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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.

Image from this question https://physics.stackexchange.com/questions/207810/mach-zehnder-interferometer-two-output-interference-pattern-question

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 $c$)?

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  • $\begingroup$ 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? $\endgroup$ – Emilio Pisanty Jan 22 at 9:39
  • $\begingroup$ 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 $\frac{1}{\sqrt{2}}(|1\rangle_b |0\rangle_c + i |0 \rangle_b |1 \rangle_c)$ $\endgroup$ – The Bosco Jan 22 at 9:41
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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:

enter image description here

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.

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