# Expressing the Schrödinger equation in terms of spinors

I appreciate that the Dirac equation can be thought of in terms of spinors, as it directly implies the presence of spin, in addition to initiating the concept of treating fields as operators.

From memory alone, (and I may be incorrect in this) the appendex to Penrose's "Cycles of Time" expresses the Schrödinger equation in spinor notation. I do realise the motivation behind the Dirac equation, (the fusion of Special Relativity and Quantum Mechnics), giving a mathemathetical basis to particles of spin 1/2, and also that the Schrödinger equation does not incorporate the concept of spin.

My question is, can anybody give me a quick deriviation / translation of the transformation of the Schrödinger equation (for, say a free particle) from it's conventional notion, into spinor notation? (Assuming my memory of seeing it expressed in spinor notion is correct, I don't have the book any more to check).

Schrodinger equation just says that $$\hat{H}\left| \psi(t) \right\rangle = i\hbar \frac{\partial}{\partial t} \left| \Psi(t) \right\rangle$$ It is an operator equation, in order to transform it into a PDE of a c-number function (the wave function) you have to project the Hilbert space vector $\left| \Psi(t) \right\rangle$ into its components. Now the hilbert space for a spineless particle in an infinite dimensional space, one whose bases is $\left| \vec{x} \right\rangle$. So we project to get $$\int d^3 x' \left\langle \vec{x} \right |\hat{H}\left| x'\right\rangle\left\langle x'\right.\left| \psi(t) \right\rangle = i\hbar \frac{\partial}{\partial t} \left\langle \vec{x} \right |\left. \Psi(t) \right\rangle$$ which for local potentials becomes $$\hat{H}_x\Psi(\vec{x},t) = i\hbar \frac{\partial}{\partial t} \Psi(\vec{x},t)$$
Where $\hat{H}_x\equiv\left\langle H \right\rangle_x$ Now if the particle has spin, or any new quantum number that provides the hilbert space with additional dimension "on top of" that of $\left| \vec{x} \right \rangle$, i.e. $\mathcal{H}_{\text{free}}\otimes\mathcal{H}_n$, you have to project onto that as well $$\sum_{n'} \int d^3 x' \left\langle \vec{x},n \right |\hat{H}\left| x',n'\right\rangle\left\langle x',n'\right.\left| \psi(t) \right\rangle = i\hbar \frac{\partial}{\partial t} \left\langle \vec{x},n \right |\left. \Psi(t) \right\rangle$$ which again for local potentials becomes $$\sum_{n'} \hat{H}_{x,nn'} \Psi_{n'}(\vec{x},t) = i\hbar \frac{\partial}{\partial t} \Psi_n(\vec{x},t)$$ In the case of spin 1/2 particle, $\mathcal{H}_n$ is two dimensional and therefor the above equation represents a vector equation, or two equations for each spin component. And $\hat{H}_{x,nn'}$ is just a $2\times2$ matrix, where it's usually the free hamiltonian multiplied by $\mathbb{1}_2$ plus some combination of the $2\times 2$ pauli matrices
Notice that $\Psi_n(\vec{x},t)$ is a two component vector (a spinor) with index $n$ in the same way as it is an infinite component vector with "index" $x$.