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In Quantum Mechanics, the Heisenberg evolution of an observable $\hat{o}$ is defined as

$$ \hat{o}(t) = U(t,0)^{\dagger} \hat{o} U(t,0) $$

where $U(t,0)$ is the unitary time-evolution operator from time $0$ to time $t$. This satisfies the Heisenberg equations of motion

$$ i\hbar \frac{d}{dt} \hat{o}(t) = [\hat{o}(t),H(t)],$$

But is there a standard name for the "reverse Heisenberg evolution"

$$ \hat{o}_R(t) = U(t,0) \hat{o} U(t,0)^{\dagger} $$

which satifies the differential equation $$ i\hbar \frac{d}{dt} \hat{o}_R(t) = [H(t),\hat{o}_R(t)],$$

and in which circumstances should one consider it?

It came up because I was thinking about a state $|\psi\rangle$ which is defined to be the (unique, say) eigenstate of some observable $\hat{o}$ with eigenvalue $\lambda$. Then we see that the time-evolved state $|\psi(t)\rangle = U(t,0) |\psi\rangle$ can be characterized as the eigenstate of the operator $\hat{o}_R(t)$.

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  • $\begingroup$ $U(t,0)^\dagger = U(0,t)$, the unitary time-evolution operator from time t to time 0. $\endgroup$ – octonion Sep 25 '19 at 23:41
  • $\begingroup$ @octonion Yes, but I'm not sure how that addresses my question? $\endgroup$ – Dominic Else Sep 25 '19 at 23:48
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    $\begingroup$ I'm not sure what your question is then. What you wrote with flipped conjugation signs is no different than the standard time evolution. Why does it deserve a new name? $\endgroup$ – octonion Sep 25 '19 at 23:52
  • $\begingroup$ @octonian It's not the same time evolution. Maybe this is clearer in differential form (which I added in the question). The two differential equations are evidently defining two different time-dependent families of operators if you start from the same operator at time $t=0$. $\endgroup$ – Dominic Else Sep 26 '19 at 0:14
  • $\begingroup$ The first differential equation also defines operator times earlier than $t=0$. The only difference with the second equation is you flipped the sign of t. It represents the same evolution $\endgroup$ – octonion Sep 26 '19 at 0:19
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It came up because I was thinking about a state $|\psi\rangle$ which is defined to be the (unique, say) eigenstate of some observable $\hat{o}$ with eigenvalue $\lambda$. Then we see that the time-evolved state $|\psi(t)\rangle = U(t,0) |\psi\rangle$ can be characterized as the eigenstate of the operator $\hat{o}_R(t)$

This property is necessary for the derivation of the path integral in quantum mechanics and QFT. In the context I saw it, it was used in reverse, but it's the same idea. We had operators which evolve normally $$X(t) = U^\dagger (t)X U(t)$$ $$P(t) = U^\dagger (t)P U(t)$$

but states were then defined with backwards evolution: $$|x,t\rangle \equiv U^\dagger (t)|x\rangle$$

So that for all time, they remain eigenstates:

$$X(t)|x,t\rangle=U^\dagger (t)X U(t) U^\dagger(t)|x\rangle$$ $$=U^\dagger (t)x |x\rangle$$ $$=x |x,t\rangle$$ (and same with $P$)

If you want to read more, I learned this from Weinberg's The Quantum Theory of Fields, the part he does this is in Chapter 9.1, page 379. His derivation is for Quantum Mechanics but with n degrees of freedom. Taking the $n\to \infty$ would give the QFT path integral.

You asked if this evolution has a name; I don't know of any special name for it. Reverse evolution seems to get the point right.

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