Superposed photon hitting a polarizing beamsplitter

[jelrod45]

Hey everyone,

This is my first post, but I have been here a lot looking up things for homework problems in some math and science courses. I love the community here, so here I am! I am a computer science major, and know a bit about physics, but my expertise is in software engineering, so please excuse any incorrect assumptions / elementary mistakes I may have made/thought. I have a wonderful idea, and have asked professors/colleagues and have not been able to find a definite answer. So I am turning here.

Dealing with polarity of photons, take horizontal polarity to be <1,0> and vertical polarity to be <0,1> (looking at the photon moving straight at/away from you).

Say you have a polarizing beamsplitter, which to my understanding will transmit one polarization of a photon and reflect the opposing polarization. Given this definition of a polarizing beamsplitter, it seems like what would logically happen if a photon with polarity <1,1>/√2 were to hit a polarizing beamsplitter, the photon would have 1/2 chance of being transmitted and 1/2 chance of being reflected.

If this is correct, then it also seems like this case would be analogous to the famous double slit experiment. Would it be correct to interpret the activity of the photon as being both transmitted and reflected until measured after the beamsplitter? More explicitly, would the activity of the beamsplitter collapse the superposition of the polarity of the photon immediately, or would subsequent activity applied to the photon not be "applied" until a measurement decides whether the photon is transmitted or reflected.

I am considering writing a paper on optical computing that is dependent on such activity, and definitely need to know the ins and outs of all of this before choosing this as my topic.

I will be glad to clarify any ambiguities in the question and am looking forward to hearing what you all think!

Thanks for your help!

 

[Fightfish]

Hi, Welcome to PF :)

Would it be correct to interpret the activity of the photon as being both transmitted and reflected until measured after the beamsplitter?

Yup, in fact the photon can be made to interfere with itself!

This is the quantum version of the Mach-Zehner inferometer.

 

[jelrod45]

That is excellent! I thought that this would be the case, but a few people in one of my classes weren't sure about it, so I needed to get a more qualified opinion. This is super exciting. Thank you for your answer!