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I'm currently consuming a course on QFT where we need to define the unitary time-evolution to get the time evolution of the wave function in the interaction picture:

$\hat{U}(t_1,t_0) = \exp\left(\frac{i}{\hbar}\hat{H}_0t_1\right)\exp\left(-\frac{i}{\hbar}\hat{H}(t_1-t_0)\right)\exp\left(-\frac{i}{\hbar}\hat{H}_0t_0\right)$ .

Now one can show that this operator follows a Schrödinger equation by simply taking the derivative to time:

$i\hbar\frac{d}{dt}\hat{U}(t,t_0) = \hat{H}_1^I(t)\hat{U}(t,t_0)$ .

Where $\hat{H}_1^I$ is the perturbation to our free-field Hamiltonian $\hat{H}_0$.

Now I started wondering whether $\hat{U}(t_1,t_0)$ shouldn't also follow a Heisenberg equation since it's an operator.

I believe it shouldn't since $\hat{U}(t_1,t_0)$ gives a unitairy time-evolution which is a transformation, while the Heisenberg equation applies to observables. I was woundering if someone could confirm my reasoning or disprove it?

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  • $\begingroup$ If $t_1$ and $t_0$ are constants, then $\hat{U}(t_1,t_0)$ is the operator evaluated at those particular instants, not a function. Think about the function $\sin(x)$ and the function evaluated at $x_0$ $\sin(x)|_{x=x_0}=\sin(x_0)$ $\endgroup$
    – Jorge Lavín
    Jun 24, 2013 at 14:30
  • $\begingroup$ Yes but if I were to take $t_1$ as my variable, then it might follow some equation governing the time-evolution. That it follows Schrödinger can be proven. For the Heisenberg-case this doesn't work (I wouldn't know how to take the commutator of $\hat{H_1^I}$ and $\hat{H_0}$). My reasoning for this is that the Heisenberg-equation only applies to observables, not the transformation (what the $\hat{U}(t_1,t_0)$ in principle is. $\endgroup$
    – Nick
    Jun 24, 2013 at 14:54

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Yes, OP is right. The Heisenberg evolution equation in the interaction picture applies to an operator $A_I(t)$, that depends on a single moment of time $t$. On the other hand, the evolution operator $\hat{U}(t_f,t_i)$ depends in principle on the whole intermediate time interval $[t_i,t_f]$, and is in particular not a local operator of a single time.

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  • $\begingroup$ Ok, that might explain my conceptual problems upon trying to derive the Heisenberg-like equation. Thanks! Now I was woundering if it was purely because of the fact that $\hat{U}(t_f,t_i)$ is depending on the interval $[t_i,t_f]$. Or also because of the fact that it is a transformation and Heisenberg doesn't apply to transformations ? $\endgroup$
    – Nick
    Jun 24, 2013 at 15:20
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    $\begingroup$ @Nick: Well, I'm not sure what class of transformations you allow, but a sufficiently well-behaved local transformation operator that is only associated with a single time should satisfy the Heisenberg evolution equation. $\endgroup$
    – Qmechanic ♦
    Jun 24, 2013 at 15:41
  • $\begingroup$ @Nick: An other way to see that is that the evolution operator have to be expressed as a time ordered function of the perturbation hamiltonian. $\endgroup$
    – Trimok
    Jun 24, 2013 at 17:00
  • $\begingroup$ Why would they violate Heisenberg if I'd have a time-ordered product ? $\endgroup$
    – Nick
    Jun 24, 2013 at 18:05
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    $\begingroup$ @Nick : Yes. Be careful, it is a partial derivative. (Minor remark : if you want your message appears in the mail box of somebody, you have to do a @ (for instance @Trimok). Sometimes it is automatic, but not in this case.) $\endgroup$
    – Trimok
    Jun 25, 2013 at 18:21

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