# Time translation in the heisenberg picture?

1. Apr 27, 2008

### pellman

A fundamental quantity that we calculate with QM is $$\langle \Phi|\Psi\rangle$$-- the probability amplitude for observing a system to be in state $$|\Phi\rangle$$ given that it is in state $$|\Psi\rangle$$. In the Schrodinger picture the states are time-dependent and we can ask, "What is the probability amplitude for observing the system to be in state $$|\Phi\rangle$$ at time t2 given that it is in state $$|\Psi\rangle$$ at time t1. And of course the answer is $$\langle \Phi(t_2)|\Psi(t_1)\rangle$$

We can also refer them to a fixed time t=0, by writing $$|\Psi(t_1)\rangle=e^{-iHt_1}|\Psi_0\rangle$$. Then the probability amplitude for observing the system to be in state $$|\Phi\rangle$$ at time t2 given that it is in state $$|\Psi\rangle$$ at time t1 may be written

$$\langle \Phi_0|e^{+iH\Delta t}|\Psi_0\rangle$$
where $$\Delta t=t_2-t_1$$

But $$|\Psi_0\rangle$$ and $$|\Phi_0\rangle$$ are the very same states we use to describe the system from the Heisenberg-picture approach. That said we finally come to my question:

What does the quantity $$\langle \Phi_0|e^{+iH\Delta t}|\Psi_0\rangle$$ mean in the Heisenberg picture? The answer should avoid referring to "state at time t" because in the Heisenberg picture states are not time dependent; only operators are time dependent.

2. Apr 27, 2008

### masudr

Quite right, but note that your expression is not a state, rather it is an amplitude. And amplitudes most certainly can be time-dependent.

The meaning of your expression is the same in either picture.

3. Apr 28, 2008

### pellman

Thanks but they cannot be the same.

In the Schrodinger picture $$\langle \Phi_0|e^{+iH\Delta t}|\Psi_0\rangle$$ means "The probability amplitude of finding the system in state $$\Phi$$ at time t2 given that it is in state $$\Psi$$ at time t1."

What can it mean in terms other than states at specific times?

4. Apr 28, 2008

### f95toli

Of course they mean the same thing!
What you are calculating is a number (the amplitude) and since the two "pictures" are just two ways of looking at the same thing they must always agree once you start calculating amplitudes.

Note that the Heisenberg and Schroedinger pictures are just the two "extremes" of the interaction picture, in the latter both operators and states are time dependent.
You can use whatever picture you want, it really is just a question of mathematical convenience.

5. Apr 28, 2008

### Ken G

It can't mean that because it only depends on $$\Delta t$$. So it just tells you how the overlap amplitude of two states varies with time, in general. I don't see how that requires that the states themselves are associated with any time dependence.

6. Apr 28, 2008

### lbrits

The question is best answered by appealing to the Schrodinger picture, since the two pictures are mathematically equivalent. But your question is essentially "what is the meaning of the time evolution operator in the Heisenberg picture". Note that the expression you wrote gives essentially matrix elements of the time evolution operator. Thus, if we know the matrix elements of that operator, in say, the basis of states $$\left|x\right\rangle$$, we can compute, say $$\left\langle \psi \right| x^2(t) \left| \psi \right\rangle$$ more easily. Thus it is useful to know $$\left\langle x' \right| e^{-i H t} \left|x\right\rangle$$. This object is also the Green's function in the schrodinger picture.

As has been said, the two pictures are equivalent, so use whatever appeals to you at any particular instant =)

7. Apr 28, 2008

### nrqed

Good question.
The key point I think is that the state of a system does change in time no matter in what picture you are. This is clear. If you consider, say, a spin in a magnetic field, the state of the system at different times is obviously different. what the Heisenberg picture implies is not that the state of a system does not change in time, only that this time dependence is not tagged on the wavefunctions. In other words, you have that at some time t_0 the state of the system is $$| \Psi>$$ and at some later time t_1 it is now a different state $$| \Phi >$$. There is no explicit time dependence in the states but the system is described by two different wavefunctions at the two different times. This is th ekey point to your question, I think. So in fact, there is an implicit time dependence on the state of a system even in the Heisenberg picture (otherwise, the Heisenberg picture would imply that everything is static!).

Consider an operator A which does not commute with H (so that the eigenstates in the Schrodinger picture are time dependent). In the Heisenberg picture, you could find the eigenstates of that operator at some time t_0. Those eigenstates would have no time dependence. But if you consider a later time t_1, you would have to find A(t_1) and computes again its eigenfunctions. You get a new set of states (with no time depence) which represent the eigenstates of that operator at that specific time only. And so on. This is why even in the Heisenberg picture you often see a specific time assigned to some set of eigenstates (like $$|a(t_0)>$$ for the eigenstates of our operator A). This seems strange at first since states are supposed to have no time dependence in the Heisenberg picture. But that simply means that those are the eigenstates at that specfic time.

