Pauli matrices represent vectors.
Mulitplying two Pauli matrices results in a bivector.
In three dimensions there are 3 bivectors.
A bivector can only rotate a vector 90 degrees.
A scalar plus a bivector can rotate a vector any angle.
Using Pauli matrices the scalar plus all 3 bivectors looks like this:
$\begin{pmatrix}x_0 +ix_3&ix_1+x_2\\ix_1-x_2&x_0-ix_3\end{pmatrix}$
So a scalar plus 3 bivectors is a quaternion which is evidently some sort of rotation matrix (in the sense that it "rotates" one Pauli matrix into another).
This leads naturally to the question of what do you get if you add a pseudoscalar to the 3 vectors. (Note that I am reusing the variable x even though they are different from the four above)
$\begin{pmatrix}ix_0+x_3&x_1-ix_2\\x_1+ix_2&ix_0-x_3\end{pmatrix}$
I have no idea at all. I am hoping someone here either knows the answer or recognizes the matrix above.
Thanks in advance. 🙂
Best Answer
What you've written out as an explicit 4x4 matrix, can be written in the more compact "sigma" notation as $$x_0 \sigma_0 + \boldsymbol{\sigma} \cdot \mathbf{x}.$$ It isn't appropriate to describe this multivector as a quaternion, nor as a rotation. Quaternions may be represented in the Pauli algebra as multivectors with scalar+bivector components, such as $$x_0 \sigma_0 + i \boldsymbol{\sigma} \cdot \mathbf{x}.$$ (recall that $ i \sigma_0 = \sigma_1 \sigma_2 \sigma_3 $ is the pseudoscalar in the Pauli representation, and that a pseudoscalar and vector product is a bivector.) In particular, by setting $ \mathbf{i} = \sigma_2 \sigma_3, \mathbf{j} = \sigma_1 \sigma_3, \mathbf{k} = \sigma_1 \sigma_2 $, it is easy to see that the usual quaternion multiplication table can be recovered.
As for rotations in the Pauli algebra, rotating a vector about a normal $ \mathbf{n} $ can be accomplished by sandwiching the vector between two conjugate rotors, exponentials with bivector arguments $$\sigma \cdot \mathbf{x} \rightarrow e^{-i \boldsymbol{\sigma} \cdot \mathbf{n}/2} \sigma \cdot \mathbf{x} e^{ i \boldsymbol{\sigma} \cdot \mathbf{n}/2 }$$ This works as a rotation since $ e^{-i \boldsymbol{\sigma} \cdot \mathbf{n}/2} $ commutes with any component of $ \sigma \cdot \mathbf{x} $ that lies in the normal direction, and conjugate commutes with any component that lies in the plane of the rotation (i.e. the plane represented by the bivector $i \mathbf{n}$.) For example, if $ \sigma \cdot \mathbf{x} = \sigma \cdot \mathbf{x}_\parallel + \sigma \cdot \mathbf{x}_\perp $, where $ \mathbf{x}_\parallel \cdot \mathbf{n} = 0 $, then we have $$\begin{aligned}\sigma \cdot \mathbf{x} &\rightarrow e^{-i \boldsymbol{\sigma} \cdot \mathbf{n}/2} \left( { \sigma \cdot \mathbf{x}_\parallel + \sigma \cdot \mathbf{x}_\perp} \right) e^{ i \boldsymbol{\sigma} \cdot \mathbf{n}/2 } \\ &=\left( { \sigma \cdot \mathbf{x}_\parallel } \right) e^{ i \boldsymbol{\sigma} \cdot \mathbf{n} }+ \sigma \cdot \mathbf{x}_\perp.\end{aligned}$$ The component that is perpendicular to the plane of rotation is left untouched, while we have a complex-exponential style rotation of any component of the vector that lies in the plane. For example, with $ i \mathbf{n} = i \theta \sigma_3 = \sigma_1 \sigma_2 $, we have a rotation in the x-y plane. Each of the rotors is a multivector with scalar+bivector components $$e^{ \theta \sigma_1 \sigma_2 /2 }=\sigma_0 \cos \theta/2 + \sigma_1 \sigma_2 \sin\theta/2.$$
So, if a scalar+vector is not a rotation, nor a quaternion, what is it? I don't have a good answer for that in general, but there are some interesting special cases of multivectors of this type. One is the projector, an example of which is $$P = \frac{1}{{2}} \left( { \sigma_0 + \sigma_3 } \right).$$ This squares to itself, and eats any factor of $ \sigma_3 $ $$\begin{aligned} P^2 &=\frac{1}{{4}} \left( { \sigma_0 + \sigma_3 } \right) \left( { \sigma_0 + \sigma_3 } \right) \\ &=\frac{1}{{4}} \left( { \sigma_0 + 2 \sigma_3 + \sigma_3^2 } \right) \\ &=\frac{1}{{4}} \left( { 2 \sigma_0 + 2 \sigma_3 } \right) \\ &= P,\end{aligned}$$ and $$ P \sigma_3 = \frac{1}{{2}} \left( { \sigma_0 + \sigma_3 } \right) \sigma_3=\frac{1}{{2}} \left( { \sigma_3 + \sigma_3^2 } \right)= P.$$ Projectors of this form show up as factors in (multivector electromagnetic wave) solutions of Maxwell's equation in charge and current free regions.
As for the remaining part of your question, what is a pseudoscalar + vector? In the sigma notation, such a sum has the form $$ i \sigma_0 \alpha + \boldsymbol{\sigma} \cdot \mathbf{x}.$$ Instead of answering that "what is" question, I'll cheat and give an example where we see such a sum. In particular, in the rotation above, we had products like $ \boldsymbol{\sigma} \cdot \mathbf{x} e^{ i \boldsymbol{\sigma} \cdot \mathbf{n}/2 } $. Let $ \mathbf{n} = \theta \hat{\mathbf{n}} $, so this product expands to $$\boldsymbol{\sigma} \cdot \mathbf{x}\left( { \cos \theta/2 + i \boldsymbol{\sigma} \cdot \hat{\mathbf{n}} \sin\theta/2} \right).$$ The multivector coefficient of the sine is $ i (\boldsymbol{\sigma} \cdot \mathbf{x})( \boldsymbol{\sigma} \cdot \hat{\mathbf{n}} ) $. If $ \mathbf{x} $ and $ \hat{\mathbf{n}} $ are perpendicular, such a product is a vector, and if they are parallel, such a product is a pseudoscalar. However, in general, such a product is a sum of a vector and a pseudoscalar. As it happens, all the pseudoscalar products that occur in the expansion of the rotation, cancel out in the end, leaving just a vector.
Another example (albeit somewhat contrived), of a vector+pseudoscalar sum, can be found by considering the generalization of Maxwell's equation found in Engineering antenna theory that includes fictional magnetic sources. One can construct a multivector potential field that includes both a vector potential and a magnetic scalar potential, like so: $$ A = c \boldsymbol{\sigma} \cdot \mathbf{A} - \eta i \phi_m \sigma_0.$$ This potential has both vector and pseudoscalar components, but in general would have scalar and bivector components as well.