Is CPT symmetry still valid for macroscopic physics?

[Mordred]

Is the optical isolator you have there applying the Faraday rotator ? If so your going to need to account for the 45 degree polarity shift.

Edited June 23, 2023 by Mordred

[swansont]

8 hours ago, Duda Jarek said:

However, in perspective after CPT symmetry the targets and equations would switch - hence emission formula should also symmetrically apply to the target on the left - stimulate its deexcitation (if satisfying condition of being excited).

The stimulated de-excitation can happen with the laser causing the excitation. It’s why a two-level system can’t have more than 50% of the atoms in an excited state.

And since de-excitation requires that the atom be in an excited state, the situations aren’t the same.

[Duda Jarek]

The polarization might be interesting to try to use it to stimulate deexcitation of one polarization of atoms, what might allow e.g. for CPT analogue of state preparation for quantum computing.

The basic test is for nonpolarized target: if the formula on the left applies also to the target on the left, as suggested by CPT symmetry switching the targets. The main difficulties of such test is requirement of high N_2/N excited population, and that the asymmetry is imperfect - some percentage of photons travel in the opposite direction, exciting our target.

 

1 minute ago, swansont said:

The stimulated de-excitation can happen with the laser causing the excitation. It’s why a two-level system can’t have more than 50% of the atoms in an excited state.

And since de-excitation requires that the atom be in an excited state, the situations aren’t the same.

You are writing about the central target (pumped crystal) - to which both equations apply. The question is about the left/right targets - CPT symmetry suggests both should have corresponding formula.

Edited June 23, 2023 by Duda Jarek

[swansont]

54 minutes ago, Duda Jarek said:

You are writing about the central target (pumped crystal) - to which both equations apply. The question is about the left/right targets - CPT symmetry suggests both should have corresponding formula.

There’s nothing to put the target in an excited state. How does stimulated emission happen?

[Duda Jarek]

To test application of the emission formula for the target on the left, it needs to be initially excited - e.g. a lamp.

In this case it would deexcite in isotropic way - the question to test is if there is additional deexcitation caused by the laser, what would be seen by detectors around: in this case getting a lower power of light.

[swansont]

17 minutes ago, Duda Jarek said:

To test application of the emission formula for the target on the left, it needs to be initially excited - e.g. a lamp.

In this case it would deexcite in isotropic way - the question to test is if there is additional deexcitation caused by the laser, what would be seen by detectors around: in this case getting a lower power of light.

Then you could get stimulated emission, but there’s no need to apply CPT to get it, and you’re changing the conditions of the experiment, so you’ve removed the existing symmetry - it’s a different experiment. Basic examples of CPT I’ve seen don’t do this.

It’s possible an issue here is that AFAIK time-reversal symmetry is not the same thing as the thermodynamic arrow of time; an entropy increase does not violate CPT, even though such a process is not reversible.

[Duda Jarek]

Indeed stimulate the target to emit/deexcite, like pulling photons out of it - suggested by symmetry (T/CPT) ... if possible could lead to lots of new applications, e.g.:

- new calculation possibilities e.g. for quantum computing,

- low probability nuclear transitions - if they produce some characteristic gammas, then they could be stimulated by such caused deexcitation - these high energy photons are available e.g. in free electron lasers, wigglers/undulators in synchrotron ... also in standard laser setting using above Einstein's equation. Which nuclear transitions would be the most interesting, practical?

- similarly for chemistry - probably useful for many technological processes ...

[Mordred]

While I still feel its highly unlikely to get any CPT violations with the device. Its a nice change to discuss some serious physics. A couple of details to consider. CPT of photons tie into U(1) gauge symmetry. However it also ties in rather closely to Lorentz invariants. In answer to the question "has macroscopic tests been performed for CPT violations" then the answer in light of Lorentz symmetry and its connections with CPT would become Yes we have. 

This article is an example of tests performed on the International space station utilizing atomic clocks.

https://arxiv.org/abs/hep-ph/0306190

this is the reference 14 of the above paper

https://arxiv.org/pdf/hep-ph/0306190.pdf

I chose this one simply due to its uniqueness. More commonly known methods involve synchrotrons, Penning traps, etc

Here for example is the Zeus detector results. 

https://arxiv.org/pdf/2212.12750.pdf

A synchrotron test 

https://arxiv.org/abs/0905.4346

You will likely find the information handy if anything it may grant some further ideas on how to modify your planned detector. As you can now include Lorentz invariants. Mathematically these papers use SME which which is the standard model extension to include violating terms.

this article has a decent listing of the Langrangian for the major ones

https://arxiv.org/pdf/hep-ph/0506054.pdf

hope this helps, if anything its informative to other readers as well. Granted that also opens a few doors in what to look for with your experiment. 

