Spin Operators: Axial for QM, Polar in Clifford Algebra?

[Elemental]

Hello folks! New to this forum, so hoping I'm not retreading old ground. The Pauli matrices are spin angular momentum operators in quantum mechanics and thus are axial vectors. But in Clifford algebra in three dimensions they are odd basis elements and thus polar vectors. Hestenes, Baylis, other geometric algebraists have reformulated quantum mechanics hinting at fundamental reinterpretations. So, the same mathematical objects have even parity in one formalism, and odd parity in others. Nowhere, as far as I can see, is this explained. Any insights will be greatly appreciated!

 

[Jazzdude]

Can you define the objects you refer to more precisely? In exterior algebra / geometric algebra there are no such things as polar or axial vectors to start with, only multivectors (and their Hodge duals). So I think what you claim is probably based on a misunderstanding of the algebraic structure.

Cheers,

Jazz

 

[Elemental]

Jazz,

I'm referring to $\sigma_1=\left[<br /> \begin{array}{ c c }<br /> 0 & 1 \\<br /> 1 & 0<br /> \end{array} \right],\sigma_2=\left[<br /> \begin{array}{ c c }<br /> 0 & -i \\<br /> i & 0<br /> \end{array} \right],\sigma_3=\left[<br /> \begin{array}{ c c }<br /> 1 & 0 \\<br /> 0 & -1<br /> \end{array} \right]$. They are the grade 1 elements (acknowledge that the matrices are just one particular representation) of a Clifford algebra for real three-space. Together with the grade 0 element, $\sigma_0=I_{2x2},$, the grade 2 elements $\sigma_2\sigma_3, \sigma_3\sigma_1,\sigma_1\sigma_2,$ and the grade 3 element $\sigma_1\sigma_2\sigma_3,$ they form the multivector basis for that algebra.

As far as polar or axial vectors, Hestenes states the Pauli matrices are polar in geocalc.clas.asu.edu/pdf-preAdobe8/CAIQM.pdf (as do others), and he states that the grade two elements are axial. Baylis, in web4.uwindsor.ca/users/b/baylis/main.nsf/.../$FILE/cainphys.pdf, shows that such a grade 1 vector, say $v=\sigma_3$ can be reflected in the 2-3 plane orthogonal to it with the transformation $\sigma_1\sigma_2$v$\sigma_1\sigma_2$ = $\sigma_1\sigma_2\sigma_3\sigma_1\sigma_2=-\sigma_3$. Hence, $\sigma_3$ (and so on for the other grade 1 elements) has odd parity as does a polar vector.

So, with this interpretation, I don't understand why the same quantities are axial vectors with even parity, representing spin angular momentum in quantum mechanics.

Elemental

• http://webcache.googleusercontent.com/search?q=cache:vsIvjZJ4vZIJ:geocalc.clas.asu.edu/pdf-preAdobe8/CAIQM.pdf+&cd=1&hl=en&ct=clnk&gl=us

 

[Elemental]

Typo: I meant to say, "... can be reflected in the 1-2 plane orthogonal to it".

 

[Jazzdude]

Ok, I understand now where your confusion comes from. The spin vector is the Hodge dual of the spin bivector. That means the basis vectors used to expand the spin vector in are vectors, not bivectors. Let me give you an example.

Consider the exterior algebra $\Lambda(\mathbb{R}^3)$ with a positively oriented vector basis $\{ \mathbf{x},\mathbf{y},\mathbf{z} \}$. The bivector Hodge duals of these basis vectors are then
$$\begin{eqnarray}
\star \mathbf{x} & = & \mathbf{y} \wedge \mathbf{z} \nonumber\\
\star \mathbf{y} & = & \mathbf{z} \wedge \mathbf{x} \nonumber\\
\star \mathbf{z} & = & \mathbf{x} \wedge \mathbf{y} \nonumber
\end{eqnarray}$$
Now the general bivector $\mathbf{v}$ is a linear combination of these bivectors and can be written as $$\mathbf{v}=v_1 \cdot \mathbf{y} \wedge \mathbf{z} + v_2 \cdot \mathbf{z} \wedge \mathbf{x} \nonumber + v_3 \cdot \mathbf{x} \wedge \mathbf{y} \nonumber$$

That's the equivalent to your spin bivector. And as a bivector it behaves like a axial vector. But it's best to avoid polar and axial altogether in the context of geometric algebra, because it leads to confusion like yours.

