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.

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    $\begingroup$ LaTeX tip: use, e.g. \tag{1} to label your equations. For other configurations, it may be \eqno(1). $\endgroup$ – JamalS Jul 1 '14 at 21:53

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.

  • $\begingroup$ Does Stone's theorem in the above mentioned form apply in the general case, where $\hat H$ is time-dependent where the Hamiltonians at different times need not commute? Also, are trace class operators that do not depend explicitly on time guaranteed to be self-adjoint (so that they really satisfy (3))? (This goes beyond my current knowledge, apologies if my questions are rubbish.) $\endgroup$ – Kubav Jul 2 '14 at 10:09
  • $\begingroup$ Or, is there a general result saying that if (3) is satisfied by all observables, then it satisfied by all trace class operators with no explicit time dependence? $\endgroup$ – Kubav Jul 2 '14 at 10:29
  • $\begingroup$ No, Stone's theorem does not hold for time dependent generators; however also your equation (1) is not precise when the generators are time dependent. In particular you would then obtain a two-parameter group, that is well defined only with some assumptions on the aforementioned generators, and you have a right derivative different from the left derivative. Everything gets very very technical in that case, but a similar reasoning could be replied in principle. $\endgroup$ – yuggib Jul 2 '14 at 10:32
  • $\begingroup$ As you wrote, the trace class operators are general, and everything holds in particular for self-adjoint trace class operators. Since the solution is self-adjoint, provided the initial $x$ is self-adjoint, you have a closed result that preserves self-adjointness. $\endgroup$ – yuggib Jul 2 '14 at 10:32
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    $\begingroup$ Well, the evolution can be rigorously defined by means of the Dyson expansion for bounded $H(t)$. When, as usual in QM, $H(t)$ are not bounded, a lot of problems may arise in defining a two-parameter unitary group generated by them. Most of the problems are linked to the domains of definition of $H(t)$ for each $t$. For example it is a priori possible that $D(H(t_1))\cap D(H(t_1+1))=\emptyset$. In this case you cannot apply in succession $H(t_1+1)H(t_1)$ because it is not defined, and so you see that it would not be possible to define the time-ordered exponential. $\endgroup$ – yuggib Jul 2 '14 at 11:13

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 $\hat U^\dagger(t)$.

  • $\begingroup$ Thank you for your answer. Unfortunately, I don't see how it follows from $-\hat O^\dagger(t) \hat A_S^\dagger(t) = \hat A_S(t) \hat O(t)$ that $2(\hat A_S(t)\hat O(t))^\dagger = 0$ unless $\hat A_S(t)\hat O(t)$ is Hermitian, which I don't think is the case in general. Could you please clarify? Thanks. $\endgroup$ – Kubav Jul 2 '14 at 1:03
  • $\begingroup$ @Kuba You're completely right; that was sloppy. Let me think about this for a moment. If I can't salvage the answer, then I will delete it. $\endgroup$ – joshphysics Jul 2 '14 at 1:07
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    $\begingroup$ @joshphysics Thanks a lot for responding (and I see now that you edited the answer accordingly as well). We all know how scary it is not knowing for sure whether one is asking a smart question or a really stupid one! I would certainly advise against deleting; it is quite interesting to me already! $\endgroup$ – Danu Jul 2 '14 at 1:53
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    $\begingroup$ @joshphysics A much better argument: $\widehat O^\dagger A + A\widehat O = 0$ for every Hermitian operator $A$ implies (for $A = I$) that $\widehat O$ is skew hermitian, and the condition becomes $[\widehat O, A] = 0$ for every Hermitian $A$. Hence the skew-Hermitian $\widehat O$ has a common basis of eigenvectors with every Hermitian $A$. Since we are free in our choice of $A$, every vector is an eigenvector of $\widehat O$, so $\widehat O$ has to be a scalar, purely imaginary because it is skew-Hermitian, and corresponds to a constant shift in potential energy in the Hamiltonian. $\endgroup$ – doetoe Jul 2 '14 at 16:19
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    $\begingroup$ @doetoe Now that is a nice argument. I was stuck at having used the $A=I$, never thinking to plug the resulting skew-hermitian condition back into the original equation. This also neatly explains what's going on in dimension 1 where clearly any imaginary scalar satisfies the equation. Well done. $\endgroup$ – joshphysics Jul 2 '14 at 22:11

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).

  • $\begingroup$ It does not seem sufficient, because a priori you do not know that $U=e^{-itH}$ (if else is trivial). And $[H,U]$ is zero whenever $U$ is a function of $H$, as defined by the spectral theorem. $\endgroup$ – yuggib Jul 2 '14 at 8:06
  • $\begingroup$ But then eqn. (1) and (3) aren't connected, as long as I don't choose U to be defined like that. Or how else would you find (3)? $\endgroup$ – PeMa Jul 2 '14 at 8:26
  • $\begingroup$ You assume (3) to hold for all observables, without knowing the form of $U(t)$. $\endgroup$ – yuggib Jul 2 '14 at 8:46
  • $\begingroup$ This is not an assumption. This you can proof. That's what the OP called "straight forward" and in this proof you already use $\partial_t\hat{U}$ $\endgroup$ – PeMa Jul 2 '14 at 9:20
  • $\begingroup$ This is one side of the proof, namely (1)$\Rightarrow$(3). If you want to prove the converse, i.e. (3)$\Rightarrow$(1), you have to assume (3) without knowing (1)!! $\endgroup$ – yuggib Jul 2 '14 at 9:22

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!)

  • $\begingroup$ As I explained in my answer, the unique solution of (3) (for trace class operators, but actually in any Banach space) has to be the unitary group generated by $H$, it is not necessary to assume the group property a priori. $\endgroup$ – yuggib Jul 2 '14 at 9:54
  • $\begingroup$ Only now I've fully understood your answer, that seems decisive to me now. Do you think that my answer answers to the question (sorry for the repetition) "How to obtain (1) from Heisenberg picture, i.e. A_h(t) = U(t) A_h(0) U^{dagger}(t) with some physical assumptions: H does not evolve, U is a group"? This is not the OP question that instead is "how to obtain (1) in Heisenberg picture if we assume (3) (and nothing else)", right? $\endgroup$ – giulio bullsaver Jul 2 '14 at 10:41
  • $\begingroup$ Well, there is a one to one correspondence between groups and self-adjoint operators that generate them (Stone's Theorem). So asking $U$ is a group it means that $U$ satisfies (1) only with its generator on the r.h.s. In some sense then postulating the Heisenberg picture let you know that the sought generator is effectively $H$. $\endgroup$ – yuggib Jul 2 '14 at 11:07

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