# When should we consider “reverse Heisenberg” evolution of operators?

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 $$\hat{o}_R(t)$$.

• $U(t,0)^\dagger = U(0,t)$, the unitary time-evolution operator from time t to time 0. – octonion Sep 25 '19 at 23:41
• @octonion Yes, but I'm not sure how that addresses my question? – Dominic Else Sep 25 '19 at 23:48
• 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? – octonion Sep 25 '19 at 23:52
• @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 $t=0$. – Dominic Else Sep 26 '19 at 0:14
• 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 – octonion Sep 26 '19 at 0:19

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.