# 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.

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The equations have complex solutions. What are the restrictions on the solutions in this context? Not everyone here is familiar with Quantum Mechanics. – Ayman Hourieh Jun 16 '12 at 16:38
@Ayman: My apologies, I have updated the question. AFAIK, the only remarkable thing about this system that I didn't mention is that there can be complex solutions. – Xodarap Jun 16 '12 at 18:35

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 $\Delta = -1 \lt 0$.

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I believe amplitudes can be complex so showing non-real solutions doesn't satisfy, unfortunately. – Xodarap Jun 16 '12 at 15:20
But complex solutions are still solutions depending on the context, aren't they? – Bolutife Ogunsola Jun 16 '12 at 15:27
@Xodarap - I'm not familiar with this topic in Quantum Mechanics and thought that the values were supposed to be real. Of course, the equations have complex solutions and they all satisfy $|\alpha|^2 \le 1$. Are there other restrictions on the values? – Ayman Hourieh Jun 16 '12 at 16:32
You're right, there are complex solutions. My problem is probably at the physics level, not the math level, so I will ask there. – Xodarap Jun 16 '12 at 18:41

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.

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$${1\over 2}(\alpha_0-\beta_0)^2 + {1\over 2}(\alpha_1 - \beta_1)^2 = {1\over 2}(\alpha_0^2 + \beta_0^2 + \alpha_1^2 + \beta_1^2) - \alpha_0\beta_0 - \alpha_1\beta_1 = 1 - {2\over \sqrt{2}} = 1 - \sqrt{2}.$$

Since the left-hand side is a sum of squres, you are now done.

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Nice solution!${}{}{}{}{}$ – André Nicolas Jun 16 '12 at 15:11
Sorry, could you expand? If it is a sum of squares then...? The amplitudes can be complex – Xodarap Jun 16 '12 at 15:21
You did not specify this is a complex system. – ncmathsadist Jun 16 '12 at 17:21

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.

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