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Let us consider nonrelativistic quantum mechanics, wherein Galiliean relativity reins supreme. This question is motivated by what I think is a misunderstanding I'm having in reading Fonda's Symmetry Principles in Quantum Physics.

Consider two different observers $\overline{O}$ and $O$ of a given quantum system $S$. These observers differ only in their time coordinate. Their time coordinates are related by $\overline{t} = t - \tau$; that is, $\overline{O}$'s time coordinate is delayed with respect to $O$'s.

Now Fonda suggests that in terms of how the two observers view the quantum system, the "translation" between the two frames of reference in their vectors (neglecting the ray technicality) used to describe a given state must obey $$|(\phi_{\overline{O}}(\overline{t}),\psi_{\overline{O}}(\overline{t}))|^2=|(\phi_{O}(t),\psi_{O}(t))|^2.$$ My question is how do we justify this requirement on the translation map? Do we argue that since $\overline{t} = t - \tau$ (since these are the same absolute times), we must have $\psi_{\overline{O}}(\overline{t}) = \psi_{O}(t-\tau))$ and so, trivially, $$|(\phi_{\overline{O}}(\overline{t}),\psi_{\overline{O}}(\overline{t}))|^2 = |(\phi_{O}(t-\tau),\psi_{O}(t-\tau))|^2 .$$ Then by the unitary evolution of non-relativistic quantum mechanics, one has (if we evolve forward in time by $\tau$, $$|(\phi_{O}(t-\tau),\psi_{O}(t-\tau))|^2 = |(\phi_{O}(t),\psi_{O}(t))|^2$$ and so we arrive at the required form.

The reason I am not sure about this is that Fonda says "In fact, the state of the system $S$ that $\overline{O}$ is observing at his time $\overline{t}$ is the evolved, through the time interval $\tau$, of the state observed by $O$ at his own time $t$", whereas my interpretation of what we have done is just opposite! I seem to have concluded that the state of the system $S$ that $\overline{O}$ is observing at his time $\overline{t}$ is the evolved backwards, through the time interval $\tau$, of the state observed by $O$ at his own time $t$ (since $\psi_{\overline{O}}(\overline{t}) = \psi_{O}(t-\tau))$).

The discussion in question is related to what Fonda discusses near equation 1.9 below.

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Eq (1.3) alluded to is about the unitarity of time evolution (a hypothesis in nonrelativistic QM, I suppose).

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  • $\begingroup$ What is e.q. 1.3? $\endgroup$ Commented Aug 13, 2023 at 20:58
  • $\begingroup$ Will update OP, one moment. Basically, unitary time evolution. @TobiasFünke $\endgroup$
    – EE18
    Commented Aug 13, 2023 at 21:01
  • $\begingroup$ @TobiasFünke I don't think it's quite that. Like I say in the OP, I think there's a two step argument. In the first step, one uses that the two observers are seeing things at the same absolute time, and in the second step one uses that unitary evolution preserves inner products. But my problem, as per the answer by fulis which I still don't understand, is why in that first step we should be led to $\phi'(t) = \phi(t+\tau)$ rather than $\phi'(t) = \phi(t-\tau)$ (which is what I wrote in my OP). $\endgroup$
    – EE18
    Commented Aug 13, 2023 at 22:12
  • $\begingroup$ Below (1.9), Fonda discusses two cases "in ordinary quantum mechanics". First, he says, two different observers are there who watch the system at the same objective (absolute) time and, I assume, the same position etc. Then, trivially, we have $\psi_{\overline{O}}(t) =\psi_{O}(t)$ so that (1.9) is trivially true. Fonda then refers to observers who differ in terms of their time only, but are otherwise at the same location. Suppose $\overline{O}$ has their time retarded by $\tau$: $\overline{t} = t - \tau$. One then has (and here I think is where I make my mistake according to fulis)... $\endgroup$
    – EE18
    Commented Aug 13, 2023 at 23:14
  • $\begingroup$ ... that since they only disagree on time, we should have $\psi_{\overline{O}}(t) =\psi_{O}(t-\tau)$, so that trivially $|(\phi_{\overline{O}}(\overline{t}),\psi_{\overline{O}}(\overline{t}))|^2 = |(\phi_{O}(t-\tau),\psi_{O}(t-\tau))|^2 .$. If one then employs (1.3) on the RHS, we arrive at $|(\phi_{O}(t-\tau),\psi_{O}(t-\tau))|^2 = |(\phi_{O}(t),\psi_{O}(t))|^2$ which establishes (1.9). But my problem is that apparently $\psi_{\overline{O}}(t) =\psi_{O}(t-\tau)$ is wrong and it should be $\psi_{\overline{O}}(t) =\psi_{O}(t+\tau)$ according to fulis. This doesn't change (1.9), since one... $\endgroup$
    – EE18
    Commented Aug 13, 2023 at 23:17

