# Schmidt rank

1. Jun 23, 2015

### valesdn

Hi guys. I'm studying an article on the measurements of entanglement in a pure bipartite state.
I don't understand the definition of the Schmidt rank. It is equal to the rank of the reduced density matrix, isn't it?
Could you help me?

2. Jun 23, 2015

### Strilanc

Are you referring to the schmidt decompositionhttps://en.wikipedia.org/wiki/Schmidt_decomposition)? [Broken]

Maybe this blog post will be helpful?

Suppose you have a quantum system involving two parties, Alice and Bob. Alice has an $n$-level quantum system, and Bob has an $m$-level quantum system. The combined system can be described by $n m$ amplitudes. Arrange those amplitudes into a grid with $n$ rows and $m$ columns, where each row corresponds to one of Alice's system's levels and each column corresponds to one of Bob's system's levels. For example, if they both have a qubit then the grid would be laid out like:

Code (Text):

| Alice qubit OFF | Alice qubit ON
--------------+-----------------+----------------------
Bob qubit OFF | amplitude_00    | amplitude_01
Bob qubit ON  | amplitude_10    | amplitude_11
Which happens look a whole lot like a matrix:

$\begin{bmatrix} a_{00} & a_{01} \\ a_{10} & a_{11} \end{bmatrix}$

and if you treat it like a matrix, and perform a singular value decomposition to turn it into $U \cdot S \cdot V$, then you'll find that Alice's operations affect $U$ and Bob's operations affect $V$ but neither of them can affect $S$. Because $S$ is a pretty simple matrix, with nothing but real non-negative entries in descending order along its diagonal, it makes sense to think of the entries on that diagonal as a measure of the entanglement between Alice and Bob.

The terms on the diagonal of $S$ might be what you mean by "schmidt rank"?

Fully entangled systems have an equal value along the whole diagonal, making the system act a whole lot like a unitary matrix. Non-entangled systems have a single non-zero entry in the top left cell, and act like the tensor product of two vectors.

Last edited by a moderator: May 7, 2017
3. Jun 23, 2015

### atyy

Where did you see the term, and how is it used?

4. Jun 23, 2015

### valesdn

Thank you Strilanc for your explanation. I don't know if the terms on the diagonal are what I mean "Schmidt rank". I think so. I have just read that for all quantum system, the sum of all eigenvalues is essentially finite and the whole entangled system becomes finite dimensional. I don't know if it is useful to consider "not continuos" the Schmidt rank ( however it is possibile use continuous variables and define an infinite Schmidt rank, cit. Lewenstein). I should demonstrate that the rank is neither continuos nor additive.
Hi atyy. I read an article ( maybe lecture notes) where the Schmidt rank is defined as the number of terms in the Schmidt decomposition, and it is also equal to the dimension of the support of the reduced density matrix. That's all.
Could the Schmidt rank be a measurement of entanglement ( being neither continuos nor additive)? For systems with local dimensional 2, I think that it is not possible. I don't know how to explain it...