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How can I show explicitly that the bell state $$|\psi^{-}>=\frac{1}{\sqrt{2}}(|0>|1>-|1>|0>)$$ is invariant under local unitary transformations $U_{1}\otimes U_{2}$ ?

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closed as off-topic by jinawee, Frederic Brünner, Brandon Enright, Waffle's Crazy Peanut, Dan Dec 18 '13 at 21:04

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It is not invariant: Try $U_{1}\otimes U_{2} = \sigma_x \otimes \mathbb{Id}$ – Trimok Jul 2 '13 at 7:38
I think that you meant that it is invariant (up to a phase) under $U\otimes U$ in which the same unitary transformation is performed on each qubit separately, right? That's true because $U\otimes U$ is the action of an $SU(2)$ rotation and the singlet state is invariant under the total $SU(2)$, the group generated by the total angular momentum. But it is not invariant under the $SU(2)_1$ and $SU(2)_2$ groups separately. For example, a rotation of the first qubit only, a la Trimok, gives you states like $00+11$ etc. – Luboš Motl Jul 2 '13 at 7:42
For demonstration by "brute force", you may use that a unitary operator $U$ of $U(2)$has the form: $U = e^{i\alpha} S$, where $\alpha$ is a real. and $S$ belongs to $SU(2)$ $S$ can be written $S = e^{i~\theta~ \vec n.\vec \sigma} = (cos \theta ~\mathbb{Id} + i sin\theta~\vec n.\vec \sigma)$, where $\theta$ is a real, $\vec n$ is a normed vector ($(\vec n)^2 = 1$), and $\vec \sigma$ are the Pauli matrices. And you have : $S|0> = (cos \theta + i sin\theta~n^z)|0> + i sin\theta(~n^x + in^y)|1>$, $S|1> = i sin\theta(~n^x - in^y) |0> + (cos \theta - i sin\theta~n^z) |1>$ – Trimok Jul 2 '13 at 10:19