Wolfram alpha disagrees with the answer in question (b).

Wolfram alpha link to the solution

I have also worked out the example and got similar results to the answer in the second link above.

Am I missing something?

Wolfram alpha disagrees with the answer in question (b).

Wolfram alpha link to the solution

I have also worked out the example and got similar results to the answer in the second link above.

Am I missing something?

Hey @Nicolas_Hadjittoouli! Our answer is correct, but so is the one on Wolfram (sort ofâ€¦)! Let me show you:

Wolframâ€™s eigenvectors are

\begin{align*}
\vert \tilde{v}_1 \rangle & = -0.780776 i \vert 0 \rangle + \vert 1 \rangle \\
\vert \tilde{v}_2 \rangle & = 1.28078 i \vert 0 \rangle + \vert 1 \rangle \
\end{align*}.

I put the â€śtildeâ€ť (~) sign over those vectors because they arenâ€™t *normalized*! After normalizing them, we get

\begin{align*}
\vert v_1 \rangle & = -0.61541221 i \vert 0 \rangle + 0.780776 \vert 1 \rangle \\
\vert v_2 \rangle & = 0.780776 i \vert 0 \rangle + 0.61541221 \vert 1 \rangle
\end{align*}.

So, at this stage, \vert v_1 \rangle matches what we have in the codebook. But \vert v_2 \rangle still looks a little different. Hereâ€™s what we can do to make it â€ślookâ€ť like ours:

\begin{align*}
\vert v_2 \rangle & = 0.780776 i \vert 0 \rangle + 0.61541221 \vert 1 \rangle \\
\vert v_2 \rangle & = i \left( 0.780776 \vert 0 \rangle - 0.61541221 i \vert 1 \rangle \right) \\
\end{align*}

In the second line, I factored out the factor of i, which essentially acts like a global phase to the state. We can simply â€śdo awayâ€ť with the global phase and redefine \vert v_2 \rangle as

\vert v_2 \rangle =0.780776 \vert 0 \rangle - 0.61541221 i \vert 1 \rangle,

which is the answer that we have in the codebook.

So, two things to remember when finding eigenvectors:

- Always normalize them
- Global phases donâ€™t matter

1 Like

Just to illustrate this with a more familiar-feeling example, the Pauli Y matrix

\hat{\sigma}^y = \begin{pmatrix}
0 & -i \\
i & 0
\end{pmatrix}

has eigenvalues / eigenvectors

1, \sqrt{\frac{1}{2}}\left(\vert 0 \rangle + i \vert 1 \rangle\right)

and

-1, \sqrt{\frac{1}{2}}\left(\vert 0 \rangle - i \vert 1 \rangle\right).

Try factoring out factors of i and playing the same game that we did with the codebook example and you will find that they donâ€™t affect the eigenequations!

Oh, right! I couldnâ€™t see, how I could get there. I totally forgot about the normalization.

Thanks a lot for the help!

1 Like

Thank you for posting here! I forget about normalizing every once and a while too