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I'm reading a book where quantum interference is explained with a common experiment, where photons are thrown to some apparatus formed by two beam splitters:

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Describing the problem using linear algebra, we could assign some vectors for the two possible paths that the photon could take after the first beam splitter, for example

$$ \text{Red path} \equiv \left(\begin{array}{c}1\\0\end{array}\right) \quad \text{Green path} \equiv \left(\begin{array}{c}0\\1\end{array}\right) $$

Then, if we say that the photon starts in the state of the red path, the quantum state after the first beam splitter is

$$ \lvert\psi\rangle = \dfrac{1}{\sqrt{2}} \left(\begin{array}{cc}1&i\\i&1\end{array}\right)\left(\begin{array}{c}1\\0\end{array}\right) = \dfrac{1}{\sqrt{2}}\left(\begin{array}{c}1\\i\end{array}\right) = \dfrac{1}{\sqrt{2}}\left(\begin{array}{c}1\\0\end{array}\right) + \dfrac{i}{\sqrt{2}}\left(\begin{array}{c}0\\1\end{array}\right) $$

Mathematically, this is nice because it gives 50% of probabilities for the photon to be found in each path since

$$ \left|\dfrac{1}{\sqrt{2}}\right|^2 = \dfrac{1}{2} \quad \text{and} \quad \left|\dfrac{i}{\sqrt{2}}\right|^2 = \dfrac{1}{2} $$

However, I don't see why the book represents the beam splitter using the imaginary number $i$. Wouldn't it be the same physically speaking (in terms of the resultant probability) if we represent the beam splitter with the usual Hadamard's gate

$$ \dfrac{1}{\sqrt{2}} \left(\begin{array}{cc}1&1\\1&-1\end{array}\right) $$

Is there any reason (historical, physical, ...) that could make the author choose the matrix with the $i$ components?


marked as duplicate by glS, GiorgioP, Jon Custer, ZeroTheHero, Aaron Stevens Apr 1 at 13:13

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