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In this Review of modern physics, section "II. A.", optical homodyne detection is briefly reviewed. In equation (9), it is stated that the fields after the beam splitter are

$$E_{1}=\frac{E_\text{L}+E_\text{S}}{\sqrt{2}},$$ $$E_{2}=\frac{E_\text{L}-E_\text{S}}{\sqrt{2}}.$$

However, since the setup is perfectly symmetrical (see Fig. 1), I don't understand why the fields are not

$$E_{1}=\frac{-E_\text{L}+E_\text{S}}{\sqrt{2}},$$ $$E_{2}=\frac{E_\text{L}-E_\text{S}}{\sqrt{2}}.$$

Can anyone shed some light on it? As I understand it, the whole point of homodyne detection is this difference in sign.

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  • $\begingroup$ physics.stackexchange.com/questions/387731/… Does this solve your question? $\endgroup$ – Maxim May 7 at 15:55
  • $\begingroup$ @Maxim Yes, for beam splitters like this! What about fiber beam splitters? $\endgroup$ – nabla May 7 at 19:31
  • $\begingroup$ The beam splitter needs to be represent by a unitary matrix because it maintains that normalization of the input state. That still allows one different ways to represent it, because one can always multiply the individual input or output ports by arbitrary phase factors. One way to represent it is $\endgroup$ – flippiefanus May 9 at 4:12
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The beam splitter needs to be represented by a unitary matrix because it maintains that normalization of the input state. That still allows one different ways to represent it, because one can always multiply the individual input or output ports by arbitrary phase factors. One way to represent it is $$ U_{bs} = \frac{1}{\sqrt{2}} \left(\begin{array}{cc} 1 & 1 \\ -1 & 1 \end{array}\right) $$ which would correspond to the first set of equations you provide.

If you prefer a more symmetric matrix, then you can use $$ U_{bs} = \frac{1}{\sqrt{2}} \left(\begin{array}{cc} 1 & i \\ i & 1 \end{array}\right) .$$ The second set of equations you provide does not correspond to a unitary process.

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