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.


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.

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.

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  • $\begingroup$ Using spatial-translation (and classically, sure), consider (a) a system S at position $x_1$ (from an observer at that position) or an identical system at position $x_2$, transforming between them is an "active" transformation, (b) a system S at position $x_1$ but from an observer at $x_2$, i.e. a standard coordinate transformation, a "passive" transformation, and (c) how system S changes spatially, which is what $\frac{d}{dx}|\psi \rangle$ would signify in analogy to time evolution. I can sort-of see how (a) and (b) are "mirror images" of each other, perhaps, but not how they relate to (c). $\endgroup$ – PhysicsTeacher Jan 30 '19 at 19:48

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