How does time-translation symmetry morph into evolution in time?

Question

I am reading Ballentine's textbook "Quantum Mechanics: A Modern Development". In it he transitions from discussing time-symmetry to discussing evolution (of the state) in time. I'm finding it difficult to understand why this move is justified. I suppose one can just take it as a hyposthesis, but I feel there is something deeper here, an implication that evolution has to follow the form of the symmetry, which I don't see.

In a bit more detail: After laying-down the postulates of quantum mechanics, Ballentine discusses continuous single paramter symmetries of the laws of nature and argues that they must preserve the quantum amplitudes and hence must correspond to a unitary operator $U(s)$.

Ballentine emphazises that he is invoking the "active point of view, in which the object ... is transformed relative to a fixed coordinate system". I take this to mean that the meaning of time-translation here is that I actually change when I'm considering the system. If I do the experiment today or I translate it in time and do it tomorrow, the experiment will have the same results.

After some effort, Ballentine further shows that time-translation is given by the operator $U(t)=exp(i t H)$.

Critically, after showing that for dynamics he needs to consider the change in time of the state $\frac{d}{dt}|\psi(t)\rangle$, Ballentine argues that "corresponding to the time displacement [] there is a vector space transformation of the form" $exp(i t H)$ from which he formally derives the dynamic equation for time evolution in quantum mechanics $\frac{d}{dt}|\psi(t)\rangle=-iH|\psi(t)\rangle$ (eq. 3.38).

But it appears to me that the $t$ in the time-translation operator is the amount of time-translation that I engage in, rather than the physical time that passes. I don't see how Ballentine can move from the idea that the symmetry of the laws of nature in time implies that time-translation can be described by $U(t)=exp(i t H)$, to the idea that this operator can be applied to describe the evolution of the state in time.

Answer 1

This has nothing to do with quantum mechanics. Exactly the same things can be said in classical Hamiltonian mechanics, where the Hamiltonian is the generator of time translation in the sense that the Poisson bracket $\{H,A\}$ (quantum analogue: the commutator $[H,A]$) is the infinitesimal evolution of $A$ and the flow of its associated Hamiltonian vector field $X_H$ (quantum analogue: the exponential $\mathrm{e}^{\mathrm{i}Ht}$) is finite time translation.¹

That time translation is the same as time evolution is a tautology. There is no difference between "time passes of its own accord" and "an object moves in time", at least not in the formalism. Maybe try out the analogy with position to see that there really is nothing here to be discussed. You could say:

But it appears to me that the $x$ in the spatial translation operator is the amount of spatial translation that I engage in, rather than the physical distance I cover . I don't see how Ballentine can move from the idea that the symmetry of the laws of nature in space implies that spatial translation can be described by $T_x(x)=exp(ixp)$, to the idea that this operator can be applied to describe the change of position of the state in space.

In particular, does the first sentence of the above make sense to you? It does not to me, "spatial translation that I engage in" and "physical distance I cover" are the same things, and so are "time translation that I engage in" and "physical time that passes". The passage of time is the same as all objects translating in time, rather by definition of what we mean when we say that time passes. When distance passes you by, you are engaging in spatial translation, when time passes you by, you are engaging in time translation.

Finally, let me address another confusion that seems to shine through when the question mentions "symmetry": For this idea of operators generating one-parameter groups through exponentiation, it is wholly irrelevant whether or not the operator/group is a symmetry or not. The Hamiltonian generates time-translation whether the system is time-symmetric or not. The momentum generates spatial translation no matter whether the system is Galilean (or Poincaré) invariant or not. It is not symmetry that implies the exponential of an operator $A$ with $[A,B] = \mathrm{i}\hbar$ describes translations in $B$, but pure algebra that doesn't care whether $\exp(A)$ or $\exp(B)$ are symmetries of anything.

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¹As an aside, the vector field and the exponential are not as different as one might think. In both cases there is a Lie algebra - of vector fields classically and of Hermitian operators quantumly - that has an exponential map to a (possibly infinite-dimensional) Lie group - the group of diffeomorphisms (perhaps symplectomorphisms) classically and the group of unitary operators quantumly.

Answer 2

I'm also reading Ballentine and had this exact question. I added a bounty, but I'm now satisfied with the below explanation. Will happily award the bounty if someone explains it more clearly though!

