How to determine which irrep the Hilbert space of states carry?

Is the following statement ([1]) correct?

[1] If the universe has a symmetry under a group $$G$$, does this mean the Hilbert space carries a unitary representation formed by taking the direct product of all the irreps of the group $$G$$?.

When considering symmetries in a non-relativistic theory, often we look for operators $$\hat{S}$$ such that the statement $$\Big[\hat{H},\hat{S}\Big]=0$$ is true for a Hamiltonian $$\hat{H}$$. My confusion is that this Hamiltonian is usually formulated to only describe a certain type of particle e.g. the Pauli Hamiltonian for spin $$\frac{1}{2}$$, so the overarching general symmetry principles are lost.

Should there exist one-particle Hamiltonians $$\hat{H}_0, \;\hat{H}_\frac{1}{2},\;\hat{H}_1,\dots\;$$

where the $$n$$ in $$\hat{H}_n$$ refers to the spin of the particle? Also, I'm assuming that the list goes on indefinitely leading me to think that statement [1] is correct. Does this have something to do with all unitary representations on $$\mathcal{H}$$ are infinitely dimensional for $$SO(1,3)$$? i.e. Is there a proof that all the unitary representations on $$\mathcal{H}$$ of $$SO(1,3)$$ can be decomposed and written as a direct sum of all the irreps.

Another example is that non-relativistically, there exists an $$SO(3)$$ symmetry that leads to angular momentum. The state-space seems to carry a representation of the direct product of all the irreps of this group. Could someone confirm if what I've said is true or detail where I've gone wrong?

An example for which the above seems not to be true is when talking about internal symmetries however I'm not quite sure why. Is this because the theory is set up for the state space to carry a very specific unitary representation of the internal symmetry group?

• What $SU(2)$ symmetry is the first paragraph referring to? Aug 21, 2021 at 23:25
• @ChiralAnomaly Non-relativistic spin Aug 21, 2021 at 23:27
• Surely I can if the Hamiltonian has a $2$ dimensional state space and only admits one-particle states. Aug 22, 2021 at 11:25
• edited to add to my answer Aug 22, 2021 at 11:30
• @ZeroTheHero Thanks for your answer. How do I know which irreps a Hilbert space carries? Aug 22, 2021 at 11:46

The actual irrep is not linked to the number of particles. It is true that you can construct any irrep (of a semi-simple Lie group or algebra) by taking suitable tensor products of basic representations but this does not imply that these basic representations are “single particle representations”, while others are “multi-particle”.

For instance spherical harmonics of angular momentum $$\ell$$ span a vector space that carries a $$2\ell+1$$-dimensional irrep, irrespective of the number of particles involved.

Moreover, the tensor product of two $$\ell=1$$ irreps decomposes into $$L=2,1,0$$ so the singlet can hardly be thought as a 0-particle irrep, and the (antisymmetric) $$L=1$$ piece is a 2-particle irrep, despite being isomorphic to the original 1-particle $$\ell=1$$ pieces.

Summarizing (and repeating): there is no link between the representation labels and the number of particles. The same applies to $$SU(2)$$.

If a Hilbert space carries a representation of a group, you cannot say anything about this irrep in general as it could be any irrep.

The Hilbert space does not have to carry every representation of the group, but presumably it will carry one, which may be reducible, but does not need to contain every representation. For instance, the even harmonic oscillator states $$\{\vert 2n\rangle\}$$ transform by a single representation of $$\mathfrak{sp}(2,\mathbb{R})$$, the odd states by one other, but there are plenty of other representations that we don’t see.

Spin is treated no different than anything other group or algebra. If there are no elementary particles with $$s=7/2$$, this is because of the physics of the model, not because of the group theory behind the model. Conversely, simply knowing $$s$$ isn’t enough to say anything about the number of particles.

Consider for instance \begin{align} \vert \textstyle\frac{1}{2};\frac{1}{2}\rangle&=\frac{1}{\sqrt{2}}\left(\vert +\rangle_1\vert -\rangle_2-\vert -\rangle_1\vert +\rangle_2\right)\vert +\rangle_3\, ,\\ \vert \textstyle \frac{1}{2};-\frac{1}{2}\rangle&=\frac{1}{\sqrt{2}}\left(\vert +\rangle_1\vert -\rangle_2-\vert -\rangle_1\vert +\rangle_2\right)\vert -\rangle_3\, . \end{align} These are clearly a 3-particle states, yet both of them span the Hilbert space of a 2-dimensional irrep of $$\mathfrak{su}(2)$$. The matrix representations of the spin operators would be exactly the $$2\times 2$$ Pauli matrices.

The job of the theorist is often to explain why this irrep does or does not appear: representation theory cannot tell you this no more than the theory of differential equations can tell you why you have this rather than that boundary condition to determine your solution.

• Sorry I meant $s=1/2$. I've changed the question. So in reference to the above comment, what is preventing all the other $SU(2)$ irreps for a one-particle state? Aug 21, 2021 at 23:40
• Why do you think something prevents them? Spin-1 particles exist.
– fqq
Aug 21, 2021 at 23:48
• @fqq How do I form a spin $2$ one particle state using the Pauli equation though? Aug 21, 2021 at 23:57
• @DIRAC1930 I don't know what you mean by "form a state using an equation", but obviously if you start with an equation for spin-1/2 particles you get spin-1/2 particles.
– fqq
Aug 22, 2021 at 0:00
• @fqq So how would I go about forming a non-relativistic wave equation for spin $1$ particles? Aug 22, 2021 at 0:06