# Troubles with stimulated emission of single photons and the no cloning theorem

I think I don't understand something basic about stimulated emission: it seems to violate the no cloning theorem, and applying its arguments give me 2 different results when trying to understand a basic experiment.

Before I present my question, I will use the following notations: $$|0\rangle = \operatorname{one photon with horizontal polarization}$$ $$|1\rangle = \operatorname{one photon with vertical polarization}$$ and the two states of a photon in a diagonal polarization: $$|+\rangle = \frac{|0\rangle +|1\rangle}{\sqrt2} , |-\rangle = \frac{|0\rangle -|1\rangle}{\sqrt2}$$ Now for the question: I know that if you enter a state $$|0\rangle$$ to a stimulated emission medium you get $$|00\rangle$$ and if you enter $$|1\rangle$$ you get $$|11\rangle$$. But what happens if you enter a state $$|+\rangle$$? Well, according to the no cloning theorem you can't get $$|++\rangle$$, but instead you will get $$\frac{|00\rangle +|11\rangle}{\sqrt{2}}$$ But in the diagonal basis this is just: $$\frac{(|+\rangle + |-\rangle)\otimes (|+\rangle + |-\rangle) + (|+\rangle - |-\rangle)\otimes (|+\rangle - |-\rangle)}{2} = \frac{|++\rangle +|--\rangle}{\sqrt{2}}$$ Which means that although I entered a single photon with a $$|+\rangle$$ polarization, I could measure 2 photons with $$|-\rangle$$ polarization.

What am I missing?

Thank you

Note: I saw that there is a similar question regarding the issue about the no cloning theorem and stimulated emission but the answers don't address this problem so please don't mark this as duplicate.

Note2 : I discussed the problem with a professor. The problem is that I neglected spontaneous emission and with one photon it's just as strong as stimulated... Using the real Hamiltonian of the interaction (with electric field operators and light Fock states) there is no problem.

• I don't understand how you are "measuring" photons in a $|-\rangle$ polarization when you explicitly indicated that this polarization is not a pure state and measurements are implicitly made only in the horizontal and vertical directions. From the point of view of horizontal and vertical measurements, it's impossible to distinguish between the $|+\rangle$ and $|-\rangle$ states because both would give an equal distribution of H and V measurements – Jim Oct 17 '19 at 13:33
• It is a pure state, not a mixed one. You can just put a polarizer in 45 degrees (or a polarizing beam splitter) – Ofek Gillon Oct 17 '19 at 21:18
• If you do that, you effectively make 45 degrees the new vertical and then recover the original case with no issues. In order to have this problem with your measurements, you need to ensure that "diagonal" is actually diagonal w.r.t. your measurement axis. Cylindrical symmetry is awful, eh? – Jim Oct 18 '19 at 11:35
• @Jim , that is exactly what I'm saying. If the process is symmetric, I should get another photon in the $|+\rangle$, but as you can see, I don't. – Ofek Gillon Oct 20 '19 at 13:53
• Hey every one, after discussing with a professor I figured out what I was missing: stimulated emission with one photon is just as strong as spontaneous emission so I can't neglect that effect. When you count that in the calculation you see that everything works! – Ofek Gillon Oct 20 '19 at 13:58

This is of no concern to the no-cloning theorem. As you correctly point out, if $$|0\rangle\mapsto|00\rangle$$ and $$|1\rangle\mapsto|11\rangle$$ then you cannot have $$|+\rangle\mapsto|++\rangle$$. However, there is no problem with mapping $$|+\rangle$$ to some state which has some overlap with $$|++\rangle$$ and/or $$|--\rangle$$, which is indeed what happens, as you observe.
• @OfekGillon well no, it doesn't. In all of these processes the momentum is conserved, so the output photons can't really be identical. But regardless of this, no-cloning still doesn't apply because if you have $|0\rangle\mapsto|00\rangle$ through stimulated emission you do not have $|+\rangle\mapsto|++\rangle$ via the same process – glS Oct 20 '19 at 14:42