\[|V > =\frac{1+\sqrt{6}}{2\sqrt{6}}|00> + \frac{1-\sqrt{6}}{2\sqrt{6}}|01>+\frac{\sqrt{2}-\sqrt{3}}{2\sqrt{6}}|10>+\frac{\sqrt{2}-\sqrt{3}}{2\sqrt{6}}|11>\]
http://prntscr.com/8kds95

- dan815

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- dan815

@Empty

- Empty

cool

- Empty

pellet why did I speak it breaks this terrible site's latex :'(

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## More answers

- Empty

Maybe we should go to the superior Peer Answer

- dan815

lol

- Empty

- Empty

What would we do without wio's add on to make this site less trash? Like literally users are fixing the site haha.

- dan815

tkhunny would have your head for such talk

- Empty

I'll ban him idgaf

- Empty

ok so what do you wanna do with this shiz son? This is a matrix basically. I'll write it out too one sec.

- dan815

this tensor product |ij> = |i> * |j>, 1<=i,j<=2
whats that about

- dan815

does this first mean that i and j are 2 by 2 matrces and we take tensor product of these 2 matrices?

- dan815

or are they 2 vectors of dimension 2

- dan815

|dw:1443201862130:dw|

- dan815

like that?

- Empty

Nah not quite.

- Empty

\[|ij \rangle = |i \rangle \otimes |j \rangle \] i and j can each take on 0 or 1 to give you entries of a matrix.
It might be easier to imagine taking the product of two vectors to make a matrix:
\[A_{ij} = x_iy_j^\dagger\]
Notice that this is like the dot product but flipped backwards so that it creates a matrix. This is called the outer product as opposed to the dot product which is an inner product. :P

- dan815

wait an outer product always produces an operator right

- dan815

|dw:1443202519738:dw|

- Empty

Yeah, all you're doing is this: |dw:1443202706547:dw|

- dan815

ok i gotcha

- dan815

would that thing you did there be the tensor product of the 2 vectors i and j defined above

- dan815

where that matrix can be rewritten as
ac|11> + ad|12>+bc|21> +bd |22>

- dan815

i dont think that matrix can be rewritten like this nevermind

- dan815

se this would be like one of those column vectors they keep using for 4 states

- dan815

okay so basically a tensor product is just this diract notation for a matrix

- Empty

well ok so really all this does is take the hermitian conjugate:
\[|i \rangle^\dagger = \langle i| \]
Also I realize that this stuff is pretty confusing since it has a lot of overlap of notation for stuff.
Here we're looking at:
\[| ij \rangle = \left[\begin{matrix}a & b \\ c & d\end{matrix}\right] = a|00\rangle + b |01 \rangle +c |10 \rangle + d | 11 \rangle\]

- Empty

At least I think so, I might be mixing the notation up possibly.

- dan815

ya im looking through the book they are reseving this tensor product strictly for like a matrix and that way of writing http://prntscr.com/8ke94r for like your column matrices only like if u look at page 25 example 2.2.4

- dan815

http://prntscr.com/8keczo

- dan815

wait okay before that question so supppose i just write it out like this 2 by 2 matrix

- dan815

|dw:1443204170570:dw|

- dan815

http://prntscr.com/8kejen

- dan815

i better go eat

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