# Compton-scattering = Entanglement ?

1. Feb 12, 2005

### Edgardo

Hi all,

1) does the Compton-scattering produce an entangled state?
That is, if I measure the energy of the photon, the energy of the
electron is immediately known and vice versa.

2) Can the photon after scattering be considered as a superposition
of energy-states?

-Edgardo

2. Feb 12, 2005

### dextercioby

Since the 4momentum is conserved,i would say that knowing one energy,automatically determines the other...
As for second point,why...?The energy is known precisely from the energy-momentum conservation law...

Daniel.

3. Feb 19, 2005

### Edgardo

My thought was the following:
Before scattering, 'everything' is known, that is photon energy (wavelength).
But right after scattering, we don't know the photon's energy. There's a certain probability that it will be scattered at $\theta_{1}$
with energy $E_{1}$, scattered at $\theta_{2}$
with energy $E_{2}$ etc.

a) So could I describe the photon's state like the following?

$|\Psi \rangle = \sum c_{k} |k \rangle$

b) When is the photon's energy determined?
Right after scattering or when we measure the energy?

4. Feb 19, 2005

### dextercioby

But let's not forget the V-th principle of QM:we measure the energy of the outgoing photon and force his state vector to collapse.So even if,a priori,you can speak of a entangled state,one the measurement is performed,things are definitive...

Daniel.

5. Feb 22, 2005

### Edgardo

Yes yes, I know that QM-axiom.
But in your opinion, is the energy (for the Compton-scattering) determined before measurement?

6. Feb 22, 2005

### dextercioby

Of course not.

Daniel.

7. Feb 22, 2005

### Edgardo

Would you then say that the photon's state can be written
as a superposition of energy-states as I described above?

8. Feb 22, 2005

### dextercioby

Sure,before the measurement it is in an entangled state...I thought we had this cleared.

Daniel.

9. Feb 23, 2005

### Edgardo

Hello dextercioby,

do you know other examples where the photon's state
can be written as a superposition of energy-states?
and the question came up whether such a superposition for a single photon exists.

10. Feb 23, 2005

### dextercioby

We have pure quantum states and mixed quantum states.A pure state wan be seen as an uniparticle state (photon with momentum "k" and helicity eigenvalue Delta),or a multiparticle state (a state in the tensor product of Hilbert spaces of uniparticle states).As for mixed states,they are (normalized) combinations of pure quantum states (for a photon a (normalized) combination of uniparticle states).This last part (statistical enesemble of photons) assumes entanglement and measurement of energy/helicity eigenvalue would collapse the state vector into the one corresponding to the measured eigenvalue/spectral value.

Daniel.

11. Feb 23, 2005

### Edgardo

I am not that expert in QM :rofl:

I was rather asking for simple examples like Compton-effect.

12. Feb 23, 2005

### dextercioby

In told u,up until measurement,the energy of the photon could be any possible value allowed by the energy conservation law for the $e^{-}\gamma$ scattering.Once u know the final energy of the (outgoing) photon,then u can compute physical observables,as the differential (and then integral) cross-section.

Daniel.