# Ways in which $\frac{1}{\sqrt 2} (|00\rangle + |11\rangle)$ can be expressed as $\frac{1}{\sqrt 2} (|uu\rangle + |vv\rangle)$

I want to find out what values $$|u\rangle$$ and $$|v\rangle$$ can take if I want to write $$\frac{1}{\sqrt 2} (|00\rangle + |11\rangle)$$ as $$\frac{1}{\sqrt 2} (|uu\rangle + |vv\rangle).$$

Say $$|u\rangle = a|0\rangle + b|1\rangle$$

$$|v\rangle = c|0\rangle + d|1\rangle.$$

Now, $$\frac{1}{\sqrt 2} (|uu\rangle + |vv\rangle)$$

= $$(a^2 + b^2)|00\rangle + (ab + cd)(|01\rangle + |10\rangle) + (c^2 + d^2)|11\rangle.$$

We have

$$(a^2 + b^2)e^{i\theta} = 1$$

$$(c^2 + d^2)e^{i\theta} = 1$$

(for the same $$\theta$$)

$$ab + cd = 0$$

We also know that:

$$|a|^2 + |b|^2 = 1 \implies a^*a + b^*b = 1$$

$$|c|^2 + |d|^2 = 1 \implies c^*c + d^*d = 1$$

How do I find the relation between $$a, b, c, d$$ as rigorously as possible?

I think the clearest way to specify this is $$|u\rangle,|v\rangle\in\mathbb{R}^2$$ such that $$\langle u|v\rangle=0$$ (up to the same global phase shared by both states).
To see how this corresponds to what you wrote: We start with $$b^2=e^{-i\theta}-a^2$$ and take the mod-square: $$|b|^4=1+|a|^4-a^2e^{i\theta}-{a^*}^2e^{-i\theta}$$ and we can compare this to the square of the normalisation condition: $$|b|^4=1+|a|^4-2|a|^2$$ Hence, we require $$a^2e^{i\theta}+{a^*}^2e^{-i\theta}=2|a|^2,$$ which must mean that $$a=|a|e^{-i\theta/2}$$. Putting this back in $$b^2=e^{-i\theta}-a^2$$ gives that $$b=e^{-i\theta/2}|b|$$. Hence, we might as well take $$a,b\in\mathbb{R}$$, with $$\theta/2$$ just being a global phase. We can apply an identical argument to $$c,d$$.
Now that we know $$a,b,c,d$$ are real, we see that $$\langle u|v\rangle=ab+cd=0$$, i.e. $$|u\rangle$$ and $$|v\rangle$$ are orthogonal.
Actually, I should probably point out that you're presupposing that $$\langle u|v\rangle=0$$ because otherwise your state $$(|uu\rangle+|vv\rangle)/\sqrt{2}$$ wouldn't be normalised.
I've also just noticed that you've switched somewhere in the middle between b being $$\langle 1|u\rangle$$ and $$\langle 0|v\rangle$$. I don't think that affects the calculation, but it would be better to have it consistent!
The state $$|\omega\rangle=\tfrac{1}{\sqrt{2}}(|00\rangle+|11\rangle)$$ is invariant under transformations of the form $$U\otimes \bar{U}$$, with $$U$$ unitary (or more generally, $$X\otimes {\bar X^{-1}}$$): $$|\omega\rangle=(U\otimes \bar U)|\omega\rangle\ .$$ Thus, \begin{align} |\omega\rangle &= \tfrac{1}{\sqrt{2}}(U|0\rangle\otimes \bar U|0\rangle+U|1\rangle\otimes \bar U|1\rangle) \\ &= \tfrac{1}{\sqrt{2}}(|u\rangle\otimes |\bar u\rangle+|v\rangle\otimes |\bar v\rangle)\ , \end{align} with $$|u\rangle=U|0\rangle$$, $$\bar u=\bar U|0\rangle$$, etc. Thus, if you want $$|\bar u\rangle=|u\rangle$$, $$|\bar v\rangle=|v\rangle$$, you need to choose $$U$$ real. In that case, \begin{align} |u\rangle = \cos\phi|0\rangle + \sin\phi|1\rangle\ , |v\rangle = -\sin\phi|0\rangle + \cos\phi|1\rangle\ . \end{align}