When a Hilbert space's state vector becomes a spinor?

Question

Given a Hilbert space $\mathcal{H}$, we pick up some state vector $| \psi\rangle$ which lives in the Hilbert space $\mathcal{H}$.

The $| \psi\rangle$ is a vector of the Hilbert space satisfies the rule like $\langle\psi| \psi\rangle =1$.

My question is about when a Hilbert space's state vector $| \psi\rangle$ can become a spinor? Of course I need to clarify what I meant and possible interpretations.

1. can the state vector $| \psi\rangle$ be a spinor as a spinor representation of the spacetime symmetry? (the projective representation of the rotational group $SO(d,1)$)? If so, if this state vector $| \psi\rangle$ is a ground state of the Hamiltonian system, we can transform the state vector $| \psi\rangle$ as the spinor under the rotational group $SO(d,1)$ or more precisely the spin group $Spin(d,1)$. Does this imply that the system have finite or infinite many ground states? [Well, if the spinor representation is finite dimensional, it looks that we still have a finite dimensional ground state subspace of the Hilbert space.] In any case, is this possible?

2. can the state vector $| \psi\rangle$ be a spinor as a spinor representation of the flavor or internal symmetry, of say some $Spin$ group? (For example, we can have a solid state spin system with $Spin(3)=SU(2)$, symmetry, and one of the state vector $| \psi\rangle$ is a spinor representation (spin-1/2) of the $Spin(3)=SU(2)$.) If so, what are the implications of the properties and dynamics of the system?

This post imported from StackExchange Physics at 2020-12-03 17:31 (UTC), posted by SE-user annie marie heart

Comments

You seem to confuse the „state vectors” and „operators”. In QFT the spinors are the spinorial fields, not the abstract state vectors.

This post imported from StackExchange Physics at 2020-12-03 17:31 (UTC), posted by SE-user DanielC

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Not confused - I am only asking the situation when state vector is in a spinor rep of some Spin group (what implications can be)

This post imported from StackExchange Physics at 2020-12-03 17:31 (UTC), posted by SE-user annie marie heart

Answer 1

Perhaps this question can be answered more generally, but this is what comes to mind. Any vector in a representation of the $\text{Spin}(d,1)$ group can be considered a "spinor" but its interpretation depends entirely on the Hamiltonian. It will only be able to be ground state if the spin operator $\vec{\sigma}$ conmutes with the Hamiltonian, and whether it is a "real" spinor or an internal symmetry depends on something like, say, how it couples semiclassically to the electromagnetic field under $\vec{p}\mapsto \vec{p}-q\vec{A}$. If you obtain a term of the sort $\vec{\sigma}\cdot\vec{A}$ in the Hamiltonian, you may interpret the spin in the traditional sense of giving rise to a magnetic moment.

This post imported from StackExchange Physics at 2020-12-03 17:31 (UTC), posted by SE-user Gaston Barboza

Comments

thanks+1 for your attempt!

This post imported from StackExchange Physics at 2020-12-03 17:31 (UTC), posted by SE-user annie marie heart