Does Heisenberg equation of motion imply the Schrodinger equation for evolution operator?

Question

Let us choose to postulate (e.g. considering the analogy of the Hamiltonian being a generator of time evolution in classical mechanics) $$ i\hbar \frac{d\hat{U}}{dt}=\hat{H}\hat{U}\tag{1} $$ where $\hat{U}$ is the (unitary, linear) evolution operator and $\hat{H}$ the Hamiltonian (the most general version of which; i.e. time-dependent with instances at different times non-commuting).

In S-picture, one can easily show that (1) is equivalent to $$ i\hbar \frac{d}{dt}|\psi_S(t)\rangle=\hat{H}_S|\psi_S(t)\rangle\tag{2} $$ where $\psi$ is a state, $|\psi_S(t)\rangle = \hat{U}(t)|\psi_S(0)\rangle \equiv \hat{U}(t)|\psi_H\rangle$ and $\hat{H}_S:=\hat{H}$.

In H-picture, it is straightforward to show that (1) implies $$ \frac{d\hat{A}_H}{dt} = \frac{\partial\hat{A}_H}{\partial t}+\frac{1}{i\hbar}[\hat{A}_H,\hat{H}_H]\tag{3} $$ where $A$ is an observable and $\hat{A}_H(t)=\hat{U}(t)^\dagger \hat{A}_H(0)\hat{U}(t)\equiv\hat{U}(t)^\dagger \hat{A}_S\hat{U}(t)$ (and also $[\hat{H}_{H},\hat{H}_H]=0$ implying that the time dependence of $\hat{H}_H$ is purely explicit, i.e. $\hat{H}_H=\hat{H}_S\equiv \hat{H}$ with $[\hat{H},\hat{U}]=0$).

My question: is it possible to obtain (1) from (3), i.e. to show that (1) is equivalent to (3)?

Some thoughts on this: it is extensively mentioned in literature that both pictures yield same answers. Therefore, it should be possible to obtain (1) from (3) since (1) and (2) are equivalent. Assuming (3), the best I can get is that given an observable $A$, the operator $$ \hat{C}:= \hat{A}_S\left(\frac{d\hat{U}}{dt} \hat{U}^\dagger - \frac{\hat{H}}{i\hbar}\right) $$ must be skew-Hermitian.

Answer 1

The result can be proved in a general way using, well, math. In particular the theory of semigroups of linear operators on Banach spaces (I know that seems advanced and maybe not physical, but it is an elegant way of proving the result you seek ;-) ).

Define the Banach space $\mathscr{L}^1(\mathscr{H})$ of trace class operators over a separable Hilbert space the set of all bounded operators $u$ such that $\mathrm{Tr}\lvert u\rvert<\infty$. The fact that (3) holds for all observables (I will not discuss about domains here for the sake of simplicity) implies that it holds also for trace class operators that does not depend explicitly on time. In this case (3) is equivalent to the following Cauchy problem on the Banach space $\mathscr{L}^1(\mathscr{H})$: $$\frac{du(t)}{dt}=L u(t)\;, \; u(0)=x$$ where $\mathscr{L}^1(\mathscr{H})\ni x\equiv \hat{A}_0$ that does not depend on time, $u(t)\equiv \hat{A}_H(t)$ and $L$ is the linear operator that acts as $i[\hat{H},\cdot]$ (I am assuming $\hslash=1$). If $x\in D(L)$, i.e. $x$ such that $\mathrm{Tr}\lvert [\hat{H},x]\rvert <\infty$, the solution of the Cauchy problem above is unique, because $H$ is self-adjoint. We know by Stone's theorem an explicit solution, namely $$u(t)=\hat{U}(t)x \hat{U}^\dagger(t)\; ,$$ where $\hat{U}(t)=e^{-it\hat{H}}$ is the unitary group generated by $\hat{H}$, that satisfies (1) on $D(\hat{H})$. That solution is unique, so assuming (3) for all observables implies that the operator $\hat{U}$ you used to define Heisenberg picture operators has to be exactly the group generated by $\hat{H}$, i.e. satisfy (1).

Just for the sake of completeness: once you have solved the Cauchy problem for $x\in D(L)$, you can extend the solution to trace class operators or compact operators or bounded operators; also to unbounded operators (provided $\hat{U}(t)x \hat{U}^\dagger(t)$ makes sense on some dense domain). This is what is called a mild solution of the Cauchy problem, because we don't know a priori if we are allowed to take the derivative. However uniqueness is usually proved under general assumptions and for mild or even weak solutions, so I think it is quite safe to conclude that $\hat{U}(t)x \hat{U}^\dagger(t)$ is the unique solution of (3) in a suitable sense.

Answer 2

CAUTION - ANSWER INCOMPLETE There is a gap in my argument (see the send); it relies on the claim that \begin{align} - \hat O^\dagger \hat A = \hat A\hat O \end{align} for all hermitian $\hat A$ implies $\hat O = 0$ which may not be true. Please comment if you know how to prove this or know of a counterexample.

Update. Actually the claim above is definitely false in one dimension, so the ensuing argument is certainly incomplete.

---

Some notational clarifications.

Let me first that (3) as you wrote it, although very much standard, is really a rather severe abuse of notation.

