# Schmidt decomposition - example

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.

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.)
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 $$|\psi\rangle$$.