What is the Schmidt Rank and Its Role in Measuring Entanglement?

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Discussion Overview

The discussion revolves around the concept of Schmidt rank in the context of measuring entanglement in pure bipartite quantum states. Participants explore its definition, properties, and implications for entanglement, including questions about continuity and additivity.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant seeks clarification on whether the Schmidt rank is equal to the rank of the reduced density matrix.
  • Another participant introduces the Schmidt decomposition and discusses how it relates to the entanglement between two parties, Alice and Bob, using a matrix representation of their quantum states.
  • A participant questions the continuity and additivity of the Schmidt rank, suggesting that it may not be continuous or additive and referencing a source that discusses infinite Schmidt rank.
  • There is mention of the Schmidt rank being defined as the number of terms in the Schmidt decomposition and its relation to the dimension of the support of the reduced density matrix.
  • One participant speculates on whether the Schmidt rank can serve as a measurement of entanglement, particularly for systems with local dimension 2, expressing uncertainty about how to explain this idea.

Areas of Agreement / Disagreement

Participants express uncertainty regarding the properties of the Schmidt rank, particularly its continuity and additivity. There is no consensus on these issues, and multiple viewpoints are presented without resolution.

Contextual Notes

Participants reference various definitions and properties of the Schmidt rank, but there are unresolved questions about its continuity and additivity, as well as its role as a measure of entanglement.

valesdn
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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?
Is the Schmidt rank continuous and/or additive? I have no found on the net any articles or information about this.
Could you help me?

Thanks in advance.
 
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Are you referring to the schmidt decompositionhttps://en.wikipedia.org/wiki/Schmidt_decomposition)?

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

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