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

[Lindsayyyy]

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

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

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:

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

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

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

 

[Fightfish]

Basically, the tensor product is simply a product of two separate subspaces. So,
\mid \uparrow \downarrow \rangle can be more explicitly written as \mid \uparrow \rangle_{1}\mid\downarrow \rangle_{2}

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

You just have to expand (r_1 - r_2)^2 and then "act them" on the appropriate subspaces.

 

[Lindsayyyy]

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?

 

[Fightfish]

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: r_{1} = r \otimes I and r_{2} = I \otimes r where I is identity. The subscripts 1 and 2 refer to the particle number.

Let me work out the trickier part explicitly: r_{1}r_{2} = r \otimes r
Lets act it on the state:
(\langle \uparrow \mid \otimes \langle \downarrow \mid)(r \otimes r)(\mid \uparrow \rangle \otimes \mid \downarrow \rangle)
Now, operations on each subspace are independent of each other ie.
(A \otimes B) (C \otimes D) = AC \otimes BD
So, the previous expression simplifies to
(\langle \uparrow \mid r \mid \uparrow \rangle) \otimes (\langle \downarrow \mid r \mid \downarrow \rangle)
But these are just c-numbers. So the tensor product becomes a normal product and we arrive at
\langle \uparrow \mid r \mid \uparrow \rangle\langle \downarrow \mid r \mid \downarrow \rangle

 

[Lindsayyyy]

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.