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Show that H is hermitian.

  1. Oct 20, 2012 #1
    1. The problem statement, all variables and given/known data
    Show that the unitary time evolution time operator requires that the Hamiltonian
    be hermitian.
    And then it tells us to use the infinitesimal time evolution operator.
    3. The attempt at a solution
    [itex] U(dt)=1-\frac{iHdt}{\hbar} [/itex]

    so now we take [itex] U^tU=(1+\frac{iH^tdt}{\hbar})( 1-\frac{iHdt}{\hbar}) [/itex]
    When I multiply this out it still is not obvious to me why H need to be hermitian.
    although I could make an argument about the cross terms with i's in them
    if H is not hermitian then I would have an imaginary number in my expansion and that would
    not be equal to 1. But then my last term with H^tH might have some i's in it to cancel the i's from the cross terms but if this were the case H would need to have i's in it and the cross terms would not have i's after they were multiplied to H.
    What do you guys think.
     
  2. jcsd
  3. Oct 20, 2012 #2
    We expand your expression out in full:
    [tex]U^{\dagger}U = (1+ i H^{\dagger} dt) (1 - i H dt ) = 1 - i H dt + i H^{\dagger} dt + i H^{\dagger} H (dt)^{2} = 1 + i (H^{\dagger} - H) dt[/tex]
    neglecting the [itex](dt)^{2}[/itex] term as per usual, and setting [itex]\hbar[/itex] to 1.
    I think you are too preoccupied with whether [itex] i (H^{\dagger} - H) dt[/itex] is real or imaginary. It doesn't matter! For [itex]U^{\dagger}U [/itex] to be equal to 1, that cross term has to vanish totally.
     
  4. Oct 20, 2012 #3
    I dont see why that cross term has to vanish though.
     
  5. Oct 20, 2012 #4

    dextercioby

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    You already have the unit operator in the RHS, it comes from multilplying the unit operators in each bracket. So you must have

    [tex] \hat{1} + i\left(\hat{H}^{\dagger}-\hat{H}\right) dt \equiv \hat{1} [/tex]
     
  6. Oct 20, 2012 #5
    ok dex I could see how your expression in post # 4 would force H to be hermitian but
    where did the [itex] iH^tH{dt}^2 [/itex] term go.
     
  7. Oct 20, 2012 #6

    dextercioby

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    Well, you're interested in the linear approximation of the product of U and U^dagger, since both U and U^dagger are linear to start with. You must discard the quadratic term.
     
  8. Oct 20, 2012 #7
    ok thanks for everyones response, it cleared things up.
    so we are not just neglecting the dt^2 term because it is really small.
     
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