Are Input Photons #2 and #3 Still Entangled After Going Through a Beam Splitter?

[StevieTNZ]

In regards to the attached image;

If the input photons were, for input b = photon #2, and for input c = photon #3, where:
photon #1 and #2, and #3 and #4 are entangled as |H>|V> + |V>|H>,

for the outputs:|H>(b’)|H>(b’) - |V>(b’)|V>(b’) + |H>(c’)|H>(c’) - |V>(c’)|V>(c’) + H(b’)|V>(c’) - |V>(b’)|H>(c’) (Not sure if I got the + and – signs correct?)

in the case of getting the outputs |H>(b’)|H>(b’) - |V>(b’)|V>(b’) + |H>(c’)|H>(c’) - |V>(c’)|V>(c’), would photons #2 and #3 be entangled with each other? Or would they remain entangled with their original partner?

So up until the actual detection of H(b’)|V>(c’) - |V>(b’)|H>(c’) (one specific bell-state) - or once that basis state is created when the photons entangle with the final detectors at the end output - the photons remain entangled with the original partner?

would the above be the same if photons #1 and #2, and #3 and #4 were entangled as |H>|V> - |V>|H>?

 

[mfb]

You send one photon of each entangled pair through a (non-polarizing) beam splitter?
Why don't you get parts like H(b’)>|V(b’)>?

Why don't you look at each photon individually? If they are not entangled in some way, I would not expect anything special from using both at the same time.

 

[StevieTNZ]

You send one photon of each entangled pair through a (non-polarizing) beam splitter?

And they reach the beam splitter at zero delay.

Why don't you get parts like H(b’)>|V(b’)>?

Well, I don't know. Perhaps you can shed some light on this?

 

[mfb]

Well, you left them out in your state, and I ask why you did this.

 

[StevieTNZ]

My guess is they get canceled out.