Relativistic quantum mechanics and causality

[Haelfix]

"I don't find anything "trivial" or "clear" in these statements."

Weinberg in his book links the original paper by Bohr and Rosenfeld
"translation by R Cohen and J Stachel" phys review. 78, 794 (1950).

I like reading papers from that time as the connection with experiment was much closer and transparent, so I'd recommend checking it out.

Absent a blackboard its hard to flush out what I mean mathematically in this setting, and its getting to the point where I can no longer english speak to explain what I mean, (which is fine).

 

[meopemuk]

Weinberg in his book links the original paper by Bohr and Rosenfeld
"translation by R Cohen and J Stachel" phys review. 78, 794 (1950).

Haelfix,
Thank you for the reference. I'll check it out.

Eugene

 

[reilly]

When a photon leaves its mark on the photographic plate, we measure an observable "photon's position". The probability for such a measurement is determined by the square of photon's wave function in the position space. I'm afraid that photon's quantum field $A_{\mu}(x)$ has nothing to do with this probability.


..........

Then, what does?
Regards,
Reilly Atkinson

 

[meopemuk]

When a photon leaves its mark on the photographic plate, we measure an observable "photon's position". The probability for such a measurement is determined by the square of photon's wave function in the position space. I'm afraid that photon's quantum field $A_{\mu}(x)$ has nothing to do with this probability.

..........

Then, what does?
Regards,
Reilly Atkinson

Below I will briefly describe how I understand quantum field theory and its relationship to QM. I understand that my views do not look like the mainstream, but I believe they are in full accord with the way QFT presented in S. Weinberg, "The quantum theory of fields", vol. 1. So, here it goes...


Why quantum mechanics is no good enough? In quantum mechanics we normally deal with systems having fixed number of particles (N). The Hilbert space is built as a tensor product of N 1-particle Hilbert spaces where operators of observables, state vectors, dynamics, etc are defined. In relativistic physics energy can be converted to mass due to Einstein's $E=mc^2$, so we must also take into account the possibility of changing the number of particles (emission, absorption, annihilation, etc.). This can be achieved by switching from fixed-number-of-particles Hilbert spaces to the Fock space, which is simply a direct sum of N-particle Hilbert spaces, where N varies from 0 to infinity. Fundamentally, QFT is nothing else but quantum mechanics in the Fock space.

Operators of observables and wave functions in the Fock space One can define operators of particle observables in the Fock space in exactly the same way as in QM. For example, in each N-particle sector we have well-defined operators (of position, momentum, spin, etc.) for each of the N particles described there. These operators have usual common bases of eigenvectors, and an arbitrary N-particle state vector can be projected on these bases to obtain N-particle wave functions in different representations. The new feature is that one also has state vectors with non-zero projections on sectors with different N. Wave functions of states with such undefined particle numbers are just sets of N-particle wave functions, each with its own coefficient. The square of the coefficient is the probability to find N particles in this general state.

Unitary representation of the Poincare group. The most general way to construct a relativistic quantum theory is by defining an unitary representation of the Poincare group in the Hilbert space (in our case this is the Fock space) of the system. Basically, it is sufficient to construct interaction operators $V$ and $\mathbf{W}$ in the generators of time translations (the Hamiltonian) and boost

$$H = H_0 + V$$
$$\mathbf{K} = \mathbf{K}_0 + \mathbf{W}$$

so that the Poincare commutation relations between all 10 generators (including the total momentum $\mathbf{P}_0$ and the total angular momentum $\mathbf{J}_0$) remain preserved. This is a very non-trivial problem, and currently there is only one known solution which satisfies all requirements, such as the cluster separability, the possibility to describe particle-number-changing interactions, etc. [Please note that the fact that it is the only known solution does not imply that it is the only possible solution.] This is where quantum fields come into play:

1. Define particle creation and annihilation operators in the Fock space.
2. For each particle type build a certain linear combination (the free quantum field) $\psi(\mathbf{r},t)$ of the creation and annihilation operators with two major properties
2a. (anti)commutativity at space-like separations
2b. manifestly covariant transformation laws with respect to the non-interacting representation of the Poincare group in the Fock space
3. Then operator $V(t)$ can be constructed as an integral on $\mathbf{r}$ of field products (or polynomials)

$$V(t) = \int d^3r \psi(\mathbf{r},t) \phi(\mathbf{r},t) A(\mathbf{r},t) \ldots$$

(for brevity I omit possible indices of operators $\psi, \phi, A$ and summations over these indices). The "boost interaction" $\mathbf{W}(t)$ is given by a similar formula.

It is important to note that we don't need to give any physical interpretation to quantum fields $\psi, \phi, A, \ldots$. They are just abstract mathematical quantities, whose role is to facilitate the construction of interaction operators $V$ and $\mathbf{W}$. Once this construction is completed, we have a full interacting quantum theory in the Hilbert space with particle operators, wave functions, the time evolution operator

[tex] U(t) = \exp(\frac{i}{\hbar} Ht) [/itex]......(1)

etc, i.e., everything one would need to solve any kind of physical problem.

