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Beam splitter energy conservation

  1. Oct 1, 2011 #1

    If we describe a beam splitter as follows:

    [tex]e^{ikx} -> \sqrt{T}e^{ikx} + \sqrt{R}e^{i\theta}e^{iky}[/tex]
    [tex]e^{iky} -> \sqrt{T}e^{iky} + \sqrt{R}e^{i\theta'}e^{ikx}[/tex]

    then [itex]\theta+\theta'=\pi[/itex] is a condition to ensure conservation of energy according to my text.

    I tried working this out by taking [itex]Ae^{ikx}+Be^{iky}[/itex] incident on a beam splitter. The incident energy is [itex]A^{2}+B^{2}[/itex].

    The output is

    [itex]A\sqrt{T}e^{ikx} + A\sqrt{R}e^{i\theta}e^{iky} +B\sqrt{T}e^{iky} + B\sqrt{R}e^{i\theta'}e^{ikx}[/itex]

    Its energy is [itex]A^{2} + B^{2} + AB\sqrt{TR}(e^{i\theta}+e^{-i\theta})+AB\sqrt{TR}(e^{i\theta'}+e^{-i\theta'})[/itex]

    So, to preserve conservation, we must have [itex]2cos(\theta)+2cos(\theta')=0[/itex]

    That gives [itex]\theta+\theta'=\pi[/itex] or [itex]\theta -\theta'=\pi[/itex]. But I never see this second result anywhere. Why is it there and how is it eliminated?

    Thank you
  2. jcsd
  3. Oct 2, 2011 #2
    Anyone? Strange thing is that I have the answer but an extra solution which should be invalid. Thank you
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