# What does a beam splitter look like in path-encoded notation for at most one photon?

Assume a Hilbert space that is (i) truncated to at most one photon, and (ii) is path-encoded such that $$(1,0)^T$$ and $$(0,1)^T$$ represent the photon in two separate optical modes, respectively. Here, these could be the upper $$|u\rangle$$ and lower $$|l\rangle$$ arms incident on a beam splitter. If this beam splitter is symmetric, it would be written as $$$$\hat{B} = \frac{1}{\sqrt{2}} \begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix}.$$$$

My problem is with the representation of the vacuum $$|0\rangle$$. I could write it as $$(0,0)^T$$ and the effect of the beam splitter would indeed be to leave it unchanged. However this can't be a valid representation of the vacuum state since it's not normalized. I then thought of extending the Hilbert space by including an extra dimension at the beginning of the state vector such that $$(a,b,c)^T$$ would be equivalent to $$a|0\rangle + b|u\rangle + c|l\rangle$$.

The question is then, how would the beam splitter look like in this three-dimensional Hilbert space? In order to capture the fact that the beam splitter leaves the vacuum unchanged, I tried $$$$\hat{B} = \frac{1}{\sqrt{2}} \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 1 \\ 0& 1 & -1 \end{pmatrix},$$$$ but this doesn't satisfy the unitarity of the beam splitter. What am I missing?

• Why is the vacuum not normalized? Isn't it just a problem with the mathematical representation? Commented Nov 8, 2023 at 3:46
• It is. My question is then how to represent the vacuum and the beam splitter in such a way that they are normalized and unitary, respectively. Commented Nov 8, 2023 at 11:35
• You could make the matrix $\hat{B}$ unitary by only having the factor of $1/\sqrt{2}$ in the lower-right block. I will add though that generally you don't want your description of an optical element to depend on the number of photons. One therefore typically uses the transformation of the creation/annihilation operators and not states in the Fock space. Commented Nov 8, 2023 at 14:18
• @fulis Agreed, but the ladder operators are unwieldy when one needs to use matrix operations. Commented Nov 8, 2023 at 14:51

To leave the vacuum unchanged, you need to put a $$\sqrt{2}$$ in the top left corner.