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I have a problem minimizing this norm with respect to $\alpha$:

$\min_{\alpha}||e^{i\alpha}|\psi\rangle-|\phi\rangle ||^2$ (1)

The result is that this achieves min when $\alpha=-\measuredangle \langle\psi|\phi\rangle$

and the above is equal to $2-2\langle\psi|\phi\rangle$.

But when I take the derivative of (1) and set it to 0, I get $e^{i\alpha} = \langle\psi|\phi\rangle$, which is not the right desired result.

Can anyone help me with this, please? thanks in advance.

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  • $\begingroup$ You wouldn’t be solving for the maximum probability of distinguishing those two qubits with a measurement by any chance, would you? $\endgroup$ Mar 18, 2023 at 0:54

1 Answer 1

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For unit vectors $|\psi\rangle, |\phi\rangle \in \mathbb{C^n}$ we have: $$ ||e^{i\alpha}|\psi\rangle-|\phi\rangle ||^2 = \left(e^{-\alpha i}\langle\psi|-\langle \phi| \right) \left(e^{i \alpha} |\psi\rangle - |\phi\rangle \right) = 2 - 2Re(e^{i \alpha}\langle \phi|\psi \rangle). $$ Since the inner product of $\langle \phi|\psi \rangle$ might be a complex number, we can write it as $\langle \phi|\psi \rangle = a + ib$. We can set $b=0$ if the inner product is real. Then we have $$Re(e^{i \alpha}\langle \phi|\psi \rangle) = Re((\cos\alpha+i\sin\alpha)(a + ib)) = a \cos \alpha - b \sin \alpha.$$ Hence, the quantity we want to minimize is $$ \min_{\alpha} \left \{2 - 2(a \cos \alpha - b \sin \alpha) \right\}. $$ Take the derivative of $2 - 2(a \cos \alpha - b \sin \alpha)$ w.r.t. $\alpha$ and set it to zero. Then we have $$ a \sin \alpha = b \cos \alpha. $$ Equivalently, $$ \tan \alpha = \frac{b}{a}, \textrm{ for } a \neq 0. $$

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