# Dot product of two pauli matrices

In some text, I read something like this

$$\vec{S}_i\cdot\vec{S}_j$$

where $$\vec{S}_i$$ and $$\vec{S}_j$$ are "vectors" with each components be the pauli matrices $$S_x, S_y, S_z$$ individularly. My question is: if all components of this kind of vector are a 3x3 matrix, so how do you carry out the dot product? So the dot product will give another matrix instead of a number?

Matterwave
Gold Member
"Vectors" are not matrices. Matrices are tensors of rank 2, while vectors are tensors of rank 1. I don't see how you can have "vectors" which are Pauli-Matrices.

Matrix multiplication (2 matrices multiplied together, not a matrix and a vector) has specific rules and will result in another matrix with the same number of columns and rows.

"Vectors" are not matrices. Matrices are tensors of rank 2, while vectors are tensors of rank 1. I don't see how you can have "vectors" which are Pauli-Matrices.

Matrix multiplication (2 matrices multiplied together, not a matrix and a vector) has specific rules and will result in another matrix with the same number of columns and rows.

In the text of linear space, it read: matrices can be a special kind of "vectors" (that's why we add qutoation mark). In abstract linear vector space, anything satisfying the RULES of linear space can form a vector. And in many QM text, the author like to form a "vector" with three pauli matrices Sx, Sy, Sz. What I am really confuse is how to carry out the dot product b/w such "vectors"?

Matterwave
Gold Member
Hmmm, I don't know if the dot product is the same as matrix multiplication (I'm not sure how the inner product is defined in a vector space that includes matrices), but if it is then:

http://en.wikipedia.org/wiki/Matrix_multiplication

That explains it pretty well.

Oh, I was thinking of only square matrices before, if the matrix is not square then the resulting matrix won't have the same dimensions as before. Sorry, it's been a while since I've really had to do Matrix multiplication. :P

A logical way to define a dot product when using pauli matrixes as basis vectors would be to use the anticommutator

$$a \cdot b = \frac{1}{2} \{ a, b \} = \frac{1}{2} (a b + b a )$$

EDIT: latex in PF doesn't appear to be working right now. That was:
a \cdot b = \frac{1}{2} \{ a, b \} = \frac{1}{2} (a b + b a )

Such a dot product won't be in the span of the pauli matrixes themselves, but will be your typical vector dot product multiplied by the identity matrix. If you wanted a dot product that produced a true scalar instead of scalar times identity, then you should be able to couple the above with a trace operation:

$$a \cdot b = \frac{1}{4} TR \{ a, b \} = \frac{1}{4} TR (a b + b a )$$

EDIT: latex in PF doesn't appear to be working right now. That was:
a \cdot b = \frac{1}{4} TR \{ a, b \} = \frac{1}{4} TR (a b + b a )

You can similarily and naturally define the cross product or wedge product using the commutator. Some playing with these ideas can be found here:

( as written this assumes some clifford algebra background ).

Thanks a lot. It helps.
A logical way to define a dot product when using pauli matrixes as basis vectors would be to use the anticommutator

$$a \cdot b = \frac{1}{2} \{ a, b \} = \frac{1}{2} (a b + b a )$$

EDIT: latex in PF doesn't appear to be working right now. That was:
a \cdot b = \frac{1}{2} \{ a, b \} = \frac{1}{2} (a b + b a )

Such a dot product won't be in the span of the pauli matrixes themselves, but will be your typical vector dot product multiplied by the identity matrix. If you wanted a dot product that produced a true scalar instead of scalar times identity, then you should be able to couple the above with a trace operation:

$$a \cdot b = \frac{1}{4} TR \{ a, b \} = \frac{1}{4} TR (a b + b a )$$

EDIT: latex in PF doesn't appear to be working right now. That was:
a \cdot b = \frac{1}{4} TR \{ a, b \} = \frac{1}{4} TR (a b + b a )

You can similarily and naturally define the cross product or wedge product using the commutator. Some playing with these ideas can be found here: