I understand the derivation to get

$$\frac{d}{dt}⟨\psi|Q|\psi⟩ = -\frac{1}{i\hbar}⟨\psi|[H,Q]\psi⟩ + ⟨\psi|\frac{d}{dt}Q|\psi⟩$$

and that setting $[H,Q] = 0$ results in

$$\frac{d}{dt}⟨\psi|Q|\psi⟩ = ⟨\psi|\frac{d}{dt}Q|\psi⟩$$

However I don't understand the next part. It just says "if $Q$ is not explicitly dependent on time" , i.e.

$$\frac{dQ}{dt} = 0,$$ then

$$\frac{d}{dt}⟨Q⟩ = 0.$$

I don't see how or why we would just assume $\frac{dQ}{dt} = 0$ after all that effort. I thought that was what we were trying to prove in the first place! It seems circular/tautological to me.

  • $\begingroup$ Consider an operator $\cos(\omega t)\hat{x} + \sin(\omega t)\hat{p}$. That thingy has explicit time dependence aside from whatever time dependence there is in $\langle x \rangle$ coming from the evolution of the system. Is that what you're puzzled about? $\endgroup$ – DanielSank Jan 9 '17 at 21:17
  • $\begingroup$ Sorry I don't understand what you mean .. $\endgroup$ – dain Jan 9 '17 at 21:23
  • $\begingroup$ "if Q is not explicitly dependent on time" means that $\frac{\partial Q}{\partial t}=0$. It does not mean that $\frac{\mathrm dQ}{\mathrm dt}=0$. But recall that HEO implies that $\frac{\mathrm dQ}{\mathrm dt}=\frac{\partial Q}{\partial t}+i[H,Q]$. $\endgroup$ – AccidentalFourierTransform Jan 9 '17 at 21:25
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    $\begingroup$ I think someone needs to give two mutually exclusive examples: one where $\langle O \rangle$ depends on time explicitly and one where it doesn't, and show how those cases differ. $\endgroup$ – DanielSank Jan 9 '17 at 21:32
  • $\begingroup$ Possible duplicate: physics.stackexchange.com/q/9122/2451 $\endgroup$ – Qmechanic Jan 9 '17 at 21:35

So there are a bunch of "pictures" (that's the technical term!) of quantum mechanics, agreeing in the broad perspective that:

  1. There is some vector space $\{|\phi\rangle\}$ over the field $\mathbb C$ and its canonical dual space $\{\langle\phi|\},$ such that the dual operation $\mathcal D$ maps $$\mathcal D\Big(a |\alpha\rangle + b |\beta\rangle\Big) = \langle \alpha|a^* + \langle\beta|b^*$$ and there is an inverse mapping the other way and so on; we usually write this dualizing operation with a superscript $\dagger$ so that $\big(c~|\alpha\rangle\langle\beta|\big)^\dagger = c^* |\beta\rangle\langle\alpha|.$
  2. Observable quantities are represented by Hermitian operators $\hat O^\dagger = \hat O,$ or in other words you have expressions like $$\mathcal D\Big(\hat O |\phi\rangle\Big)~|\psi\rangle = \langle\phi| \hat O|\psi\rangle,$$or what mathematicians will sometimes write $\langle \hat O\phi,~\psi\rangle = \langle \phi,~\hat O\psi\rangle.$ The point is that they are their own conjugate transpose, in the sense that they play nice with this dualizing operation.
  3. The central prediction of QM is: "you observe the eigenvalues of the Hermitian operators, but we only predict the averages of these eigenvalues over many measurements. The average always takes the form $\langle O \rangle = \langle \psi|\hat O|\psi\rangle,$ where $|\psi\rangle$ is a vector we regard as the state of the system."

In one of these pictures in particular, the Schrödinger picture, all of the operators $\hat p$ and $\hat x$ and so on are generally formally independent of time, and the state $|\psi\rangle$ changes explicitly with time according to the Schrödinger equation, $$i \hbar |\partial_t \Psi(t)\rangle = \hat H |\Psi(t)\rangle,$$ where $\hat H$ is an observable for the total energy in the system. Of course we could still define time-dependent observables like $\hat O = \hat x ~ \cos(\omega t) + \hat p/\hbar ~ \sin(\omega t)$ if we wanted, and then we would have something that we'd call maybe $d \hat O\over d t,$ but the basic point is that the theory is made out of basic things which are not fundamentally time-dependent, and you can do the time dependence if you want to. So $\hat p = -i\hbar \partial_x$ as an operator, it does not change over time.

One nevertheless gets that the actual observable change in the average value is given by the formula you gave, which includes the possibility of explicit time dependence. Explicit time dependence is unusual in the Schrödinger picture, but we can handle it by saying $$\frac{d\langle A\rangle}{dt} = \frac{i}{\hbar} \langle [H, A]\rangle + \langle \frac{dA}{dt} \rangle,$$ where all of the stuff inside brackets is fundamentally some operator expression first and foremost, so $\langle dA/dt\rangle$ means, "first figure out what operator $d\hat A/dt$ is, then its average appears above."

