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meBigGuy

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- Thread starter meBigGuy
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meBigGuy

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meBigGuy

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Office_Shredder

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https://www.physicsforums.com/forumdisplay.php?f=62

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meBigGuy

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The "broadest" vector state product is the Cartesian product, and it includes the "new product" terms that one might try to associate with classical multiplying. The tensor product somehow eliminates those terms.

But classical multiplying is all about mixing, and its application expresses an interaction between elements. The tensor product provides a type of vector stae multiplication that works such that associative and communicative laws hold, but so does classical multiplication in its domain. The subtleties of what the tensor product operation actually represents and the nature of what is being excluded escapes me.

Saying that it represents all the possible states of two fully independent electron's spin vectors is well and fine, and I see the value of it in subsequent manipulations, but I don't get the why of it.

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naima

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(a Phi1,b phi2) and (ph1, phi2) are mapped to states that are collinear viz that have

the same physical content

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meBigGuy

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The idea behind the tensor product, as applied to quantum mechanics, is pretty simple: If you have one particle in state [itex]|\psi\rangle[/itex] and another particle in state [itex]|\varphi\rangle[/itex], then the composite system, made up of those two particles, is in the state [itex]|\psi\rangle \otimes |\varphi\rangle[/itex]. The technical details of what makes it a tensor product are how the operation [itex]\otimes[/itex] works on superpositions.

If [itex]|\psi\rangle = a |\psi_1\rangle + b |\psi_2\rangle[/itex] and [itex]|\varphi\rangle = c |\varphi_1\rangle + d |\varphi_2\rangle[/itex], where [itex]a, b, c, [/itex] and [itex]d[/itex] are complex numbers, then

[itex]|\psi\rangle \otimes |\varphi\rangle

= a c (|\psi_1\rangle \otimes |\varphi_1\rangle)

+ a d (|\psi_1\rangle \otimes |\varphi_2\rangle)

+ b c (|\psi_2\rangle \otimes |\varphi_1\rangle)

+ b d (|\psi_2\rangle \otimes |\varphi_2\rangle[/itex])

I don't think that there is anything else you really need to know about tensor products. Are you wondering why the tensor product is used for composite systems?

- #8

meBigGuy

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Maybe, but not why in the simple sense that it is communicative and associative and therfore usefull, but why at a higher level. Why it turns out to represents the unentangled system states. Is it like "we want an operation that can work in this equation and we will call it a tensor product"Are you wondering why the tensor product is used for composite systems?

I think I read that the tensor product is the cartesian product with certain terms removed so it becomes communicative and associative, but that is still kind of abstract.

Is there a classical system analogue to tensor product? What are other physical systems where it is useful? I didn't find any in Arfken.

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