Time evolution operator in terms of Hamiltonian

  • Thread starter dEdt
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  • #1
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Homework Statement


"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]

Homework Equations


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

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?
 

Answers and Replies

  • #2
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What is the derivative of [itex]\ln U[/itex]?
 
  • #3
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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.
 
  • #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.
 
  • #5
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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.

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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