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?


2 Answers 2


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).


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.

  • $\begingroup$ 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 $\rho_{AB}$? $\endgroup$ Dec 7, 2018 at 21:29
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    $\begingroup$ @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. $\endgroup$ Dec 7, 2018 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. $\endgroup$ Dec 7, 2018 at 22:14

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