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Time evolution operator in terms of Hamiltonian

  1. Aug 28, 2012 #1
    1. The problem statement, all variables and given/known data
    "Show that if the Hamiltonian depends on time and [itex][H(t_1),H(t_2)]=0[/itex], the time development operator is given by
    [tex]U(t)=\mathrm{exp}\left[-\frac{i}{\hbar}\int_0^t H(t')dt'\right]."[/tex]

    2. Relevant equations
    [itex]i\hbar\frac{d}{dt}U=HU[/itex]
    [itex]U(dt)=I-\frac{i}{\hbar}H(t)dt[/itex]

    3. The attempt at a solution
    The first thing I tried was to rearrange the first of the relevant equations:
    [tex]\left(\frac{d}{dt}U\right)U^{-1}=-\frac{i}{\hbar}H(t).[/tex]
    I can then integrate both sides; if the LHS could turn into an expression like [itex]\ln{U}[/itex] I'd be done, but that didn't work out. Any hints?
     
  2. jcsd
  3. Aug 28, 2012 #2
    What is the derivative of [itex]\ln U[/itex]?
     
  4. Aug 28, 2012 #3
    If U was an ordinary function I would say [itex]U^{-1}\frac{d}{dt}U[/itex], but considering that U is an operator function, I'm not sure. I don't know how [itex]\ln{U}[/itex] is even defined, much less how to apply chain rule to it.
     
  5. Aug 28, 2012 #4

    vela

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    Try expanding the expression you have for U(t) using a Taylor series and show that it satisfies the Schrodinger equation.
     
  6. Aug 28, 2012 #5
    Like any function of an operator, [itex]\ln{U}[/itex] is defined in terms of it's Taylor series. More specifically in this case, it's the inverse series of the exponential function. It exists iff [itex]U^{-1}[/itex] exists.

    So, working along the lines of your attempted solution, you would get something like:
    [tex]U^{-1} \partial_t U = -\frac{i}{\hbar}U^{-1}HU \;.[/tex]
     
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