Quick Question About Tensor Products

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Discussion Overview

The discussion revolves around the properties of unitary operators acting on tensor products of quantum states, specifically whether such operators are distributive over the tensor product. The context is primarily within quantum mechanics and quantum computing.

Discussion Character

  • Technical explanation
  • Debate/contested
  • Conceptual clarification

Main Points Raised

  • One participant questions if the unitary operator U is distributive over the tensor product or behaves like a dot product with scalars.
  • Another participant suggests that the behavior of U depends on its definition and context.
  • A participant clarifies that U is a unitary matrix, which may influence its action on tensor products.
  • It is proposed that there are multiple ways to define the action of a linear map on tensor products, depending on the context.
  • One participant emphasizes the need for clarity regarding the definition of "context" and the specific action of U on the tensor product.
  • Another participant notes that the action is typically on the right, expressing surprise at the lack of standard properties for linear operators over tensor products.
  • A later reply indicates that while one expression is correct, the context often dictates how the operator acts on the individual kets.
  • It is mentioned that for distributivity to hold, a coalgebra structure or comultiplication map would be necessary.
  • One participant expresses understanding after the discussion.

Areas of Agreement / Disagreement

Participants express differing views on the distributive properties of operators over tensor products, with no consensus reached on a definitive answer. The discussion remains unresolved regarding the standard properties of linear operators in this context.

Contextual Notes

Participants highlight the importance of context in defining the action of operators on tensor products, indicating that assumptions about operator behavior may vary based on specific definitions and applications.

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In the expression

[tex]U(|a \rangle \otimes |b \rangle)[/tex]

is U distributive over the tensor product or does it behave like the dot product with scalars, i.e.

[tex]U(|a \rangle \otimes |b \rangle) = U|a \rangle \otimes |b \rangle = |a \rangle \otimes U|b \rangle[/tex]
 
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It depends on what U is?
 
Right. U is a unitary matrix.
 
I hope I can see through the unimportant, and confusing, QM notation to say that you can define the action of a linear map on V on V\otimes V in at least 3 ways. It is up to the context to determine which you mean.
 
This is in the context of quantum computing.
 
That is not the meaning of the word "context" I was aiming for. What is the definition given in the source this occurs? That is an elementary but necessary question. The action could be on the left, the right or diagonal. (But none of these is precisely what you wrote: a left and right action will not in general coincide as you implied.) None of these is necessarily canonical.
 
From what I'm reading, the action is on the right. I thought there would be some standard properties that linear operators obey over tensor products. Weird.
 
Actually the only right thing then is

[tex]U(|a \rangle \otimes |b \rangle)[/tex]

but as the others say, it depends on the context, because often the operator is defined in such a way that it is clear that it is only working on the first ket, or the second ket, and then one would write it just like above and then it would be implicit that

[tex]U(|a \rangle \otimes |b \rangle) = U|a \rangle \otimes |b \rangle[/tex]

or in the second case

[tex]U(|a \rangle \otimes |b \rangle) = |a \rangle \otimes U|b \rangle[/tex]

but this is because one actually mean

[tex]U = 1 \otimes U[/tex]

or

[tex]U = U \otimes 1[/tex]

which is a ***** to write all the time so the shorthand notation is used. But in general a operator isn't distributive over the tensor product.
 
In order for it to be distributive, one would need something like a coalgebra structure, or at least a comultiplication map from X to X\otimes X.
 
  • #10
I think I understand now. Thanks.
 

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