This article will help get a handle on polarizations which you will need with your setup

https://scholar.harvard.edu/files/schwartz/files/lecture14-polarization.pdf

As it is informative I will add some mathematical relations that you may or may not choose to use but are of use.

Useful relations 

Parity Many textbooks describe this as a mirror reflection its not quite accurate but its a useful analogy. Its more useful to use an inversion of coordinates for parity difference described below

reflection of the x-y plane

\[(x,y,z)\longrightarrow(z,-y,z)\]

inversion (combines the reflection with a 180 degree rotation) the reason this is easier to work with is you don't need to choose a mirror plane.

\[(x,y,z)\longrightarrow (-x,-y,-z)\]

\(\mathbb{P}:(t,x,y,z)\longrightarrow(t,-x-y-z)\Rightarrow \mathbb{P}(\vec{V}=-\vec{V}\)

gives

\[\mathbb{P}((\vec{V})\cdot(\vec{W})=\mathbb{P}(\vec{V})\cdot(\vec{W})=(-\vec{V})-\cdot(\vec{W})=\vec{V}\cdot\vec{W})\]

\[\mathbb{P}((\vec{V})\times(\vec{W})=\mathbb{P}(\vec{V})\times(\vec{W})=(-\vec{V})\times(\vec{W})=\vec{V}\times\vec{W})\]

above denotes changes from left to right hand coordinate systems.

note two types of vectors those that reverse signs under parity (vector V and those that don't pseudovector A

gives

\(\mathbb{P}:\vec{V}\times\vec{A}=-\vec{V}\times\vec{A}=-\vec{V}\times\vec{A}=\) a vector

\(\mathbb{P}:\vec{V}\cdot\vec{A}=-\vec{V}\cdot\vec{A}=-\vec{V}\cdot\vec{A}=\) a pseudovector

with twice applied parity \(\mathbb{P}^2=\mathbb(I)\) eugenvalues of parity plus or minus 1. scalars and pseodovectors have eugenvalue +1, while pseudoscalars and vectors have eugenvalue -1.

with GR parity has the sign convention (+1,-1,-1,-1) 

\[(\Lambda_\mathbb{P})^\mu_\nu=\begin{pmatrix}1&0&0&0\\0&-1&0&0\\0&0&-1&0\\0&0&0&-1\end{pmatrix}\]

under Lorentz transformation

\[g_{\alpha\beta}(\Lambda_\mathbb{P})^\alpha_\nu (\Lambda_\mathbb{P})^\beta_\mu=g_{\mu\nu}\]

by examining the photon in a hydrogen atom we can find the parity of a photon.

this is too lengthy so you can get the gist here

https://en.wikipedia.org/wiki/Spherical_harmonics  of course QED rules further include the above. The photon for energy conversation has parity -1

 

Edited June 23, 2023 by Mordred

[Duda Jarek]

CPT violation would be not seeing this looking unknown effect required by CPT symmetry - that as we "push photons" to the target on the right, we should also symmetrically "pull photons" from the target on the left - because from perspective after CPT transform, we would "push photons" to this target.

I also suspect this effect is true, but it needs experimental confirmation - if positive, it would open lots of new possibilities, applications. E.g. in case of low probability nuclear transition producing characteristic gammas, we could "pull them" this way e.g. with free electron laser - hopefully stimulating this nuclear transition.

In case of negative result: preparing setting able to see effect on expected level but not seeing it, would mean violation of CPT symmetry on macroscopic level - that macroscopic setting does not work the same in perspective after CPT symmetry. Such credible negative result should also inspire further experiments - improving on potential limitations, searching for the scale where CPT starts being violated and its mechanisms.

Edited June 23, 2023 by Duda Jarek

[Mordred]

For photon beams you would want the current densities of the polarity related wavefunctions for both both quantum and macroscale regimes. LOL you also need these relations for your readouts on your test equipment truth be told. We do push the umbrella at looking for CPT at respectable energy levels and are always pushing for test methodologies at higher. In point of detail its also looked for in our highest energy level test equipment such as our particle accelerators. Lots of ongoing research is still looking for new methods to test for it of course.