The difference between the "total" and "partial" derivatives in the equation is that the partial derivative term is supposed to reference the time-dependence carried by the Schrodinger picture operator itself, while the total derivative refers to that plus the additional time-dependence introduced by conjugating the operator by $\hat U(t)$.

To see this, note that if as usual we define \begin{align} \hat A_H(t) = \hat U^\dagger(t) \hat A_S(t) \hat U(t) \tag{$\star$} \end{align} then differentiation with respect to time on both sides and invoking (1) yields \begin{align} \frac{d\hat A_H}{dt}(t) = U^\dagger(t)\frac{d \hat A_S}{dt}(t)\hat U(t) + \frac{1}{i\hbar} [\hat U^\dagger(t)\hat A_S(t) \hat U(t), \hat H] \tag{$\star\star$} \end{align} so if we feel like good physicists who like using partial derivative symbols in rather odd ways and define \begin{align} \frac{\partial \hat A_H}{\partial t}(t) = U^\dagger(t)\frac{d \hat A_S}{dt}(t)\hat U(t) \end{align} then we get precisely your equation (3).

Proof that (3) $\implies$ (1).

All right, so now that we know what that equation is really saying. Let's try to use it to prove (1) as you desire. We start with $(\star)$ and $(\star\star)$ and try to prove (1). In fact, plugging the $(\star)$ into $(\star\star)$ and canceling the common term yields \begin{align} \frac{d \hat U^\dagger}{dt}(t) \hat A_S(t)\hat U(t) + \hat U^\dagger(t) \hat A_S(t) \frac{d\hat U}{dt}(t) = \frac{1}{i\hbar} [\hat U^\dagger(t)\hat A_S(t) \hat U(t), \hat H] \end{align} Expanding out the commutator, and multiplying both sides by $\hat U(t)$ on the left, and $\hat U^\dagger(t)$ on the right, we find that \begin{align} \hat U(t) \frac{d\hat U^\dagger(t)}{dt} \hat A_S(t) + \hat A_S(t) \frac{d \hat U}{dt}(t) \hat U^\dagger(t) = -\frac{1}{i\hbar}\hat U(t)\hat H \hat U^\dagger (t) \hat A_S(t) + \frac{1}{i\hbar}\hat A_S(t) \hat U(t) \hat H \hat U^\dagger (t). \end{align} which, upon a some rearrangement gives \begin{align} \left(\hat U(t) \frac{d\hat U^\dagger(t)}{dt} + \frac{1}{i\hbar}\hat U(t)\hat H \hat U^\dagger (t)\right)\hat A_S(t) =\hat A_S(t)\left(\frac{1}{i\hbar}\hat U(t) \hat H \hat U^\dagger (t)-\frac{d \hat U}{dt}(t) \hat U^\dagger(t)\right) \end{align} Now let the term in parentheses on the right be called $\hat O(t)$, then using the fact that $\hat A_S(t)$ is hermitian, this equation can be written as \begin{align} -\hat O^\dagger(t) \hat A_S(t) = \hat A_S(t) \hat O(t) \end{align} This holds for all hermitian $\hat A_S(t)$, so $\hat O(t) = 0$, which is to say that \begin{align} \frac{1}{i\hbar}\hat U(t) \hat H \hat U^\dagger (t)-\frac{d \hat U}{dt}(t) \hat U^\dagger(t) =0 \end{align} and (1) follows upon multiplying both sides on the left by $\hat U^\dagger(t)$.

Answer 3

Just calculate (3) for $\hat{A}_H=\hat{U}_H$. With $\hat{U}_H(t)=\hat{U}^{\dagger}(t)\hat{U}(t)\hat{U}(t)=\hat{U}(t)$ and $\hat{H}_H=\hat{H}$ (as you said) you get:

$\frac{d\hat{U}_H}{dt}=\frac{\partial \hat{U}}{\partial t}+\frac{1}{i\hbar}\underbrace{[\hat{H},\hat{U}]}_{=0 \text{(as you said)}}\stackrel{\text{$\hat{U}=e^{\hat{H}t/(i\hbar)}$}}{=}\frac{1}{i\hbar}\hat{H}\hat{U}$

In words: $\hat{U}$ commutes with $\hat{H}$ and thus the total derivative of $\hat{U}$ equals it's partial derivative, which I can calculate to be $\frac{1}{i\hbar}\hat{H}\hat{U}$. So (1) is implied in (3).

Answer 4

I think you need to add a postulate to (3) in order to obtain (1). This postulate it's actually assumed in every picture of the dinamic and it is that the evolution operator is a one group parameter. It is such a natural assumption that it comes no harm in taking it for true. Even in classical hamiltonian mechanics the flow in the coordinate space form a group.

Eq(3) on its own, as prievously stated, substantially tells you that $[U,H] = 0$ and then you can put $U(t) = f(\hat{H},t)$ for an opportune function $f$, via spectral theorem. This of course may be expressed in full mathematical rigour at will.

Now if you ask $U$ to be a group (and take $H$ time independent for sake of semplicity), it should be easy to prove that $f( - , t)$ must be $exp(i (-) t)$ (the i comes from the request of unitarity). After all, if you write your problem in a eigenbasis for $U$ and $H$, you are looking for a function $f(E,t)$ such that $f(E,t) f(E,t') = f(E,t+t')$, for each eigenvalue $E$ of $\hat{H}$ and each $t,t'$. (Note that now the product is the one in $\mathbb{R}$ not the composition of operators!)