# I Pair production proximity

1. Oct 11, 2016

### God Plays Dice

Does anyone know how close a photon needs to be to a nucleus (an ion really, no shielding from electrons) for pp to occur? I assume it's a probability as a function of distance, any ideas/equations?

Thanks

2. Oct 11, 2016

### Staff: Mentor

Photons do not have a well-defined position.

There is a cross-section that depends on the nucleus and the photon energy, but don't interpret that as classical area.

3. Oct 12, 2016

### snorkack

Does a photon have a definable (and quantized) angular momentum of its translational angular momentum past a target?

4. Oct 12, 2016

### Zypheros_Knight

Did you say pp on a bare nucleus with a photon? Good luck getting a photon of >2 GeV energy!!! and for the second part, charge conjugation occurs when the photon ( I seriously suggest a high-energy proton for this) enters the coulomb field of a nucleus (or another proton). The probability of pp increases with increasing energy from the threshold energy level, the equation that governs this : σ(pair E)≈αZ2r(electron)2InEγ

5. Oct 12, 2016

### vanhees71

Photons are special since they are massless. There's a unique gauge invariant definition of total angular momentum only. The split into orbital and spin-angular momentum is arbitrary and gauge dependent and thus unphysical.

There are different complete one-photon bases. One is the momentum-helicity basis. These states are the usual plane-wave states of circular-left ($\lambda=-1$) and circular-right ($\lambda=+1$) polarization. Of course you could choose for each momentum any other polarization basis like linear polarized states etc.

Another one is in terms of the usual multipole expansion, which is a energy-total-angular-momentum eigenbasis.

This is all pretty similar to the free classical electromagnetic waves, only that in the corresponding Fourier decomposition the coefficients be come annihilation and creation operators for the corresponding photon states, with help of which you can build the Fock states (states with definite photon number).

As a massless particle of spin 1 a photon by definition has no position in the literal sense, because there's no position operator that has the usual Heisenberg commutation relations with momentum.

6. Oct 12, 2016

Staff Emeritus
This is not a problem.

All these messages are focusing on the same thing - getting a well-defined photon position is problem, and it becomes a bigger problem when you impose innocuous-seeming additional conditions (like 'exactly one'). What is well-defined, however, is the intersection of the electron and positron trajectories. That might be a sensible proxy.

7. Oct 12, 2016

### Staff: Mentor

I read "pp" as "pair production", not "proton antiproton", but it does not matter. Multi-GeV photons (and even a few photons with more than 1 TeV, see e.g. this recent ATLAS publication) are routinely produced at colliders like the LHC.
What does "charge conjugation occurs" mean? Charge conjugation is a mathematical operation. When does "addition of two numbers" occur in collision processes?
I would be surprised if that works for hadron production.

8. Oct 13, 2016

### Zypheros_Knight

Its okay , and yes the equation is for pair-production of electron which, if I remember distinctly ,I mentioned in the equation. And yes again I forgot to mention both the equations for Hadron pp and that charge conjugation operator changes the signs of all quantum charges (dammit). Nevertheless I will get better, I'm still in the process of healing.
PS: That article is nice! Keep me posted on latest stuff like this!

9. Oct 13, 2016

### Staff: Mentor

10. Oct 13, 2016

### ChrisVer

11. Oct 14, 2016

### Zypheros_Knight

Thank you very much! I've read about ALICE before, oh and have you read about AWAKE? Its a new method for accelerating charged particles!!( I'm talking about Wakefield Acceleration).

12. Oct 15, 2016

### snorkack

So, in a frame where angular momentum is defined with respect to the nucleus, you can define and measure:
spin angular momentum of photon (from state of polarization)
spin angular momentum of initial state of nucleus

If the end result is a single nucleus (excited state), can the distance between photon and nucleus thereby be defined?

13. Oct 16, 2016

### vanhees71

There's no way to split the total angular momentum of the em. field into spin and orbital angular momentum. Only total angular momentum is well-defined and gauge invariant. There is also no way to define a position observable for a photon (or any massless particle with spin $\geq 1$).

14. Oct 16, 2016

### ShayanJ

You mean its possible for massless spin-$\frac 1 2$ particles?

15. Oct 16, 2016

### vanhees71

16. Oct 16, 2016

### Staff: Mentor

Unfortunately we don't know any massless particles with spin < 1. Well, one neutrino mass eigenstate could be massless, but that would be odd.