Edited June 23, 2023 by Mordred

[Duda Jarek]

From perspective after CPT symmetry, you would cause excitation of atoms of target on the left, what means causing deexcitation in perspective without CPT - sure polarized beam would mean pumping/unpumping one polarity, there should be also unpolarized settings, e.g. using free electron laser, wiggler/undulator, synchrotron radiation.

[swansont]

10 hours ago, Duda Jarek said:

Indeed stimulate the target to emit/deexcite, like pulling photons out of it - suggested by symmetry (T/CPT) ... if possible could lead to lots of new applications

Seems to me the time reversal of a photon absorption would be an excited atom emitting a photon. Not stimulated emission. 

[Duda Jarek]

Mathematically, we have the two equations (from https://en.wikipedia.org/wiki/Stimulated_emission#Mathematical_model ), after CPT they would be switched: the right one (absorption equation) would apply to the target on the left, what without CPT means the emission equation should apply to the target on the left.

In other words, while laser causes absorption by target on the rights ("push photons" there"), CPT symmetry says it has to also cause emission by target on the left ("pull photons" from there) - if only N_2 > 0: satisfied condition of being excited in the first place, e.g. lamp.

To my knowledge, existence of such effect was not tested yet (?) - negative result would mean macroscopic violation of CPT symmetry, positive would lead to lots of new possibilities/applications.

Edited June 24, 2023 by Duda Jarek

[swansont]

11 hours ago, Duda Jarek said:

Mathematically, we have the two equations (from https://en.wikipedia.org/wiki/Stimulated_emission#Mathematical_model ), after CPT they would be switched: the right one (absorption equation) would apply to the target on the left, what without CPT means the emission equation should apply to the target on the left.

In other words, while laser causes absorption by target on the rights ("push photons" there"), CPT symmetry says it has to also cause emission by target on the left ("pull photons" from there) - if only N_2 > 0: satisfied condition of being excited in the first place, e.g. lamp.

How does the lamp appear in the problem? Specifically, how do you get it from a time reversal, charge or parity transform?

 

 

 

 

[Duda Jarek]

To observe the dN_2/dt = -B rho N_2 evolution, we need N_2 > 0 initial excitation of the target - e.g. gas discharge lamp.

Its continuously excited atoms normally deexcite in isotropic way, the laser should additionally increase probability of directional deexcitation - what would be seen by detectors around, watching isotropic radiation, by reduction of seen light intensity.

[swansont]

45 minutes ago, Duda Jarek said:

To observe the dN_2/dt = -B rho N_2 evolution, we need N_2 > 0 initial excitation of the target - e.g. gas discharge lamp.

Yes, but what does that have to do with CPT symmetry, when the setup isn’t symmetric?

 

[Duda Jarek]

The CPT theorem says that looking at setting from perspective after CPT symmetry, it should be governed by the same physics.

For the discussed setting with asymmetric light source, in perspective after CPT symmetry the two target would be switched, laser would increase the number of excited atoms in the target on the left - toward negative time, what means decrease toward positive time.

In other words, it should stimulate emission from this target - not only when it is the central pumped crystal, but also when it is shifted left.

[swansont]

9 hours ago, Duda Jarek said:

The CPT theorem says that looking at setting from perspective after CPT symmetry, it should be governed by the same physics.

For the discussed setting with asymmetric light source, in perspective after CPT symmetry the two target would be switched, laser would increase the number of excited atoms in the target on the left - toward negative time, what means decrease toward positive time.

In other words, it should stimulate emission from this target - not only when it is the central pumped crystal, but also when it is shifted left.

CPT is a set of transforms. x —> -x, q —> -q and t —> -t

A photon causes an excitation in an atom would become an excited atom emitting a photon.

If you have to add lamp, you don’t even have the same system, so any application of CPT is irrelevant. 

[Duda Jarek]

The T transform here reverses photon direction, making the laser cause excitation of target on the left, what means causing deexcitation without T symmetry.

Look at the two equations (from https://en.wikipedia.org/wiki/Stimulated_emission#Mathematical_model ) - to which of 3 targets they apply?

The equation on the right applies to the central and right target.

