# Special relativity and QM

1. Jun 26, 2008

### jordi

There is an "inconsistency" between SR and QM: the collapse of the wavefunction happens at a given point in time, but SR implies that different observers at different speeds could see either the collapsed wavefunction or the not collapsed wavefunction.

However, this is not a problem: QM is not-relativistic.

QFT "is" a relativistic quantum theory. I have heard (but not understood) that there are no wavefunctions in QFT. In fact, I am embarrassed to admit that I do not understand how the framework of QM (collapse of the wavefunction, ...) fits into QFT.

How does QFT, as a relativistic quantum theory, deal with different observers seeing or not seeing the collapse of a "state" (or wavefunction), in a given observation by an observer?

2. Jun 26, 2008

### Icosahedron

Observers at different speeds in SR have the Lorentz transformation to compare their mearsurements made in each coordinate system.
No inconsistencies here.

3. Jun 26, 2008

### Icosahedron

Well, of course they disagree o simultaneity, so they also disagree on simultaneity of two distant-apart measurements/ wave collapses. But what inconsistencies should arise from that?

4. Jun 27, 2008

### jordi

Definitely, I am talking about "inconsistency" in an intuitive way, not in a rigorous way. What I mean by inconsistent is: if I understand correctly lorentz or poincaré invariance, embedded in SR, it says that all observers moving at constant speed with each other should see the same physical laws, and final observation in a given frame can be predicted by taking the final observation in another frame and lorentz-rotating the coordinates in the observed solution.

I cannot imagine that a collapsed wavefunction and a non-collapsed wavefunction can be obtained one from the other just by lorentz-rotating the coordinates.

5. Jun 27, 2008

### jordi

But of course, there is something in SR that I do not understand properly. I do not understand the typical example of two men shooting at themselves while in a train. They are both at the extremes of the wagon. In the middle, there is a light that will turn on when an observer in the train decides (T). There is another observer outside the train (L) (that is moving at a relative speed with respect to the land). The two men can shoot once the see the light.

According to T, once the light is turned on, the light arrives to both men at exactly the same time. Then, they both shoot at the same time, and die at the same time.

According to L, one observer sees the light first (because it is moving towards the light), so this observer shoots first. I have not done the calculations, but it seems intuitive that according to L, the observer who shoots first will die the latest.

Well, this is what SR books explain: SR excludes the concept of simultaneity.

But if we add a new rule (there is a shield that covers the other shooter once the other shooter is dead), the first observer should see both shooters dying, and the second one, only one. Intuitively, this does not look a Lorentz-invariant solution.

Of course, there is a lot of blablabla and handwaiving in my argument, and there is something wrong in it. Probably what I miss is what I miss in my first question.

Can anybody help me with my misunderstandings?

6. Jun 27, 2008

### jordi

Note: the previous question should not be in QF, but in SR subforum. I have put it there just for the sake of continuity with my initial question.

7. Jun 27, 2008

### Icosahedron

Observers in relative motion to each other in SR disagree duration of time between two events, the length of objects and the simultanity of two events.

One event, in your case a measurement, does not lead to disagreement.

8. Jun 27, 2008

### jordi

Exactly, that is what I thought: spacetime (ie events) is unique. Different observers are different ways to "slice" the spacetime. Measures of coordinates (slice-dependent) will change for different observers, but events will be the same.

However, my intuitive arguments lead me to think that there can be different events (eg, one observer sees two men dying, the other observer sees one man dying, or also one observer sees a collapsed wavefunction, the other a non-collapsed wavefunction). Of course I am wrong, because SR is known to be true, but I do not know where my handwaiving argument fails.

9. Jun 27, 2008

### Fredrik

Staff Emeritus
The homogeneous Lorentz transformations can be described as "rotations", but the full set of Lorentz transformations (the Poincaré group) includes translations in space and time. In this case the relevant Lorentz transformation is a translation in time, and those are just as relevant to non-relativistic QM as to relativistic QM. So the problem here has nothing to do with SR. It sounds like what you're talking about (in this post, not in the OP) is the measurement problem of quantum mechanics, which is the fact that there are really two different (i.e. contradictory) kinds of time evolution in QM: one that applies when nothing is being measured, and one that applies when something is being measured. The first is deterministic and the second is not.

10. Jun 27, 2008

### Fredrik

Staff Emeritus
Don't bother with calculations (unless you think it's fun), but draw a space-time diagram. Suppose that the train is moving to the right according to L, who's standing on the ground. Then the man on the left will see the light first. This is easy to see in a space-time diagram showing the events from T's point of view.

The word "once" in this context refers to a set of simultaneous events, but simultaneous in what frame? You probably took that frame to be T's frame when you tried to figure out what T would see, and L's frame when you tried to figure out what L would see.

11. Jun 27, 2008

### Fredrik

Staff Emeritus
The two observers would disagree about on which slice of space-time the collapse occured (because they disagree about simultaneity). The reason this isn't a problem is that the wave function isn't measureable, so neither is its collapse.