2 Answers 2

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I believe you're are using the wrong sign in the time-translation. If we write

$$ \phi'(t) = \phi(t-\tau) $$

then $\phi'$ is translated forward in time with respect to $\phi$. This is because it's a passive transformation.

Thus, in the notation of the book (which only uses a single time coordinate) we have

$$ \mathbf{T}\mathbf{\Phi}_O(t) = \mathbf{\Phi}_{\overline{O}}(t) = \mathbf{\Phi}_O(t+\tau) $$

and

$$ \left|\bigl( \phi_{\overline{O}}(t),\psi_{\overline{O}}(t)\bigr) \right| = \left| \bigl( \phi_O(t+\tau),\psi_O(t+\tau) \bigr)\right| $$

which is consistent with the statement in the book.

Note that it works the same for other coordinate translations. For example, for a wavefuction $\psi(x)$ we have the position basis state

$$ |\psi(x)\rangle = \int \psi(x) |x\rangle dx. $$

Let $T|x\rangle = |x+a\rangle$. Then

$$ T|\psi(x)\rangle = \int \psi(x) |x + a\rangle dx. $$

The displaced wavefunction is

$$ \langle x' | T|\psi(x)\rangle = \int \psi(x) \langle x'|x + a\rangle dx = \int \psi(x) \delta(x+a-x') dx = \psi(x'-a) $$

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  • $\begingroup$ So, if I am understanding you correctly, you are saying (I now use the same time coordinate like the book does so that it makes sense) that the fact that $t$ in $\overline{O}$'s frame is the same as the time $t-\tau$ in $O$'s frame (or, as you use, $t + \tau$ in $\overline{O}$'s frame is the same as the time $t$ in $O$'s frame) means that $\phi_\overline{O}(t+\tau) = \phi_O(t)$? $\endgroup$
    – EE18
    Commented Aug 11, 2023 at 13:39
  • $\begingroup$ If so, does the rest of the argument make sense? That is, that in Step 1 we argue as above (that these describe the same system/same absolute time) and in Step 2 we argue using unitarity? $\endgroup$
    – EE18
    Commented Aug 11, 2023 at 13:40
  • $\begingroup$ You're still using a forward translation. The backwards translation is $\phi_{\overline{O}}(t) = \phi_O(t+\tau)$. I believe this is what the author means with the quoted sentence, but I think it's worded in a confusing way. If your clock is retarded by $\tau$ units of time then that is the same as letting the system evolve for $\tau$ units of time while you pause your own clock. That's how I would phrase it. And yes, using the unitarity is correct. $\endgroup$
    – fulis
    Commented Aug 11, 2023 at 14:06
  • $\begingroup$ I find myself still confused, I'm sorry. If the time in $\overline{O}$'s frame is the same as the time $t-\tau$ in $O$'s frame then we have $\overline{t} = t-\tau \implies \overline{t} +\tau = t$, right? This is what "retarded by $\tau$" means I thought? Then since what observers call their own times is immaterial, and they are examining the same system at some absolute time, we must have $\phi_{\overline{O}}(t) \equiv\phi_{\overline{O}}(\overline{t}) =\phi_O(t - \tau)$, where $\equiv$ is because our notation for variable is irrelevant and $=$ is because what... $\endgroup$
    – EE18
    Commented Aug 11, 2023 at 16:25
  • $\begingroup$ ... $\overline{O}$ sees at his $t$ is exactly what $O$ sees at his $t- \tau$? $\endgroup$
    – EE18
    Commented Aug 11, 2023 at 16:26
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There is a subtle confusion about your argument below

