Does this equation mean anything to you?

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Izaac
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Joined: Tue Mar 20, 2012 2:24 am

Does this equation mean anything to you?

Postby Izaac » 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...

actrask
Posts: 19
Joined: Tue Feb 12, 2013 7:52 pm

Re: Does this equation mean anything to you?

Postby actrask » 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.

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Izaac
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Re: Does this equation mean anything to you?

Postby Izaac » 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.

bfollinprm
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Joined: Sat Nov 07, 2009 11:44 am

Re: Does this equation mean anything to you?

Postby bfollinprm » 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).




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