Mrthdrb

I'm trying to compute the Schmidt decomposition of $left| psi right> = (left| 00 right> + left| 01 right> + left| 10 right>)/sqrt3$. This should be possible by first computing the reduced density matrices $rho_A=Tr_Bleft| psi right> left< psi right|=sum_i p_i left| a_i right> left< a_i right|$ and $rho_B=Tr_Aleft| psi right> left< psi right|=sum_i p_i left| b_i right> left< b_i right|$, and then by identifying $left| psi right>=sum_i sqrtp_ileft| a_i right> otimes left| b_i right>$.

However, somewhere in the computation I'm doing a mistake: I find that
$$rho_A=rho_B=left(2left| 0 right> left< 0 right| + left| 0 right> left< 1 right| + left| 1 right> left< 0 right|+left| 1 right> left< 1 right|right)/3,$$
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
$$ left| psi right> = sqrtlambda_1 left| a_1 right> otimes
left| a_1 right> + sqrtlambda_2 left| a_2 right> otimes
left| a_2 right>.$$
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 $1/sqrt3$.

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

Thanks in advance.

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 $|a_irangle$. Then, rewrite
$$
|psirangle = sum_i |a_irangleotimes |b_irangle .tag*
$$
($|b_irangle$ can be determined e.g. as $|b_irangle = langle a_i|psirangle$.) Then, the $|b_irangle$ are orthogonal with $langle b_i|b_irangle=lambda_i$ (cf. Preskill), i.e., the form $(*)$ above is the Schmidt decomposition (upon normalizing the $|b_irangle$).

(The latter can be seen by computing the reduced density matrix of A from $(*)$, which yields
$$
sum |a_irangle langle a_j | ; langle b_j|b_jrangle = sum lambda_i |a_irangle langle a_i| ,
$$
which yields $langle b_j|b_jrangle = lambda_idelta_ij$ as the $|a_iranglelangle a_j|$ are linearly independent.)

Norbert Schuch's answer is correct. Just for fun, here's the exact decomposition:
$$
|psiranglepropto
(3+sqrt5)|Arangleotimes |Arangle
-(3-sqrt5)|Brangleotimes |Brangle
$$
with
beginalign
|Arangle &= 2|0rangle+(sqrt5-1)|1rangle \
|Brangle &= 2|0rangle-(sqrt5+1)|1rangle.
endalign
Using these equations, we can verify that the coefficient of $|11rangle$ is zero and that the coefficients of $|00rangle$, $|01rangle$, and $|10rangle$ are all equal to each other, and
$$
langle A|Brangle = 0.
$$
Unnormalized vectors $A,B$ are used here to simplify the coefficients in the overall expression for $|psirangle$.