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I suppose this question ultimately boils down to: when we speak of a time translation (in nonrelativistic mechanics, so that the Galilean group is the apporpiate symmetry group under which the physics of a system must not change) what do we mean? In particular, do we mean that the value we assign to the current moment does not matter (of course) or that if we actually actively translate the system in time, then nothing changes (how can this be true? The state may evolve in time!).

This question is motivated by the following excerpt from Ballentine's quantum text on page 77:

Corresponding to the time displacement $t \to t' = t + s$, there is a vector space transformation of the form (3.8) [i.e. effected by the unitary operator for time translations which was earlier seen to be $e^{isH}$ in this context],$|\psi(t)\rangle \to |\psi'(t')\rangle \equiv e^{isH}|\psi(t)\rangle \stackrel{(?)}{=} |\psi(t-s)\rangle$

where it's the equality marked with (?) which I can't follow and which I think has to do with my lack of understanding of time translations. If I am doing $t \to t' = t + s$ on the system then shouldn't my system get mapped to $|\psi'(t')\rangle=|\psi(t+s)\rangle$? If $s>0$ I am ahead in time after the active transformation.

This is related to my not understanding the end of Saoirse's answer here.

Edit on my understanding after answers from Prof. Moretti and ACuriousMind:

A summary of my lack of understanding: If $e^{isH}|\psi(t)\rangle$ propagates a state $|\psi(t)\rangle$ forward in time (to $t+s$) then why do I have from earlier quantum mechanics courses that $|\psi(t+s)\rangle = e^{-isH}|\psi(t)\rangle$. Is there a different meaning to propagating forward in time to the two cases? In the first case, do we mean keeping the state still somehow while things evolve around it while in the latter we actually evolve the state? I generally don't have trouble understanding these transformations when it comes to any other aspect of the Galilean group (I just think in terms of an active transformation) but here I am struggling.

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  • $\begingroup$ After 30seconds of searching I've found e.g. this...Have you checked any other source - e.g. Google, Wikipedia, any other text book? $\endgroup$ Commented Jan 26, 2023 at 23:02
  • $\begingroup$ @TobiasFunke Yes I have looked at e.g. Fonda’s book (too advanced) and am looking at Corry’s book right now but to no avail. I don’t think the linked answer is detailed enough to really help. $\endgroup$
    – EE18
    Commented Jan 27, 2023 at 1:16
  • $\begingroup$ What about the discussion around eq. 3.8? $\endgroup$ Commented Jan 27, 2023 at 7:09
  • $\begingroup$ From that I deduce $|\psi '(t) \rangle = |\psi(t-s) \rangle$ which isn't quite right either. I think the crux of my problem is not following what's meant by "time translation". Does it mean evolving in time or does it mean picking a new time coordinate (the latter of which doesn't seem like it would do anything to the system)? @TobiasFünke $\endgroup$
    – EE18
    Commented Jan 27, 2023 at 14:10
  • $\begingroup$ I don't know what the author means exactly either... BTW: Your link does not work, consider to edit the question accordingly. $\endgroup$ Commented Jan 27, 2023 at 14:11

3 Answers 3

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This is nothing specific to time translations:

When you have a function $f : X \to Y$ and you act on the $X$ with some operator $O : X\to X$, then the new function $f'$ ("after applying $O$") is defined so that $f'(O(x)) = f(x)$, and thus $f'(x) = f(O^{-1}(x))$.

What you have here is just this general notion of how a function transforms applied to functions of time and time translation, i.e. $X=\mathbb{R}$ and $O: t\mapsto t + a$ and $f(t) = \lvert \psi(t)\rangle$.

