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When I first learned group theory in the context of quantum mechanics, I got it into my head that a group was the unification of different "transformation actions" that ultimately have the same behaviour. This seemed to make sense when I looked at an example such as $Z_4$. I can define a group action, $$ g_1 \equiv i, $$ and get the behaviour $$ 1 \longrightarrow i \longrightarrow -1 \longrightarrow -i \longrightarrow 1, $$ but I can also define a group action, $$ g_2 \equiv \begin{pmatrix} -1 & 1\\\ 0 & -i \end{pmatrix} $$ and get the "same" behaviour $$ \begin{pmatrix} 1 & 0\\\ 0 & 1 \end{pmatrix} \longrightarrow \begin{pmatrix} -1 & 1\\\ 0 & -i \end{pmatrix} \longrightarrow \begin{pmatrix} 1 & -1-i\\\ 0 & -1 \end{pmatrix} \longrightarrow \begin{pmatrix} -1 & i\\\ 0 & i \end{pmatrix} \longrightarrow \begin{pmatrix} 1 & 0\\\ 0 & 1 \end{pmatrix}. $$ My thinking was then that group theory helped us identify that, in the end, $g_1$ and $g_2$ are just the same mathematical objects in this context.

This intuition is then shattered when we define different particles as different representations of the same group. I take this to mean that, in some weird alternate universe, $g_1$ and $g_2$ could correspond to two different particles (neglecting to check if they are irreducible.) So I know my intuition is wrong, I'm just not sure how it is wrong, and I still struggle to see the logic behind that definition of particles.

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    $\begingroup$ I think you are confusing the representation of the group elements with the vector space on which they act. In particle physics, elementary particles transform under irreducible representations of the symmetry groups. These are the Poincare group + all the internal (gauge) symmetries which we have discovered. $\endgroup$ Mar 13 at 21:15
  • $\begingroup$ You are comparing the 1D to the 2D representation of your finite group. Analogous infinite group representation matrices act on particles in different representations of the same Lie group. What, exactly, is the question? $\endgroup$ Mar 13 at 22:42
  • $\begingroup$ Yea okay, but what is the right way to think about it then? What is the difference between transforming under one representation and another, if the group has the same actions in all? $\endgroup$
    – Depenau
    Mar 13 at 22:43
  • $\begingroup$ The argument I've read states that you can start a treatment of QFT by defining particles as transforming under irreducible unitary representations of the Poincaré group. I know why they need to be unitary, I think I know why they need to be irreducible, but I fail to understand how this is a good axiom. So my question is, how can we know we can exhaust all possibilities of different particles by just looking at different irreducible representations of the Poincaré group? What even is the difference between transforming under on group as opposed to another? $\endgroup$
    – Depenau
    Mar 13 at 23:22
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    $\begingroup$ Why do you neglect to check irreducibility in your example? It might help to look at other simpler examples, how about considering Parity $\mathbb{Z}_2$ (subgroup of full Poincare group), which has two irreducibles, the trivial representation where Parity sends particle to itself, or the sign representation where Parity collects a minus sign. Is it clear how these two representations should correspond to two different behaviours under Parity (and hence, the full Poincare group)? $\endgroup$ Mar 14 at 3:05

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The idea that different particles 'are' irreducible representations of a group, or that they 'are' the group elements themselves, is a shorthand for a more complicated situation. Particles are excitations of a field that obeys a differential equation; something like the Schrodinger equation. If the differential equation has certain symmetries, expressed by a symmetry group, then its solutions must in some sense have the same symmetries. We can deduce a lot about the solutions and how they are interrelated by studying the symmetry group and its representations, to the point where we never even bother to work out the differential equations and its solutions explicitly - we only look at the group and its properties.

I'm going to argue by analogy with the situation we have for the Schrodinger equation of a Hydrogen atom. I think the truth is probably more complicated than this, but this gives the gist.

So suppose we have a differential equation of the form $H\psi=\epsilon \psi$, where $H$ is a differential operator, $\epsilon$ is a constant number, and $\psi$ is the wavefunction. The possible solutions for $\epsilon$ are the eigenvalues, and for each eigenvalue there are one or several possible solutions $\psi$ which we call eigenvectors or eigenfunctions or eigenstates. For the Schrodinger equation, $H$ is the Hamiltonian, $\epsilon$ is the energy, and the wavefunction evolves in time according to the differential equation $H\psi=i\hbar \partial \psi / \partial t$.

