## dan815 one year ago $|V > =\frac{1+\sqrt{6}}{2\sqrt{6}}|00> + \frac{1-\sqrt{6}}{2\sqrt{6}}|01>+\frac{\sqrt{2}-\sqrt{3}}{2\sqrt{6}}|10>+\frac{\sqrt{2}-\sqrt{3}}{2\sqrt{6}}|11>$ http://prntscr.com/8kds95

1. dan815

@Empty

2. Empty

cool

3. Empty

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

4. Empty

Maybe we should go to the superior Peer Answer

5. dan815

lol

6. Empty

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

7. Empty

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

8. dan815

tkhunny would have your head for such talk

9. Empty

I'll ban him idgaf

10. 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.

11. dan815

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

12. dan815

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

13. dan815

or are they 2 vectors of dimension 2

14. dan815

|dw:1443201862130:dw|

15. dan815

like that?

16. Empty

Nah not quite.

17. 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

18. dan815

wait an outer product always produces an operator right

19. dan815

|dw:1443202519738:dw|

20. Empty

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

21. dan815

ok i gotcha

22. dan815

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

23. dan815

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

24. dan815

i dont think that matrix can be rewritten like this nevermind

25. dan815

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

26. dan815

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

27. 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: $| ij \rangle = \left[\begin{matrix}a & b \\ c & d\end{matrix}\right] = a|00\rangle + b |01 \rangle +c |10 \rangle + d | 11 \rangle$

28. Empty

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

29. 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

30. dan815
31. dan815

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

32. dan815

|dw:1443204170570:dw|

33. dan815
34. dan815

i better go eat