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Time-dep Hamiltonian commute with time evolution?

  1. Dec 27, 2014 #1
    Hi all,
    I'm attempting to prove that [tex] i \frac{d \xi (t)}{dt}=[\xi(t),H(p,q ; t)] [/tex] where the Hamiltonian is explicitly time-dependent, in general. We also have some unitary U(t) which generates time-evolution. I wrote up a quick proof but realized afterward that I had assumed that H and U(t) commute. But this isn't true is it? My reasoning is this: U(t) flows the system though time, and since H depends explicitly on time, it is no longer a constant of the motion. Therefore, H is not conserved and [U,H] does not vanish. Is this right?
     
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  3. Dec 27, 2014 #2

    Matterwave

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    The unitary time evolution operator is basically built up of the Hamiltonian. For a time-independent Hamiltonian, U(t) is simply e^iHt so in that case it certainly commutes with the Hamiltonian. However, in a time-dependent Hamiltonian, the Hamiltonian at one time may not commute with the Hamiltonian at some other time. And so you are right that U(t',t) and H(t) will not in general commute.
     
  4. Dec 28, 2014 #3
    Great, thanks. So with the statement I'm trying to prove, I simply take a time-derivative of [itex] \xi(t)=U^{\dagger} \xi U [/itex] to get: [tex] i \frac{d \xi(t)}{dt}=U^{\dagger}\xi H U-U^{\dagger} H \xi U .[/tex]
    Then I simply commuted H and U to complete the proof, but this doesn't work, like you said. So where is my logic flawed? Or is this just the wrong way to approach the proof?
     
  5. Dec 29, 2014 #4

    Matterwave

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    I'm guessing you are using ##U(t)=e^{iHt}## when you are taking your derivatives? It looks that way to me. Like I said in my previous post, this is no longer true if ##H=H(t)##. In addition, even if the above equality held for a time-dependent Hamiltonian, you missed those terms with derivatives of the Hamiltonian. Anyways, in the case of a time dependent Hamiltonian, the unitary evolution operator is not so simple. It can be obtained perturbatively via e.g. the Dyson series, but doesn't really have a closed form solution like the above that I am aware of.
     
  6. Dec 29, 2014 #5
    No, I understand that isn't true in our case. But I did prove that [itex] HU=i\frac{dU}{dt} [/itex] follows from the Schrodinger equation. And I think I didn't assume anything silly in proving this. So I used this in evaluating derivatives; I never used [itex] U(t)=e^{iHt} [/itex] that I know of.
     
  7. Dec 29, 2014 #6

    Matterwave

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    Can you show how you derived ##HU=i\frac{dU}{dt}##? Because this would seem to show that ##U=e^{i\int H(t) dt}## which seems to be a too simple expression to me.
     
  8. Dec 29, 2014 #7
    If we're given some unitary that brings states to motion [itex] | \Psi,t\rangle=U(t) | \Psi \rangle [/itex], then the Schrodinger Equation turns into [itex] HU=i \frac{dU}{dt} [/itex], I believe.
     
  9. Dec 30, 2014 #8

    Matterwave

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    I'm not sure I can guide you to the final answer but here's a couple of comments. 1) I think we are getting confused because we are omitting the time dependencies in most of these expressions. Restoring time dependencies your post #3 should read $$i\frac{d\xi(t)}{dt}=U^\dagger (t,t_0)\xi(t_0)H(t)U(t,t_0)-U^\dagger (t,t_0)H(t)\xi(t_0)U(t,t_0)$$

    2) It seems that you are mixing the Schrodinger picture with the Heisenberg picture in some of your calculations. In the Schrodinger picture the kets evolve in time, the operators do not (unless they are explicitly time-dependent), while it's in the Heisenberg picture that ##\xi(t)=U^\dagger (t,t_0)\xi(t_0) U(t,t_0)##, but in that picture the kets are time-independent. Usually, for a time-dependent Hamiltonian we work in the interaction picture, which is a mix of the two, but we must be explicit which picture we are working in.
     
  10. Dec 30, 2014 #9
    I agree with the equation you write. However, to get to the result I'm trying to prove, we'd need to commute [itex] H(t) [/itex] and [itex] U(t,t_{0}) [/itex] in the first term and [itex] U^{\dagger}(t,t_{0}) [/itex] and [itex]H(t) [/itex] in the second. But this isnt allowed. Do you think this is just not the right way to approach the proof? Thanks a lot!
     
