Write the state of the system after r photons leave the |nϕ> state

[Malamala]

Homework Statement

(Note: this is the last part of a longer problem, so I hope I won't miss anything important)

We have a laser with 2 energy levels, so the emitted photons will have the energy ##\omega##. For this given energy there are ##m## states in which the photon can be, denoted ##|\psi_i>##. Assume we have ##n## photons in state ##|\psi_1>## and one more in the state ##|\phi>=\frac{1}{\sqrt{m}}(|\psi_1>+|\psi_2>+...+|\psi_m>)## and denote the state as ##|n\phi>##. From a previous part, this can be written as $$|n\phi>=\sqrt{\frac{m}{(m+n)(n+1)}}(|11..1\phi>+|11..1\phi1>+...|\phi1..1>)$$ where, for example ##|11..1\phi>## stands for ##|\psi_1>|\psi_1>...|\psi_1>|\phi>## and the above wave function is normalized and symmetrized (for bosons). Now for the last part we are asked to write the state of the system after ##r## photons leave the ##|n\phi>## state.

Homework Equations

The Attempt at a Solution

I'll try to do it for a simpler case (which should be easily generalized). So for n=3 we have $$|3\phi>=N(|111\phi>+|11\phi1>+|1\phi11>+|\phi111>)$$ where N is a normalization. If we remove a photon, this can be the ##|\psi_1>## or the ##|\phi>## So after one removal the state would be a liner combination of these 2 possibilities, something like: $$N_1(|11\phi>+|1\phi1>+|\phi11>+|111>)$$ After one more, you get $$N_2(|1\phi>+|\phi1>+|11>)$$ after one more: $$N_3(|\phi>+|1>)$$ And of course this can be generalized for arbitrary ##n## and ##r##. But I am not sure if my logic is correct. I am mainly confused about the fact that ##|\phi>## is not an eigenstate of this system, but a linear combination, so if you would think in terms of annihilation operators I am not sure how you would write this removal. Any suggestion would be greatly appreciated. Thank you!

 

[mark726]

Hello there,

Thank you for sharing your attempt at solving this problem. Your approach seems to be on the right track. Let's break it down step by step.

First, we have the state ##|n\phi>##, which is a combination of ##n## photons in state ##|\psi_1>## and one photon in state ##|\phi>##. As you correctly pointed out, this state is not an eigenstate of the system, but a linear combination. In terms of annihilation operators, we can write this state as follows:

$$|n\phi>=\sqrt{\frac{m}{(m+n)(n+1)}}(\hat{a_1}^\dagger)^n \hat{a_\phi}^\dagger|0>$$

Here, ##\hat{a_i}^\dagger## represents the creation operator for state ##|\psi_i>## and ##\hat{a_\phi}^\dagger## represents the creation operator for state ##|\phi>##. The factor in front is the normalization constant.

Next, we want to remove ##r## photons from this state. As you correctly pointed out, there are two possibilities - either we remove a photon from the ##|\psi_1>## state or from the ##|\phi>## state. Let's consider each case separately.

Case 1: Removing a photon from the ##|\psi_1>## state

In this case, we can write the state after removal as follows:

$$|n\phi>_1=\sqrt{\frac{m}{(m+n)(n+1)}}(\hat{a_1}^\dagger)^{n-1} \hat{a_\phi}^\dagger|0>$$

Here, we have simply removed one creation operator for state ##|\psi_1>##. The factor in front remains the same.

Case 2: Removing a photon from the ##|\phi>## state

In this case, we can write the state after removal as follows:

$$|n\phi>_2=\sqrt{\frac{m}{(m+n)(n+1)}}(\hat{a_1}^\dagger)^n \hat{a_\phi}^\dagger(\hat{a_\phi}^\dagger)^{-1}|0>$$

Here, we have removed one creation operator for state ##|\phi>## and added one