8. Apr 28, 2008

### pellman

Are you saing that the quantity in question in the Heisenberg picture means "the probability amplitude of finding the system in state $$\Phi$$ at time t2 given that it is in state $$\Psi$$ at time t1." even though in the Heisenberg the states are taken to be time independent?

9. Apr 28, 2008

### pellman

I agree. I think the problem with the Heisenberg picture--at least as applied by many recent writers--is just to obfuscate that time in quantum theory must be handled differently than position. We can consider the abstract quantity $$|\Psi(t)\rangle$$ where the wavefunction is $$\Psi(x,t)=\langle x|\Psi(t)\rangle$$ because we can just as well work in the momentum representation $$\langle p|\Psi(t)\rangle$$. Position and momentum are dynamical variables.

We can formally pull out the time dependence using $$e^{-iHt}$$ and so that we can pretend that we are manifestly Lorentz covariant, but it's not really. Time is a parameter not an operator. See the recent thread on quantum myth: time energy uncertainty. https://www.physicsforums.com/showthread.php?t=230693

This question was prompted by reading Chap 3 of Weinberg's Quantum Theory of Fields. He is setting up the discussion for the S-matrix and stresses that the 'in' state and 'out' state are not time dependent and that we are "maintaining manifest Lorentz invariance", etc, etc, but then immediately starts applying the $$e^{-iHt}$$ operation to them to calculate amplitudes.

Sounds like BS to me. Might as well write $$\Psi(t)$$ and stop trying to confuse me.

10. Apr 28, 2008

### f95toli

Not really, "the state of the system" usually means something like "an electron at position x", in a time dependent problem this electron can move to a new position -i.e. state- and this must obviously happen in the same way regardless of which picture you decide to use.
After all, if the state was truly time independent the electron wouldn't move!
Rhe fact that the state vectors (mathematical notation) are time independent doesn't mean that the state of the system (physical property) can't change.

11. Apr 28, 2008

### Ken G

To me, the purpose of the Heisenberg picture is simply to draw out the fact that if both the "in" and "out" states share a similar-in-form time evolution, that similarity can be capitalized on to simplify how we think of time evolution. If the two wave functions had completely different time dependences, the Heisenberg approach would have no value, but as you showed, the similarity in their form allows for a unified form for the time dependence. So the key is not whether you think of it as an "operator" or a "wave function", it is essentially whether you want to picture "time dependence" as happening in two places, or just one.

Personally, I like the Heisenberg picture, because it shows that the "states" are somewhat arbitrary-- they are of our making because we choose the preparation and the measurement of the system. What we don't control is the dynamical object that maps between the preparation and measurement, so thinking of that mapping as a time-dependent operator makes perfect sense. It also gets us away from the erroneous idea that "the system itself has a wave function", rather than our choices for how to treat it, but that's a controversial issue that gets into the Copenhagen interpretation.

Last edited: Apr 28, 2008
12. Apr 28, 2008

### reilly

The Schrodinger and Heisenberg (and Interaction) pictures are completely equivalent, as they must give the same physics, and related as they are by a unitary transformation. Simply work it out for eigenstates of the exact Hamiltonian and all will be clear.
Regards,
Reilly Atkinson
(This is discussed in virtually every QFT book written after 1950.)

13. Apr 29, 2008

### pellman

I am sure they are. But please humor me. I cannot see it for this specific question.

I understand the mathematical difference. It is just a matter of where we put the $$e^{-iHt}$$. An expectation value of observable Q can be written

$$\langle Q\rangle=\langle\Psi(t)|Q|\Psi(t)\rangle=(\langle\Psi_0|e^{+iHt})Q(e^{-iHt}|\Psi_0\rangle)$$ - Schrodinger
or
$$\langle Q\rangle=\langle\Psi_0|Q(t)|\Psi_0\rangle=\langle\Psi_0|(e^{+iHt}Qe^{-iHt})|\Psi_0\rangle$$ - Heisenberg

Same answer either way.

In the Schrodinger picture we have time dependent states which obey the Schrodinger equation of motion and (typically) time independent operators. In the Heisenberg picture (supposedly) we treat states as if they are time independent; the operators carry all of the time dependence. There is no equation of motion for the states but observables obey the Heisenberg equations of motion.

So consider the quantity $$\langle \Phi_0|e^{+iH(t_2-t_1)}|\Psi_0\rangle$$. This is the answer to the question posed in the Schrodinger picture: "What is the probability amplitude for finding the system to be in state $$\Phi$$ at time t2 given that we prepared it in state $$\Psi$$ at time t1?" In the Heisenberg picture the time-dependence is wholly in the observables; states do not have time-dependence. The reference "state $$\Psi$$ at time t1" does not apply.

