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?


2 Answers 2


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}


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!


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