12. Jun 27, 2008

### jordi

"Once" I think is observer-independent: it means that when one observer dies (an event, observer-independent) a light signal is sent to the other shooter, and a shield appears. I do not think this is observer-dependent.

13. Jun 27, 2008

### Icosahedron

Here is a real inconsistency of quantum mechanics and SR, which makes it necessary to bring quantum fields in.

Measurements of quantum observables, the collapse of the state vector happens instantaneous, faster than light, clear violation of SR.

So when you naively add SR to QM faster than light information results, which leads to violation of causality.

Root of the problem is that measurements in non-relativistic qm are made by operators that are not attached to space-time points, there is just the position operator, the momentum operator, etc.

That's why we introduce quantum fields: operator-valued functions of spacetime whose dynamics are purely local.

14. Jun 27, 2008

### Fredrik

Staff Emeritus
You didn't mention light signals, so I assumed that you meant that raising the shield in front of gunfighter B would happen simultaneously with the death of gunfighter A.

This is what happens:

According to T: Everything is perfectly symmetrical. Whatever happens to gunfighter A also happens to B at the same time.

According to L: Gunfighter A shoots first. gunfighter A dies first. When the light-signal carrying the information about A's death reaches gunfighter B, gunfighter B has been dead for a while.

(I'm assuming that L is standing on the ground watching the train move to his right, and that gunfighter A is to the left of gunfighter B)

Last edited: Jun 27, 2008
15. Jun 27, 2008

### Fredrik

Staff Emeritus
This is wrong. When you construct the operators from quantum fields, you integrate out the space-time dependence.

The collapse of the wave function isn't a "clear violation of SR", since the wave function isn't measurable. No information is transmitted from one measurement to the other.

I also don't like the description of QFT as something different than QM, but this depends on how you define "quantum mechanics" so I can't say that you're wrong about that. I would define QM as "states are represented by the rays of a Hilbert space, observables are represented by hermitian linear operators on that space,..." and so on. None of that changes when we go from non-relativistic to relativistic QM. What changes is the explicit construction of the Hilbert space and the operators. Each QFT defines a way to do the construction. So if your definition of QM includes that the Hilbert space must be specifically the one used in non-relativistic QM, then I suppose what you said is correct. It would be a strange definition of QM though, since then we don't have a term for what I defined as QM.

16. Jun 27, 2008

### Icosahedron

Collapse of wave function= measurement.

Construct operators from quantum fields?? Quantum fields are operators, operator-valued function of spacetime.

Last edited: Jun 27, 2008
17. Jun 27, 2008

### Fredrik

Staff Emeritus
The measurement that collapses the wave function is a measurement of the particle's spin, position, momentum, or whatever, but it's not a measurement of the wave function. Wave functions are vectors in a Hilbert space. Observables are operators on that Hilbert space. So a wave function is a different kind of object than the ones that can be measured.

Yes, in quantum field theories you construct the generators of Lorentz transformations (e.g. the Hamiltonian and the spin operators) by integrating certain combinations of the fields and their derivatives over $\mathbb R^3$. Noether's theorem tells us which combinations to use.

That's sort of the point of the quantization procedure. It's a relatively easy way to construct useful representations of the Poincaré group. You can take the solutions of the field equations to be the Hilbert space, and then Noether's theorem tells you how to explicitly construct Lorentz transformation operators.

18. Jun 27, 2008

### jordi

Let me state the situation: the train is moving to the right. A is on the left of the wagon, B on the right. According to L, A sees first the light, since A is moving towards the light.

So, A shoots first, and I would expect gunfighter B to die first, not A (argument: A shoots first!).

Intuitively and without writing formulas, if the speed of the train is very high, according to L the initial light signaling the start of the shooting should arrive to A (assuming A being on the left of the wagon, and the train moving to the right) quite "soon" (because A is moving at almost light-speed towards a ray of light moving towards her).

Then, if I assume the speed of the bullet is very slow, I think that the bullet sent from A to B will arrive earlier than the one sent from B to A (argument: the bullet will start travelling from A to B much earlier). Then, if the bullet is very slow, the light ray sent once B is dead, towards A, to shield A, could possibly arrive to A before the bullet arrives.

Of course, lorentz-velocity composition could change things.

Any inputs?

Last edited: Jun 27, 2008
19. Jun 28, 2008

### Fredrik

Staff Emeritus
Don't bother with formulas. Draw a space-time diagram showing the events from T's point of view. Then draw the world line of L, in the same diagram, and also draw L's simultaneity lines. (If the slope of the world line is 1/v, the slope of the simultaneity lines is v). It's quite obvious from the diagram that any two events that are simultaneous in T's frame are not in L's frame, and that in L's frame it's always the event on the left that happens first.

20. Jun 28, 2008

### Icosahedron

Frederick, please open any QFT text of your choice and read that quantum fields are operator-valued functions of space-time.

Also, a wave function made collapsed by a non-localized quantum mechanical operator is a clear violation of special relativity.