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.


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.)

  • $\begingroup$ A link or source to Preskill’s notes would nicely complement your answer. $\endgroup$ – ZeroTheHero Mar 27 '19 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 '19 at 9:45

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 $|\psi\rangle$.


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