So for what question in the Heisenberg picture is $$\langle \Phi_0|e^{+iH(t_2-t_1)}|\Psi_0\rangle$$ the answer? I don't need a lengthy explanation of how the two pictures are equivalent. I just need a single question: "What is the probability amplitude .... ?" whose answer is $$\langle \Phi_0|e^{+iH(t_2-t_1)}|\Psi_0\rangle$$ and that is phrased such that states are not associated with times and the times t2 and t1 refer to operators only.

Last edited: Apr 29, 2008
14. Apr 29, 2008

### masudr

The probability for a transition in the Heisenberg picture is

$$\langle \psi | U(t, t_0) | \psi \rangle$$

The expression for $U(t_2, t_1)$ is precisely what you have sandwiched between your bra and ket.

I don't wish to trivialize the issue, but is part of the problem that you have written the two states with different symbols?

15. Apr 29, 2008

### pellman

I wrote them with different symbols because I meant them to signify two different states.

16. Apr 29, 2008

### Ken G

It is the answer to the question "what is the probability amplitude for the observable state $$\Psi_0$$ to project onto the observable state $$\Phi_0$$ after time $$t_2 - t_1$$." The fact that it depends on $$t_2$$ and $$t_1$$ in a way that does not need to reference "state $$\Phi(t_2)$$" etc., shows that it is merely referring to the meaning of a projection operator as time elapses. If $$t_2$$ and $$t_1$$ had some special association with states $$\Phi(t_2)$$ and $$\Psi(t_1)$$ then you could not think of is simply as a type of time-advance operation. Indeed, here it depends only on the combination $$t_2 - t_1$$ and not on either time explicitly, so it becomes especially clear that a type of "time evolution operator" is a more general way to think of it than as time evolution of individual states.

Last edited: Apr 29, 2008
17. Apr 30, 2008

### pellman

Thanks, K G.

But it seems to me more and more that when an author says that we're going to work in the Heisenberg picture to "maintain manifest Lorentz invariance" or the like, (e.g., PCT, spin and statistics, and all that or Weinberg's Quantum Theory of Fields), that it is just sweeping the troubling difference in the way time and position are necessarily handled in quantum theory under the rug. I'll try to keep an open mind though. I am still not far along in this subject.

18. Apr 30, 2008

### Ken G

I don't understand the connection with Lorentz invariance but it may be that they are trying to unite all the explicit time and space dependence in the same place, the evolution operator, so they can test its Lorentz invariance in one fell swoop, rather than having to treat the time dependence of the states separately and check the invariance of the combination. If you get any insight into this issue, please explain it.

19. Apr 30, 2008

### pellman

Ha! Maybe I can check back with you in a year or two.

Actually, though I'm not that far along, what I suspect is that they "demote" position to the kind of parameter status that time has in non-relativistic QM, so that Schrodinger-like picture and Heisenberg picture states would be related by $$|\Psi(x,t)\rangle=e^{i\hat{p}_{\mu}x^{\mu}}|\Psi_0\rangle$$ (for a one particle state). Of course, then we lose the position operator and position representation basis states $$\langle x|$$, which is probably just as well since I don't think wave-functions are defined for relativistic theories. I know for certain they are not for spin 0. But we would keep momentum eigenstates just as we have energy eigenstates in regular QM.

Actually, maybe I am beginning to understand this after all. A little.

20. Apr 30, 2008

### reilly

It was the Interaction Representation that made it possible to do covariant computations in QED. Suppose We've got a differential equation,

idX/dt = -wX + F(X,t,...)

F is quite arbitrary. The usual trick is to set X=exp(iwt) Y. Then the w term goes away, and

idY/dt = exp(iwt) ∫ exp(-iwt')F (X(Y),t'….)dt'.

Translate this to the Schrodinger Eq., and , lo and behold, you have the interaction rep emerging. That is, in the IP, the "free" motion is removed.

Assigning a time to a Heisenberg state is against the rules. So, how can a system evolve? Well, we know from evolution in the Schrodinger picture. For a time independent H, the evolution of a Schrodinger state is exp(-iEt)|S0>, and |S0> is the initial state at time t=0. It is thus equivalent to a Heisenberg state, since it is time independent. All the evolution is carried by the exponential operator of time translation -- not dissimilar to the various coordinate systems used in fluid mechanics, one of which travels along with the stream, sort'a like the Heisenberg picture.

Pines and colleagues, in solid state physics, used the Heisenberg rep. with great success. They worked with the operator equations of motion, which don't exist in the Schrodinger picture. As usual, it all depends…... .

To some degree this is a summary of what's been said before.

Regards,
Reilly