$$A(t)=e^{+iHt/\hbar}Ae^{iHt/\hbar}$$
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...
Does this equation mean anything to you?
Re: Does this equation mean anything to you?
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.
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?
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.

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Re: Does this equation mean anything to you?
$$A = A(0)$$
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 timetranslation invariant (e.g. is the same for all time).
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 timetranslation invariant (e.g. is the same for all time).