# 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 $$c$$)?

• The identity you've used, $\hat{a}^\dagger \longrightarrow \frac{1}{\sqrt{2}}(\hat{b}^\dagger + i \hat{c}^\dagger) \hat{a}^\dagger$, is incorrect - it should read $\hat{a}^\dagger \longrightarrow \frac{1}{\sqrt{2}}(\hat{b}^\dagger + i \hat{c}^\dagger)$. Where did you see that version? Jan 22 '19 at 9:39
• I thought it was that way because we first act on the state $| 0 \rangle$ with the $a$ operator to get $|1\rangle_a$ and then with the splitter 1 operator to get $\frac{1}{\sqrt{2}}(|1\rangle_b |0\rangle_c + i |0 \rangle_b |1 \rangle_c)$ Jan 22 '19 at 9:41

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.