Next, let's take this bivector to its vector Hodge dual.

$$ \star\mathbf{v}=v_1\cdot\mathbf{x}+v_2\cdot\mathbf{y}+v_3\cdot\mathbf{z}$$

As you can see, it has been expanded into the basis vectors that each transform like polar vectors. So where is the axial character of the spin in this representation? The answer is, it is hidden in the Hodge dual and absorbed in the sign of the expansion coefficients. Had I not stated in the beginning that the basis is positively oriented in the order given, but rather negatively oriented, you'd have a factor of $-1$ with every basis vector. So if you apply a spatial inversion, the signs of the coefficients will change.

The very same thing happens for your spin operator
$$S = s_1 \sigma_1 + s_2 \sigma_2 + s_3 \sigma_3$$
where the $\sigma$ are your grade one (axial) Clifford algebra basis elements. Spatial inversion will again generate a factor of $-1$ for all the duals which will be absorbed in your coefficients.

But this is really all superfluous, because you can work with the grade-2 elements (bivectors) of the clifford algebra directly without using the Hodge dual, and things will become more obvious. The fact that spin is treated as a pseudo-vector and not a bivector has mostly historical reasons, so let's fix that.

The grade-2 elements of the spin Clifford algebra are explicitly
$$\begin{eqnarray}
\sigma_1 \cdot \sigma_2 & = & \mathrm{i} \sigma_3 \nonumber \\
\sigma_2 \cdot \sigma_3 & = & \mathrm{i} \sigma_1 \nonumber \\
\sigma_3 \cdot \sigma_1 & = & \mathrm{i} \sigma_2 \nonumber \\
\end{eqnarray}$$

So the canonical bivector basis is $\{ \mathrm{i} \sigma_1,\mathrm{i} \sigma_2,\mathrm{i} \sigma_3 \}$ and you can check that these transform correctly under spatial inversion.

This gives us a spin operator of $$S = s_1\cdot \mathrm{i} \sigma_1 + s_2\cdot \mathrm{i} \sigma_2 + s_3\cdot \mathrm{i} \sigma_3$$

which is very convenient if you use this spin for generating rotations. For the spin vector, the lie algebra of spatial rotations was generated by $R = \exp(\mathrm{i}S)$. And for the spin bivector representation it becomes just $R = \exp(S)$.
If we also adopt the bivector basis as the Lie algebra basis we get a mildly different lie algebra structure:
$$[\mathrm{i}\sigma_k,\mathrm{i}\sigma_l]=-[\sigma_k,\sigma_l]=[\sigma_l,\sigma_k]$$

So in total, not much changes if you keep spin a bivector instead of Hodging it, at least for a cartesian basis. I'll leave the details of the non-cartesian case to you, if you're interested.

Hope this helps!

Jazz

 

[Elemental]

Jazz,

So had we expressed the spin operator as a bivector in the first place, it would explicitly retain its sign under spatial inversion. But if instead we express it as the Hodge dual of a bivector, the operator reverses sign under inversion, and one must take into account the fact that taking the Hodge dual yields a plus or minus sign depending on the orientation of the space.

But in typical quantum mechanical expressions spin is not explicitly shown as the dual of a bivector; e.g., in the Pauli equation σ is not multiplied by i, and seemingly one must deduce the axial character from context (I'm probably missing something here). So reverting to physicist language (I'm new to geometric algebra) we have the eigenvalue equation $\sigma_3|\uparrow\rangle=|\uparrow\rangle$. The reflection operator for $\sigma_3$ is the unitary operator $U=\sigma_1\sigma_2$. So we reflect $\sigma_3$ in this equation using the transformation and get $$U^†\sigma_3U|\uparrow\rangle=-\sigma_2\sigma_1\sigma_3\sigma_1\sigma_2|\uparrow\rangle=\sigma_1\sigma_2\sigma_3\sigma_1\sigma_2|\uparrow\rangle=-\sigma_3|\uparrow\rangle=-|\uparrow\rangle.$$ This is certainly the right answer, the spin does not change sign under reflection. If we interpret $\sigma_3$ as the measurement of spin in the positive z direction, then $-\sigma_3$ must be the measurement of spin in the negative z direction, and an eigenstate yielding 1 for spin in the positive z direction must yield -1 for a measurement of spin in the negative z direction.

