Solution to a system of quadratics I am learning about a Bell State, and am trying to show that they are entangled. I believe that the required proof is to show that the system
$$\alpha_0^2+\alpha_1^2=1$$
$$\beta_0^2+\beta_1^2=1$$
$$\alpha_0\beta_0=1/\sqrt{2}$$
$$\alpha_1\beta_1=1/\sqrt{2}$$
has no solutions. I have tried various ways of rewriting each variable in terms of others etc. without success. Any hints?
EDIT: I apologize, I should have made clear. In QM probabilities can be complex - what exactly this means intuitively is unclear to me, but algebraically it means that $\alpha_i,\beta_i$ can be complex.
 A: By the first equation we may write $$ \alpha_0 = \sin(x), \alpha_1=\cos(x)$$ and by the second $$\beta_0 = \sin(y), \beta_1 = \cos(y).$$
Multiplying the third and fourth equations then gives
$$(\sin(x)\cos(x))(\sin(y)\cos(y)) = \left(\frac{1}{2}\sin(2x)\right)\left(\frac{1}{2}\sin(2y)\right) = \frac{1}{2} $$
$$\iff \sin(2x)\sin(2y) = 2 $$
which is not possible since the product of two numbers with absolute value at most 1 has absolute value at most 1.
A: $${1\over 2}(\alpha_0-\beta_0)^2 + {1\over 2}(\alpha_1 - \beta_1)^2
= {1\over 2}(\alpha_0^2 + \beta_0^2 + \alpha_1^2 + \beta_1^2) - \alpha_0\beta_0
- \alpha_1\beta_1 = 1 - {2\over \sqrt{2}} = 1 - \sqrt{2}.$$
Since the left-hand side is a sum of squres, you are now done.
A: Another trigonometric approach: let $x$ and $y$ be such that $\alpha_0 = \sin x$ and $\beta_0 = \sin y$, as in nullUser's answer.
Then use the following identities on the third and fourth equations:
$$\sin s \sin t = \frac{\cos (s-t) - \cos (s+t)}{2}$$
$$\cos s \cos t = \frac{\cos (s-t) + \cos (s+t)}{2}$$
This gives the following system of equations:
$$\cos (s-t) - \cos (s+t) = \sqrt{2}$$
$$\cos (s-t) + \cos (s+t) = \sqrt{2}$$
From this it is clear that $\cos (s-t) = \sqrt{2}$, but that is impossible.
A: From (3) and (4), we have:
\begin{align*}
\beta_0^2 &= \frac{1}{2\alpha_0^2} \\
\beta_1^2 &= \frac{1}{2\alpha_1^2}
\end{align*}
Plug into (2):
\begin{align*}
\frac{1}{2\alpha_0^2} + \frac{1}{2\alpha_1^2} &= 1 \\
\frac{\alpha_1^2}{\alpha_0^2\alpha_1^2} + \frac{\alpha_0^2}{\alpha_0^2\alpha_1^2} &= 2 \\
\frac{\alpha_0^2 + \alpha_1^2}{\alpha_0^2\alpha_1^2} &= 2
\end{align*}
Use (1) in the numerator to get:
$$
\alpha_0^2 \alpha_1^2 = \frac{1}{2}
$$
Therefore:
$$
\alpha_1^2 = \frac{1}{2\alpha_0^2}
$$
Plug into (1) and multiply both sides by $\alpha_0^2$:
$$
\alpha_0^4 - \alpha_0^2 + \frac{1}{2} = 0 
$$
This is a quadratic equation for $\alpha_0^2$ with no real solutions, as $\Delta = -1 \lt 0$.
