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?

  • $\begingroup$ What is |n,0>? If |n> is a Fock state, how is this a two-mode squeezed state? $\endgroup$ – Norbert Schuch Jan 4 '19 at 15:47
  • $\begingroup$ $|n, 0)$ 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! $\endgroup$ – Heisenberg Jan 4 '19 at 15:49
  • $\begingroup$ By "squeezed state", people usually mean a squeezed coherent state. Your state isn't even gaussian. $\endgroup$ – Norbert Schuch Jan 4 '19 at 16:17
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    $\begingroup$ Are you happy with a destructive measurement, or do you want a non-destructive measurement? $\endgroup$ – Norbert Schuch Jan 4 '19 at 16:19
  • $\begingroup$ 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 $\endgroup$ – 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.

  • $\begingroup$ Thanks, How come removing the squeezing operation deos not make any difference? $\endgroup$ – Heisenberg Jan 4 '19 at 16:29
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    $\begingroup$ What do you mean by "does not make any difference"? It changes the measurement! $\endgroup$ – Norbert Schuch Jan 4 '19 at 16:31
  • $\begingroup$ 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. $\endgroup$ – Heisenberg Jan 4 '19 at 16:37
  • $\begingroup$ @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! $\endgroup$ – Norbert Schuch Jan 4 '19 at 16:38
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    $\begingroup$ @Heisenberg: Some experiments do essentially this for $n=1$. $\endgroup$ – Frédéric Grosshans Jan 8 '19 at 16:21

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