  11. Dec 30, 2014 #10

    vanhees71

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    That's a big mess in this thread :-(. First of all you should work in one picture of time evolution. You started to argue in the Heisenberg picture and then invoked the Schrödinger picture. These are opposite extremes of puting the time dependence either to the state kets or the self-adjoint operators that represent observables.

    Here's the derivation in the Heisenberg picture: Let ##\hat{A}(t)## be an operator that's not explicitly time dependent. Then (in the Heisenberg picture!) it obeys the equation of motion (setting ##\hbar=1##):
    $$\mathrm{d}_t \hat{A}(t)=\frac{1}{\mathrm{i}} [\hat{A}(t),\hat{H}(t)]. \qquad (1)$$
    On the other hand the time-evolution should be described by a unitary transformation
    $$\hat{A}(t)=\hat{U}(t) \hat{A}(0) \hat{U}^{\dagger}(t). \qquad (2)$$
    This implies
    $$\mathrm{d}_t \hat{A}(t)=\mathrm{d}_t \hat{U}(t) \hat{A}(0) \hat{U}^{\dagger}(t) + \hat{U}(t) \hat{A}(0) \mathrm{d}_t \hat{U}^{\dagger}(t). \qquad (3)$$
    Now we use (2):
    $$\mathrm{d}_t \hat{A}(t)=\mathrm{d}_t \hat{U}(t) \hat{U}^{\dagger}(t) \hat{A}(t) \hat{U}(t) \hat{U}^{\dagger}(t) + \hat{U}(t) \hat{U}^{\dagger}(t) \hat{A}(t) \hat{U}(t) \mathrm{d}_t \hat{U}^{\dagger}(t) = \mathrm{d}_t \hat{U}(t) \hat{U}^{\dagger}(t) \hat{A}(t) + \hat{A}(t) \hat{U}(t) \mathrm{d}_t \hat{U}^{\dagger}(t). \qquad (4)$$
    Now we have
    $$\hat{U}(t) \hat{U}^{\dagger}(t)=\hat{1} \; \Rightarrow \mathrm{d}_t \hat{U}(t) \hat{U}^{\dagger}(t)=-\hat{U}(t) \mathrm{d}_t \hat{U}^{\dagger}(t). \qquad (5)$$
    Using this in (4) you get
    $$\mathrm{d}_t \hat{A}(t)=[\mathrm{d}_t \hat{U}(t) \hat{U}^{\dagger}(t),\hat{A}(t)]. \qquad (6)$$
    From this you get by comparing with (1)
    $$\mathrm{d}_t \hat{U}(t) \hat{U}^{\dagger}(t)=\mathrm{i} \hat{H}(t) \; \Rightarrow \mathrm{d}_t \hat{U}(t) =\mathrm{i} \hat{H}(t) \hat{U}(t). \qquad (7)$$
    Now you can, with the initial condition ##\hat{U}(0)=\hat{1}##, rewrite this operator-differential equation as an integral equation
    $$\hat{U}(t)=\hat{1} + \mathrm{i} \int_0^{t} \mathrm{d} t' \hat{H}(t') \hat{U}(t'). \qquad (7)$$
    Now you can iteratively solve this equation to obtain the formal Dyson series. It is imporant to realize that in general neither the Hamiltonians at different times nor the Hamiltonian with ##\hat{U}## commutes.

    The final result is the formal solution
    $$\hat{U}(t)=\mathcal{T}_c \exp(\mathrm{i} \int_0^t \mathrm{d} t' \hat{H}(t'), \qquad (8)$$
    where ##\mathcal{T}_c## is the time-ordering symbol. The meaning becomes clear when expanding the exponential into its power series. For the ##n^{\text{th}}## order term you get
    $$\hat{U}^{(n)}(t)=\frac{\mathrm{i}^n}{n!} \int_0^{t} \mathrm{d} t_1' \cdots \int_0^{t} \mathrm{d} t_n' \mathcal{T}_c \hat{H}(t_1') \hat{H}(t_2') \cdots \hat{H}(t_n').$$
    The time-ordering operator demands to order the Hamiltonians such that the time arguments are ordered from right to left in ascending order.

    For a detailed derivation for this result, see

    http://fias.uni-frankfurt.de/~hees/publ/lect.pdf

    Sect. 1.3. Note that there I discuss a general picture of time evolution with an arbitrary initial time ##t_0##, and instead of ##\hat{U}(t)## I wrote ##\hat{A}(t,t_0)## and instead of ##\hat{H}(t)## we have an arbitrary self-adjoint operator ##\hat{X}(t)##.
     
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