note-math

$\mathrm{\varphi }$ induces the co-vector of #link(<Hermitian-tensor>)[]

Since ${𝑝}_{\text{spin }}\in {⨀}^{\ast 2}{\mathrm{ℂ}}^2$ is a Hermitian matrix, then ${\left({\mathrm{\varphi }}^{†}{𝑝}_{\text{ spin }}\mathrm{\varphi }\right)}^{†}={\mathrm{\varphi }}^{†}{𝑝}_{\text{ spin }}\mathrm{\varphi }$, so ${\mathrm{\varphi }}^{†}{𝑝}_{\text{ spin }}\mathrm{\varphi }\in \mathrm{ℝ}$

Because ${𝑝}_{\text{spin}}$ is Hermitian or self-adjoint with respect to the inner product of ${\mathrm{ℂ}}^2$, ${𝑝}_{\text{spin}}$ can be considered to act symmetrically on the two slots ${\mathrm{\varphi }}^{†}\left({𝑝}_{\text{spin }}\mathrm{\varphi }\right)={\left({\mathrm{\varphi }}^{†}\left({𝑝}_{\text{spin }}\mathrm{\varphi }\right)\right)}^{†}={\left({𝑝}_{\text{spin }}\mathrm{\varphi }\right)}^{†}\mathrm{\varphi }$

The base of the vector space ${𝜎}^0,{𝜎}^1,{𝜎}^2,{𝜎}^3\in {\mathrm{ℝ}}^{1,3}$ gives the coefficients of the co-vector ${\mathrm{\varphi }}^{†}{𝜎}^{𝜇}\mathrm{\varphi }\in \mathrm{ℝ}$

The action of $\left(\frac{1}{2},0\right)$ on the co-vector $\mathrm{\varphi }$ is

It also corresponds to the transformation of the dual base, i.e., the base of the co-vector space ${\left({\mathrm{ℝ}}^{1,3}\right)}^{⊺}$, which is ${𝑓\left(𝐴\right)}^{⊺}$

#link(<parity>)[] dual and $\left(0,\frac{1}{2}\right)$ induced action

Similarly, co-vectors can also be defined for anti-Hermitian tensors

For complex conjugate two-tensors

• formally corresponds momentum $𝑝$ to gradient momentum $\text{i }\cdot \left({\partial }_0,-{\partial }_1,-{\partial }_2,-{\partial }_3\right)=\text{ i }\cdot {\left({\partial }_0,{\partial }_1,{\partial }_2,{\partial }_3\right)}^{◊}$, and ${𝑝}_{\text{spin}}$ to #link(<spacetime-momentum-spinor-representation>)[]

  $$\begin{array}{ccc}\text{i }\cdot \left(\begin{array}{cc}{\partial }_0+\left(-{\partial }_1\right)& -\left({\partial }_2+{\partial }_3\right)\text{ i}\\ -\left({\partial }_2-{\partial }_3\right)\text{ i}& {\partial }_0-\left(-{\partial }_1\right)\end{array}\right)& =& \text{ i }\cdot {\left(\begin{array}{cc}{\partial }_0+{\partial }_1& {\partial }_2+{\partial }_3\text{ i}\\ {\partial }_2-{\partial }_3\text{ i}& {\partial }_0-{\partial }_1\end{array}\right)}^{◊}\\ & =& \text{ i }{𝜎}^{𝜇}{\partial }_{𝜇}^{◊}=\text{ i }{𝜎}^{𝜇◊}{\partial }_{𝜇}\\ & ≕& \text{ i }{{\partial }_{\text{ spin}}}^{◊}\end{array}$$

• formally used for co-vector generated by $\mathrm{\varphi }$ to obtain $\mathrm{ℂ}$ field

  $$\begin{array}{cc}& {\mathrm{\varphi }}^{†}\text{ i }{{\partial }_{\text{ spin }}}^{◊}\mathrm{\varphi }\end{array}$$

• action + product rule + divergence + zero boundary + integral quadratic form ==> self-adjoint operator $\text{i }{{\partial }_{\text{ spin}}}^{◊}$

  $${⟨\mathrm{\varphi },\text{ i }{{\partial }_{\text{ spin }}}^{◊}𝜓⟩}_{{𝐿}^2}={⟨\text{i }{{\partial }_{\text{ spin }}}^{◊}\mathrm{\varphi },𝜓⟩}_{{𝐿}^2}$$

or ${\mathrm{\varphi }}^{†}\left({𝜎}^{◊}\cdot \text{ i }\partial \right)\mathrm{\varphi }$ or ${\mathrm{\varphi }}^{†}\left({𝜎}^{𝜇◊}\text{ i }{\partial }_{𝜇}\right)\mathrm{\varphi }$

where ${𝐿}^2$ is integrated using ${\mathrm{ℝ}}^{1,3}$ + quadratic form of ${\mathrm{ℂ}}^2$ as ${\mathrm{\varphi }}^{†}𝜓$