Here's my understanding of the logic used to arrive at equation 3.38: to model time dependencies of the observable value $\langle \mathrm{A} \rangle = \langle \Psi | A | \Psi \rangle$, we have the freedom to either make the state vector $\langle \Psi |$ time-dependent and $A$ time-independent (the Schrodinger picture), or vice versa (the Heisenberg picture). In the Schrodinger picture, we define $|\Psi(t)\rangle$ to be the vector that can be used to make predictions for time $t$ according to the formula $\langle \mathrm{A}(t) \rangle = \langle \Psi(t) | A | \Psi(t) \rangle$. The vector $|\Psi(t)\rangle$ should not be interpreted as physically existing at time $t$! Hence there is no "physical time" passing, hopefully resolving your confusion. By convention we also define the observable $A$ in a way that if we analyze a system relative to a particular frame of reference and obtain a state vector $|\Psi\rangle$, then $\langle \mathrm{A} \rangle = \langle \Psi | A | \Psi \rangle$ will be a prediction for time $0$ of that frame.

Therefore, by definition, the state vector obtained from analyzing the system with respect to the current frame is $|\Psi(0)\rangle$. If we consider an observer in another frame whose time $0$ corresponds to time $t$ in the current frame, the state vector they obtain will necessarily be $|\Psi(t)\rangle$, since it can be used to obtain predictions for time $t$ of the original frame. Because the second frame is simply a time translation of the first by $-t$, the two state vectors are related by $|\Psi(t)\rangle = e^{-iHt}|\Psi(0)\rangle$, and differentiating, we get equation 3.38.

Answer 3

You are right about the pictures of quantum mechanics and the argument beyond Ballentine's derivation can be restated in the way you did. The only thing one could object here is when you say that $t$ is not physical time.

A thorough discussion about the role of time in physics foundations is found in Arnold's Mathematical Methods of Classical Mechanics, because as it was mentioned before, the role of time in non-relativistic quantum mechanics is the same as in classical mechanics: just a parameter. You can see this also in Ballentine's book, just below equation (3.2) the author considers a family of operators $U(s)$ where $s$ is any continuous parameter. Later, just above equation (3.38) it says that $t=s$. So that is $t$, a continuous parameter.

Why then it turns out to be hard to deal with the notion of time in quantum mechanics? Because in classical mechanics one relies on daily physical experience and time is just there (as in the foundations of classical mechanics). Even though it could be hard to define time, everyone knows what you mean when you talk about time in classical mechanics. So, the difference are the limitations that we have in quantum mechanics to rely on daily experience because everything is macroscopic.

In conclusion, if you want you can regard $t$ as something non-physical, just a mathematical entity, a continuous parameter that defines an unitary transformation $e^{isH}=e^{itH}$. Doing so, all your reasoning is correct, no objections. However, I don't find anything wrong in regarding this $t$ as physical time, even though we don't see microscopic systems in our daily experience, the time-translation that you defined at the end, is the very same translation that we experience in classical mechanics.

With that clarification, I will just make some comments about other things you refer and finally, rephrase Ballentine's argument with all the intermediate details:

1. Both pictures of quantum mechanics can be used as pleased. Therefore, it should be a situation where they coincide and it is at $t=0$. We have (superscripts stand for Heisenberg and Schrödinger pictures):

$$A^{H}(t=0)=A^{S},$$ for operators, and

$$|\psi^{S}(t=0)\rangle=|\psi^{H}\rangle,$$ for states.

What is important is that expectation values are the same regardless of the picture you chose (that is why I said that they can be used arbitrarily), so then we have:

$$\langle\psi(t)|A|\psi(t)\rangle=\langle\psi|A(t)|\psi\rangle$$

For more information about the pictures I suggest you the chapter 2 of Sakurai's Modern Quantum Mechanics.

2. Now say that in frame $O$ you know the physics at time $t$ (or parameter $s=t$ if you want). For a second frame $O'$ the same physics occurs at $t'$ which turns to be time $t+r$ with respect to $O$. Now you should go to Fig. 3.1 in Ballentine's book and the discussion below it. Since the physics is the same for each frame:

$$|\psi'(t')\rangle=|\psi(t)\rangle$$

But the state for $O'$ is just a time translation with respect to $O$, so:

$$e^{isH}|\psi(t')\rangle=|\psi(t)\rangle,$$

where it seems weird to have the same $t'$ in the left but you must remember that you are working in Schrödinger's picture, what changes are the states ($\psi$ to $\psi'$) and not the parameter.

Finally, $t'=t+r$ as mentioned before, so $t=t'-r$ and:

$$e^{isH}|\psi(t')\rangle=|\psi(t'-r)\rangle.$$

At this point Ballentine says "putting $s=t'$" and we get:

$$e^{it'H}|\psi(t')\rangle=|\psi(0)\rangle.$$

But you want to invert this to get:

$$|\psi(t')\rangle=e^{-it'H}|\psi(0)\rangle,$$

or just

$$|\psi(t)\rangle=e^{-itH}|\psi(0)\rangle.$$

Those are all the intermediate states for Ballentine's derivation, I only show you the rigorous way but your argument is essentially the same thing.

Finally, equation (3.38) follows from this by taking the derivative with respect to time as you said.