Quantum beam-splitter matrix I have seen the matrix for the action on a quantum beam splitter described in one of two ways: 
$$\begin{pmatrix}
t_1 & r_2 \\
r_1 & t_2 \end{pmatrix}$$
(this appears in Quantum Optics by Girish Agarwal page 104)
And also like this: 
$$\begin{pmatrix}
r_1 & t_2 \\
t_1 & r_2 \end{pmatrix}$$
(this appears in this http://physics.gu.se/~ostlund/kvant/ajpbs02-2.pdf)
Both of these sources then go onto say that for a 50:50 beam splitter the matrix $$\frac{1}{\sqrt{2}}\begin{pmatrix}
1 &i\\
i & 1 \end{pmatrix}$$
Is one of the sources wrong or are we free to chose which coefficient (i.e. the tranmission or reflection) we include the $i$ with, without it affecting the results? 
 A: In this case, this particular definition tacitly assumes that the beamsplitter's transfer matrix is invariant with respect to a swap in the roles of the ports, which is why you swap the roles of reflexion and transmission, yet get the same result. This seemingly odd assumption simply arises from an assumption that the beamsplitter has a certain spatial symmetry.
To understand this, let's look at the general case and understand that there's actually a great deal of leeway in specifying a beamsplitter, even a lossless 50:50 one. Look up your favorite reference to find a general expression for a general $2\times 2$ unitary matrix $U$; one possibility is:
$$U = \left(\begin{array}{cc}a&b\\-b^\ast\,e^{i\,\varphi}&a^\ast\,e^{i\,\varphi}\end{array}\right) = \left(\begin{array}{cc}e^{i\,\alpha}\,\cos\theta&e^{i\,\beta}\,\sin\theta\\-e^{i\,(\varphi-\beta)}\,\sin\theta&e^{i\,(\varphi-\alpha)}\,\cos\theta\end{array}\right)$$
where $a,\,b\in\mathbb{C}$ and $|a|^2+|b|^2=1$. So, equivalently, our general unitary matrix can be specified by putting $a=e^{i\,\alpha}\cos\theta$ and $b = e^{i\,\beta}\,\sin\theta$ for $\alpha,\,\beta,\,\theta\in\mathbb{R}$. In general, therefore, a unitary matrix, once its splitting ratio has been specified by fixing $\theta$, has three real parameters $\alpha,\,\beta,\,\varphi$ that can be arbitrarily chosen. Now, an even power splitting condition is simply the condition that $\theta \in \left\{\frac{\pi}{4}+2\,k_1\,\pi,\,\frac{3\,\pi}{4}+2\,k_2\,\pi,\,\frac{5\,\pi}{4}+2\,k_3\,\pi,\,\frac{7\,\pi}{4}+2\,k_4\,\pi \right\}$ where $k_i\in\mathbb{Z}$. The phase factors $\alpha,\,\beta,\,\varphi$ are varied by choosing different reference points as "ports" where the phase of the wave is defined. We can arbitrarily slide a "port" along its beam; as long as we stick with a position during a calculation, any position in the beam will do.
The stated matrices actually arise from the further condition that the beamsplitter is symmetric and that the reference ports are symmetrically placed with respect to the beamsplitter. That is, the transfer matrix must be invariant under the similarity transformation defined by $\left(\begin{array}{cc}0&1\\1&0\end{array}\right)$, which swaps the roles of transmission and reflexion ports. So we get:
$$\left(\begin{array}{cc}0&1\\1&0\end{array}\right)\left(\begin{array}{cc}e^{i\,\alpha}\,\cos\theta&e^{i\,\beta}\,\sin\theta\\-e^{i\,(\varphi-\beta)}\,\sin\theta&e^{i\,(\varphi-\alpha)}\,\cos\theta\end{array}\right)\left(\begin{array}{cc}0&1\\1&0\end{array}\right)^{-1}=\left(\begin{array}{cc}e^{i\,\alpha}\,\cos\theta&e^{i\,\beta}\,\sin\theta\\-e^{i\,(\varphi-\beta)}\,\sin\theta&e^{i\,(\varphi-\alpha)}\,\cos\theta\end{array}\right)$$
or:
$$\left(\begin{array}{cc}e^{i\,\alpha}\,\cos\theta&e^{i\,\beta}\,\sin\theta\\-e^{i\,(\varphi-\beta)}\,\sin\theta&-e^{i\,(\varphi-\alpha)}\,\cos\theta\end{array}\right)=\left(\begin{array}{cc}e^{i\,(\varphi-\alpha)}\,\cos\theta&-e^{i\,(\varphi-\beta)}\,\sin\theta\\e^{i\,\beta}\,\sin\theta& e^{i\,\alpha}\,\cos\theta\end{array}\right)$$
so that now we must have $\alpha = \frac{\varphi}{2}\mod 2\,\pi$ and $\beta = \frac{\varphi}{2}+\frac{\pi}{2}\mod 2\,\pi$ so that:
$$U = e^{i\frac{\varphi}{2}}\left(\begin{array}{cc}\cos\theta&i\,\sin\theta\\i\,\sin\theta&\cos\theta\end{array}\right)$$
and the most convenient choice is $\phi=0$. We can vary $\phi$ because we can still shift the reference ports, as long as we do so symmetrically.
A: I think that it in this case is a bit of semantics. In an ideal beam-splitter there is really no "true" reflection. Rather the beam is split into two output directions. Which one of the two directions you call transmitted and deflected is a matter of taste.
Naively I would say the example with transition $t$ on the diagonal looks more natural. In this case the identity matrix just means that light is transmitted, as if the beam splitter was not present.
