Creation and Annihilation operators on photons

[Malamala]

Homework Statement

The possible (normalized) eigenstates of a photon in a given system are written as: $$|\psi_1>,|\psi_2>,...|\psi_m>$$ Let another state be $$|\phi> = \frac{|\psi_1>+|\psi_2>+...+|\psi_m>}{\sqrt{m}}$$ and denote: $$|n>=|\psi_1>|\psi_1>...|\psi_1>$$ which represent a state containing n photons in $$|\psi_1>$$ If we add $|\phi>$ to $|n>$ we get the totally symmetric (for bosons) state $|n\phi>$. For $n=1$ this would be $$|1\phi>=N(|\phi>|1>+|1>|\phi>)$$ where N is a normalization constant. I showed in a previous part of the problem that: $$|n\phi>=N(a_1^\dagger+a_2^\dagger++...a_m^\dagger)|n>$$ where $a_i^\dagger$ is the creation operator for a photon in the state $i$. The question I have problems with is: Starting in the state $|n\phi>$ what is the state of the system after $r$ photons leave? The photons that leave are the one in the state $|\psi_1>$

Homework Equations

The Attempt at a Solution

I'll do this for $r=1$ for now. Intuitively, if one photon in the state $|\psi_1>$ leaves, the final state would be $|(n-1) \phi>$ (I will ignore normalization constants for now). However, if I apply the annihilation operator to get rid of a photon in $|\psi_1>$ i.e. $a_1$ and using $$a_1a_1^\dagger|n>=(n+1)|n>$$ and for $i \neq 1$ $$a_1a_i^\dagger|n>=\sqrt{n}|i(n-1)>$$ I get in the end $$a_1|n\phi>=(n+1)|n>+\sum_{i=2}^m \sqrt{n}|i(n-1)>$$ which is equal to $$a_1|n\phi>=(n+1)|n>-\sqrt{n}|n>+\sqrt{n}|(1+2+...m)(n-1))>$$ $$a_1|n\phi>=(n+1)|n>-\sqrt{n}|n>+\sqrt{nm}|\phi(n-1))$$ so in the end I get an extra term of $|n>$ and I am not sure why is it there. It looks as if $|\phi>$ itself was annihilated, but I am not sure how. Did I do a mistake? Can someone help me? Thank you!

 

[mark726]

Thank you for your detailed explanation and attempt at solving the problem. It seems like you are on the right track, but there are a few small mistakes in your calculations.

Firstly, when applying the annihilation operator $a_1$ to the state $|n\phi>$, you should get $$a_1|n\phi>=(n+1)|n>+a_1a_2^\dagger|n>+\dots+a_1a_m^\dagger|n>.$$ This is because the state $|n\phi>$ contains $n$ photons in the state $|\phi>$, and the annihilation operator $a_1$ will only act on one of these photons. So, your final result should be $$a_1|n\phi>=(n+1)|n>+\sqrt{n}|\phi(n-1)>.$$

Secondly, when considering the case $r=1$, the final state will indeed be $|(n-1)\phi>$, as you correctly stated. However, your calculations seem to suggest that this state would also contain an extra term $|n>$. This is not the case, as the photons leaving the system are in the state $|\psi_1>$, and not in the state $|\phi>$. So, the final state should be $$|(n-1)\phi>=(n+1)|n>-\sqrt{n}|\phi(n-1)>.$$

I hope this helps clarify your doubts. Keep up the good work!