Trace of the spin matrix of spin-1

[Chenkb]

Spin-1 matrix Sx, Sy, Sz are traceless 3*3 matrix, and have the property $[S_i, S_j] = i\epsilon_{ijk}S_k$, and we know that $Tr(S_i^2) = 1^2+0^2+(-1)^2=2$.
All of the above are independent of representation, of course, the trace of a matrix is representation-independent.

so, if we want to know $Tr(S_xS_z)=?$, we can use $Tr(S_yS_z^2)=Tr[(S_zS_y+iS_x)S_z]=Tr(S_yS_z^2)+iTr(S_xS_z)$, thus $Tr(S_xS_z)=0$.

And my question is, what about $Tr(S_xS_yS_xS_y)=?$ Using the similar method mentioned above.

Regards!

 

[samalkhaiat]

... we know that $Tr(S_i^2) = 1^2+0^2+(-1)^2=2$.

What is this? Why should the sum (of squares of the eigenvales of $S_{ z }$) gives the trace of $S_{ i }^{ 2 }$? For spin one matrices, you have
$$\sum_{ 1 }^{ 3 } S_{ i }^{ 2 } = \left( \ \begin{array} {rrr} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \end {array} \right) .$$
So, the trace of this is 6 not 2.

All of the above are independent of representation, of course, the trace of a matrix is representation-independent.

No, who told you this? The trace depends on the dimension of the representation. In any representation $\rho$, the Casmir matrix is given by
$$\sum_{ i = 1 }^{ 3 } J_{ i }^{ 2 } = j ( j + 1 ) \mbox{ I }_{ \rho ( j ) \times \rho ( j )} ,$$
where $\rho ( j ) = 2 j + 1$ is the dimension of the representation and I is the identity matrix in that representation. So, clearly
$$\mbox{ Tr } \left( \sum_{ i = 1 }^{ 3 } J_{ i }^{ 2 } \right) = j ( j + 1 ) ( 2 j + 1 ) .$$
So, for spin-1/2 the trace is $( 1/2 ) ( 3/2 ) ( 2 ) = ( 3/2 )$. And, for spin one you get $( 1 ) ( 2 ) ( 3 ) = 6$.

so, if we want to know $Tr(S_xS_z)=?$, we can use $Tr(S_yS_z^2)=Tr[(S_zS_y+iS_x)S_z]=Tr(S_yS_z^2)+iTr(S_xS_z)$, thus $Tr(S_xS_z)=0$.

And my question is, what about $Tr(S_xS_yS_xS_y)=?$ Using the similar method mentioned above.

Regards!

Why do you need to do all this? The spin-1 matrices are simple enough to determine the trace of products of any number of them. However, you can use the identity
$$( S_{ x } S_{ y } )^{ 2 } = S_{ x }^{ 2 } S_{ y }^{ 2 } - i S_{ x } S_{ z } S_{ y }$$
Now, take the trace and use the fact that
$$\mbox{ Tr } ( S_{ x }^{ 2 } S_{ y }^{ 2 } ) = 1 , \ \mbox{ and } \ \ \mbox{ Tr } ( S_{ x } S_{ z } S_{ y } ) = - i$$

Sam

 

[Chenkb]

Sam

Maybe my notations could cause ambiguity, for $Tr(S_i^2)$, I didn't mean to sum from 1 to 3, $i$ stands for $x$ or $y$ or $z$. And we can go to the representation which makes $S_i$ diagonal, and the diagonal elements are the eigenvalues, i.e. 0, 1, -1, so I got $Tr(S_i^2)=2$.

I know that we can choose a specific representation(usually Sz diagonal), and do the matrix product explicitly, then get the traces. But I want to do it in another way, using the properties of spin operators like the commutation relations. I think this is a good homework. ^_^
I had proved that $Tr(S_xS_zS_y)=-i$, but got some trouble for $Tr(S_x^2S_y^2)=1$. Is there any chance you could possibly show me the details of $Tr(S_x^2S_y^2)=1$?

Thanks a lot!

 

[Avodyne]

$$S_x^2+S_y^2+S_z^2=2I$$$$S_x^2+S_y^2=2I-S_z^2$$$$(S_x^2+S_y^2)^2=(2I-S_z^2)^2$$$$S_x^4+S_y^4+S_x^2S_y^2+S_y^2S_x^2=4I-4S_z^2+S_z^4$$Take the trace of both sides, and use
$${\rm Tr}\,S_i^2={\rm Tr}\,S_i^4=2$$$${\rm Tr}\,I=3$$

 

[Chenkb]

$$S_x^2+S_y^2+S_z^2=2I$$$$S_x^2+S_y^2=2I-S_z^2$$$$(S_x^2+S_y^2)^2=(2I-S_z^2)^2$$$$S_x^4+S_y^4+S_x^2S_y^2+S_y^2S_x^2=4I-4S_z^2+S_z^4$$Take the trace of both sides, and use
$${\rm Tr}\,S_i^2={\rm Tr}\,S_i^4=2$$$${\rm Tr}\,I=3$$

Wow! That's amazing, thank you so much!