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


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: \begin{equation} \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}. \end{equation}

  • $\begingroup$ It is really just the Discrete Fourier Transform. $\endgroup$ Commented Nov 24, 2023 at 14:15

1 Answer 1


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}

BTW this matrix has a number of "nice" properties. See for instance:

Mehta, M. L. (1987). Eigenvalues and eigenvectors of the finite Fourier transform. Journal of Mathematical Physics, 28(4), 781-785.

  • $\begingroup$ It's worth remarking that for systems with dimension $d=2^k$ you can also use a Hadamard matrix to only have real entries. $\endgroup$
    – fulis
    Commented Nov 24, 2023 at 12:58
  • $\begingroup$ @fulis ... to model what? $\endgroup$ Commented Nov 24, 2023 at 14:15
  • $\begingroup$ @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 $\endgroup$
    – fulis
    Commented Nov 24, 2023 at 14:22
  • $\begingroup$ @fulis Hm, this is puzzling. How can both be correct? How do those beamsplitters differ? $\endgroup$ Commented Nov 24, 2023 at 17:10
  • $\begingroup$ @NorbertSchuch even for a 2-mode BS the transformation matrix is not uniquely determined because the phases are not measurable, see this question. To go between different the 4-mode descriptions you need 2-mode mixing operations in addition to phase shifts, but that isn't too surprising if you consider that any N-mode path unitary can be decomposed into 2-mode unitaries, for example with the Clements decomposition. $\endgroup$
    – fulis
    Commented Nov 25, 2023 at 1:45

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