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Creation and Annihilation operators on photons

  • Thread starter Malamala
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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!
 

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