# What is a general expression for a symmetric $N$-mode beam splitter?

BACKGROUND

A symmetric beam splitter can be represented as $$$$\hat{B}^{(2)} = \frac{1}{\sqrt{2}}\begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix},$$$$ and according to Zukowski et al., the three-mode equivalent is $$$$\hat{B}^{(3)} = \frac{1}{\sqrt{3}}\begin{bmatrix} 1 & 1 & 1 \\ 1 & \alpha_3 & \alpha_3^2 \\ 1 & \alpha_3^2 & \alpha_3^4 \end{bmatrix},$$$$ where $$\alpha_3 = e^{i\frac{2\pi}{3}}$$.

QUESTION

I would like to generalize the above to a symmetric $$N$$-mode beam splitter. I understand that there has to be a unitarity requirement. Based on the two-mode and three-mode examples, I also notice that there is some kind of pattern whereby the matrix elements might be powers of $$\alpha_N = e^{i\frac{2\pi}{N}}$$. However, I'm not really sure how to assemble them into a matrix. I'm looking for something that would look like this: $$$$\hat{B}^{(N)} = \frac{1}{\sqrt{N}}\begin{bmatrix} 1 & 1 & 1 & ⋯ & 1\\ 1 & \alpha_N^? & \alpha_N^? & ⋯ & \alpha_N^? \\ ⋮ & ⋮ & ⋮ & ⋱ & ⋮ \\ 1 & \alpha_N^? & \alpha_N^? & ⋯ & \alpha_N^? \end{bmatrix}.$$$$

• It is really just the Discrete Fourier Transform. Nov 24, 2023 at 14:15

It seems the matrix you want is just the Fourier matrix. Thus, if $$\alpha=e^{2\pi i/N}$$ so that $$\alpha^N=1$$, then \begin{align} \frac{1}{\sqrt{N}}\left( \begin{array}{ccccc} 1&1&1&\ldots&1\\ 1&\alpha&\alpha^2&\ldots &\alpha^{N-1}\\ 1&\alpha^2&\alpha^4&\ldots&\alpha^{2(N-1)}\\ \vdots&\vdots &\vdots &\ddots&\vdots\\ 1&\alpha^{N-1}&\alpha^{2(N-1)}&\ldots&\alpha^{(N-1)(N-1)} \end{array}\right) \end{align}
• It's worth remarking that for systems with dimension $d=2^k$ you can also use a Hadamard matrix to only have real entries. Nov 24, 2023 at 12:58
• @NorbertSchuch to model a symmetric d-mode lossless beam splitter (obviously with a $1/\sqrt{d}$ factor in front). See this for example doi.org/10.1364/OPTICA.388912 and doi.org/10.1103/PRXQuantum.2.010320 Nov 24, 2023 at 14:22