The only one that works is $\text{Re }{\mathrm{\varphi }}^{†}\left({𝜎}^{𝜇◊}\text{ i }{\partial }_{𝜇}\right)\mathrm{\varphi }=-\text{ i }\text{ Im }{\mathrm{\varphi }}^{†}\left({𝜎}^{𝜇◊}\text{ i }{\partial }_{𝜇}\right)$, because $\text{Im }{\mathrm{\varphi }}^{†}\left({𝜎}^{𝜇◊}\text{ i }{\partial }_{𝜇}\right)\mathrm{\varphi }=\text{ i }\text{ Re }{\mathrm{\varphi }}^{†}\left({𝜎}^{𝜇◊}{\partial }_{𝜇}\right)\mathrm{\varphi }=\text{ i }{\partial }_{𝜇}\left({\mathrm{\varphi }}^{†}{𝜎}^{𝜇◊}\mathrm{\varphi }\right)=\text{ i }\text{div}\left({\mathrm{\varphi }}^{†}𝜎\mathrm{\varphi }\right)$ is a divergence quantity, using Stokes' theorem + zero boundary

variation gives linear part $\int 2\text{ Re }{\left(\mathrm{\Delta }\mathrm{\varphi }\right)}^{†}\text{ i }{{\partial }_{\text{ spin }}}^{◊}\mathrm{\varphi }$

or $\left(𝜎\cdot {\partial }^{◊}\right)\mathrm{\varphi }=0$ or $\left({𝜎}^{𝜇}{\partial }_{𝜇}^{◊}\right)\mathrm{\varphi }=0$

Similar to ${\mathrm{ℂ}}^2\simeq {\mathrm{ℝ}}^2$ via $𝑥\pm 𝑦\text{ i}$, varying with respect to $\mathrm{ℂ}$-valued $\mathrm{\varphi }$ is equivalent to varying with respect to $\mathrm{ℝ}$-valued $\mathrm{\varphi },{\mathrm{\varphi }}^{†}$

${\partial }_{\text{spin }}{\mathrm{\varphi }}^{◊}=\left(𝜎\cdot {\partial }^{◊}\right)\mathrm{\varphi }=\left({𝜎}^{𝜇}{\partial }_{𝜇}^{◊}\right)\mathrm{\varphi }$ can be interpreted as the gradient momentum of the field (after metric-dual) $\text{i }\cdot \left({\partial }_0,-{\partial }_1,-{\partial }_2,-{\partial }_3\right)\mathrm{\varphi }\in {\mathrm{ℝ}}^{1,3}\otimes {\mathrm{ℂ}}^2$, compounded to, the multiplication of momentum and spinor $\left({𝑝}_{\text{spin }}\otimes \mathrm{\varphi }⇝{𝑝}_{\text{ spin }}\mathrm{\varphi }\right)\in {\mathrm{ℝ}}^{1,3}\otimes {\mathrm{ℂ}}^2\to {\mathrm{ℂ}}^2$

parity dual action uses $\left(0,\frac{1}{2}\right)$ spinor

parity dual eq

or $\left(𝜎\cdot \partial \right)\mathrm{\varphi }=0$ or $\left({𝜎}^{𝜇}{\partial }_{𝜇}\right)\mathrm{\varphi }=0$

Plane wave solution $\mathrm{\Phi }{𝑒}^{-\text{ i }𝑝𝑥}$ with ${𝑝}^2={𝑚}^2$ and ${𝑝}_{\text{spin }}\mathrm{\Phi }=0$

A linear equation with $\text{det }{𝑝}_{\text{ spin }}={𝑝}^2=0$ indicates non-zero solutions. The solution space is one-dimensional, and the solution can be written as $\mathrm{\Phi }={{𝑝}_{\text{ spin }}}^{◊}𝜉$ with $\text{dim}\left(\text{im}\left({{𝑝}_{\text{spin }}}^{◊}\right)\right)=1$

couple Weyl spinors and their parity $\left(\frac{1}{2},0\right),\left(0,\frac{1}{2}\right)$ to $\left(\frac{1}{2},0\right)\oplus \left(0,\frac{1}{2}\right)$

