1
$\begingroup$

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.

  • 0
    @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.2012-06-16

4 Answers 4

2

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 $\Delta = -1 \lt 0$.

  • 0
    You're right, there are comple$x$ solutions. M$y$ problem is probabl$y$ at the ph$y$sics level, not the math level, so I will ask there.2012-06-16
5

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.

3

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

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

  • 0
    You did not specify this is a complex system.2012-06-16
3

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.