Consider a pure state $\boldsymbol{\eta} \in \mathcal{H}_{AB}$. There exist orthonormal sets $\{\alpha_1, \alpha_2 \dots \alpha_i\} \subset \mathcal{H}_A$ and $\{\beta_1, \beta_2 \dots \beta_i\} \subset \mathcal{H}_B$, and real numbers $\lambda_k > 0$ such that \begin{equation*} \boldsymbol{\eta} = \sum_{i=1}^d \lambda_i \alpha_i \otimes \beta_i \end{equation*} My question is if it is it possible (if so, how?) to find $d$ without using the decomposition above.

What I have done so far is that I have set $\text{dim}\mathcal{H}_A=m$ and $\text{dim}\mathcal{H}_B = n$. This means that $d \leq \min(m,n)$. Besides this, I do not know what to do. I know about a theorem called Caratheodory's theorem, but I am not sure if it will help me here. Can I use any of this to show that $d$ only depends on $\boldsymbol{\eta}$? Thanks!

  • $\begingroup$ Also $\mathcal{H}_{AB}=\mathcal{H}_{A}⊗\mathcal{H}_{B}$ is a pure state (Hilbert space) $\endgroup$
    – Annonymus
    Commented Jun 2, 2020 at 23:06
  • 1
    $\begingroup$ "to show that $d$ only depends on $\boldsymbol\eta$" what else could it depend on? $\endgroup$
    – glS
    Commented Jun 3, 2020 at 19:21

2 Answers 2


A priori, the only thing you can know is, as you say, $d\leq\min(m,n)$. To get more information, you're going to have to do a state-dependent calculation.

Let's say you're told $|\eta\rangle$ but not its Schmidt decomposition. So, you possibly have $$ |\eta\rangle=\sum_{i,j}\eta_{ij}|i\rangle_A|j\rangle_B, $$ and you want to know how many non-zero Schmidt coefficients it has. There are several ways which are all variants on making a start to finding the Schmidt decomposition. For example, you could calculate $$ \rho_A=\text{Tr}_B|\eta\rangle\langle\eta|. $$ In this case, $d=\text{rank}(\rho_A)$, so you just have to find the number of non-zero eigenvalues of $\rho_A$.

Equally, just write the coefficients $\eta_{ij}$ as an $m\times n$ matrix and find the rank (i.e. number of non-zero singular values).


It is important to refer the the concept of Schmidt rank to understand the minimum number of terms required in this summation expression. The Schmidt rank of vectors (pure states) and Schmidt number of states in a bipartite finite dimensional Hilbert space are measures of entanglement.

The concept of a subspace that does not contain any vector of Schmidt rank is quite useful to understand the bounds. It is proved by T. Cubitt, A. Montanaro, and A. Winter that the dimension of any subspace with Schmidt rank is bounded.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.