# Projective measurement using two mode squeezed state?

Let me define two mode squeezed states as $$\left | \xi \right>_n=\exp\left(\xi \hat{a}^\dagger \hat{b}^\dagger-\hat{a} \hat{b} \xi^\star\right)\left | n,0 \right>$$ where $$\left|n,0\right>$$ is the number state with $$n$$ photons in mode 1 and none in mode 2. Now, let me define a projective measurement $$\{\left | \xi \right>_n\left<\xi\right|,\mathbb{I}-\left | \xi \right>_n\left<\xi\right|\}$$. Is it possible to implement this projective measurement in the lab? have you seen any paper which has been discussed such a measurement?

• What is |n,0>? If |n> is a Fock state, how is this a two-mode squeezed state? – Norbert Schuch Jan 4 '19 at 15:47
• $|n, 0)$ is the number state with n photons in mode 1 and none in mode 2 and I applied two mode squeezing operation on this state! – Heisenberg Jan 4 '19 at 15:49
• By "squeezed state", people usually mean a squeezed coherent state. Your state isn't even gaussian. – Norbert Schuch Jan 4 '19 at 16:17
• Are you happy with a destructive measurement, or do you want a non-destructive measurement? – Norbert Schuch Jan 4 '19 at 16:19
• You are write, its not gaussian. Some older papers call this state two mode squeezed state. For example look at this one: iopscience.iop.org/article/10.1088/0253-6102/32/3/471/meta – Heisenberg Jan 4 '19 at 16:20

If you are happy with destructive measurements, this can be done in principle in the lab. Let $$\rho$$ be the input state, and write $$|\xi\rangle = S|n,0\rangle$$.

1. Undo the squeezing transformation $$S$$, i.e., transform $$\rho'=S^\dagger\rho S$$. Note that $$S^\dagger$$ is just another squeezing transformation which can in principle be implemented in the lab. Correspondingly, this means that now you need to perform the measurement $$\{|n,0\rangle\langle 0,n|,I-|n,0\rangle\langle 0,n|\}$$.

2. Use photon number resolving detectors to measure both modes. Again, this is something which can be done in principle (to a certain accuracy). Count the outcome $$(n,0)$$ towards the first outcome, and all others towards the second.

Formally speaking, what you are looking for is the probability $$p_\xi$$ for the first outcome, which we can rewrite as $$p_\xi = \langle \xi|\rho|\xi\rangle =\langle n,0|S^\dagger \rho S|n,0\rangle = \langle n,0| \rho'|n,0\rangle\ ,$$ where the last term is exactly what the proposed measurement determines.

• Thanks, How come removing the squeezing operation deos not make any difference? – Heisenberg Jan 4 '19 at 16:29
• What do you mean by "does not make any difference"? It changes the measurement! – Norbert Schuch Jan 4 '19 at 16:31
• Thanks. If it changes the measurement how does it answer my question? I want a measurement that projects any state $\rho$ to $\left|\xi \right>_n$ with some probablity and by performing the measurement over and over I aim to somehow calculate the fidelity between the $\rho$ and $\left|\xi \right>_n$. I do not know if the scheme you proposed is able to solve my problem. – Heisenberg Jan 4 '19 at 16:37
• @Heisenberg It does! The trick is that it first transforms the state to a different state, and then carries out a different measurement on the new state, which is just the same as the original measurement on the original state! Basically it's just Schrödinger vs. Heisenberg picture -- given your username, you should understand that ;) --- I'll add a formula! – Norbert Schuch Jan 4 '19 at 16:38
• @Heisenberg: Some experiments do essentially this for $n=1$. – Frédéric Grosshans Jan 8 '19 at 16:21