And then, you have all of the other pictures. It turns out that we can think about solving the equation $i\hbar |\partial_t \psi(t)\rangle = \hat h |\psi(t)\rangle$ for an arbitrary Hermitian $\hat h$, and we get that $\psi(t) = \hat u(t) |\psi(0)\rangle$ for some "unitary operator" $\hat u(t)$, meaning that $\hat u \hat u^\dagger = \hat u^\dagger \hat u = 1.$ One particular one of these, $\hat U(t)$, corresponds to the case where $\hat h = \hat H.$

We can insert these into the expectation value given by the Schrödinger picture to do a sort of quantum coordinate transform,$$\langle A \rangle = \langle \psi_0|\hat U^\dagger \hat u ~ \hat u^\dagger \hat A \hat u ~ \hat u^\dagger \hat U |\psi_0\rangle.$$

The point is that now instead of $|\psi\rangle =\hat U |\psi_0\rangle$ we think about $|\psi'\rangle = \hat u^\dagger \hat U |\psi_0\rangle$ which we can derive evolves according to $$i\hbar |\partial_t \psi'\rangle = (\hat H' - \hat h') |\psi'\rangle.$$ Above you can also see primes on the operators; see now it is also more typical for operators to have explicit time dependence, since we are also replacing $\hat A$ with $\hat A' = \hat u^\dagger \hat A \hat u$ and finding $$i \hbar \frac{d\hat A'}{dt} = -\hat u^\dagger \hat h \hat A \hat u + \hat u^\dagger \hat A \hat h \hat u = [\hat A',~\hat h'].$$ In the most extreme form of this, the Heisenberg picture, we choose $\hat h = \hat H$ so that the state does not evolve at all and remains at $|\psi_0\rangle$ in perpetuity. Instead all of the operators evolve in time. This was the basis for the original "matrix mechanics" form of quantum mechanics before Schrödinger discovered his wave equation.

It is also very common to have "interaction pictures" where we divide $\hat H$ into a nice easy noninteracting part $\hat H_0$ plus whatever complications exist in the interactions $\hat H_I.$ Then we choose $\hat h = \hat H_0$ which usually just throws some $e^{i \omega t}$ terms on all of the operators we're analyzing, and then we can make various approximations for the remaining dynamics now that the easy part is "out of the way."

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    $\begingroup$ I think I get it now. I was getting confused between the operator $\hat{Q}$ and the quantity $Q$. Obviously $\hat{Q}$ shouldn't change with time, but $Q$ might or might not, depending on what $\hat{Q}$ is. $\endgroup$ – dain Jan 9 '17 at 22:00
  • $\begingroup$ @dain yeah, that's about right. $\endgroup$ – CR Drost Jan 9 '17 at 22:01
  • $\begingroup$ @CRDrost I don't think calling $Q$ a time dependent quantity but $\hat{Q}$ time independent is standard/sensible at all. Usually, $Q$ and $\hat{Q}$ mean exactly the same thing, i.e. an operator, be it time dependent or not. $\endgroup$ – DanielSank Jan 9 '17 at 22:17
  • $\begingroup$ In fact, between @dain's comment here and the ones on the OP, I'm pretty sure he/she doesn't understand what's going on yet. $\endgroup$ – DanielSank Jan 9 '17 at 22:18
  • $\begingroup$ @DanielSank: I suppose. I am assuming that if dain has found a distinction to be made, then it is the only distinction that can be made, namely $Q$ as the nonlinear operator from the Hilbert space $\mathcal H$ to $\mathbb R$, a shorthand for $Q = Q[\psi] = \langle \psi|\hat Q|\psi\rangle,$ as distinct from $\hat Q$, the linear operator from $\mathcal H \to \mathcal H.$ Possibly I should be more cynical about such things but I am hoping that the above answer gets enough "in the weeds" that it reminds people that there's a difference between those two. $\endgroup$ – CR Drost Jan 9 '17 at 22:25

You could have set $dQ/dt=0$ at the beginning, but at the end you would not have known if $$ \frac{d}{dt}\langle \psi |Q |\psi\rangle=0 $$ depends on that assumption or not.

If, on the other hand, you are asking about the difference between $$ \frac{d}{dt}\langle \psi |Q |\psi\rangle \qquad \text{and} \qquad \langle \psi |\frac{d}{dt}Q |\psi\rangle, $$ note that they are not the same, they differ by the quantity $$ i \langle \psi | [H,Q]| \psi \rangle $$ as your calculation shows. The state $|\psi\rangle$ depends on time, that is where the derivative acts in evaluating $$\frac{d}{dt}\langle \psi |Q |\psi\rangle$$

This was all in Schroedinger's picture, where states evolve with time.


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