How are time translations understood in non-relativistic quantum mechanics?

Question

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.

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

Answer 1

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

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

Answer 2

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.