# Scalar product of many-particle states?

1. Nov 19, 2009

### Gerenuk

How do you find the scalar product of two non-orthogonal many particle states?
For example $<\leftarrow,\rightarrow|\uparrow,\downarrow>$
I wanted to express both as a 4-vector in the up/down basis, but this seems weird, since then a state $|\uparrow\downarrow+\downarrow\uparrow>$ is like $|\uparrow\uparrow+\downarrow\downarrow>$.

2. Nov 24, 2009

### peteratcam

What you said is correct. Express both as a vector in the 4-dimensional space of states.
A sensible basis for the hilbert space of two spin halfs is
uu,ud,du,dd
Your last sentence doesn't make sense though. (ud+du) is obviously different from (uu + dd)
(0,1,1,0) vs. (1,0,0,1);

Your first example is
<(u-d)(u+d)/2|ud>=(1/2)<uu-du-dd+ud|ud> = 1/2.
where I've used the fact that downx = (u-d)/root(2) and upx = (u+d)/root(2) for a single spin half.

3. Nov 24, 2009

### Gerenuk

Thanks for the answer. I used to do calculations right, but I wasn't aware that some other weird approach I tried was something incorrect. I had some misconception about entanglement in mind.
Today I wrote a python program that seems to get it right and so I can play around with it :)

4. Nov 24, 2009

### Fredrik

Staff Emeritus
$|\uparrow,\downarrow\rangle$ is a shorthand notation for the tensor product $\left|\uparrow\rangle_1\otimes\left|\downarrow\rangle_2$. I'm actually not sure if the notation $\langle\leftarrow,\rightarrow|$ usually puts particle 1 first, or if the order is reversed in the bras. I'm guessing that the order is the same. (Check your book to make sure). If that's the case, you're looking for the scalar product of $\left|\leftarrow\rangle_1\otimes\left|\rightarrow\rangle_2$ and $\left|\uparrow\rangle_1\otimes\left|\downarrow\rangle_2$, which is defined as

[tex]_2\langle\rightarrow|\otimes\ _1\langle\leftarrow|\ (\left|\uparrow\rangle_1\otimes\left|\downarrow\rangle_2)={}_1\langle\leftarrow|\uparrow\rangle_1\ _2\langle\rightarrow|\downarrow\rangle_2[/itex]

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