3
$\begingroup$

I'm trying to compute the Schmidt decomposition of $\left| \psi \right> = (\left| 00 \right> + \left| 01 \right> + \left| 10 \right>)/\sqrt{3}$. This should be possible by first computing the reduced density matrices $\rho_A=Tr_B\left| \psi \right> \left< \psi \right|=\sum_i p_i \left| a_i \right> \left< a_i \right|$ and $\rho_B=Tr_A\left| \psi \right> \left< \psi \right|=\sum_i p_i \left| b_i \right> \left< b_i \right|$, and then by identifying $\left| \psi \right>=\sum_i \sqrt{p_i}\left| a_i \right> \otimes \left| b_i \right>$.

However, somewhere in the computation I'm doing a mistake: I find that $$\rho_A=\rho_B=\left(2\left| 0 \right> \left< 0 \right| + \left| 0 \right> \left< 1 \right| + \left| 1 \right> \left< 0 \right|+\left| 1 \right> \left< 1 \right|\right)/3,$$ which has eigenvalues $\lambda_1=0.87$ and $\lambda_2=0.13$ with corresponding eigenvectors $$ \left| a_1 \right>=0.85 \left| 0 \right> + 0.53 \left| 1 \right>,\quad \left| a_2 \right>=-0.53 \left| 0 \right> + 0.85 \left| 1 \right>. $$ This means that $\left| \psi \right>$ should be given by $$ \left| \psi \right> = \sqrt{\lambda_1} \left| a_1 \right> \otimes \left| a_1 \right> + \sqrt{\lambda_2} \left| a_2 \right> \otimes \left| a_2 \right>.$$ This, however, is not true. If you check for example the numerical value in front of $\left| 00 \right>$, you find that it is not equal to $1/\sqrt{3}$.

I would appreciate if someone could help me to see where I made the mistake.

$\endgroup$
3
$\begingroup$

The phase of your eigenvectors is not correct (or rather, it is not determined, so you need to make a judicious choice). If you put a minus sign in front of the 2nd term, it works out.


To elaborate on that: If you want to find the Schmidt decomposition, you can proceed e.g. as in Preskill's lecture notes: Diagonalize the reduced state of A, which yields eigenvalues $\lambda_i$ and eigenvectors $|a_i\rangle$. Then, rewrite $$ |\psi\rangle = \sum_i |a_i\rangle\otimes |b_i\rangle\ .\tag{*} $$ ($|b_i\rangle$ can be determined e.g. as $|b_i\rangle = \langle a_i|\psi\rangle$.) Then, the $|b_i\rangle$ are orthogonal with $\langle b_i|b_i\rangle=\lambda_i$ (cf. Preskill), i.e., the form $(*)$ above is the Schmidt decomposition (upon normalizing the $|b_i\rangle$).

(The latter can be seen by computing the reduced density matrix of A from $(*)$, which yields $$ \sum |a_i\rangle \langle a_j | \; \langle b_j|b_j\rangle = \sum \lambda_i |a_i\rangle \langle a_i|\ , $$ which yields $\langle b_j|b_j\rangle = \lambda_i\delta_{ij}$ as the $|a_i\rangle\langle a_j|$ are linearly independent.)

$\endgroup$
  • $\begingroup$ A link or source to Preskill’s notes would nicely complement your answer. $\endgroup$ – ZeroTheHero Mar 27 at 9:37
  • $\begingroup$ @ZeroTheHero I think the answer is self-contained. And if I add a link, I would have to link a specific section, and Preskill occasionally updates his notes, which could both make the link and an exact reference within the notes obsolete, so people might complain about that. $\endgroup$ – Norbert Schuch Mar 27 at 9:45
2
$\begingroup$

Norbert Schuch's answer is correct. Just for fun, here's the exact decomposition: $$ |\psi\rangle\propto (3+\sqrt{5})|A\rangle\otimes |A\rangle -(3-\sqrt{5})|B\rangle\otimes |B\rangle $$ with \begin{align} |A\rangle &= 2|0\rangle+(\sqrt{5}-1)|1\rangle \\ |B\rangle &= 2|0\rangle-(\sqrt{5}+1)|1\rangle. \end{align} Using these equations, we can verify that the coefficient of $|11\rangle$ is zero and that the coefficients of $|00\rangle$, $|01\rangle$, and $|10\rangle$ are all equal to each other, and $$ \langle A|B\rangle = 0. $$ Unnormalized vectors $A,B$ are used here to simplify the coefficients in the overall expression for $|\psi\rangle$.

$\endgroup$

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.