The equation on the left applies to the central target - my point is it also symmetrically applies to the target on the left - because in perspective after CPT symmetry the equations are switched.

[Mordred]

I mentioned earlier that the tests for CPT and Lorentz invariance are extensive as well as to high precision.

To give a better idea here is a collective of the relevant datatables. Tables D15 and D16 apply to photons. These tables are a bit tricky as there is 44 total parameters 20 for CPT related. These additional parameters involve extensions to describe mathematically the relevant violation.

https://arxiv.org/abs/0801.0287

1 hour ago, Duda Jarek said:

The T transform here reverses photon direction, making the laser cause excitation of target on the left, what means causing deexcitation without T symmetry.

Look at the two equations (from https://en.wikipedia.org/wiki/Stimulated_emission#Mathematical_model ) - to which of 3 targets they apply?

The equation on the right applies to the central and right target.

The equation on the left applies to the central target - my point is it also symmetrically applies to the target on the left - because in perspective after CPT symmetry the equations are switched.

Changing the photon direction is not the same thing as time reversal symmetry involved in the photon/antiphoton symmetry relations. Stimulated emissions involve momentum transfer. Yes you should have Lorentz symmetry in the momentum transfer however the time symmetry in Lorentz has distinctions from CPT symmetry which involve the Dirac equations.

Lets do a simple thought experiment take a photon/antiphoton pair. The only difference is the polarity relations and the photon is its own antiparticle. Left hand,  right polarizations. They both have the same energy/momentum so the transfer of momentum should be the same when interacting with an atom.

We should be clear which time reversal symmetry are we testing for 

Edited June 25, 2023 by Mordred

[swansont]

3 hours ago, Duda Jarek said:

The T transform here reverses photon direction, making the laser cause excitation of target on the left, what means causing deexcitation without T symmetry.

Look at the two equations (from https://en.wikipedia.org/wiki/Stimulated_emission#Mathematical_model ) - to which of 3 targets they apply?

The equation on the right applies to the central and right target.

 

I don’t see an equation “on the right” and I see more than two equations in that link

If you’re referring to the Einstein rate equations, that look like dN/dt = BN, please show what time reversal does to these equations

 

Quote

The equation on the left applies to the central target - my point is it also symmetrically applies to the target on the left - because in perspective after CPT symmetry the equations are switched.

The target on the left doesn’t get any photons. Why do you think it gets photons under time reversal?

[Duda Jarek]

Turns out there is considered "negative radiation pressure" - predicted for solitons, searched e.g. for mechanical waves on graphene: https://scholar.google.pl/scholar?q=negative+radiation+pressure ... so the question is if it could be realized with EM waves, photons using lasers?

On 6/25/2023 at 8:18 PM, swansont said:

The target on the left doesn’t get any photons. Why do you think it gets photons under time reversal?

Because believing in CPT symmetry, from its perspective photon trajectories would be reversed, the target on the left would be "standard target" to which we "push photons" (absorption equation), what from perspective without CPT would mean emission equation, "pulling photons", "negative radiation pressure".

 

[Duda Jarek]

Also, pulling with photons is done e.g. by https://en.wikipedia.org/wiki/Optical_tweezers

EM radiation pressure is <E x H>/c ( https://en.wikipedia.org/wiki/Radiation_pressure#Radiation_pressure_from_momentum_of_an_electromagnetic_wave ) - doesn't have to be positive.

[swansont]

6 hours ago, Duda Jarek said:

Because believing in CPT symmetry, from its perspective photon trajectories would be reversed, the target on the left would be "standard target" to which we "push photons" (absorption equation), what from perspective without CPT would mean emission equation, "pulling photons", "negative radiation pressure".

You have asserted this, but I don’t see how you get there from transforming x —> -x

That transform would flip the experiment - its mirror image. IOW, I don’t think you’ve done the P transform properly. Repeating your conjecture is not sufficient.

[Duda Jarek]

From perspective after CPT symmetry the photons would travel toward CPT(target on the left) - if physics works the same after this symmetry, toward "minus time" this target would be excited accordingly to equation on the right, what toward "plus time" means it would be deexcited accordingly to the equation on the left ... if only it was excited in the first place like lamp: N_2 > 0.

As ring laser causes "positive radiation pressure" in one direction, from perspective after CPT it means causing "negative radiation pressure".