1
$\begingroup$

On page 547 of N&C, for $|\psi_{0}\rangle=|0\rangle$ and $|\psi_{1}\rangle=(|0\rangle+|1\rangle)/\sqrt{2}$ and for $|\tilde{0}\rangle=\cos(\pi/8)|0\rangle+\sin(\pi/8)|1\rangle$ and $|\tilde{1}\rangle=-\sin(\pi/8)|0\rangle + \cos(\pi/8)|1\rangle$, that $|\langle\tilde{0}|\psi_{k}\rangle| = \cos(\pi/8)$ and $|\langle\tilde{1}|\psi_{k}\rangle| = \sin(\pi/8)$ for k = $\{0,1\}$. I just don't see how this can be the case for either of them.

I get $\cos(\pi/8)$ and $\cos(\pi/8)/\sqrt{2}+\sin(\pi/8)\sqrt{2}$ for $|\langle\tilde{0}|\psi_{k}\rangle|$ and for $|\langle\tilde{1}|\psi_{k}\rangle|$ I get $-\sin(\pi/8)$ and $-\sin(\pi/8)/\sqrt{2}+\cos(\pi/8)/\sqrt{2}$.

The inner product only produce these for $|\psi_{0}\rangle$. Is this a typo, and what they mean to say is that the inner product $|\langle\tilde{0}|\psi_{k}\rangle|$ is much larger than $|\langle\tilde{1}|\psi_{k}\rangle|$. But even in this case, that isn't completely true, as it's only just larger in the case of $|\langle\tilde{1}|\psi_{k}\rangle|$

For context, $|\tilde{0}\rangle$ and $|\tilde{1}\rangle$ come from the spectral decomposition of the density operator representing the source that generates $|\psi_{0}\rangle$ and $|\psi_{1}\rangle$ with probability a half for each.

What am I missing here? It seems like a simple inner product should be used but I can't get their results.

$\endgroup$
2
$\begingroup$

You're missing a bit of algebraic trickery. Remember that $\frac{1}{\sqrt{2}}=\sin(\pi/4)=\cos(\pi/4)$. Thus, $$ \cos(\pi/8)/\sqrt{2}+\sin(\pi/8)/\sqrt{2}=\cos(\pi/8)\cos(\pi/4)+\sin(\pi/8)\sin(\pi/4)=\cos(\pi/4-\pi/8)=\cos(\pi/8) $$ by the double angle formula.

Also, be careful of signs. It might be an amplitude is $\pm\sin(\pi/8)$, but when you take the modulus, that becomes $\sin(\pi/8)$.

$\endgroup$
1
  • $\begingroup$ That is...... extremely frustrating. Yeah the -sin was typo on my part, but the rest of it, I made the mistake of checking it using a calculator and forgot to convert into radians. Jesus every time I think I have something down some small part like this knocks me on my ass. $\endgroup$ Jul 13 '20 at 15:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.