Expanding over R and L polarisations

[StevieTNZ]

We expand over, in the H/V basis, |RR> + |LL>, likewise |RR> - |LL>. Because R = (H + iV) and L = (H - iV), how do we factor in the complex number to obtain the final result?

So:
1. (H + iV)(H + iV) + (H - iV)(H - iV) and
2. (H + iV)(H + iV) - (H - iV)(H - iV)

Thanks,
Stevie

 

[StevieTNZ]

Anyone?

 

[netheril96]

What difficulties have you encountered in expanding this tensor product?

 

[StevieTNZ]

I guess my confusion lies with |HH> - |VV>, which allows RR combinations. That's why I'm asking what to do in cases involving complex numbers.

 

[netheril96]

I guess my confusion lies with |HH> - |VV>, which allows RR combinations. That's why I'm asking what to do in cases involving complex numbers.

I don't see how your second sentence (complex number problem) logically follows your first one.

 

[StevieTNZ]

|HH> - |VV>
=(R + iL)(R + iL) - (R - iL)(R - iL)
= RR - RR + iLL - iLL + RiL + RiL ??

 

[netheril96]

|HH> - |VV>
=(R + iL)(R + iL) - (R - iL)(R - iL)
= RR - RR + iLL - iLL + RiL + RiL ??

R=H+iV
H=(R+L)/2

 

[StevieTNZ]

So H = (R + L) and
V = -i (R - L)

=|H>|V> - |V>|H>
=(R + L) -i(R - L) - -i(R - L)(R + L)

-i = +1? (-i^2)

How do we expand over that?

 

[Delta Kilo]

What seems to be the problem exactly? Just use the bilinearity of ⊗:
(cA)⊗B = A⊗(cB) = c(A⊗B), (A+B)⊗C = A⊗C+B⊗C, A⊗(B+C) = A⊗B+A⊗C

Spoiler

R⊗R + L⊗L=(H + iV)⊗(H + iV) + (H - iV)⊗(H - iV) =
=(H⊗(H + iV) + iV⊗(H + iV)) + (H⊗(H - iV) - iV⊗(H - iV)) =
=H⊗H + H⊗iV + iV⊗H + iV⊗iV + H⊗H - H⊗iV - iV⊗H + iV⊗iV =
=H⊗H + i(H⊗V) + i(V⊗H) -V⊗V + H⊗H - i(H⊗V) - i(V⊗H) - V⊗V =
= 2(H⊗H - V⊗V)
Similarly, RR - LL= 2i (HV + VH).