Scalar product of many-particle states?

[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>.

 

[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.

 

[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 :)

 

[Fredrik]

|\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

_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]