Deriving a vector identity using Pauli spin matrices

[quasar_4]

Homework Statement

I'm supposed to derive the following:

[tex] \left({\bf A} \cdot {\bf \sigma} \right) \left({\bf B }\cdot {\bf \sigma} \right) = {\bf A} \cdot {\bf B} I + i \left( {\bf A } \times {\bf B} \right) \cdot {\bf \sigma} [/tex]

using just the two following facts:

Any 2x2 matrix can be written in a basis of spin matrices:

[tex] M = \sum{m_\alpha \sigma_\alpha} [/tex]

which means that the beta-th component is given by

[tex] m_\beta = \frac{1}{2}Tr(M \sigma_\beta) [/tex]

Homework Equations

listed above...

The Attempt at a Solution

It should just be a left side= right side proof.

I started by saying [tex] \left({\bf A} \cdot {\bf \sigma} \right) \left({\bf B }\cdot {\bf \sigma} \right) =

\left(\sum_\alpha a_\alpha \sigma_\alpha \cdot \sigma \right) \left(\sum_\alpha b_\alpha \sigma_\alpha \cdot \sigma \right) =

\left( \sum_\beta a_\alpha \delta_{\alpha \beta} \right) \left( \sum_\gamma b_\alpha \delta_{\alpha \gamma} \right) =

\sum_\beta a_\beta \sum_\gamma b_\gamma =

\frac{1}{2} Tr(A \sigma_\beta) \frac{1}{2} Tr(B \sigma_\gamma) [/tex]

Not sure if this is even the right way to start, and I can't see at all where I would go from here to get the appropriate RHS of the identity I'm proving. Any ideas?

 

[nickjer]

You can't expand A or B into Pauli matrices because neither of them are 2x2 matrices. They are both vectors of scalars.

Also, are you able to use any other info, such as the product of two Pauli matrices?

 

[quasar_4]

Thanks for the reply. I can use the usual commutation relations between the Pauli matrices. Also, I know the Pauli spin matrices are traceless... not sure if that's helpful or not.

 

[nickjer]

Yes, that helps a lot. First you will need to make a 2-D matrix from the original equation. You had the right idea originally but expanded too much early on. You should do something like:

[tex](A\cdot\sigma)(B\cdot\sigma) = \sum_{i} a_i \sigma_i \sum_{j} b_j \sigma_j
= \sum_{ij}a_i b_j \sigma_i \sigma_j
=M[/tex]

That last term is now a 2d matrix. You will expand that in spin matrices including the identity. I suggest solving for the [tex]m_0[/tex] (identity piece) first, since that will give you your dot product. Be sure to use properties of the trace. For example, a_i and b_j are scalars. And you can break the sum of matrices in a trace up into a sum of traces of the individual matrices.

 

[paranormal]

Your approach is on the right track. To continue, you can use the fact that the trace operation is linear, meaning that it distributes over addition and scalar multiplication. This means that you can rewrite the last line of your attempt as:

\frac{1}{2} Tr(A \sigma_\beta) \frac{1}{2} Tr(B \sigma_\gamma) = \frac{1}{4} Tr(A \sigma_\beta B \sigma_\gamma)

Next, we can use the fact that the Pauli spin matrices satisfy the anti-commutation relation:

\{\sigma_\beta, \sigma_\gamma\} = 2 \delta_{\beta \gamma} I

where {\cdot} denotes the anti-commutator. Using this relation, we can rewrite the previous line as:

\frac{1}{4} Tr(A \sigma_\beta B \sigma_\gamma) = \frac{1}{4} Tr(A B \sigma_\beta \sigma_\gamma) = \frac{1}{4} Tr(A B (\sigma_\beta \sigma_\gamma + \sigma_\gamma \sigma_\beta))

Finally, we can use the fact that the Pauli spin matrices are traceless, meaning that their trace is zero, except for the identity matrix. This allows us to simplify the previous line as:

\frac{1}{4} Tr(A B (\sigma_\beta \sigma_\gamma + \sigma_\gamma \sigma_\beta)) = \frac{1}{4} Tr(A B (2 \delta_{\beta \gamma} I)) = \frac{1}{2} Tr(A B \delta_{\beta \gamma} I)

which is equal to:

\frac{1}{2} Tr(A B \delta_{\beta \gamma} I) = \frac{1}{2} Tr(A B I) \delta_{\beta \gamma} = \frac{1}{2} Tr(A B I) \delta_{\beta \gamma} = \frac{1}{2} (A \cdot B) \delta_{\beta \gamma} = \frac{1}{2} (A \cdot B) \delta_{\beta \gamma} = \frac{1}{2} (A \cdot B) I

Therefore, we have shown that:

\left({\bf A} \cdot {\bf \sigma} \right) \left({