Help with the matrix representation of <-|+|->. Does "+"=|+>?

[yoyopizza]

Homework Statement

I'm working on problem 1.10 in Sakurai (1.8 in version 1 or 2), where I'm supposed to prove the commutation relation on [S_i,S_j]=i\epsilon_{ijk}h_bar*S_k

Relevant Equations

S_x=hbar/2(|+><-|+|-><+|)
S_y=ihbar/2(-|+><-|+|-><+|)
S_z=hbar/2(|+><+|-|-><-|)

Trying to use <+|+>=1=<-|-> and <-|+>=0 to prove each iteration of the equation, so I have 6 different versions to prove. But the part I'm currently stuck on is understanding how to simplify any given version. I've written out [S_x,S_y]=S_xS_y\psi-S_yS_x\psi and expanded it in terms of the |+>,|-> kets and bras. Then using associativity I've broken up some of the three term components like <-|+|-> into (<-|)(+|->) which I'm assuming will allow me to remove some terms, however now I don't really understand what + or - means in the absence of being inside a ket or bra. I can't imagine they equal their usual matrix representation because then +=|+> which makes no sense.

 

[PeroK]

Some of Sakurai's notation is a bit wild and woolly in my opinion. What we have here, for example, is:
$$S_x = \frac \hbar 2[(|+\rangle \langle -|) + (|-\rangle \langle +|)]$$ I.e. it is the sum of two operators, each formed by the outer product of a ket with a bra.

Does that help?

PS The $+$ in the middle is the plain old symbol for (in this case) operator addition.

 

[yoyopizza]

Ahhh, that does help. So to expand on that, $S_y=\frac{i\hbar}{2}(-|+><-|)+(|-><+|)$ i assume. And this would imply that the first negative sign there just means negative $|+>$ Thanks

 

[PeroK]

Ahhh, that does help. So to expand on that, $S_y=\frac{i\hbar}{2}(-|+><-|)+(|-><+|)$ i assume. And this would imply that the first negative sign there just means negative $|+>$ Thanks

Yes, I don't know why he didn't write: $$S_y=-\frac{i\hbar}{2}[(|+><-|)-(|-><+|)]$$

 

[yoyopizza]

Yes, I don't know why he didn't write: $$S_y=-\frac{i\hbar}{2}[(|+><-|)+(|-><+|)]$$

Yeah that would have been much more clear. Thanks a ton.