# Elements of vector algebra and particle physics

What does scalar, vector, pseudovector and pseudoscalar particles have to do with the concept of scalar, polar vectors and axial vectors in vector algebra?

• Similar rules under parity transform. – Turgon Jul 20 '18 at 6:27
• Vectors etc play a role in all physics. – marshal craft Jul 20 '18 at 6:43
• The terms are with respect to how they change under proper and improper Lorentz transformations and the language is synonymous with how ordinary 3D vectors transform under proper/improper rotations. – CAF Jul 20 '18 at 10:53

The elementary particles of the standard model of particle physics have been discovered and modeled with a complex mathematical model and can be described with various vector spaces. The basic way we know of their existence is by measuring the momentum and energy in various and numerous experiments, use conservation laws (energy , momentum, angular momentum) and define them.

The classifications on the particle's list as vector , pseudovector , scalar have to do with the intrinsic angular momentum vector characterizing them, called spin.This was found and identified with each particle by measurement and the use of conservation laws. This classification also follows resonances and composite particles, and in addition spinors are introduced (again necessary because of the use of conservation laws), which are an extension of vector algebra.

Example the rho meson, which is a vector meson, spin 1 from the way the constituent quarks define the meson.

First of all, axial vector is not the same as pseudo-vector but in 4D they will coincide. I will explain why this is so at the end of the answer.

Now, these terms are referring to the transformation properties under the Lorentz symmetry of the field theory.

Let me tell the generic picture first, without the contents of the objects in question. Any term, $\mathbf{F}(x)$, of rank 1, regardless of its constituents, transforms as $$\mathbf{F} (x) \rightarrow \mathbb{U}(x) \, \mathbf{F}(x)$$ where $\mathbb{U} (x)$ is an element of Lorentz algebra, $\mathfrak{so} (1,3)$, and it forms according to the representations of the algebra, i.e., spinor, vector, etc.. So, for example, if $\mathbb{U} (x)$ is $1$, i.e. the term doesn't change, then it is called a scalar. It can be a spinor field's mass term, $m \bar\psi (x) \psi (x)$ or the kinetic term of a scalar field, $\partial_\mu \phi (x) \partial^\mu \phi (x)$, etc., but it will not change under the transformations.

We can also write the transformation of a term with higher ranks (like tensors and matrices), but let's keep it simple.

Suppose we do a Lorentz transformation as follows: $$x^\mu \rightarrow \Lambda_\nu^\mu x^\nu$$ where $\Lambda_\mu^\nu$ are the coefficients of the Lorentz transformation. For spinors the transformation would be a 4x4 matrix, denoted as $S_a^b (\Lambda)$, depending on $\Lambda$, where small Latin indices represent the spinor indices while Greek indices denote spacetime ones.

It is trivial to see that a free Greek index in the term will imply a vector tranformation, and a free Latin index will imply a spinor transformation. However, pseudo-scalar or pseudo-vector applies when $\gamma^5$ matrix is present, which is defined as follows: \begin{align} \gamma^5 &= i \epsilon_{\mu\nu\alpha\beta} \gamma^\mu \gamma^\nu \gamma^\alpha \gamma^\beta \\ &= i \gamma^0 \gamma^1 \gamma^2 \gamma^3 \end{align} where $(\gamma^\mu)^a_b$ are the Dirac matrices living in the spinor space (on the formula, I omitted the spinor indices), and $\epsilon_{\mu\nu\alpha\beta}$ is the Levi-Civita symbol (not the tensor).

Since $\gamma^5$ is a matrix, under the transformation it will produce a determinant of the Jacobian of the transformations, even if the matrix multiplication gives a scalar at the end. Because the definition consists the Levi-Civita symbol, which is just a bunch of 0's, 1's and -1's, so that the Greek indices of the multiplied gamma matrices can not form a true scalar, i.e. their transformation matrix would not cancel out by its inverse (since there are no inverses produced there, the symbol does not transform). \begin{align} \gamma^5 &\rightarrow i \epsilon_{\mu\nu\alpha\beta} \Lambda_\bar\mu^\mu \, \gamma^\bar\mu \Lambda_\bar\nu^\nu \, \gamma^\bar\nu \Lambda_\bar\alpha^\alpha \, \gamma^\bar\alpha \Lambda_\bar\beta^\beta \, \gamma^\bar\beta \\ & = i \epsilon_{\mu\nu\alpha\beta} \Lambda_\bar\mu^\mu \, \Lambda_\bar\nu^\nu \, \Lambda_\bar\alpha^\alpha \, \Lambda_\bar\beta^\beta \, \gamma^\bar\mu \gamma^\bar\nu \gamma^\bar\alpha \gamma^\bar\beta \\ & = \epsilon_{\mu\nu\alpha\beta} \Lambda_\dot\mu^\mu \, \Lambda_\dot\nu^\nu \, \Lambda_\dot\alpha^\alpha \, \Lambda_\dot\beta^\beta \, \epsilon^{\dot\mu\dot\nu\dot\alpha\dot\beta} \epsilon_{\bar\mu\bar\nu\bar\alpha\bar\beta} i\gamma^\bar\mu \gamma^\bar\nu \gamma^\bar\alpha \gamma^\bar\beta \\ &=\det |\Lambda| \, \gamma^5 \end{align} where determinant is defined in terms of the contraction of two Levi-Civita symbols with four copies of a matrix.

