Calculating Photon Polarization States with PBS & Wave Plates

[StevieTNZ]

I have hopefully what is regarded as simple and straightforward questions.

If we have the attached set up (comprising a source for photons entangled as |H>|V> - |V>|H>), polarizing beam splitters (PBS) and a wave plate that converts |H> to |45> and |V> to |135>.

How does one calculate the state of the partner photon if we measure one photon in the 45/135 basis (via inference using the wave plate and PBS), and the partner in the H/V basis.

My guess is we start with the initial state
|H>|V> - |V>|H>
Convert one photon into the 45/135 basis from the use of the wave plate, thus |45>|V> - |135>|H>

And then have those photons reach the beam splitters: convert |45> back to |H> (which means the photon was measured, in effect, by a 45 degree orientated polarizer and passed it). In this case if we measure the partner photon in the H/V basis, we’d always find it vertically polarized. Is this correct?I’m wondering, though, as even using polarizers instead of combined wave plates and PBS we could have one photon |135> and the other |V>, which isn’t reflected my calculation above. I might be restricting polarizations photons can take on based on the initial state (e.g. one must be horizontal and the other vertical).I believe it is the case that with the photon state |H>|H> - |V>|V>, it is re-written in the 45/135 basis as -|45>|135> - |135>|45>. By using wave plates and PBS (for both photons) to measure in that 45/135 basis, one can come up with results like |H> and |V> - indicating one was |135> and one was |45>. So not necessarily restricted to both needing to be |H>|H> or |V>|V> as per the initial state. Is that right?Many thanks
Stevie

 

[DrClaude]

Is this correct?

Up to this point, yes. (Although I should pedantically point out that you should use normalized states, so add a factor $1/\sqrt{2}$.)

I’m wondering, though, as even using polarizers instead of combined wave plates and PBS we could have one photon |135> and the other |V>, which isn’t reflected my calculation above. I might be restricting polarizations photons can take on based on the initial state (e.g. one must be horizontal and the other vertical).I believe it is the case that with the photon state |H>|H> - |V>|V>, it is re-written in the 45/135 basis as -|45>|135> - |135>|45>. By using wave plates and PBS (for both photons) to measure in that 45/135 basis, one can come up with results like |H> and |V> - indicating one was |135> and one was |45>. So not necessarily restricted to both needing to be |H>|H> or |V>|V> as per the initial state. Is that right?

I don't understand what you are asking. Can you please explain exactly the setup you are considering here?

 

[StevieTNZ]

Thanks for your reply.

The set up in relation to the first part of the second half is using polarizers and photon detectors, really to illustrate how we can come up with a solution (using the same method as my first calculation) that allows for one photon to be |135> polarized, while the other is |V> (rather than find one photon as |45> (inferred from the use of the wave plate and PBS) and the other as |V>).

Regarding the situation with the initial state 1/√2 |H>|H> - |V>|V>, both photons are measured using wave plates and PBS', in the 45/135 basis. In that basis, one photon can be |135> while the other is |45>. By using wave plates and PBS, to indicate such a state, we'd find the photons after the PBS in the state |H>|V> or |V>|H>.

I wonder if that clarifies?

 

[vanhees71]

In this setup the measurement is described by a unitary operator (assuming loss-less wave plates and polarizing beam splitters). Just apply the corresponding operators to the initial biphoton state $1/\sqrt{2} (|HH \rangle - |VV \rangle)$ to see what's the outgoing state after the wave plates and PBSs.