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PH 401

PH 401. Dr. Cecilia Vogel. Review. Sx, Sy, Sz eigenstates spinors, matrix representation states and operators as matrices multiplying them. Time dependent perturbations approximations perturbation symmetry . Outline. Time Dependence. We have looked at stationary states

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PH 401

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  1. PH 401 Dr. Cecilia Vogel

  2. Review • Sx, Sy, Sz eigenstates • spinors, matrix representation • states and operators as matrices • multiplying them • Time dependent perturbations • approximations • perturbation symmetry Outline

  3. 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?

  4. Unperturbed System • Let’s assume that we have solved the TISE • for the UNperturbed syste • with UNperturbed Hamitonian Ho. • Ho|yn> = En|yn> • If our system begins in state • |y>=Scn|yn> • where cn=amplitude for state |yn> • then at time t, it will evolve to • |y>=Scne-iEnt/|yn> • Note that the probability of any state |yn> is still |cn|2. The unperturbed system will not change energy levels.

  5. 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=Ho+Vpert(t). • (Note: what I call Vpert(t), your text calls H1)

  6. Perturbed System • If our system begins in state • |y>=Scn(0)|yn> • where cn(0)= initial amplitude for state |yn> • then at time t, it will evolve to • |y>=Scn(t)e-iEnt/|yn> • Because the perturbation depends on time, the amplitudes also depend on time. • The probability of any state |yn> changes with time. The perturbed system can indeed change energy levels. • Let’s see how….

  7. TDSE • The state of our system • |y>=Scn(t)e-iEnt/|yn> • must obey the TDSE • not the TISE, because it is not a stationary state of the full Hamiltonian • Plug into TDSE: • H|y>=i∂|y>/∂t • (Ho+Vpert(t)) Scn(t)e-iEnt/|yn>= i∂/∂t Scn(t)e-iEnt/|yn> • S(En+Vpert(t)) cn(t)e-iEnt/|yn>= i[ S(-iEn/)cn(t)e-iEnt/|yn>+ Sc’n(t)e-iEnt/|yn>] • Sooo… • S cn(t)e-iEnt/Vpert(t)|yn>= i Sc’n(t)e-iEnt/|yn>

  8. Diff Eqn for Amplitudes • Sooo… • S cn(t)e-iEnt/Vpert(t)|yn>= i Sc’n(t)e-iEnt/|yn> • One more step to give us numbers rather than states: • Take an overlap of this with some final state |yf> • we can do this for any and all possible |yf>, so we don’t lose generality. • < yf|S cn(t)e-iEnt/Vpert(t)|yn>= < yf|i Sc’n(t)e-iEnt/|yn> • i c’f(t)e-iEft/=S cn(t)e-iEnt/< yf|Vpert(t)|yn> • c’f(t)=(-i/) S cn(t)ei(Ef-En)t/< yf|Vpert(t)|yn>

  9. Time Dependence of c(t) • Thus the amplitudes evolve according to • c’f(t)=(-i/) S cn(t)ei(Ef-En)t/< yf|Vpert(t)|yn> • 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.

  10. 1st 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/) ei(Ef-Ei)t/< yf|Vpert(t)|yi> • In this case, the derivative = computable ftn of t, that can in principle be integrate to get c(t).

  11. Matrix Element • The amplitudes evolve according to • c’f(t)=(-i/) ei(Ef-Ei)t/< yf|Vpert(t)|yi> • The evolution depends on the quantity • < yf|Vpert(t)|yi> • 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 cf(t) remains zero for all time • that state does not get excited by the potential

  12. Symmetry and Matrix Element • Under what circumstances is • < yf|Vpert(t)|yi>=0? • and thus incapable of exciting that state? • If Vpert commuteswith Q, then • |yf> and |yi> cannot have different eigenvalues of Q • and Vpert cannot change the eigenvalue of Q • Proof: • Consider initial and final states with initial and final eigenvalues of Q = qi and qf • < yf|QVpert|yi>= < yf|VpertQ|yi> since they commute. • qf < yf|Vpert|yi>= qi< yf|Vpert|yi> • So… either qi = qf, or < yf|Vpert|yi>=0

  13. Symmetry Example • If Vpert commuteswith Q, then • |yf> and |yi> cannot have different eigenvalues of Q • and Vpert cannot change the eigenvalue of Q • Example: • If Vpert = Vpert (y,z) is not a function of x. • then Vpert commutes with px. • 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 Vpert = Vpert (r,q) is not a function of f. • then Vpert commutes with Lz. • and this perturbation cannot change the particle’s z-component of angular momentum

  14. PAL

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