Proving the Pauli Matrix Identity with Ordinary Vectors: A Simplified Approach

[ognik]

I'm not sure I have the right approach here:

Using the three 2 X 2 Pauli spin matrices, let $ \vec{\sigma} = \hat{x} \sigma_1 + \hat{y} \sigma_2 +\hat{z} \sigma_3 $ and $\vec{a}, \vec{b}$ are ordinary vectors,

Show that $ \left( \vec{\sigma} \cdot \vec{a} \right) \left( \vec{\sigma} \cdot \vec{b} \right) = \vec{a} \cdot \vec{b} I_2 + i \sigma \cdot \left( \vec{a} \times \vec{b}\right)$

I'm not sure how to go about this - the $i$ in the last term suggests to me that I have to laboriously multiply both sides ...

But the Pauli matrices are 2 X 2 , ex. $ \sigma_1 = \begin{bmatrix}0&1\\ 1&0\end{bmatrix}$ and $\vec{\sigma}$ appears to be Cartesian 3-D.

So I tried $ \vec\sigma \cdot \vec{a} = \hat{x} \sigma_1 \cdot \vec{a} + ... = \begin{bmatrix}1\\ 0 \\ 0 \end {bmatrix} \left( \begin{bmatrix}0&1\\1&0\end{bmatrix} \cdot \begin{bmatrix}a_1\\ a_2 \end {bmatrix} \right) + ...$

Not possible to multiply out like this, so keep the unit vectors as $\hat{x}$ etc. ...

I then get $ \vec\sigma \cdot \vec{a} = \hat{x} \begin{bmatrix}a_2 \\ a_1 \end {bmatrix} +\hat{y} \begin{bmatrix} -a_2\\ a_1 \end {bmatrix} +\hat{z} \begin{bmatrix}a_1\\ -a_2 \end {bmatrix} $ and for $\vec{\sigma} \cdot \vec{b}$ a very similar eqtn by symmetry.

But again I won't be able to multiply out $ \left( \vec\sigma \cdot \vec{a} \right) \left( \vec{\sigma} \cdot \vec{b} \right) $ because of different matrix ranks? Must be a better way to do this (one that also works :-))

 

[topsquark]

I'm not sure I have the right approach here:

Using the three 2 X 2 Pauli spin matrices, let $ \vec{\sigma} = \hat{x} \sigma_1 + \hat{y} \sigma_2 +\hat{z} \sigma_3 $ and $\vec{a}, \vec{b}$ are ordinary vectors,

Show that $ \left( \vec{\sigma} \cdot \vec{a} \right) \left( \vec{\sigma} \cdot \vec{b} \right) = \vec{a} \cdot \vec{b} I_2 + i \sigma \cdot \left( \vec{a} \times \vec{b}\right)$

I'm not sure how to go about this - the $i$ in the last term suggests to me that I have to laboriously multiply both sides ...

But the Pauli matrices are 2 X 2 , ex. $ \sigma_1 = \begin{bmatrix}0&1\\ 1&0\end{bmatrix}$ and $\vec{\sigma}$ appears to be Cartesian 3-D.

So I tried $ \vec\sigma \cdot \vec{a} = \hat{x} \sigma_1 \cdot \vec{a} + ... = \begin{bmatrix}1\\ 0 \\ 0 \end {bmatrix} \left( \begin{bmatrix}0&1\\1&0\end{bmatrix} \cdot \begin{bmatrix}a_1\\ a_2 \end {bmatrix} \right) + ...$

Not possible to multiply out like this, so keep the unit vectors as $\hat{x}$ etc. ...

I then get $ \vec\sigma \cdot \vec{a} = \hat{x} \begin{bmatrix}a_2 \\ a_1 \end {bmatrix} +\hat{y} \begin{bmatrix} -a_2\\ a_1 \end {bmatrix} +\hat{z} \begin{bmatrix}a_1\\ -a_2 \end {bmatrix} $ and for $\vec{\sigma} \cdot \vec{b}$ a very similar eqtn by symmetry.

But again I won't be able to multiply out $ \left( \vec\sigma \cdot \vec{a} \right) \left( \vec{\sigma} \cdot \vec{b} \right) $ because of different matrix ranks? Must be a better way to do this (one that also works :-))

To start with you need to figure out what \(\displaystyle \vec{ \sigma } \cdot \vec{a}\) is. \(\displaystyle \vec{\sigma} = \sigma _x ~ \hat{x} + \sigma _y ~ \hat{y} + \sigma _z ~ \hat{z}\). This is a "vector" of matrices. You are "dotting" it with a 3-vector \(\displaystyle < a_x,~a_y,~a_z >\). So:
\(\displaystyle \vec{ \sigma } \cdot \vec{a} = \sigma _x ~ a_x + \sigma _y ~ a_y + \sigma _z ~ a_z = \left ( \begin{matrix} a_z & a_x - i~a_y \\ a_x + i~a_y & -a_z \end{matrix} \right )\)

Can you take it from here? (I'd give you a really cool and elegant method but I don't have any tricks for this one.)

-Dan

 

[ognik]

Yup thanks, once you pointed out they were a vector. I really need to find a way to notice that sort of thing!

Got a sore hand now :-)