Find the number of elements in the Schmidt decomposition of a pure state

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!

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

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).