Would you concur with this interpretation? I was tempted, when I wrote the equation earlier, to conclude that the axial character of spin is imparted by the states, not the operators. Your explanation above suggests that it is a property of the operators, but I can only see that if we represent spin as iσ, not σ, but of course in doing so they would not be Hermitian and thus not observables.

Thank-you very much for your assistance,

Elemental

 

[Jazzdude]

Sorry for the late response, I was quite busy.

The reflection operator for $\sigma_3$ is the unitary operator $U=\sigma_1\sigma_2$.

I don't quite see what you mean with this. First of all, if you have a reflection, then it's not specific to one operator but acts like a reflection on all operators. And secondly, $\sigma_1\sigma_2$ is a rotation by the way it transforms. So I guess everything that follows is quite questionable.

Cheers,

Jazz

 

[Elemental]

†

Sorry for the late response, I was quite busy.



I don't quite see what you mean with this. First of all, if you have a reflection, then it's not specific to one operator but acts like a reflection on all operators. And secondly, $\sigma_1\sigma_2$ is a rotation by the way it transforms. So I guess everything that follows is quite questionable.

Cheers,

Jazz

Jazz,

My mistake for using the wrong operator and limiting it to one dimension. So I try again. The reflection operator sending $\sigma$ to $\sigma'$ by $$\sigma'=-n\sigma n,$$ is n, where n is the unit normal to the plane of reflection. If $\sigma$ is a normized vector $\sigma=s_1\sigma_1+s_2\sigma_2+s_3\sigma_3$ (scalar coefficients satisfying $s_1^2+s_2^2+s_3^2=1$) , and the plane of reflection is the 1-2 plane, this becomes $$\sigma'=-\sigma_3(s_1\sigma_1+s_2\sigma_2+s_3\sigma_3)\sigma_3=s_1\sigma_1+s_2\sigma_2-s_3\sigma_3.$$ As expected, only the component normal to the plane changes.

Now suppose $|\uparrow_\sigma\rangle$ is an eigenvector of $\sigma$ satisfying $\sigma|\uparrow_\sigma\rangle=|\uparrow_\sigma\rangle$. Let us reflect $\sigma$ in the plane orthogonal to it to obtain $\sigma"=(-\sigma)\sigma(\sigma),$ where now the reflection operator is just $\sigma$, and substitute it into the eigenvalue equation. My argument above runs $$\sigma"|\uparrow_\sigma\rangle=-\sigma\sigma\sigma|\uparrow_\sigma\rangle=-\sigma|\uparrow_\sigma\rangle=-|\uparrow_\sigma\rangle.$$ Well, the reflection changed the sign of the spin operator (which was what was bothering me since spin has even parity), but nevertheless the eigenvalue equation yields the expected answer: if a measurement in the positive $\sigma$ direction yields spin angular momentum in that direction, then a measurement in the negative $\sigma$ direction still yields spin angular momentum whose direction is positive $\sigma$. Think of this as measuring electrons prepared in a spin-up date with a Stern Gerlach apparatus resulting in all the electrons being deflected upwards. Now if you rotate the apparatus 180 degrees about the beam axis, but don't change how the states are prepared, the electrons will be deflected downward. You still interpret the electrons as spin-up per the original convention.

This makes it seem to me like the axial character of spin is imparted, in quantum mechanical formalism, by the states, not the operators. In that formalism, spin 1/2 operators are $\hbar/2$ times the the Pauli matrices $\sigma_1,\sigma_2,\sigma_3$. I think what you're saying is that in quantum mechanical formalism, these $\sigma s$ are not vectors, but Hodge duals of bivectors, and their axial character is hidden in the formalism. And furthermore, had we avoided the Hodge duals by staying explicitly with the bivectors (retaining the factor i), the commutation relations are preserved but the axial character would become explicit in the operators. But I'm not sure how to do that since it would make the operators non-Hermitian while nevertheless they're interpreted as observables.

Another thing bothering me. The reflection operator is unitary (but not special unitary). I've always been able to stick with special unitary operators, and the transformation looks like $U^†\sigma U$. But here, the reflection operator, n, is also Hermitian, so we'd have $U^†=n^†=n=U$, and we can't write this as $-n\sigma n$, which is how I got tripped up in my earlier post.

Well, I've probably really wrapped myself around the axle this time, but so be it.

Cheers,

Elemental