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Substituting the definitions of $\ket{+}$ and $\ket{-}$ in the standard basis,

$\frac{1}{\sqrt{2}}(\ket{0}\ket{+} + \ket{1}\ket{-}) = \frac{1}{2}( \ket{00} + \ket{01} + \ket{10} - \ket{11})$

The state is not entangled if there are $a_1, b_1, a_2, b_2$ such that

$(a_1\ket{0} + b_1\ket{1}) \otimes (a_2\ket{0} + b_2\ket{1}) = \frac{1}{2}( \ket{00} + \ket{01} + \ket{10} - \ket{11})$

which would mean that the following system of equations has a solution:

$\begin{array}{lcl} a_1a_2 = \frac{1}{2} \\ a_1b_2 = \frac{1}{2} \\ a_2b_1 = \frac{1}{2} \\ b_1b_2 = -\frac{1}{2} \end{array}$

If a solution exists, $b_1$ and $b_2$ must have opposite signs. The second and third equations require that $a_1$ has the same sign as $b_2$ and $a_2$ has the same sign as $b_1$, therefore $a_1$ and $a_2$ also have opposite signs. But the first equation requires $a_1$ and $a_2$ to have the same sign. Therefore, the system has no solutions and the state is entangled.

ReplyOptionsThe talk of “opposite signs” seems to assume that $a_1$, $a_2$, $b_1$ and $b_2$ are real, but in fact they can be complex.

Here is an alternative argument. We see that $a_1 a_2 = a_1 b_2$, and these are non-zero; hence $a_2 = b_2$. Similarly $a_1 = b_1$. But then $b_1 b_2$ must be equal to [[$a_1 a_2]], and we can see that this is not the case. Thus the system has no solutions and the state is indeed entangled.

ReplyOptions