Evolution of the operators in Mach-Zehnder interferometer I have trouble understanding how the operators that change the initial state, change when adding extra splitters. As an example, I will use an idealized Mach-Zehnder interferometer where the splitters have no thickness.

I am using the picture from Mach-Zehnder Interferometer: two output interference pattern question
I will use the labeling of the modes as follows:


*

*purple = $a$

*red between splitter 1 and splitter 2 = $b$

*green between splitter 1 and splitter 2 = $c$

*between splitter 2 and screen 1 = $d$

*between splitter 2 and screen 2 = $e$
Now, I know that the creation operator $\hat{a}^\dagger$ creates the photon at the very beginning of the diagram.
Then, it changes with splitter 1 as follows
$$\hat{a}^\dagger \longrightarrow \frac{1}{\sqrt{2}}(\hat{b}^\dagger + i \hat{c}^\dagger) \hat{a}^\dagger$$
since now apply the $\hat{a}^\dagger$ operator first, and then the operator that 'creates' the other two paths after the first splitter (including the phase $i$).
Now, my problem is how do I take this and add a second splitter since I have two different paths ($b$ and $c$)?
 A: The identity you've used, 
$$\hat{a}^\dagger \longrightarrow \frac{1}{\sqrt{2}}(\hat{b}^\dagger + i \hat{c}^\dagger) \hat{a}^\dagger
\qquad \text{(wrong!)}
$$
is incorrect. Instead, what you need to do is this:
$$
\hat{a}^\dagger = \frac{1}{\sqrt{2}}(\hat{b}^\dagger + i \hat{c}^\dagger) 
$$
That is, the operator $\hat{a}^\dagger$ is exactly equal to the linear combination   of operators $\frac{1}{\sqrt{2}}(\hat{b}^\dagger + i \hat{c}^\dagger)$, or in other words the operation of creating a photon on channel $a$ is exactly identical to the superposition of creating photons on the $b$ and $c$ channels.
If you want to keep going and add the second beam splitter, you do the same but with the relevant combination of modes:
$$
\hat{b}^\dagger = \frac{1}{\sqrt{2}}(\hat{d}^\dagger + i \hat{e}^\dagger).
$$
Similarly, you should have a relationship between $\hat{c}^\dagger$, $\hat{d}^\dagger$ and $\hat{e}^\dagger$, where you should ensure that the matrix that connects the two sets is unitary.
More importantly, you need a second such relationship coming from the beam splitter, which relates the $\hat{b}^\dagger$ and $\hat{c}^\dagger$ operators with the empty port of the first beam splitter:



The first time you see this, it feels extremely weird that this empty port is so important, but it is a crucial ingredient. In short, the port looks empty to you, but it still contains vacuum fluctuations, and those vacuum fluctuations can have a strong effect on the experiments downstream from the beamsplitter.
Once you put all of this together, you will be able to construct explicit relationships between the input modes ($\hat{a}^\dagger$ and $\hat{f}^\dagger$) and the output modes ($\hat{d}^\dagger$ and $\hat{e}^\dagger$) of the interferometer, so you can


*

*change input states like $\hat{a}^\dagger|0\rangle$ into output states, by simply substituting $\hat{a}^\dagger$ for the relevant combination of $\hat{d}^\dagger$ and $\hat{e}^\dagger$, or

*Express your observable (which is likely to be some combination of the number operators $\hat{d}^\dagger\hat{d}$ and $\hat{e}^\dagger\hat{e}$, since you're detecting intensity at the screens) in terms of your input operators $\hat{a}^\dagger$ and $\hat{f}^\dagger$, which can then be used to take expectation values for different input states, and so on.


Hopefully that clarifies how the formalism is meant to be used.
