Graduate Understanding time translations in Ballentine

[EE18]

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→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],|𝜓(𝑡)⟩→$e^{isH}$|𝜓(𝑡)⟩ =(?) |𝜓(𝑡−𝑠)⟩

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→𝑡′=𝑡+𝑠 on the system then shouldn't my system get mapped to |𝜓′⟩=|𝜓(𝑡′)⟩=|𝜓(𝑡+𝑠)⟩? If 𝑠>0 I am ahead in time after the active transformation, no? I think the crux of my misunderstanding is why a $t-s$rather than $t+s$ appears in the argument of the transformed state.

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

 

[andresB]

It all boils down to equation (3.8)
$$U(\tau)\psi(x)=\psi(\tau^{-1}x).$$
Notice that transforming "forward" the wavefunctions is the same as leaving the wavefunction intact but transforming "backward" the space-time variable variables.Hence, moving forward in time, $t \rightarrow t+s$, is given by $\psi(t) \rightarrow\psi(t-s).$

 

[strangerep]

I think the crux of my misunderstanding is why a $t-s$rather than $t+s$ appears in the argument of the transformed state.

Adding to andresB's answer...

Imagine a classical observer in spacetime who, at time $t$ (as given by his wristwatch) constructs a (Cartesian, say) local coordinate system, i.e., a grid marked out in space and time numbers. To get to the point on the grid where his watch would read $t+s$ he must tardis himself forward in time to where the time coordinate on the grid says "$t+s$". Alternatively, he could grab the whole grid and slide it backwards in time by an amount $-s$.

The former is called an "active" transformation, the latter is called a "passive" transformation. In general these are usually inverses of each other.

As for complications regarding nontrivial state evolution, that can only happen we have a nontrivial Hamiltonian. We still use the Schrodinger equation (3.38) to compute what $\Psi(t)$ looks like at arbitrary times.