# Schmidt rank

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?

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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:
              | 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.

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