Compton-scattering = Entanglement ?

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

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?
 
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.
 
Yes yes, I know that QM-axiom.
But in your opinion, is the energy (for the Compton-scattering) determined before measurement?
 
Would you then say that the photon's state can be written
as a superposition of energy-states as I described above?
 
Sure,before the measurement it is in an entangled state...I thought we had this cleared.

Daniel.
 
Hello dextercioby,

do you know other examples where the photon's state
can be written as a superposition of energy-states?
I'm asking because we just recently had a discussion about white photons,
and the question came up whether such a superposition for a single photon exists.
https://www.physicsforums.com/showthread.php?t=62946&page=5&pp=15
 
  • #10
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
Argh! Please not so complicated :cry:
I am not that expert in QM :smile:

I was rather asking for simple examples like Compton-effect.
 
  • #12
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.
 
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