$\left(\begin{array}{cc}𝐴& \\ & {\left({𝐴}^{†}\right)}^{-1}\end{array}\right)\left(\begin{array}{c}\mathrm{\varphi }\\ 𝜓\end{array}\right)=\left(\begin{array}{c}𝐴\mathrm{\varphi }\\ {\left({𝐴}^{†}\right)}^{-1}𝜓\end{array}\right)$

${𝑝}_{\text{spin }}\in {⨀}^{\ast 2}{\mathrm{ℂ}}^2$, ${𝑝}_{\text{spin }}⇝\left(\begin{array}{cc}{{𝑝}_{\text{ spin}}}^{◊}& \\ & {𝑝}_{\text{ spin}}\end{array}\right)$

invariant non couple term ${\left(\begin{array}{c}\mathrm{\varphi }\\ 𝜓\end{array}\right)}^{†}\left(\begin{array}{cc}\text{i }{{\partial }_{\text{ spin}}}^{◊}& \\ & \text{i}{\partial }_{\text{spin}}\end{array}\right)\left(\begin{array}{c}\mathrm{\varphi }\\ 𝜓\end{array}\right)=\text{ i }{\mathrm{\varphi }}^{†}{{\partial }_{\text{ spin }}}^{◊}\mathrm{\varphi }+\text{ i }{𝜓}^{†}{\partial }_{\text{ spin }}𝜓$

non-couple term variation with respect to $\mathrm{\varphi }$ gives $\int 2\text{ Re }{\left(\mathrm{\Delta }\mathrm{\varphi }\right)}^{†}\text{ i }{{\partial }_{\text{ spin }}}^{◊}\mathrm{\varphi }$

invariant couple term $-{\left(\begin{array}{c}\mathrm{\varphi }\\ 𝜓\end{array}\right)}^{†}𝑚\left(\begin{array}{cc}& \mathrm{𝟙}\\ \mathrm{𝟙}& \end{array}\right)\left(\begin{array}{c}\mathrm{\varphi }\\ 𝜓\end{array}\right)=-𝑚\left({\mathrm{\varphi }}^{†}𝜓+{𝜓}^{†}\mathrm{\varphi }\right)=-2\text{ Re }𝑚{\mathrm{\varphi }}^{†}𝜓$

couple term variation with respect to $\mathrm{\varphi }$ gives $\int 2\text{ Re }{\left(\mathrm{\Delta }\mathrm{\varphi }\right)}^{†}\left(-𝑚𝜓\right)$

• overall variation with respect to $\mathrm{\varphi }$ gives $\text{i }{{\partial }_{\text{ spin }}}^{◊}\mathrm{\varphi }-𝑚𝜓=0$
• overall variation with respect to $𝜓$ gives $\text{i }{\partial }_{\text{ spin }}𝜓-𝑚\mathrm{\varphi }=0$
• when $𝑚=0$, decouples into two parity-dual massless-spinors

These two PDEs imply

and $∆={\partial }_{\text{ spin }}{{\partial }_{\text{ spin }}}^{◊}={{\partial }_{\text{ spin }}}^{◊}{\partial }_{\text{ spin}}$ as "square root of $∆$" square-root-of-spacetime-Laplacian_(tag)

Overall $\left(∆+{𝑚}^2\right)\left(\begin{array}{c}\mathrm{\varphi }\\ 𝜓\end{array}\right)=0$, square root of KG. If a field satisfy Dirac eq, then itt satisfy KG eq

All partial derivatives of the action $𝑆$ is zero $\frac{\partial 𝑆}{\partial \mathrm{\varphi }}=\frac{\partial 𝑆}{\partial 𝜓}=0$, giving massive-spinor-equation_(tag) , alias Dirac-equation_(tag)

Although it might be possible to use the eigenvalues of the Hermite matrix ${𝑝}_{\text{spin}}$ (${𝑝}_0\pm {\left({𝑝}_0^2-{𝑚}^2\right)}^{\frac{1}{2}}$), we will calculate it directly here. let ${𝑝}_{\text{spin }}={{𝑞}_{\text{ spin}}}^2$ or

==>

==> Use ${𝑝}_1^2+{𝑝}_2^2+{𝑝}_3^2={𝑝}_0^2-{𝑚}^2$

==> Quadratic equation for ${𝑞}_0^2$: ${\left({𝑞}_0^2\right)}^2-{𝑝}_0\left({𝑞}_0^2\right)+\frac{1}{4}\left({𝑝}_0^2-{𝑚}^2\right)$, solution

