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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.

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    $\begingroup$ maybe they meant spin 0? $\endgroup$
    – lineage
    Commented Apr 8 at 14:26
  • $\begingroup$ @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? $\endgroup$ Commented Apr 8 at 16:41
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    $\begingroup$ 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. $\endgroup$
    – lineage
    Commented Apr 8 at 17:42
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    $\begingroup$ (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. $\endgroup$
    – tobi_s
    Commented Apr 9 at 1:33

1 Answer 1

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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.

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  • $\begingroup$ 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? $\endgroup$ Commented Apr 8 at 17:01
  • $\begingroup$ @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). $\endgroup$
    – Hyperon
    Commented Apr 8 at 19:37
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    $\begingroup$ @SiriusBlack With respect to these space-time symmetries, the three components $\pi_i(x)$ of the pion field are transforming as scalars! $\endgroup$
    – Hyperon
    Commented Apr 8 at 19:41
  • $\begingroup$ Very good, thank you very much! $\endgroup$ Commented Apr 13 at 4:45

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