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I am confused about something in Schmidt decomposition. I remind the theorem then a possible proof for it.

Schmidt theorem: Let $H_A$ and $H_B$ be two hilbert spaces of dimension $n_A < n_B$

For any vector $|\psi\rangle$ in this space, there exists an orthonormal basis of $H_A$, $|a_i\rangle$ and an orthonormal basis of $H_B$ $|b_i\rangle$ such that:

$$|\psi\rangle=\sum_{i=1}^{n_A} \sqrt{\lambda_i} |a_i\rangle|b_i\rangle$$

I start from an arbitrary state $|\psi\rangle=\sum_{i,j} c_{i,j} |e_i\rangle|f_j\rangle$ defined on $H_A \otimes H_B$.

Where $|\psi\rangle$ is an orthogonal basis in which $\rho_A$ is diagonal and $|f_j\rangle$ is any basis of $B$.

I can rewrite:

$$|\psi\rangle=\sum_{i} |e_i\rangle |\widetilde{b}_i\rangle$$

Where:

$$|\widetilde{b}_i\rangle=\sum_j c_{i,j} |f_j\rangle$$

It remains to prove that $|\widetilde{b}_i\rangle=\sqrt{\lambda_i}|b_i\rangle$. I now use the fact that the $|e_i\rangle$ diagonalise $\rho_A$:

$$\rho=\sum_{i,j,k,l} c_{i,j}c^*_{k,l} |e_i\rangle \langle e_k | |f_j\rangle \langle f_l |$$

$$\rho_A=\sum_{i,j,p} c_{i,p}c^*_{k,p} |e_i\rangle \langle e_k |$$

Thus, because $\rho_A$ must be diagonal in this basis:

$$\sum_{p} c_{i,p}c^*_{k,p} = \lambda_i \delta_{i,k}$$

Using this, I can prove that the family $\{| \widetilde{b}_k \rangle \}_k$ is a family of orthogonal vectors). Indeed:

$$\langle \widetilde{b}_k | \widetilde{b}_l \rangle = \sum_{i,j} c_{l,i} c^*_{k,j} \langle f_j | f_i \rangle = \sum_i c_{l,i} c^*_{k,i} = \lambda_l \delta_{k,l}$$

Which finally proves:

$$|\psi\rangle=\sum_{i=1}^{n_A} \sqrt{\lambda_i} |a_i\rangle|b_i\rangle$$

My question:

I don't understand where the condition $n_A<n_B$ is encoded in this proof ? I never used it. I know that if $n_A>n_B$ we would have a contradiction because it is not possible to find more than $n_B$ orthogonal vectors in $H_B$ but I don't get where it is encoded in this derivation. This disturbs me a lot to understand...

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2 Answers 2

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Start with a state $\vert\psi\rangle = \sum\limits_{i,j}c_{ij}\vert i\rangle\vert j\rangle$. Assume $n_A > n_B$ and take the partial trace over the $B$ subsystem to get

$$\rho_A =\sum\limits_{i,k,p} c_{ip}c^*_{kp} \vert i\rangle\langle k\vert$$

The index $p$ runs from $1$ to $n_B$. Thus, $\rho_A$ is the sum of $n_B$ rank $1$ matrices so it has rank at most $n_B$. When you diagonalize $\rho_A$, you will have at least $n_A - n_B$ zeros since the $A$ subsystem is of dimension $n_A$. Hence, in your line

$$\sum_{p} c_{i,p}c^*_{k,p} = \lambda_i \delta_{i,k}$$

you must allow $\lambda_i = 0$ and then there is no contradiction.

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The condition $n_A < n_B$ can be relaxed to $n_A \leq n_B$ here without causing problems. By the way, you can gain flexibility in your statement if you talk about "orthonormal systems" on the spaces rather than "orthonormal bases". After all, the dimensions $n_A$, $n_B$ have to be at least as large as the "Schmidt rank" which is the rank of $\rho_A$.

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  • $\begingroup$ Hello. Thanks for your answer. My problem is not the inequality being strict or not. It is on the condition on the dimension in itself. At which point exactly in my proof i would be wrong if I actually had na>nb ? $\endgroup$
    – StarBucK
    Dec 10, 2019 at 16:36
  • $\begingroup$ If $n_A>n_B$ and $\rho_A$ has full rank, the $b_i$s will fail to form an orthogonal system. Your proof might fail in proving this property. $\endgroup$
    – Gisbert
    Dec 10, 2019 at 16:53
  • $\begingroup$ I know but at which line in my proof it fails exactly. What disturbs me is that all my step seems consistant to me. I cannot point out precisely the moment my reasoning is wrong. This is the point of my question. Could you help me pointing out precisely the line I make an error ? $\endgroup$
    – StarBucK
    Dec 10, 2019 at 23:05
  • $\begingroup$ I think in case $n_A>n_B$, and $\rho_A$ having full rank your proof fails right at the start. The $f_j$ in your do not form an orthonormal basis - a property you use when evaluating the partial trace as you do. $\endgroup$
    – Gisbert
    Dec 11, 2019 at 11:25
  • $\begingroup$ The $f_j$ are by definition one orthonormal basis of $H_B$ which always exists $\endgroup$
    – StarBucK
    Dec 13, 2019 at 2:52

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