How to Calculate Expectation Value of Product State in a Potential-Free System?

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Lindsayyyy
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Hi everyone

Homework Statement



I have to particles without a potential. The coordinates are r_1 and r_2 (for particle 1 and 2). Both have orthonormal states |↑> and |↓>. I shall show that the expectation value is the following, where as |↑↓> is a product state

[tex]d^2=\langle \uparrow \downarrow \mid (r_1-r_2)^2 \mid \uparrow \downarrow \rangle = \langle \uparrow \mid r^2 \mid \uparrow\rangle +\langle \downarrow \mid r^2 \mid \downarrow \rangle -2 \langle \uparrow \mid \vec r \mid \uparrow \rangle \langle \downarrow \mid \vec r \mid \downarrow \rangle[/tex]



Homework Equations



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The Attempt at a Solution


Well, my attempt so far isn't very good I think because I have many problems understanding this.

I think I can write my r_1 as:

[tex]\vec r_1 = \frac {1}{\sqrt 2} (\mid \uparrow \rangle + \mid \downarrow \rangle)[/tex]
and the 2nd one as

[tex]\vec r_2 = \frac {1}{\sqrt 2} (\mid \uparrow \rangle + \mid \downarrow \rangle)[/tex]

I can now to the tensor product, but that doesn't lead to anywhere ( I tried to use it to get my up down product state, but this term looks so complicated I can't use it to ease up my euqations)

Thanks for your help
 
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Basically, the tensor product is simply a product of two separate subspaces. So,
[itex]\mid \uparrow \downarrow \rangle[/itex] can be more explicitly written as [itex]\mid \uparrow \rangle_{1}\mid\downarrow \rangle_{2}[/itex]

Furthermore, [itex]r_{1}[/itex] and [itex]r_{2}[/itex] are really [itex]r \otimes I[/itex] and [itex]I \otimes r[/itex] respectively.

You just have to expand [itex](r_1 - r_2)^2[/itex] and then "act them" on the appropriate subspaces.
 
thanks for your help so far.

yeah I know that (r1-r2)^2 is on of the binomial theorems. But I don't know actually how, let's say (r_1)^2 acts on ∣↑↓⟩. That's where I'm stuck.

edit: actually, I don't know what r even is (without the index). I thought it might have been a typing mistake by the task given, but you posted it aswell. Or did they just leave out the indices?
 
Lindsayyyy said:
edit: actually, I don't know what r even is (without the index). I thought it might have been a typing mistake by the task given, but you posted it aswell. Or did they just leave out the indices?
My best guess is what I posted earlier: [itex]r_{1} = r \otimes I[/itex] and [itex]r_{2} = I \otimes r[/itex] where I is identity. The subscripts 1 and 2 refer to the particle number.

Let me work out the trickier part explicitly: [itex]r_{1}r_{2} = r \otimes r[/itex]
Lets act it on the state:
[tex](\langle \uparrow \mid \otimes \langle \downarrow \mid)(r \otimes r)(\mid \uparrow \rangle \otimes \mid \downarrow \rangle)[/tex]
Now, operations on each subspace are independent of each other ie.
[tex](A \otimes B) (C \otimes D) = AC \otimes BD[/tex]
So, the previous expression simplifies to
[tex](\langle \uparrow \mid r \mid \uparrow \rangle) \otimes (\langle \downarrow \mid r \mid \downarrow \rangle)[/tex]
But these are just c-numbers. So the tensor product becomes a normal product and we arrive at
[tex]\langle \uparrow \mid r \mid \uparrow \rangle\langle \downarrow \mid r \mid \downarrow \rangle[/tex]
 
thanks for your help. I'm back home tomorrow then I will try to understand it a bit better. If I have problems again I will post here.