# How can a scalar field have components and how do I interpret these components?

From lecture notes$$^\zeta$$ I've been reading that:

Consider a real three-component scalar field $$\phi=\begin{pmatrix}\phi_1 \\\ \phi_2 \\\ \phi_3\end{pmatrix}\tag{a}$$ with Lagrangian $$\mathcal{L}=\frac12\partial_\mu\phi^T\partial^\mu\phi-\frac12m^2\phi^T\phi-\frac14\left(\phi^T\phi\right)^2.\tag{1}$$ This Lagrangian is invariant under internal transformations that correspond to multiplication by a $$3\times 3$$ matrix $$M$$, $$\phi\to M\phi\tag{2},$$ provided that the transformation leaves the combination $$\phi^T\phi$$ invariant for any $$\phi$$. Because $$\phi^T\phi\to \left(M \phi\right)^TM\phi=\phi^TM^TM\phi\tag{3},$$ this is true if $$M^TM=\mathbb{I}$$. In other words, the matrix $$M$$ has to be an orthogonal $$3\times 3$$ matrix. These matrices form a group called $$\mathrm{O}(3)$$.

The notes eventually generalize to complex $$N$$-component scalar fields:

Consider a complex $$N$$-component scalar field $$\phi=\begin{pmatrix}\phi_1 \\\ \phi_2 \\\ \vdots \\\ \phi_N\end{pmatrix}\tag{b}$$ we find that the Lagrangian $$\mathcal{L}={\partial_\mu}\phi^\dagger\partial^\mu\phi-V\left(\phi^\dagger\phi\right)\tag{4}$$ is invariant under $$\mathrm{U}(N)$$ transformations. If the scalar field components are real, the symmetry group of the Lagrangian $$\mathcal{L}={\partial_\mu}\phi^T\partial^\mu\phi-V\left(\phi^T\phi\right)\tag{5}$$ is $$\mathrm{O}(N)$$.

Most of the above passages were included to provide some context; I'm not that concerned with equations $$(1)-(5)$$, it is equations $$(\mathrm{a})$$ and $$(\mathrm{b})$$ that I don't know how to interpret.

For now taking eqn. $$(\mathrm{a})$$, $$\phi=\begin{pmatrix}\phi_1 \\\ \phi_2 \\\ \phi_3\end{pmatrix}$$ I'm getting confused by the nomenclature used in the above passages of notes. This 'object', $$\phi$$ which is called a "scalar field" must actually be a vector, since it has components. But how does one interpret these components, $$\phi_1,\,\phi_2,\,\phi_3$$?

For comparison, the electric-field is a vector-field and can be written as $$\vec E=\begin{pmatrix}E_x \\\ E_y \\\ E_z\end{pmatrix}$$ where $$E_x,\,E_y,\,E_z$$ are the (Cartesian) components which are the orthogonal directions of this vector field.

But I can't write $$(\mathrm{a})$$ as $$\vec\phi=\begin{pmatrix}\phi_1 \\\ \phi_2 \\\ \phi_3\end{pmatrix}$$ and claim that $$\phi_1,\,\phi_2,\,\phi_3$$ are the 'directions' of $$\vec \phi$$, even though $$\phi$$ in eqn. $$(\mathrm{a})$$ really is a vector.

So if the components are not 'directions' perhaps they are representing something else, such as a label for each particle in the system, this would appear to make more sense especially considering eqn. $$(\mathrm{b})$$, $$\phi=\begin{pmatrix}\phi_1 \\\ \phi_2 \\\ \vdots \\\ \phi_N\end{pmatrix}$$ where $$\phi_1,\,\phi_2,\,\cdots\phi_N$$ are the field components for each of the respective particles in the field.

What is the correct way to interpret these scalar field components?

$$\zeta$$ - These are lecture notes on quantum field theory from ICL dept. of physics.

