I Beam-splitter transformation matrix

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Summary
Why is the transformation matrix of a beam splitter unitary?
The transformation matrix for a beam splitter relates the four E-fields involved as follows:

$$
\left(\begin{array}{c}
E_{1}\\
E_{2}
\end{array}\right)=\left(\begin{array}{cc}
T & R\\
R & T
\end{array}\right)\left(\begin{array}{c}
E_{3}\\
E_{4}
\end{array}\right)
\tag{1}$$

Here, the amplitude transmission and reflection coefficients are given by ##T=|T|e^{i\theta}## and ##R=|R|e^{i\varphi}##. In various textbooks we are told that for a lossless beam splitter these two quantities are subject to:

$$|T|^2 + |R|^2 =1 \tag{2}$$

I used to believe that this is because ##T^2## and ##R^2## represent transmittance and reflectanace respectively, so in the presence of loss (e.g. absorptance ##A##), one would have ##T^2 + R^2 + A = 1##.

However, according to this, the reason is because the matrix is unitary, which means that we can write:

$$\left(\begin{array}{cc}
T & R\\
R & T
\end{array}\right)\left(\begin{array}{cc}
T^{*} & R^{*}\\
R^{*} & T^{*}
\end{array}\right)=\left(\begin{array}{cc}
1 & 0\\
0 & 1
\end{array}\right) \tag{3}$$

Indeed one of the two relations you get from the above is equation (2). So, how did they know that the transformation matrix must be unitary? Did they assume equation (2) to be true in advance? Or, is equation (2) a consequence of the fact that the matrix is unitary?

Any explanation would be greatly appreciated.
 

vanhees71

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The most simple explanation is energy conservation. Note that the ##E_j## are components of the incoming and outcoming electric field, written in the usual complex notation. A lossless optical device is defined as one where there's no energy loss regarding the em. field, but the energy density is given by
$$u \propto \vec{E}^{\dagger} \cdot \vec{E}.$$
Now in order to have ##u_{\text{out}}=u_{\text{in}}##, the requirement for some optical element to be lossless, with ##\vec{E}_{\text{out}}=\hat{T} \vec{E}_{\text{in}}## you must have for any (!!!) in-field
$$\vec{E}_{\text{out}}^{\dagger} \vec{E}_{\text{out}} = \vec{E}_{\text{\in}}^{\dagger} \hat{T}^{\dagger} \hat{T} \vec{E}_{\text{in}} \stackrel{!}{=} \vec{E}_{\text{in}} ^{\dagger} \vec{E}_{\text{in}}.$$
This can hold only for all (!!!) in-fields if
$$\hat{T}^{\dagger} \hat{T}=\hat{1},$$
which means that ##\hat{T}## must be unitary to be a transfer matrix describing a linear lossless optical component.

For the special beam splitter built with a dielectric reflective coating, it's quite easy to understand also from classical electrodynamics (Fresnel equations for reflection and refraction of em. waves at dielectrica). It's built such that the reflected beam coming from one side gets a phase shift of ##\pi##, while the reflected beam coming from the other side doesn't get such a phase shift (because the phase shift occurs only if the beam is reflected on the boundary of the optically thicker medium coming from the optically thinner medium). In this case you can derive the beam-splitter transfer matrix to be
$$\hat{T}=\exp(\mathrm{i} \varphi) \frac{1}{\sqrt{2}} \begin{pmatrix} 1 & -1 \\ 1 & 1 \end{pmatrix},$$
where for an arbitrary phase shift ##\varphi## obviously ##\hat{T}^{\dagger} \hat{T}=\hat{1}## (just check it by doing the matrix multiplication explicitly).

For more details, see

https://en.wikipedia.org/wiki/Beam_splitter#Phase_shift
 
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Thank you so much @vanhees71

That makes perfect sense now.

P.S. Do symbols ##\dagger## and ##^*## really mean different things in this context, or can they be used interchangeably? I thought they both represented the conjugate transpose, but in the document I linked to, they are using both.
 

vanhees71

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Gold Member
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5,196
##\hat{T}^{\dagger}## is the usual notation for taking the (Hermitean) adjoint of a complex matrix, i.e., you transpose it (i.e., you write the columns of the original matrix as the lines of the new matrix) and take the complex conjugate of all these entries. The usual physicists' notation for complex conjugation of a number ##z=x+\mathrm{i} y## (with ##x,y \in \mathbb{R}##) is ##z^*=x-\mathrm{i} y##.
 

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