Renormalization and dressing Unfortunately, this nice theory has a serious problem. It appears that for all realistic interaction operators $V$ the scattering (S-) matrix cannot be calculated, because its matrix elements come out infinite. To solve this problem one can add (as Tomonaga, Schwinger, and Feynman did) infinite renormalization counterterms to the Hamiltonian $H$ (and to the boost operator $\mathbf{K}$ as well). Then one can get a very accurate S-matrix, but the Hamiltonian $H$ becomes infinite and useless for calculating the time evolution (1) and for finding bound states via diagonalization. This problem can be solved by applying an "unitary dressing transformation" to the Hamiltonian and all other generators of the Poincare group. Then the theory assumes the form very similar to ordinary quantum mechanics: We have a full description for the system of particles interacting with each other via instantaneous potentials, which, in addition, can change the number of particles in the system. The Poincare commutators, cluster separability and other important physical requirements are exactly satisfied. Quantum fields are not needed for the physical interpretation of this theory..

Eugene.

 

[genneth]

Actually, there is a promising line of research to build a quantum relativistic theory of particles that interact via instantaneous potentials. It is called the "dressed particle" theory, which I mentioned a few times already. It's predictive power is just as good or even better than in the traditional renormalized QFT. The "dressed particle" approach does not use Dirac or Klein-Gordon equations.

Eugene.

Whilst that may be for a QFT (or equivalents there of), I was thinking more elementarily, of a theory for just a simple free relativistic particle, without any complications of pair creation and the such.

 

[kof9595995]

I guess what meopemuk wants is a unambiguous Alice-Bob thought experiment, hopefully in which we can prove if [A,B]=0 no superluminal signal can be sent, and if [A, B]!=0, a superluminal can be sent.
And we can indeed accomplish this "proof" in QM, for example, in this link:http://everything2.com/title/Quantum+entanglement+and+faster+than+light+communication
under the "general case" the author proved Cooper can never know whether Alice has done the measurement or not, with a premise that Alice and Cooper has a simultaneous set of eigenbasis (namely [A,B]=0). Vice versa, if we [A,B]!=0, we can construct such a state so that Cooper can know whether Alice has done the measurement .

 

[schieghoven]

When a photon leaves its mark on the photographic plate, we measure an observable "photon's position". The probability for such a measurement is determined by the square of photon's wave function in the position space. I'm afraid that photon's quantum field $A_{\mu}(x)$ has nothing to do with this probability.


..........

Then, what does?
Regards,
Reilly Atkinson

Meopemuk -- what do you mean by a photon wave function? When you say it is not related to $A_{\mu}(x)$ do you refer to a representation based on EM field strength, such as the Riemann-Silberstein representation (as described in [1-3])? As far as I can see, this representation does not admit a unitary representation of the Poincare group and therefore can not be interpreted as a probability density, so it doesn't get much traction in the community. Nevertheless I like the idea.

[1-3]
Bialynicki-Birula, I. Exponential Localization of Photons, Phys. Rev. Lett., 1998, 80, 5247-
Smith, B. J. & Raymer, M. G. Two-photon wave mechanics, Phys. Rev. A, 2006, 74, 062104
Hawton, M. Photon wave functions in a localized coordinate space basis, Phys. Rev. A, 1999, 59, 3223-

 

[A. Neumaier]

The traditional way of handling properties of light seems totally inconsistent to me.

Everything is simple when we are dealing with one photon. We know that it is described by a complex-valued wave function, whose square exactly describes the probability of the photon hitting any given point on the photographic plate.

Who is the ''we'' that knows that?

There is no such wave function. The photon (i.e., the massless spin 1 representation of the Poincare group) does not admit a canonical position representation in which $$|psi(x)|^2$$ could be interpreted as a probability density. Attempted constructions of such a position representation depend on additional ingredients that change under gauge transforms. But a position probability would have to be gauge independent.

 

[meopemuk]

Who is the ''we'' that knows that?

There is no such wave function. The photon (i.e., the massless spin 1 representation of the Poincare group) does not admit a canonical position representation in which $$|psi(x)|^2$$ could be interpreted as a probability density. Attempted constructions of such a position representation depend on additional ingredients that change under gauge transforms. But a position probability would have to be gauge independent.

Thank you for reminding me about that. You are right, it is not possible to define a position operator with commuting components (X,Y,Z) in the 1-photon Hilbert space. So, strictly speaking, there are no position-space wave functions associated with photons.

However, I still believe that ordinary 1-particle quantum mechanics should be applicable to photon states. In this quantum description there should exist some approximate (or non-commutative) representation of the photon's position. It must be so, because experimentally we can measure photon's position, e.g., as dark spots of the photographic plate. This quantum theory should also provide an explanation for the Young's double-slit experiment.

The point of my comment was that there exists another (non-quantum) theory that pretends to provide a different explanation for the wave properties of light. This is Maxwell's electrodynamics, in which the light is described as an electromagnetic wave. In my opinion, it is totally intolerable when the same physical effect (e.g., the light interference) has two completely different explanations. Only one explanation must survive. I think that the quantum-mechanical explanation is the correct one.

Happy New Year!
Eugene.

