QED explanation of entanglement

1. Nov 30, 2015

zonde

I would like to understand basic construction of entanglement "explanation" in QED.
As I understand because we talk about coincidences QED explanation necessarily involves Fock states. And Fock states are expressed using annihilation and creation operators, right?

2. Nov 30, 2015

vanhees71

Right. You just use appropriate creation operators to the vacuum to construct the polarization entangled state of two photons. The usual (covariant) commutation relations for the quantum fields (particularly the commutativity for operators representing local observables at spacelike separated arguments) ensure that there is no spooky action in the measurement/registration of photons is involved.

3. Nov 30, 2015

zonde

Is polarization entangled state expressed as superposition of two Fock states? Say Fock state of two H-polarized photons minus Fock state of two V-polarized photons (with appropriate coefficient).

4. Nov 30, 2015

vanhees71

exactly! On top the formalism also ensures the correct Bose properties of the photons. For a full description see the paper

C. K. Hong, L. Mandel, Theory of parametric frequency down conversion of light, Phys. Rev. A 31, 2409 (1985)
http://dx.doi.org/10.1103/PhysRevA.31.2409

5. Nov 30, 2015

zonde

We have polarization entangled state expressed as superposition of two Fock states in certain basis but we perform measurement in different basis. So we transform this superposition from initial basis to measurement basis.
Now about this transformation, is it physical or is it just change of representation? I have impression that change of basis is viewed as not physical but then it changes physical outcome (rate of coincidences changes).

6. Dec 5, 2015

vanhees71

Of course, you can describe the vector representing a state in terms of components with respect to any basis which is most convenient for you. The change of the basis doesn't change the vector at all. It's like describing a point in space in usual geometry, once in terms of Cartesian or another time in spherical or cylinder coordinates. This is just a matter of convenience, but it doesn't change anything about the location of the object with help of either Cartesian or other coordinates.

7. Dec 7, 2015

zonde

I had impression that QED descriptions are expressed using spacetime. But Fock state has non-local description. If you say that superposition of two Fock states can be described by vector then I am at loss. In what space this vector "lives"? It's Hilbert space right? But then QED does not "live" in spacetime.

8. Dec 7, 2015

bhobba

Its a Fock space.
https://en.wikipedia.org/wiki/Fock_space

Its resides in space time. Each point of space-time contains the operator representing the quantum field

Thanks
Bill

9. Dec 7, 2015

zonde

Thanks

I see no justification for this statement. According to wikipedia Fock space is constructed from Hilbert spaces. If two individual particle states are distant in space then combination (symmetrized tensor product) of both Hilbert spaces will be non-local in respect to spacetime.

10. Dec 7, 2015

bhobba

For justification you need to go into the detail:
http://portal.kph.uni-mainz.de/T//members/wittig/talks_lecture/ral.pdf

The field operators are expanded in a Fourier Series (see equation 2.28).

The resulting creation and annihilation operators create the particle states the superposition of which requires the Fock Space.

The field resides in space-time, the Fourier transform resides in momentum space.

Thanks
Bill

Last edited: Dec 7, 2015
11. Dec 7, 2015

zonde

Fock state is single particle states plus symmetrization requirement. So if we represent superposition of Fock states in different basis symmetrization requirement determines superposition of Fock states in new basis. So the symmetrization requirement produces non-local effect in QED.

12. Dec 7, 2015

Staff: Mentor

I must confess that I'm not following this line of thinking at all? What do you mean by "superposition of Fock states in new basis"?

13. Dec 7, 2015

zonde

Look at my posts #3 and #5
Vanhees said these statements are correct.
So I am thinking with these statements on mind and this "new basis" is the one in which we perform measurement.

14. Dec 8, 2015

zonde

I would like to point out one important thing about transformation like this.
There are two ways how we can perform transformation. We can rotate the vector or we can transform the space so that vector appears rotated. It is possible that the space have such a symmetry that the two cases are mathematically equivalent and there is no way how to make them distinct. But ... if dimensions of the space are "anchored" on some physical background the distinction will become apparent and one transformation will have to be considered physical and the other one just a change of representation.
So when we the vector is expressed in different basis it is legitimate to ask if this transformation is physical or not.

If you are saying that transformation is not physical can you provide reasons why it should be considered only the change of representation?
But as I see polarizers can be physically rotated (in respect to some physical background) and if we want to represent state vector in measurement basis of polarizer then this transformation to new basis should be considered physical as well.