# Explanation of why this derivation of Schmidt decomposition works

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?

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.

• 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 $\rho_{AB}$? – user1936752 Dec 7 '18 at 21:29
• @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. – Chiral Anomaly Dec 7 '18 at 22:09
• 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. – user1936752 Dec 7 '18 at 22:14