# Relation between the spinor and wave function formalisms

1. Jul 28, 2014

### nmbr28albert

Hello everyone, this has been on my mind for a while and I finally realized I could just ask on here for some input :)

I think in general, when most people start learning quantum mechanics, they are under the impression that the wave function $\Psi$ represents everything you could possibly know about, say, an electron. If you want to know the expectation value of something, simply stick in the operator and integrate. However, when you get to spin, the spinor is introduced for spin 1/2 particles, which is a 2-D vector, and the corresponding operators are matrices. Is the spinor encoded in the wave function somehow? Or are they two distinct pieces to describing a particle?

2. Jul 28, 2014

### vanhees71

In non-relativistic quantum theory the spin operator commutes with the position operator and thus you can find common generalized eigenstates $|\vec{x},\sigma_z \rangle$. The appropriate wave function for a non-relativistic particle with spin is thus a spinor field
$$\psi_{\sigma_z}(t,\vec{x})=\langle \vec{x},\sigma_z|\Psi(t) \rangle.$$

3. Jul 28, 2014

### WannabeNewton

Wavefunction generally refers to the position representation of the state of a system. If we have a system with position and spin degrees of freedom then the state will factorize into a product state of spin and position because these two operators commute. We can then take the representation of this product state in the simultaneous eigenbasis of spin and position. This is what vanhees wrote above. The result will be a wavefunction (infinite dimensional vector) tensored into a complex
dimension 2 vector which represents the spin state of the system in the basis of spin operator.

When we write this the end result is $\psi(x)\otimes (a,b)^T$ which is often written as $(\psi_{1/2},\psi_{-1/2})^T$. This is the spinor. It is basically just a 2-component wave function for spin 1/2 particles. In general it will be an n-component wavefunction where n is the dimension of the vector space spanned by the eigenbasis of the spin operator for a given particle species (these representations are just generalizations of the Pauli matrices). The 2-component wavefunction, or spinor, can then be directly incorporated into the Schrodinger equation. See Schrodinger-Pauli equation.

4. Jul 28, 2014

### nmbr28albert

I see, so since the spin operators cannot be derived from the position and momentum operators, it seems to me that the original schrodinger equation is not the complete non-relativistic limit of the dirac equation, which is actually the pauli equation. In the historical context then, was the Schrodinger equation proposed and found to be incomplete due to spin effects evident in experiments such as the Stern-Gerlach experiment? Since the Schrodinger equation was proposed prior to the Stern-Gerlach experiment for example, someone must have noticed that the equation did not explain these anomalous effects.

5. Jul 28, 2014

### WannabeNewton

Yes this is true. You will find a derivation of the Schrodinger-Pauli equation from the non-relativistic limit of the Dirac equation in many books. See e.g. section 3.6 of Maggiore "A Modern Introduction to Quantum Field Theory".