==>

or

Still Hermite. Calculation yields

Example

$\begin{array}{ccc}{\left(\begin{array}{c}{{𝑝}_{\text{spin }}}^{\frac{1}{2}}𝜉\\ {{𝑝}_{\text{spin }}}^{◊\frac{1}{2}}𝜉\end{array}\right)}^{†}\left(\begin{array}{c}{{𝑝}_{\text{spin }}}^{\frac{1}{2}}𝜉\\ {{𝑝}_{\text{spin }}}^{◊\frac{1}{2}}𝜉\end{array}\right)& =& {𝜉}^{†}\left({𝑝}_{\text{spin }}+{{𝑝}_{\text{ spin}}}^{◊}\right)𝜉\\ & =& \left(2{𝑝}_0\right)|𝜉{|}^2\\ & =& \left(\text{tr }{𝑝}_{\text{ spin}}\right)|𝜉{|}^2\end{array}$

Inspired by the treatment of KG field quantization

We can also define a square root of a point particle complex harmonic oscillator.

Dirac plane wave $\left(\begin{array}{c}𝑢\\ 𝑣\end{array}\right){𝑒}^{-\text{ i }𝑝𝑥}$ with ${𝑝}^2={𝑚}^2$ and $\left(\left(\begin{array}{cc}{𝑝}_{\text{spin }}& \\ & {{𝑝}_{\text{spin }}}^{◊}\end{array}\right)-𝑚\left(\begin{array}{cc}& \mathrm{𝟙}\\ \mathrm{𝟙}& \end{array}\right)\right)\left(\begin{array}{c}𝑢\\ 𝑣\end{array}\right)=0$

Lagrangian $𝐿\left(\left(\begin{array}{c}\mathrm{\varphi }\\ 𝜓\end{array}\right),{\partial }_0\left(\begin{array}{c}\mathrm{\varphi }\\ 𝜓\end{array}\right)\right)={\left(\begin{array}{c}\mathrm{\varphi }\\ 𝜓\end{array}\right)}^{†}\left(\begin{array}{cc}\text{i }{\partial }_0+{𝑝}_{\text{ spin}}& -𝑚\mathrm{𝟙}\\ -𝑚\mathrm{𝟙}& \text{i }{\partial }_0-{𝑝}_{\text{ spin}}\end{array}\right)\left(\begin{array}{c}\mathrm{\varphi }\\ 𝜓\end{array}\right)$

The conditions satisfied by the $-\text{ i}$ plane wave $\left(\begin{array}{c}𝑢\\ 𝑣\end{array}\right){𝑒}^{\text{i }𝑝𝑥}$ are

Write the equation of the harmonic oscillator square root as a constant coefficient ODE

Perform $\text{exp}$ transformation. Use ${\left(\begin{array}{cc}-{𝑝}_{\text{spin}}& 𝑚\mathrm{𝟙}\\ 𝑚\mathrm{𝟙}& {𝑝}_{\text{spin }}\end{array}\right)}^2=\left({𝑝}^2+{𝑚}^2\right)\left(\begin{array}{cc}\mathrm{𝟙}& \\ & \mathrm{𝟙}\end{array}\right)={𝐸}^2\left(\begin{array}{cc}\mathrm{𝟙}& \\ & \mathrm{𝟙}\end{array}\right)$ The series rule is

The result is

Ignored some issues

Tangent projective light cone bundle is well-defined

But is a ${\mathrm{ℂ}}^2$ field or a ${\mathrm{ℂ}\mathrm{\mathbb{P}}}^1$ field needed? There are too many choices to lift a ${\mathrm{ℂ}\mathrm{\mathbb{P}}}^1$ field to a ${\mathrm{ℂ}}^2$ field (or to a ${\mathrm{ℂ}}^2\setminus 0$ field); all lifting choices form a $\text{GL}\left(1,\mathrm{ℂ}\right)={\mathrm{ℝ}}_{>0}\times \text{ U }\left(1\right)$ field

And there are only two ways to lift $\text{SO}\left(1,3\right)$ to $\text{SL}\left(2,\mathrm{ℂ}\right)$

On a curved manifold, there may not even be a global single-valued lift.

The change in the lifting method from a ${\mathrm{ℂ}\mathrm{\mathbb{P}}}^1$ field to a ${\mathrm{ℂ}}^2$ field corresponds to "changing the gauge", by multiplying the spinor $\mathrm{\varphi }\left(𝑥\right)$ by $𝑎\left(𝑥\right)=\left|𝑎\left(𝑥\right)\right|\text{exp}\left(𝜃\text{ i}\right)$ to change the gauge.

