# Dot product of two pauli matrices

1. Apr 21, 2009

### KFC

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?

2. Apr 21, 2009

### Matterwave

"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.

3. Apr 21, 2009

### KFC

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"?

4. Apr 21, 2009

### Matterwave

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

5. Apr 22, 2009

### Peeter

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: