt to time in the above equation.

Secondly, even if I could prove $\frac{\partial{\hat{H}}}{\partial{t}}$ = 0 I have absolutely no idea whatsoever what $\frac{i}{\hbar}$$[\hat{H},\hat{H}]$ means. I have no clue how to evaluate it or what its significance is.

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Ari Ben Canaan
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1 Answers1

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First, a note about the Hamiltonian and its time derivatives. I think that it is misleading to write that the Hamiltonian $$ H = i\hbar\frac{d}{dt}, $$ although the time-dependent Schrodinger equation is of course $$ H\psi = i\hbar\frac{d}{dt} \psi. $$

To evaluate e.g. $\frac{d}{dt}H$ you should consider $H=H(p, q, t)$, rather than $H = i\hbar\frac{d}{dt}$. $p$ and $q$ are the canonical coordinates in the Legendre transformation between the Hamiltonian and the Lagrangian formalisms.

The commutator is defined, $$ [A,B]=AB-BA, $$ so $[H,H]=0$ and we have $$ \frac{d}{dt}H = \frac{\partial}{\partial t}H $$ The full time derivative of $H$ is equal to the partial time derivative of $H$. What does this mean? It means that Heisenberg's equation is related to Hamilton's equations!

To see this, write the full derivative as $$ \frac{d}{dt} H = \frac{\partial H}{\partial q} \dot q + \frac{\partial H}{\partial p} \dot p + \frac{\partial H}{\partial t} $$ We must have $$ \frac{\partial H}{\partial q} \dot q + \frac{\partial H}{\partial p} \dot p = 0 $$ This is satisfied by Hamilton's equations, $$ \dot q = \frac{\partial H}{\partial q}\\ \dot p = -\frac{\partial H}{\partial p} $$

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innisfree
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