Do we argue that since $\overline{t} = t - \tau$ (since these are the same absolute times), we must have $\psi_{\overline{O}}(\overline{t}) = \psi_{O}(t-\tau))$ and so, trivially, $$|(\phi_{\overline{O}}(\overline{t}),\psi_{\overline{O}}(\overline{t}))|^2 = |(\phi_{O}(t-\tau),\psi_{O}(t-\tau))|^2 .$$ Then by the unitary evolution of non-relativistic quantum mechanics, one has (if we evolve forward in time by $\tau$, $$|(\phi_{O}(t-\tau),\psi_{O}(t-\tau))|^2 = |(\phi_{O}(t),\psi_{O}(t))|^2$$ and so we arrive at the required form.

Firstly, you are sure, there is an absolute time in the theory. in non-relativistic quantum theories, the symmetry group of spacetime is Galilean Group, and its representation. As a result, no time operator in non-relativistic quantum mechanics according to Schur's lemma, or say, the existence of absolute time.

But the crucial missing is that, $$\overline{t} = t - \tau\Rightarrow\psi_{\overline{O}}(\overline{t}) = \psi_{O}(t-\tau) $$ is wrong. The main reason is that a quantum state $\phi$ is a frame (observer)- dependent function in theory, and the probability amplitude is frame (observer)- independent which is an axiom in any quantum theory (Wigner's theorem). The only thing we can read directly from $t,\overline{t}$ corresponding to a same absolute time is $$|(\phi_{\overline{O}}(\overline{t}),\psi_{\overline{O}}(\overline{t}))|^2=|(\phi_{O}(t),\psi_{O}(t))|^2$$ If the we fix $\phi$ while running $\psi$ throughout the Hilbert space, we find $$\phi_{\overline{O}}(\overline{t})=\phi_{O}(t)$$ as expected.

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  • $\begingroup$ I'm not sure I follow. The point of my question is to prove, as Fonda says is possible, that $|(\phi_{\overline{O}}(\overline{t}),\psi_{\overline{O}}(\overline{t}))|^2=|(\phi_{O}(t),\psi_{O}(t))|^2$ obtains as a theorem in the case of time translations with Galilean invariance as the relevant symmetry group. Fonda then says that in more general theories we must assume (1.9) (i.e. postulate it). See the last sentence in the paragraph below (1.9). My question is thus about how one proves (1.9) in the time translation case in ordinary quantum mechanics, as Fonda says is possible. $\endgroup$
    – EE18
    Commented Aug 13, 2023 at 19:09
  • $\begingroup$ We can't prove (1.9) in fact. Since it is a fundamental assumption in theories, and it comes from the Galilean invariance $t\rightarrow \overline{t}$ of theory directly. The things we can do is prove something from (1.9). $\endgroup$
    – Lain
    Commented Aug 13, 2023 at 19:18
  • $\begingroup$ Comments about the invariance of the theory. If we believe the theory with some symmetries $S$, at the quantum level, the amplitude is invariant under those symmetry transformations. It is an axiom both on relativistic and non-relativistic quantum theories. A good reference is the amazon.com/Quantum-Theory-Fields-Foundations/dp/0521670535, in the chapter 2. $\endgroup$
    – Lain
    Commented Aug 13, 2023 at 19:23
  • $\begingroup$ But Fonda is explicitly saying we can prove it (in the specific cases mentioned) in what I've attached? $\endgroup$
    – EE18
    Commented Aug 13, 2023 at 19:26
  • $\begingroup$ It is a subtle point. We can prove (1.9) by symmetry. Since the theory is invariant under $t\rightarrow t+\tau, x\rightarrow x$. Meanwhile (1.9) itself is just the representation of symmetry, and thus we cant prove (1.9) from other things besides symmetry. $\endgroup$
    – Lain
    Commented Aug 13, 2023 at 19:34

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