2
$\begingroup$

I'm following Preskill's notes and he derives the Schmidt decomposition in the following way:

Let a bipartite state be $\psi_{AB} = \sum_{i,j}\lambda_{ij}\vert i\rangle\vert j\rangle = \sum_{i} \vert i\rangle\vert \tilde{i}\rangle$, where I simply choose $\sum_j \lambda_{ij}\vert j\rangle = \vert \tilde{i}\rangle$.

I choose a set of basis vectors $\vert i\rangle$ such that the partial state is diagonal, that is $\rho_A = \sum_i p_i\vert i\rangle\langle i\vert$. But I can also obtain $\rho_A = Tr_B(\rho_{AB}) = Tr_B\sum_{i,j} \vert i\rangle\langle j\vert \otimes \vert \tilde{i}\rangle\langle \tilde{j}\vert = \sum_{ij} \langle \tilde{j}\vert\tilde{i}\rangle \vert i\rangle\langle j\vert$. The last part can be computed by explicitly writing out the trace over $B$ and using the properties of an orthonormal basis.

Thus, we have $\rho_{A} = \sum_i p_i\vert i\rangle\langle i\vert = \sum_{ij} \langle \tilde{j}\vert\tilde{i}\rangle \vert i\rangle\langle j\vert$. That is $\langle \tilde{j}\vert \tilde{i}\rangle = p_i\delta_{ij}$. Suddenly, the $\vert\tilde{i}\rangle$ are all orthogonal to each other.

Why does choosing the basis where $\rho_A$ is diagonal also give you orthogonal vectors in $B$? This seemed to drop out of the sky for me although the math is clear. What is the physical meaning of this?

$\endgroup$
2
$\begingroup$

Let us start from the Schmidt decomposition $|\psi\rangle = \sum s_i |a_i\rangle |b_i\rangle$.

Now consider the reduced state of $A$: $\rho_A=\sum s_i^2 |a_i\rangle\langle a_i|$. This is, the eigenbasis of A is exactly the basis you need for the Schmidt decomposition!

Thus, if you write your state using that eigenbasis of Alice, $$ |\psi\rangle = \sum_i |a_i\rangle \Big(\sum_j \lambda_{ij}|j\rangle\Big)\ , $$ the part $|\tilde b_i\rangle=\sum_j \lambda_{ij}|j\rangle$ must be equal to $s_i|b_i\rangle$, since the Schmidt decomposition is unique (modulo degeneracies).

$\endgroup$
2
$\begingroup$

Why does choosing the basis where $\rho_A$ is diagonal also give you orthogonal vectors in $B$?

The answer is in the proof shown in the question. I'll write it out here in a slightly different way to try to help highlight what's happening:

Suppose that the state $$ \psi_{AB}=\sum_n |A_n\rangle |B_n\rangle \tag{1} $$ is such that the reduced state $$ \rho_A = \text{Trace}_{B}(\psi_{AB}) \tag{2} $$ is diagonal in the $A_n$ basis. More explicitly, the reduced state is defined by $$ \rho_A = \sum_k \big(\sum_n |A_n\rangle \langle \hat B_k|B_n\rangle\big) \big(\sum_m \langle B_m|\hat B_k\rangle \langle A_m|\,\big) \tag{3} $$ where the vectors $|\hat B_k\rangle$ are orthonormal by definition (because we're using them to compute the trace). This implies $$ \rho_A = \sum_{n,m} |A_n\rangle \langle B_m|B_n\rangle \langle A_m|. \tag{4} $$ We assumed that $\rho_A$ is diagonal in the $A_n$ basis, and the terms in the sum in (4) are all linearly independent, this is only possible if the coefficient of each individual off-diagonal term is zero: $$ \langle B_m|B_n\rangle = 0. $$ Thus equation (1) is in Schmidt form.

$\endgroup$
  • $\begingroup$ Sorry, maybe my question should have been clearer but I understood that there is a unique basis choice for $A$ (as opposed to arbitrary basis of $A$) that gives me orthogonal vectors in $B$. I just don't see why this particular choice of basis that diagonalizes $\rho_A$ is also able to give orthogonal vectors on $B$. What is the connection between diagonalizing $\rho_A$ and obtaining the Schmidt description for $\rho_{AB}$? $\endgroup$ – user1936752 Dec 7 '18 at 21:29
  • 1
    $\begingroup$ @user1936752 I replaced my answer with one that tries to make this connection more clear, although it's just a re-write of the original proof. The point is that if we look at equation (3) and assume that the off-diagonal terms are zero (which is what we're saying when we say that $\rho_A$ is diagonal in the $A_n$ basis), the conclusion that the $B_n$ are orthogonal follows immediately. I wouldn't say this has any "physical meaning"; it's just a mathematical identity. $\endgroup$ – Chiral Anomaly Dec 7 '18 at 22:09
  • $\begingroup$ Alright, thanks anyway for writing it out! I'll leave the question open for a bit to see if someone else has some intuition for why this works beyond just the mathematical proof. $\endgroup$ – user1936752 Dec 7 '18 at 22:14

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.