Dot product within the commutator

[sheelbe999]

So I am reading a book on ferromagnetism, the author writes

$$\left[S_J,S_{J-1} \cdot S_J\right] = S_J cross S_{J-1}$$

(I couldn't get the cross product x in latex code for some reason)

Where [tex]S_J[\tex] and [tex]S_{J-1}[\tex] are the spin operators for atoms j and j-1.

I was wondering how one gets from the left hand side of the equation to the right hand side.


Thanks in anticipation

 

[Meir Achuz]

Take the x component of the commutator.
Expand the dot product as L_xS_x+L_yS_y+L_zS_z.
Use [S_x,S_y]=iS_z, etc.
Recognize $$i(L\times S)_x$$.

 

[dextercioby]

I think L comes from orbital angular momentum. But it can't happen, because it's only a matter of spins in this problem.

 

[bobbytkc]

I think L comes from orbital angular momentum. But it can't happen, because it's only a matter of spins in this problem.

I think he just used L because the spin at another site is acting on another subsystem (i.e. another atom). The overall behaviour is similar to composing angular momentum and spin together, as the angular momentum and spin can be treated as two separate subsystems.

In any case, to prevent confusion, I wrote down a component to illustrate the case more clearly.

$$\begin{eqnarray*} [S_{J,x},S_{J-1,x}S_{J,x}+S_{J-1,y}S_{J,y}+S_{J-1,z}S_{J,z}] & = & S_{J-1,x}[S_{J,x},S_{J,x}]+S_{J-1,y}[S_{J,x},S_{J,y}]+S_{J-1,z}[S_{J,x},S_{J,z}]\\ & = & S_{J-1,y}[S_{J,x},S_{J,y}]+S_{J-1,z}[S_{J,x},S_{J,z}]\\ & = & S_{J-1,y}(iS_{J,z})+S_{J-1,z}(-iS_{J,y})\\ & = & i(S_{J-1,y}S_{J,z}-S_{J-1,z}S_{J,y})\\ & = & i(S_{J-1}\times S_{J})_{x}\end{eqnarray*}$$

the sub indices x, y, z refer to the components of the vector. The first thing to note is that the commutator in the question is actually a set of 3 operator equations (i.e. a vector of operator equations). So the above only broke down the x component of the commutator. This may be confusing to some people who have not done it before, so I feel like I should point it out.

Also, we also need to know that if an operator are acting on different subsystems (or the different atoms J and J-1 in this case), then they mutually commute, so you can swap the order of their relevant operators without consequence.

Otherwise, it is just an application of the algebra of spin operators.

 

[tiny-tim]

hi sheelbe999!

(I couldn't get the cross product x in latex code for some reason)

[noparse]$$\times$$[/noparse]

 

[Meir Achuz]

I think L comes from orbital angular momentum. But it can't happen, because it's only a matter of spins in this problem.

I just used L to represent an independent vector to avoid subscripts i and j. Actually L does not have to be a spin or angular momentum vector. The identity is true for any vector L.

 

[sheelbe999]

Wow! thank you so much for such helpful responses! Problem solved! :)

 

[Fredrik]

This is the type of problem that proves that it's worthwhile to get familiar with the Levi-Civita symbol, and how to use it to express cross products and the commutation relations for spin operators:

$$\big[S_J,S_{J-1} \cdot S_J\big]=e_i\big[(S_J)_i,(S_{J-1})_j (S_J)_j\big]=e_i(S_{J-1})_j\big[(S_J)_i,(S_J)_j\big]=e_i(S_{J-1})_j i\varepsilon_{ijk}(S_J)_k=iS_{J-1}\times S_J$$

 

[Meir Achuz]

This is the type of problem that proves that it's worthwhile to get familiar with the Levi-Civita symbol, and how to use it to express cross products and the commutation relations for spin operators:

$$\big[S_J,S_{J-1} \cdot S_J\big]=e_i\big[(S_J)_i,(S_{J-1})_j (S_J)_j\big]=e_i(S_{J-1})_j\big[(S_J)_i,(S_J)_j\big]=e_i(S_{J-1})_j i\varepsilon_{ijk}(S_J)_k=iS_{J-1}\times S_J$$

Is that really simpler than post #2?

 

[Fredrik]

Is that really simpler than post #2?

What I wrote is the calculation that Meir Achutz described in #2 and bobbytkc posted in #4, but for all 3 components instead of just one.

 

[bobbytkc]

Is that really simpler than post #2?

It is indeed much simpler, if you are familiar with the notation. In my elementary solution, the OP still needed to perform the same calculations for the y and z components to completely prove that it is a cross product.

 

[Meir Achuz]

It is indeed much simpler, if you are familiar with the notation. In my elementary solution, the OP still needed to perform the same calculations for the y and z components to completely prove that it is a cross product.

You don't have to. Just use cyclic substitution. That is what is nice about vectors. If you prove it for V_x, it is also true for V_y and V_z, as long as there is no preference for direction.

 

[Fredrik]

We would at least have to add a comment about it, if the calculation is to be read by someone else. It would look something like this:

The fact that the only property of the operators S_x, S_y, S_z that we have used is in the form of an equality that remains true when we make the substitutions x→y→z→x (once or twice), the result we got will also remain true when we make the same substitutions there.

 

[Meir Achuz]

I do that by writing [S_x,S_y]=iS_z, and cyclic.

 

[Fredrik]

That's of course fine in your own notes, but I doubt you would write that in a book or an article, or even in an exam.

 

[Meir Achuz]

I have seen it in a textbook.
I don't write it in my own notes because it is obvious. If there is no preferred direction, I could have called any direction x.