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

+ 1 like - 0 dislike
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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
asked Oct 4, 2020 in Theoretical Physics by annie marie heart (1,155 points) [ no revision ]
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
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

1 Answer

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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
answered Oct 7, 2020 by Gaston Barboza (20 points) [ no revision ]
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

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