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Expanding over R and L polarisations

  1. Jul 5, 2012 #1
    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
     
  2. jcsd
  3. Jul 8, 2012 #2
    Anyone?
     
  4. Jul 8, 2012 #3
    What difficulties have you encountered in expanding this tensor product?
     
  5. Jul 8, 2012 #4
    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.
     
  6. Jul 8, 2012 #5
    I don't see how your second sentence (complex number problem) logically follows your first one.
     
  7. Jul 8, 2012 #6
    |HH> - |VV>
    =(R + iL)(R + iL) - (R - iL)(R - iL)
    = RR - RR + iLL - iLL + RiL + RiL ??
     
  8. Jul 8, 2012 #7
    R=H+iV
    H=(R+L)/2
     
  9. Jul 10, 2012 #8
    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?
     
  10. Jul 11, 2012 #9
    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

    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).
     
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