Time dependent perturbation theory

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dyn
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Hi.
I have been looking at some notes for time dependent perturbation theory. The equation for the transition probability involves the matrix element < f | H | i > where f is the final state , i is the initial state and H is the perturbation switched on at t=0. If H is a constant , ie. just a number then it can be taken outside the bracket leaving H < f | i > but the final and initial states are orthonormal meaning the bracket is zero for all transitions. So a constant perturbation produces no change in state ?
Have i got all this right ? If not , where am i going wrong ?
Thanks
 

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  • #2
DrSteve
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Hi.
I have been looking at some notes for time dependent perturbation theory. The equation for the transition probability involves the matrix element < f | H | i > where f is the final state , i is the initial state and H is the perturbation switched on at t=0. If H is a constant , ie. just a number then it can be taken outside the bracket leaving H < f | i > but the final and initial states are orthonormal meaning the bracket is zero for all transitions. So a constant perturbation produces no change in state ?
Have i got all this right ? If not , where am i going wrong ?
Thanks
Looks right to me.
 
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  • #3
vanhees71
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Sure, if ##\hat{H}=h_0 \hat{1}##, ##h_0=\text{const} \in \mathbb{R}##, it doesn't do anything in the time evolution, just adding a phase factor ##\exp(-\mathrm{i} h_0 t)## to all states (in the Schrödinger picture), but that means indeed it doesn't do anything to the state at all. That's as in classical mechanics: Adding a constant term to the Hamiltonian doesn't change the equations of motion (Hamilton's canonical equations).
 
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