Quantum Computing Asked by Annonymus on January 5, 2021

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!

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

Let's say you're told $|etarangle$ but not its Schmidt decomposition. So, you possibly have $$ |etarangle=sum_{i,j}eta_{ij}|irangle_A|jrangle_B, $$ 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 $$ rho_A=text{Tr}_B|etaranglelangleeta|. $$ 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 $mtimes n$ matrix and find the rank (i.e. number of non-zero singular values).

Answered by DaftWullie on January 5, 2021

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.

Answered by Gokul Alex on January 5, 2021

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