If the conserved current of the action is to be simpler, then use $\text{U }\left(1\right)$ gauge transformation instead of $\text{GL}\left(1,\mathrm{ℂ}\right)=\text{ U }\left(1\right)\times {\mathrm{ℝ}}_{>0}$. $\text{U }\left(1\right)$ does not change the Lagrangian action, which simplifies the calculation of conserved currents (cf. the case of scalar field calculating 4-current for $\text{U }\left(1\right)$ symmetry).

Changing the gauge is not compatible with taking derivatives of tangent spaces in bundle coordinates, so an additional structure — #link(<principal-bundle-connection>)[connection] — must be introduced.

There are many possible connections. A good connection is one with the smallest curvature cf. #link(<electromagnetic-field>)[]

The ${\mathrm{ℂ}}^2$ bundle in curved spacetime can be directly defined in the bundle coordinates of the $\text{SO}\left(1,3\right)$ principal bundle (orthonormal frame bundle). Using the #link(<Lorentz-group-spinor-representation>)[$SL(1,ℂ) ↠ SO(1,3)$ correspondence], changing $\text{SO}\left(1,3\right)$ bundle coordinates automatically corresponds to changing $\text{SL}\left(2,\mathrm{ℂ}\right)$ bundle coordinates.

In curved spacetime, one needs to deal with the covariant derivative of the spinor field with respect to the metric, which is derived from the #link(<metric-connection>)[] of the tangent vector field.

For spinors, one might need to use an orthonormal frame instead of a coordinate frame, i.e., use $\text{SO}\left(1,3\right),\text{SL}\left(2,\mathrm{ℂ}\right)$ principal bundle. Does this introduce new difficulties for calculating covariant derivatives?

Even if the topology of the spacetime base manifold is non-trivial, there might exist different bundle types for gauge fields.

One problem is that, unlike spinor fields, gauge bundles do not seem to be directly related to tangent bundles.

It seems that all types of $\text{U }\left(1\right)$ bundle types based on the base manifold must be considered simultaneously.

In the homotopy sense, ${\mathrm{ℝ}}^{𝑝,𝑞}$ has only one type of $\text{U }\left(1\right)$ bundle type.

The #link(<principal-bundle>)[frame bundle] of $\text{SO}\left(1,3\right)$ derived from the tangent bundle metric and the #link(<principal-bundle-connection>)[connection] of the $\text{SO}\left(1,3\right)$ frame bundle derived from #link(<metric-connection>)[] behave as $\mathrm{\Gamma }$ is locally type ${\left({\mathrm{ℝ}}^{1,3}\right)}^{⊺}\otimes \text{so}\left(1,3\right)$, acting on ${\mathrm{ℝ}}^{1,3}$ tangent vector fields by $\partial +\mathrm{\Gamma }$

The way to derive the spin-connection is to #link(<square-root-of-Lorentz-Lie-algebra>)[map the $so(1,3)$ part of the induced metric-connection $Γ$ to $sl(2,ℂ)$] in the orthonormal-frame, yielding the connection of the $\text{SL}\left(2,\mathrm{ℂ}\right)$ bundle, locally type ${\left({\mathrm{ℝ}}^{1,3}\right)}^{⊺}\otimes {\mathrm{ℂ}}^2$, acting on the spinor field ${\mathrm{ℂ}}^2$ by $\partial +𝜔$ with $𝜔={𝜔}^{𝜇𝜈}\frac{1}{4}{\left[{𝜎}_{𝜇},{𝜎}_{𝜈}\right]}_{◊}$

Although the definition of the Pauli matrix $𝜎$ for spin representation requires $𝑔$, both $𝑔$ and Lie algebra can be expressed by the "square" of $𝜎$ thereafter.

massless-spinor-action

massless-spinor-equation

I haven't verified if this definition is conceptually reasonable. Compare with flat spacetime, try to prove or disprove it.

• $\text{i }{{\nabla }_{\text{ spin}}}^{◊}$ is self-adjoint
• Only $\text{Re }\left({𝜓}^{†}{{\nabla }_{\text{ spin }}}^{◊}𝜓\right)$ contributes to the variation of the action.
• ${{\nabla }_{\text{spin }}}^{◊}{\nabla }_{\text{ spin }}={\nabla }_{\text{ spin }}{{\nabla }_{\text{ spin }}}^{◊}=\frac{1}{2}\left({\nabla }^{†}\nabla +\nabla {\nabla }^{†}\right)$ i.e. square-root-of spacetime Laplacian (closer to the Laplacian of tangent vector fields rather than scalar fields)

massive-spinor-Lagrangian

massive-spinor-equation