So, for the following fields, the transformation properties are: \begin{align} \tag{scalar} \phi(x) &\rightarrow \phi(x) \\ \tag{pseudoscalar} \Phi(x) &\rightarrow \det |\Lambda| \, \Phi(x) \\ \\ \tag{spinor} \psi_a(x) &\rightarrow L^b_a(\Lambda) \, \psi_b(x) \\ \tag{pseudospinor} \Psi_a (x) &\rightarrow \det |\Lambda| \, L^b_a(\Lambda) \, \Psi_b(x) \\ \\ \tag{vector} A_\mu (x) &\rightarrow \Lambda^\nu_\mu \, A_\nu (x) \\ \tag{pseudovector} \Gamma_\mu (x) &\rightarrow \det |\Lambda| \,\Lambda^\nu_\mu \,\Gamma_\mu (x) \\ \\ \tag{tensor} F_{\mu\nu} (x) &\rightarrow \Lambda^\alpha_\mu \,\Lambda^\beta_\nu \, F_{\alpha\beta} (x) \\ \tag{pseudotensor} \Gamma_\mu (x) &\rightarrow \det |\Lambda| \,\Lambda^\nu_\mu \,\Gamma_\mu (x) \end{align} where $\det|\Lambda|$ is the determinant of $\Lambda_\nu^\mu$.

As you can see pseudo- prefix means that the field transforms similar to the one in subject but with a determinant of the transformation matrix, so that it becomes a density, and densities are not scalars, they carry the determinant of the Jacobian of the transformation, like a volume element.

The reason of that the pseudo-things are sometimes called axial, because $\gamma^5$ matrix, as one can see when looking to any of its representations or the definition above, is actually changes sign under parity transformations, just like axial vectors in the geometry, and you can always produce an axial vector from a vector by multiplying it with gamma-5. However the distinction is a pseudo:vector is actually a bivector where the basis is different than a vector but are in same number. Generically, a pseudo-vector is the Hodge dual of a vector, but an axial vector is a product of a vector by a vector. In field theory, since there 4 dimensions, pseudo-vectors will also be axial.

The connection is with respect to the wave function of the particles:

• Scalar particles have no spin, so their wave functions do not get any rotation under a spacial rotations. This is just like scalars in a geometrical theory do not change under coordinate rotations.

• Vector particles have spin 1, their spin vector needs to be rotated under spacial rotations. Again, this is just like a vector's components need to be rotated in a geometrical theory under a coordinate rotation.

• Spinor particles have spin 1/2, again their spin vector needs to be rotated under spacial rotations, but the way their spin vector rotates is path dependent. One common distinguishing feature people like to point out for spinor particles is that a full spacial rotation of 360 degrees generates a phase of -1 instead of acting like an identity transformation.

The prefix psuedo means that the wave function of the particle picks up an extra phase of -1 under a parity transformation, taking $(x,y,z) \rightarrow (-x,-y,-z)$. You can construct a pseudovector in a geometrical theory by taking the cross product of two regular (sometimes called polar) vectors.

$$\vec{a} \times \vec{b} = \vec{c} \\ -\vec{a} \times -\vec{b} = (-1)^2 ~\vec{a} \times \vec{b} = \vec{c}$$

$\vec{c}$ is a vector that does not change sign under parity and is therefore a pseudovector. The dot product of $\vec{c}$ with a polar vector is a pseudoscalar.

All of this is important because it helps us explain which particle decays (and productions) are possible. Decays occur through a force, if a force respects parity, like the strong and electromagnetic forces for instance, then we can rule out certain decays by working out the parity of the initial and final states and ensuring they are the same.

• A 360 degree rotation doesn't invert as spin (to the extent that that means $|\uparrow\rangle \rightarrow |\downarrow \rangle$), rather it puts a minus sign on the phase: $|\uparrow\rangle \rightarrow -|\uparrow \rangle$. – JEB Jul 20 '18 at 16:48