State Vector in the Schrodinger Equation

Question

Let us consider a nonrelativistic particle of mass $m$, spin $s$ and isospin $i$. The Schrodinger equation for the state vector $|\psi(t)\rangle$ of this particle is given by

$$i\hbar\frac{d|\psi(t)\rangle}{dt} = \hat{H}|\psi(t)\rangle \, , $$

where $\hat{H}$ is the Hamiltonian operator and the other symbols have their usual meanings.

My questions are as follows.

• Is $|\psi(t)\rangle = |\phi(t)\rangle \otimes |\chi(t)\rangle \otimes |I(t)\rangle$ ?

Here, $|\phi(t)\rangle$, $|\chi(t)\rangle$ and $ |I(t)\rangle$ are the spatial, spin and isospin state vectors of the particle.

• I have specified the time dependence of the state vectors by writing as, e.g., $ |I(t)\rangle$. Is that correct?

• Say, the particle is relativistic. Then what would be the state vector? In other words, if the particle is relativistic, then other than the spatial, spin and isospin state vectors, what else should be included in $|\psi(t)\rangle$?

If you know any relevant reference for the answer, please mention that too.

Answer 1

1. Almost. What is correct is that if you have the "spatial" space $H_S$, the spin space $H_\text{spin}$ and the isospin space $H_i$, then the total state space is $H_S\otimes H_\text{spin}\otimes H_i$. However, it is not guaranteed that any given state in this can be written as such a tensor product, since a generic vector in this space is a sum over these simple (or separable) states.

2. You include whatever is relevant for the situation you are trying to analyse. E.g. one would not ordinarily include the isospin part except if one is explicitly interested in the weak interaction. Additionally, one should not speak of a "spatial" part as such in a relativistic theory, since localization of relativistic particles is notoriously difficult.