17. Oct 16, 2016

### vanhees71

True, there's no known massless particle except the photon (note that gluons are confined in hadrons and most likely also glue balls; they never occur as asymptotic free states, as which they were also massless).

18. Oct 16, 2016

### snorkack

Does the circular polarization of electromagnetic field depend on observer?

19. Oct 17, 2016

### vanhees71

What do you mean by that? Circular polarizations are states of good helicity, and that's a Lorentz-invariant quantity (for massless fields, as the em. field is!). In that sense it's a covariant charcterization of the field and thus observer independent.

20. Oct 17, 2016

### snorkack

So, if you have a photon whose circular polarization/helicity is known, do you also know its spin?

21. Oct 18, 2016

### vanhees71

A photon has helicity, not spin. As I said, massless particles are special. One says a photon as "spin 1". This labels the representation of the rotation group for the field, i.e., $J=1$. In other words you have a vector field. However, since the photon is massless, it has not $2J+1=3$ independent but only 2 (e.g., helicity $h=\pm 1$, corresponding to right- and left-circularly polarized em. fields).

This indicates another complication. Since you want not only a theory that is rotation invariant but a local Poincare invariant theory, you need at least a four-vector field, $A^{\mu}(x)$ to describe the electromagnetic field, i.e., four field degrees of freedom. One degree of freedom is due to the fact that a four-vector field doesn't describe only spin-1 fields but also a scalar field (under rotations). You can build various scalars under rotations. A particularly convenient one is $\partial_{\mu} A^{\mu}$, and you can impose a constraint $\partial_{\mu} A^{\mu}=0$. For a massive field that would be all you have to do to get the 3 physical field degrees of freedom for a spin-1 field. However massless fields are special, and the representation theory of the Poincare group tells you that necessarily the massless spin-1 field must be represented by a gauge field. That means that a physical situation is described not only by one vector field but any vector field $A^{\prime \mu}=A^{\mu}+\partial^{\mu} \chi$ with an arbitrary scalar field $\chi$ describes exactly the same situation. This means in addition to the constraint to have only vector fields but no scalar field component you can impose another constraint. This can be achieved by "fixing the gauge completely". In order to fulfill the above Lorentz invariant constraint for both $A^{\mu}$ and $A^{\prime \mu}$ you can choose any $\chi$ fulfilling $\partial_{\mu} \partial^{\mu} \chi=\Box \chi=0$. For the free em. field a clever choice is to impose the additional constraint that $A^0=0$ (radiation gauge). Then you have
$$A^0=0, \quad \vec{\nabla} \cdot \vec{A}=0,$$
and thus you have only 2 independent field-degrees of freedom. A field with "good momentum" is a plane wave, and the corresponding field mode is $$\vec{A}=\vec{A}_0 \exp(-\mathrm{i} \omega t+\mathrm{i} \vec{k} \cdot \vec{x})$$. The radiation-gauge constraint then simplifies to $\vec{k} \cdot \vec{A}_0=0$, which tells you that there are only 2 independent modes left, namely the ones perpendicular to the momentum $\vec{k}$. As it should be the free em. field is a transverse wave.

22. Oct 18, 2016

### snorkack

There are several things that can happen when a photon interacts with a nucleus:
1. Absorption. Initial state - a nucleus of known spin and orbital momentum defined as zero, plus a photon with known helicity and uncertain location. Final state - a nucleus of known spin and orbital momentum defined as zero. Sounds like photon location is the only unknown here.
2. Photonuclear reaction. Initial state - the same. But a final state - two nuclei of known spins, but uncertain location relative to each other.
3. Pair production. Initial state - the same. But a final state - a nucleus of known spin, plus two fermions, with knowns spins, but both with uncertain location relative to each other.
In case of photon absorption, it is possible to define multipole radiation/absorption, because the orbital angular momentum of the photon is the only unknown. But due to presence of other unknown orbital angular momenta, is it possible to define a "multipole" photonuclear reaction or pair production?

23. Oct 19, 2016

### Zypheros_Knight

Yes, the circular polarization weather left-handed or right-handed will tell us about the directions of spin of a photon.

24. Oct 20, 2016

### vanhees71

Again, while helicities have a well-defined meaning for massless particles (the projection of the total (!) angular momentum to the direction of the particle's momentum), there's no well-defined way to split the total angular momentum into spin and orbital parts (see also my posting #21 above).