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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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    The square brackets denote the commutator. $[\hat{A},\hat{B}]=\hat{A}\hat{B}-\hat{B}\hat{A}$ is a measure for the difference between letting $\hat{B}$ then $\hat{A}$ operate and letting $\hat{A}$ then $\hat{B}$ operate. For operators, you usually calculate the commutator by letting the whole thing operate on a function: $[\hat{A},\hat{B}]\Psi$. But if $\hat{A} = \hat{B}$ it's clear that the commutator is zero, even without explicitly calculating it. – Wouter Apr 02 '14 at 07:12
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    Concerning the time dependence: the Hamiltonian is the Legendre transform of the Lagrangian. So if the Lagrangian has no explicit time-dependence, so does the Hamiltonian. Usually we can also identify the Hamiltonian (actually its eigenvalues) with the total energy of the system and the absence of an explicit time dependence corresponds to energy conservation. – Wouter Apr 02 '14 at 07:31
  • If my comments answered your question, let me know. Then I'll convert them into an actual answer - which I just realized they sort of are (so they shouldn't really be comments either way). – Wouter Apr 02 '14 at 07:37

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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