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Starting with the two-mode squeezing operator $ S_2(\xi) = \exp \left( \xi^* a_1 a_2 - \xi a_1^\dagger a_2^\dagger \right)$, we can factor it into [1] $$ S_2(\xi) = \frac{1}{\cosh r} \exp\left( -a_1^\dagger a_2^\dagger e^{i\varphi} \tanh r\right) \exp\left[-\left(a_1^\dagger a_1 + a_2^\dagger a_2 \right)\ln(\cosh r) \right] \exp\left( -a_1 a_2 e^{i\varphi} \tanh r\right) $$ which, when I apply to the vacuum $\vert 0\rangle _1\vert 0 \rangle_2$ I get $$ \begin{align} \vert TMSV \rangle &= \frac{1}{\cosh r} \exp\left( -a_1^\dagger a_2^\dagger e^{i\varphi}\tanh r \right) \vert 0 \rangle_1 \vert 0 \rangle_2,\\ &= \frac{1}{\cosh r}\sum_{n=0}^\infty (-1)^n(e^{i\varphi} \tanh r)^n \vert n \rangle_1 \vert n \rangle_2, \end{align} $$ which jives with what I see in the literature [2] [3]. However, in many many other places [4], I also see $$ \vert TMSV \rangle = \sqrt{1-\eta^2} \sum_{n=0}^\infty \eta^n \vert n \rangle_1 \vert n \rangle_2, $$ where $\lambda = \tanh r$ (and $\xi$ is real). For the life of me I can't figure out where the $(-1)^n$ went. This is obviously not an issue for the density matrix, but it seems like a big difference for the state itself. I don't think it comes down to convention, as the squeezing operator is the same in both cases. Am I missing something obvious?

[1] https://link.aps.org/doi/10.1103/PhysRevA.31.3093

[2] Introductory Quantum Optics, Gerry & Knight

[3] A Group-Theoretical Approach To Quantum Optics - Klimov & Chumakov

[4] E.g. How to obtain a two mode Squeezed state from two mode squeezed vacuum? and Two-mode squeezing and EPR

Qmechanic
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lionelbrits
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1 Answers1

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For me, it sounds a specific choice for the phase $\varphi$, since $e^{i\varphi}$ also disappeared. Look that if $\xi = re^{i\varphi}$, for $\varphi = \pi$ we have real $\xi$ and $(-1)^n e^{in\pi} = 1$.

Ruffolo
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