 

[A. Neumaier]

I still believe that ordinary 1-particle quantum mechanics should be applicable to photon states. In this quantum description there should exist some approximate (or non-commutative) representation of the photon's position. It must be so, because experimentally we can measure photon's position, e.g., as dark spots of the photographic plate.

The situation is actually quite complex. See
http://arnold-neumaier.at/ms/lightslides.pdf
http://arnold-neumaier.at/ms/optslides.pdf

Maxwell's electrodynamics, in which the light is described as an electromagnetic wave. In my opinion, it is totally intolerable when the same physical effect (e.g., the light interference) has two completely different explanations. Only one explanation must survive. I think that the quantum-mechanical explanation is the correct one.

The two views are closely related, and are part of the common picture called quantum electrodynamics (QED). It has both observables for the electromagnetic field and for photon number. Photons appear as the limit of geometric optics, while the Maxwell equations appear in the mean field limit.

Happy New Year!

 

[yoda jedi]

Maybe I am a bit too naive with respect to this. But isn't this simply quantum mechanics ?

commutative = simultaneously measurable = independent from each other = not causally related

or
Parameter Independence and Outcome Independence.

 

[theZ]

An interesting physical way to understand how $$[O_1(t), O_2(t') ]$$ relates to causality is through the linear response (Kubo) formalism. Suppose the system begins in the vacuum state, and you perturb the Hamiltonian with an external source $$V(t) = J(t) O_1$$. This could, for instance, be a background EM field, or an injection of particles. We wish to measure the observable $$O_2(t')$$ at a later point in time, to lowest order in J. Standard time dependent perturbation theory, carried out in the interaction or Heisenberg picture, shows that the change in the observable $$O_2(t')$$ depends on $$J(t)$$ through the stated commutator ( ie, the retarded Green's function). If it vanishes, jostling $$O_1(t)$$ leaves future measurements of $$O_2(t')$$ unaffected.

 

[A. Neumaier]

I still believe that ordinary 1-particle quantum mechanics should be applicable to photon states. In this quantum description there should exist some approximate (or non-commutative) representation of the photon's position.

Ordinary 1-particle quantum mechanics is applicable to photon states as long as you consistently work in the momentum representation, and restrict your states to be transversal to 4-momentum.

There are non-commutative position operators in the literature, but the formulas work both for photons and electrons, and in the latter case do not reduce to the Newton-Wigner position operator with commuting components that is well-known to correspond to the position representation of electrons. Thus the non-commutative position operators do not seem to have physical relevance.

It must be so, because experimentally we can measure photon's position, e.g., as dark spots of the photographic plate.

This only gives a very approximate position, consistent with the nonexistence of a position operator (that would produce sharp positions).

Note that most real measurement are not related to observables in the textbook sense but to so-called positive-operator-valued-measures (POVMs). The measurable photon position is of the latter kind.

 

[wawenspop]

If a photon wave function is spread out over a very large area then how can it collapse over that large area instantly. Remember that it is observed at one location only. So how would an identical observation at a far distant location on the same wave function area 'know' that it had already been observed.

maybe we have to break -Bohm fashion- weak causality and admit that the photon knew its destination at its outset. The instant transmission of 'knowledge' is allowed in this context (but not information).

 

[A. Neumaier]

If a photon wave function is spread out over a very large area then how can it collapse over that large area instantly. Remember that it is observed at one location only. So how would an identical observation at a far distant location on the same wave function area 'know' that it had already been observed.

You forget that measurements take time and are only approximately described by the highly idealized, instantaneous and complete measurements that figure in typical discussions of the foundations of quantum mechanics. The collapse is an idealization, too.

 

[wawenspop]

You forget that measurements take time and are only approximately described by the highly idealized, instantaneous and complete measurements that figure in typical discussions of the foundations of quantum mechanics. The collapse is an idealization, too.

We could be talking about a photo that traveled billions of years in superposition 'spread' out over billions of square miles and at one location it is decohered, so what about the symmetrically same location a billion years away where it did not decohere? Seems there is a 'knowledge' at all locations where the photon 'could be'.

 

[A. Neumaier]

We could be talking about a photo that traveled billions of years in superposition 'spread' out over billions of square miles and at one location it is decohered, so what about the symmetrically same location a billion years away where it did not decohere? Seems there is a 'knowledge' at all locations where the photon 'could be'.

A wave knows its shape locally, so there is no problem. A local measurement at a delocalized state cannot collapse the whole state but only the part within the projection to the local area.

This is enough to make your questions absurd and unanswerable.

 

[Hawkwind]

Exactly. The propagators never get outside the lightcone, neither if you simulate
the propagation on a lattice, nor with the exact analytical solutions.
...

Are you sure ?
Afaik, the propagator of a massless particle has a pole on the lightcone, but summing up the contributions due to internal lines in a diagram (by integration over 4-momenta), there are also contributions from propagators connecting events with timelike or even spacelike distance.
Afaik, causality in quantum field theories is ensured by the additional demand, that the commutators/anticommutators of bosonic/fermionic field operators have to vanish at spacelike distances. This makes sure that events with a spacelike distance cannot influence each other.
Regards,
Hawkwind