It seems that if you measure a quadrature on the first mode, let's say $\hat{q}$, on a TMSVS, the second mode will collapse into a coherent state $|q+it\rangle$ Is this true ? What is then the value of $t$ ?
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Start with a TMSV $$|TMSV\rangle=\frac{1}{\cosh r}\sum_{n=0}^\infty \tanh^n{r}|n\rangle_1\otimes|n\rangle_2.$$ Project the first mode onto a coherent state $|\alpha\rangle=e^{-|\alpha|^2/2}\sum_{n}\frac{\alpha^n}{\sqrt{n!}}|n\rangle$: $$e^{-|\alpha|^2/2}\sum_{n}\frac{\alpha^n}{\sqrt{n!}}\langle n|_1 \frac{1}{\cosh r}\sum_{n=0}^\infty \tanh^n{r}|n\rangle_1\otimes|n\rangle_2=e^{-|\alpha|^2/2}\sum_{n}\frac{\alpha^n}{\sqrt{n!}} \frac{1}{\cosh r}\tanh^n{r}|n\rangle_2.$$ Normalize, collect some terms, and you find what the new resultant coherent state is in the second mode. This can be done with a TMSV starting with a different phase, which will affect the final coherent state in the second mode.
Quantum Mechanic
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