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From unitary operator to hamiltonian

  1. Sep 16, 2012 #1

    KFC

    User Avatar

    Hi there,
    If the evolution operator is given as follows
    [tex]U(t) = \exp[-i (f(p, t) + g(x))/\hbar][/tex]

    where p is momentum, t is time. Can I conclude that the Hamiltonian is

    [tex]H(t) = f(p, t) + g(x)[/tex]

    if no, why?
     
  2. jcsd
  3. Sep 16, 2012 #2
    Well, since you have the hbar in there, that superposition of functions has units of action, so it can't be the hamiltonian, right. Maybe an idea is to taylor expand to linear order in [itex]t[/itex].
    [tex]
    U=\exp (-i (f(p,t)+g(x))/\hbar )\approx \exp (-i(f(p,0)+\dot{f}(p,0)t+g(x))/\hbar )
    [/tex]
    This kind of looks like the normal form
    [tex]
    U=e^{-iHt/\hbar}
    [/tex]
    Now you have something next to [itex]t[/itex] and divided by [itex]\hbar[/itex], which at least has to have units of energy. What do you think?
     
  4. Sep 16, 2012 #3

    KFC

    User Avatar

    Thanks jfy4. So based on your math, [tex]H=\dot{f}(p, 0)[/tex], right? Well, but if the Hamiltonian is time-dependent, should the corresponding evolution operator be
    [tex]
    U=e^{-i\int Hdt/\hbar}
    [/tex]

    If that's true, how do we find the Hamiltonian if there should be integral in the exponential ?

     
  5. Sep 16, 2012 #4
    I think it is usually quite difficult to set up a [itex]U[/itex] with a time-dependent Hamiltonian, and I agree with the expression you wrote for a time-dependent Hamiltonian, however, assuming what I wrote above in correct ( and thats a big assumption), the hamiltonian is time-independent, since it is
    [tex]
    H=\frac{\partial f(p,t)}{\partial t}\bigg|_{t=0}
    [/tex]
    thus no integral expression is required.
     
  6. Sep 16, 2012 #5

    Jano L.

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    Gold Member

    Hello KFC,

    I'll try to explain the relation between U and H as I see it.

    The first notion is the Hamiltonian operator [itex]\hat H[/itex], which plays role in the equation

    [tex]
    \partial_t \psi = -\frac{i}{\hbar} {\hat H}(t) \psi.
    [/tex]


    There is an alternative description in terms of an evolution operator. The operator [itex]\hat U(t,t_0)[/itex] is called an evolution operator, if it changes the function at time [itex]t_0[/itex] to a function at later time t:

    [tex]
    \psi(t) = \hat U(t,t_0) \psi(t_0).
    [/tex]

    The evolution operator obeys the equation

    [tex]
    \frac{\partial \hat U}{\partial t}(t) = \hat H(t) \hat U(t) ~~~(*)
    [/tex]

    In case the Hamiltonian is time-independent, Schroedinger's equation gives [itex]\hat U(t,t_0) = e^{-i\hat H (t-t_0)/\hbar}[/itex].

    In case the Hamiltonian is time-dependent, there is no simple formula but there is perturbative series called Dyson series: http://en.wikipedia.org/wiki/Dyson_series


    In your case, you seek Hamiltonian from known U. There is, as far as I know, no simple way to proceed. You can try to guess correct form of H that will recover the equation (*); if you succeed, the expression in front of U is your Hamiltonian.

    Warning: the derivative

    [tex]
    \frac{\partial U}{\partial t} \neq -i/\hbar (\dot f(p,t)) \hat U(t),
    [/tex]

    except for case when operators f(p,t) + g(x) at different times commute.
     
  7. Sep 16, 2012 #6

    KFC

    User Avatar

    Thanks Jano. It clears my doubt about the Hamiltonian and evolution operator. Actually, I found that evolution in some online materials, where they give the evolution operator first and then directly write out the corresponding Hamiltonian. But I didn't see any way to write the Hamiltonian from that evolution operator mathematically. I think they gave it, as you says, by guessing.

     
  8. Sep 16, 2012 #7

    Jano L.

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    Gold Member

    No problem. Can you post a link? I am curious what result they found for H.
     
  9. Sep 16, 2012 #8

    KFC

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    Yes, I send you the link via message. Check it out :)
     
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