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PH 401 Dr. Cecilia Vogel
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Review Outline Time dependent perturbations approximations perturbation symmetry Sx, Sy, Sz eigenstates spinors, matrix representation states and operators as matrices multiplying them
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Time Dependence We have looked at stationary states where measurable quantities don’t change at all with time We have looked at non-stationary states which develop and change with time due to the superposition of energy levels What about a system in a stationary state that changes due to some external interaction called a perturbation?
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Unperturbed System Let’s assume that we have solved the TISE for the UNperturbed syste with UNperturbed Hamitonian H o. H o | n > = E n | n > If our system begins in state | >= c n | n > where c n =amplitude for state | n > then at time t, it will evolve to | >= c n e -iE n t/ | n > Note that the probability of any state | n > is still |c n | 2. The unperturbed system will not change energy levels.
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Perturbing the System Now let’s add a perturbing potential that depends on time for example, it may be something we turn on at time t that wasn’t there before, it may be something that oscillates with time, like a light wave, etc The full Hamiltonian is then H=H o +V pert (t). (Note: what I call V pert (t), your text calls H 1 )
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Perturbed System If our system begins in state | >= c n (0)| n > where c n (0)= initial amplitude for state | n > then at time t, it will evolve to | >= c n (t)e -iE n t/ | n > Because the perturbation depends on time, the amplitudes also depend on time. The probability of any state | n > changes with time. The perturbed system can indeed change energy levels. Let’s see how….
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TDSE The state of our system | >= c n (t)e -iE n t/ | n > must obey the TDSE not the TISE, because it is not a stationary state of the full Hamiltonian Plug into TDSE: H| >=i ∂| >/∂t (H o +V pert (t)) c n (t)e -iE n t/ | n >= i ∂/∂t c n (t)e -iE n t/ | n > (E n +V pert (t)) c n (t)e -iE n t/ | n >= i [ -iE n / )c n (t)e -iE n t/ | n >+ c’ n (t) e -iE n t/ | n >] Sooo… c n (t)e -iE n t/ V pert (t)| n >= i c’ n (t) e -iE n t/ | n >
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Diff Eqn for Amplitudes Sooo… c n (t)e -iE n t/ V pert (t)| n >= i c’ n (t) e -iE n t/ | n > One more step to give us numbers rather than states: Take an overlap of this with some final state | f > we can do this for any and all possible | f >, so we don’t lose generality. = i c’ f (t) e -iE f t/ = c n (t)e -iE n t/ c’ f (t) =(-i/ ) c n (t)e i(E f -E n )t/
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Time Dependence of c(t) Thus the amplitudes evolve according to c’ f (t) =(-i/ ) c n (t)e i(E f -E n )t/ This is a differential equation relating the derivative of any of the cn’s to the values of all the (infinitely many) other cn’s. OMG We will not solve this, except in an approximate way. If we assume that the perturbation is weak and we don’t wait too long, then the state does not change much from what it was initially.
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1 st Order Approximation If we assume that the state does not change much from what it was initially. ci=1, all other cn=0 Thus the amplitudes evolve according to c’ f (t) =(-i/ ) e i(E f -E i )t/ In this case, the derivative = computable ftn of t, that can in principle be integrate to get c(t).
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Matrix Element The amplitudes evolve according to c’ f (t) =(-i/ ) e i(E f -E i )t/ The evolution depends on the quantity which is called a matrix element of the perturbing potential. The matrix element is important, because it tells us what states can and cannot be excited by the potential. If the matrix element is zero, then c’ f (t)=0, and c f (t) remains zero for all time that state does not get excited by the potential
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Symmetry and Matrix Element Under what circumstances is =0? and thus incapable of exciting that state? If V pert commutes with Q, then | f > and | i > cannot have different eigenvalues of Q and V pert cannot change the eigenvalue of Q Proof: Consider initial and final states with initial and final eigenvalues of Q = q i and q f = since they commute. q f = q i So… either q i = q f, or =0
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Symmetry Example If V pert commutes with Q, then | f > and | i > cannot have different eigenvalues of Q and V pert cannot change the eigenvalue of Q Example: If V pert = V pert (y,z) is not a function of x. then V pert commutes with p x. and this perturbation cannot change the particle’s x- component of momentum. Classical analogue: If V does not depend on x, then there is no force in the x-dir and px does not change. Example: If V pert = V pert (r, ) is not a function of . then V pert commutes with L z. and this perturbation cannot change the particle’s z- component of angular momentum
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PAL
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