The Hamiltonian for a particle in a potential well might look something like:

$$H(\mathbf{r})=-\frac{\hbar^2}{2\mu}\left( \frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\partial y^2}+\frac{\partial^2}{\partial z^2} \right)-\frac{e^2}{\sqrt{x^2+y^2+z^2}}$$

The first term on the right is the kinetic energy of the particle, the second term is the spherically symmetric potential well of the Hydrogen atom, in which an electron sits.

This potential is spherically symmetric. If we rotate the position vector about the origin, the Hamiltonian doesn't change. $H(T\mathbf{r})=H(\mathbf{r})$. The set of transformations $T \in \mathcal{G}$ for which the Hamiltonian doesn't change form a group, called the invariance group of the Hamiltonian. In this case, it is the 3D rotation group, but it doesn't have to be. Other potential well shapes will give rise to other symmetry groups.

We can apply the same transformations to the wavefunction that we just applied to the Hamiltonian operator. We write $(P(T)\psi)(\mathbf{r})=\psi(T\mathbf{r})$, where $P(T)$ is the operator that applies the coordinate transformation $T$ to the wavefunction. The operators $P(T)$ for $T \in \mathcal{G}$ form a group isomorphic to $\mathcal{G}$. We use a slightly different notation for them as they're a different mathematical object, acting on wavefunctions rather than coordinates, but it's the same set of transformations.

If we have a set of $d$ linearly-independent wavefunctions $\psi_1(\mathbf{r}),\psi_2(\mathbf{r}),\ldots,\psi_d(\mathbf{r})$, then they can be treated as the basis vectors of a $d$-dimensional representation $\Gamma$ of $\mathcal{G}$ if for every coordinate transformation $T$ of $\mathcal{G}$,

$$P(T)\psi_n(\mathbf{r})=\sum_{m=1}^d {\Gamma(T)_{mn}\psi_m(\mathbf{r})}$$

This is saying that when we transform one of our basis vectors, we get some linear combination of it and the other basis vectors as a result. The coefficients $\Gamma_{mn}$ of the linear combination for each basis vector can be arranged in a matrix, and multiplying the matrices performs the same action on any linear combination of basis vectors as applying the coordinate transformation.

And finally, going back to our eigenvalue equation $H(\mathbf{r})\psi(\mathbf{r})=\epsilon\psi(\mathbf{r})$, if we have a complete set of $d$ eigenfunctions $\psi_i(\mathbf{r})$ that all correspond to the same eigenvalue $\epsilon$, then they form a basis for a $d$-dimensional representation of the invariance group of the equation. Transforming any one of them in a way that leaves the Hamiltonian invariant gives another eigenfunction with the same eigenvalue, which is therefore in the same vector space, and which therefore has to be a linear combination of the basis eigenfunctions.

This was the reasoning that was used to help solve the Schrodinger equation for a Hydrogen atom using its spherical symmetry as a shortcut - the orbital solutions use the spherical harmonic functions as a basis. Each energy level has a different number of orbitals. The $d$ orbitals in an energy level are basis vectors of a $d$-dimensional representation of the rotation group. In the same way, we identify the internal symmetry group of the field that all particles are excitations of, which comes from the symmetries of whatever differential equation it obeys, and then the irreducible representations tell us about how the solution wavefunctions might be related. A $d$-dimensional representation corresponds to $d$ solution eigenfunctions all corresponding to the same eigenvalue.

The mental picture this ought to evoke would be to consider the group as like the way we know the Hydrogen atom is spherically symmetric, so rotations leave it unchanged, and the different particles are more like the different electron shells and orbitals in a Hydrogen atom. Each 'particle' is the universal wavefunction wobbling in a different pattern. But we don't look directly at the individual patterns, only at how they are related to the overall symmetry.

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I think my question was more basic than I let on. What I was really wondering about is how exactly can two representations correspond to two $\textit{different}$ particles, when the representations belong to the same group and thus have the same "behaviour," in a lack of better words.

I think the answer is simply that different particles have different degrees of freedom, and to model different degrees of freedom, we must put them into different objects, scalars, spinors, vectors, etc, and of course these should trasnsform under the same group but in different representations.

Sorry if my original question was poorly formulated.

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  • $\begingroup$ As it’s currently written, your answer is unclear. Please edit to add additional details that will help others understand how this addresses the question asked. You can find more information on how to write good answers in the help center. $\endgroup$
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    Mar 16 at 23:33

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