• maybe they meant spin 0? Commented Apr 8 at 14:26
• @lineage Hi there, these fields describe bosons, I'm not sure if it's spin zero though; I checked the notes and Fermions are dealt with later on. Lets suppose these fields were for spin zero, how does knowing that change anything with regards to my question? Commented Apr 8 at 16:41
• hi! in QFT, spin-0 particles are referred to as scalars, spin 1/2 as spinors and spin 1 as vectors - independent of how many component's they have. For eg. your $\phi$ would be a scalar field if $\phi_i$ are scalar fields. The fact that it has three components matters only to the internal symmetry (wrt. which it is a vector) but not symmetry under spatial transformations (wrt. which it maybe a scalar). So if the notes were referring to the spin of the field, that would explain use of the term scalar. Commented Apr 8 at 17:42
• (a) and (b) are just abbreviations. You might as well write e.g. the Lagrangian (1) on three identical lines, one for $\phi_1$, $\phi_2$, $\phi_3$. "scalar" is usually taken to mean "scalar WRT a group." When the group is not spelled out then usually the Lorentz group is meant. In this specific case, it boils down to "the $\phi_n$ are (real) numbers, so $\phi^T=(\phi_1,\phi_2,\phi_3)$ and $\phi^T\phi \equiv \sum_i \phi_i^2$. For a complex number, you'd have to complex conjugate in some place for this to make sense, for four-vectors you'd have to add (1,-1,-1,-1) in some strategic places etc. Commented Apr 9 at 1:33

The components of the electric (field) vector $$\vec{E} \,$$ "live" in ordinary three-dimensional space (where we also live in). If you place a small test charge $$q$$ at some point in space (say, the origin of your coordinate system), you can measure ("see") the direction and the magnitude of the force $$\vec{F}=q \vec{E}\,$$ exerted on the charge. With respect to a rotated coordinate system, the components of the electric field become $$E_i^\prime = R_{ij} E_j$$, where $$R$$ is an $$\rm SO(3)$$ matrix.
The components of your field $$\phi=(\phi_1,\phi_2, \phi_3)^T$$ in example (a) or $$\phi=(\phi_1, \ldots, \phi_N)^T$$ in case (b) "live" in abstract spaces, completely unrelated to our ordinary three-dimensional space we live in. Nevertheless, it turns out to be convenient using the geometric language of "vectors", "rotations", etc. also in such abstract (mathematical) spaces.
A well known physical example is the pion field $$\vec{\pi}= (\pi_1, \pi_2,\pi_3)$$ "living" in the three-dimensional representation space of the three-dimensional irreducible representation of the isospin group $$\rm SU(2)$$, where the linear combinations $$\pi^0=\pi_3$$, $$\pi^\pm =(\pi_1\mp i\pi_2)/\sqrt{2}$$ correspond to the neutral and the charged pions, respectively. The isospin group is an example of an internal symmetry group in contrast to space-time symmetries associated with the Poincare group describing rotations, Lorentz boosts and translations in ordinary space-time.
• Thanks for your answer, if I've understood you correctly the abstract space you mention is a Hilbert space. Just a couple of questions; is the pion field a vector field, (as you wrote $\vec \pi$ instead of $\pi$)? So the most specific thing I can write is that '$\pi_1,\,\pi_2,\,\pi_3$ are components of the pion field', there is really no other meaning I can assign to these components? Commented Apr 8 at 17:01
• @SiriusBlack Writing $\vec{\pi}=(\pi_1, \pi_2, \pi_3)$ or $\pi = (\pi_1,\pi_2, \pi_3)$ is simply a matter of taste without any deeper meaning. At any fixed point in space-time, the pion field is just a vector in a three-dimensional (abstract) isospin space, not related to ordinary space. It should not be called a "vector field" as this terminology is reserved for fields transforming as a four-vector $A^\mu(x)$ field with respect to Lorentz transformations (or a three-vector like $\vec{E}(\vec{x})$ with respect to rotations). Commented Apr 8 at 19:37
• @SiriusBlack With respect to these space-time symmetries, the three components $\pi_i(x)$ of the pion field are transforming as scalars! Commented Apr 8 at 19:41