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  • $\begingroup$ Please see my edit. After reading your answer, I am trying hard to understand what the requirement $f'(O(x)) = f(x)$ means physically. It seems like it corresponds to saying that $\lvert \psi'(t')\rangle$ (the new state at the new time) must equal $\lvert \psi(t)\rangle$? Or perhaps it should read "$\lvert \psi'(t)\rangle$ (the new state at the new time) must equal $\lvert \psi(t)\rangle$" (NB the prime on the $t$ is gone)? If so, then what exactly are we doing with this time translation? It seems like this would correspond to a delay in when the given state $|\psi\rangle$ occurs? $\endgroup$
    – EE18
    Commented Feb 1, 2023 at 16:54
  • $\begingroup$ It means that if you simultaneously push forwardly in time instruments (origin of time) and states you cannot see a difference with respect to the untranslated case. $\endgroup$ Commented Feb 1, 2023 at 16:59
  • $\begingroup$ @ValterMoretti If I understand...We have defined $e^{isH}$ so that with $|\psi'(t)\rangle := e^{isH}|\psi(t)\rangle$ we obey the "time invariance" requirement for all $t$ that $|\psi'(t+s)\rangle = |\psi(t)\rangle$, i.e. $e^{isH}|\psi(t+s)\rangle = |\psi(t)$. Thus $e^{isH}|\psi(t)\rangle = |\psi(t-s)\rangle$ So corresponding to our transformation of time origin (if we move ahead in time), there must be a unitary operator on our Hilbert space $e^{isH}$ which we can use on a ket in the original system so that the transformed ket is identical to the original at the corresponding time. $\endgroup$
    – EE18
    Commented Feb 1, 2023 at 17:22
  • $\begingroup$ For ACuriousMind...I think my confusion arises from the two level nature of this problem. That there is a transformation in our reference frame in spacetime (corresponding I believe to your $O$) as well as a transformation on Hilbert space (corresponding to $f \mapsto f'$). $\endgroup$
    – EE18
    Commented Feb 1, 2023 at 17:26
  • $\begingroup$ @EE18 I don't understand your problem. Really, there is nothing special about time here. A function of position $f(x)$ transforms into $f(x-a)$ under the translation $x\mapsto x+a$ in exactly the same way. $\endgroup$
    – ACuriousMind
    Commented Feb 1, 2023 at 17:46
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A visual answer is clearer. Consider the profile of a function f(t)

0 ----------x| |---------------> t axis

      t_0

the symbol | | denotes the interval where the function does not vanish. Let us suppose that this interval has length a and starts at t_0. Therefore f(t) $\neq 0$ if $t\in[t_0,t_0+a]$

Now focus on the same profile but translated forwardly of T:

0 ----------x----------| |-----------> t axis

      t_0     t_0+T

What is the analytic expression of this translated profile?

It does not vanish in the interval a+T.

This is possible only if the new function is f(t-T).

In fact: $f(t-T)\neq 0$ if $t-T \in [t_0,t_0+a]$, namely $t\in [t_0+T, t_0+a+ T]$.

In summary, pushing forward $f$ of $T$ means to replace its argument $t$ for $t-T$.

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  • $\begingroup$ Thanks for your answer. If you get the chance please take a look at my edit and at my comment on the other answer. I think my lack of understanding doesn't have to do with how functions in general transform so much as the physical meaning in this specific context with time translations/time as a parameter. $\endgroup$
    – EE18
    Commented Feb 1, 2023 at 16:56
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The space-time translational symmetry means that if you carry out the same experiment in different regions of spacetime, the experiments have similar outcomes (the outcomes are related by spacetime translations).

Both space and time translation symmetries are reflected in the Hamiltonian. The Hamiltonian is usually independent of time and depends only on the relative positions of particles $\vec{r_1} -\vec{r_2}$. This implies a spacetime translation symmetry. Keep in mind that time translation symmetry is not merely the statement that re-labelling time leaves physics unchanged. If the Hamiltonian depends on time, the universe treats different points of time differently and the same experiment yields different outcomes at different times.

Now, I do understand your confusion. Given a classical state (for simplicity) $(x, p) $, a space translation can be achieved by generating the translation by Poisson bracketing with the momentum. The new state then corresponds to the same experiment performed at a different point of space.

But, if you try to do the same thing with time translations, you'd end up generating the time evolution using the Hamiltonian. The new state $(x', p') $ will not represent the same exeriment performed at a different point of time.

It may seem that space and time are being treated asymmetrically. So here's a covariant perspective on this: Spacetime symmetry means that the same local solution to the equations of motion is permissible in any local region of spacetime. That is, the universe does not distinguish between different regions of spacetime. You can think of a local solution as the trajectory followed by the particles in an experiment, or a field solution in a small region of spacetime.

Is there a different meaning to propagating forward in time to the two cases?

Yes. In one case, you are simply dynamically evolving the system. In the other case, you are obtaining the wavefunction $|\psi ' (t) \rangle$ of the same experiment performed at a future point in time. Because the Hamiltonian is time independent, $\psi (t) '$ is simply $\psi(t)$, but offset but by some time $s$.

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