Page 1 of 1

### Does this equation mean anything to you?

Posted: Fri Jul 05, 2013 3:25 pm
$$A(t)=e^{+iHt/\hbar}Ae^{-iHt/\hbar}$$

Srednicki begins his textbook on QFT by a handful of equations ("In order to prepare for QFT, you should recognize these:"). All of them I grasp but that one.
Obviously it smells like QM, but I've never seen that exact equation, whatever the A and H mean. Anyway it seems to reduce to just $$A(t)=A$$, which is even more confusing.
So, anyone recognize something familiar in that equation? Thanks in advance.

PS: I'm pretty sure the answer is so simple as to mark me "noobie" up to the thousand and first generation...

### Re: Does this equation mean anything to you?

Posted: Fri Jul 05, 2013 5:24 pm
http://en.wikipedia.org/wiki/Heisenberg_picture

You should probably brush up on QM and Classical Mechanics before starting in on QFT. A is just a general operator, H is the Hamiltonian. It does not necessarily reduce to A(t) = A if A does not commute with H.

### Re: Does this equation mean anything to you?

Posted: Sat Jul 06, 2013 3:39 am
Oh! Thanks, now I see. Amusingly, it was discretely mentioned in the book I'm using (Griffiths) on a smallish note bottom of page, without much development.

### Re: Does this equation mean anything to you?

Posted: Sat Jul 06, 2013 9:02 pm
$$A = A(0)$$

Then

$$A(t)|{\phi}> = e^{iHt}Ae^{-iHt}|\phi>$$

means (from right to left), in the Schrodinger picture you probably learned,

1. Devolve $$|\phi>$$ to its state at time $$t = 0$$

2. Evaluate $$A$$ acting on $$|\phi>$$ there.

3. Evolve $$|\phi>$$ back to $$t = t$$. This should be the same as $$A(t)|\phi>$$

This is only trivial if $$A$$ is time-translation invariant (e.g. is the same for all time).