# Does the Schrödinger equation apply to spinors?

I was reading about Larmor precession of the electron in a magnetic field in Griffiths QM when I came across the equation $$i\hbar \frac{\partial \mathbf \chi}{\partial t} = \mathbf H \mathbf \chi,$$ where $$\mathbf\chi(t)$$ is a 2D vector that represents only the spin state and does not include information of the wave function. The Hamiltonian is $$\mathbf H = - \gamma \mathbf B \cdot \mathbf S = - \frac{\gamma B_0 \hbar}{2} \begin{pmatrix} 1 & 0 \\ 0 & -1\end{pmatrix}$$ for a uniform magnetic field $$\mathbf B = B_0 \hat k$$. Why should these spinors also obey the Schrödinger equation? The book does not provide any further information as to why this should hold.

• What is the Hamiltonian? Apr 3 at 16:38
• @JasonFunderberker I edited the question to include the Hamiltonian. Apr 3 at 16:52
• Note that your Hamiltonian is diagonal, so your Schrödinger equation are actually two independent Schrödinger equations. One for particles with spin parallel to the magnetic field, one for particles with spin antiparallel to the field. Since there is no interaction, the spin is a conserved quantity in this example and so it makes sense to treat the two cases as separate. In a realistic example you will have other terms that couple the two spin orientations as explained in the answers. Apr 4 at 13:32

The Schrödinger equation applies to any quantum system or quantum field theory, as long as you have a continuous time dimension. The equation $$\hat H\psi = i\hbar\frac{\partial \psi}{\partial t}$$ is just the statement that the Hamiltonian $$\hat H$$ is the infinitesimal translation operator in time. By Noether's theorem, it corresponds to the total energy of your system.

In order to describe a quantum system, you need to first decide what the degrees of freedom are. This decides which Hilbert space this equation is defined on. For example, for a single particle in $$\mathbb R^d$$, the Hilbert space is $$\mathcal H = L^2(\mathbb R^d)$$. And thus the wavefunction is $$\psi(\mathbf r,t)$$. Since the Hamiltonian is the total energy, we have (here for a non-relativistic particle) $$\left[-\frac{\hbar^2}{2m}\nabla^2 + V(\mathbf r)\right]\psi(\mathbf r,t) = i\hbar \frac{\partial}{\partial t}\psi(\mathbf r, t).$$

Now imagine that the particle has some internal degree of freedom, such as spin. For a spin-$$\tfrac 12$$ particle, the Hilbert space is two dimensional $$\mathbb C^2$$. Quantum mechanics tells us that the Hilbert space of the full system is a tensor product $$\mathcal H = L^2(\mathbb R^d)\otimes\mathbb C^2$$ and wavefunctions are thus $$\psi(\mathbf r,t) = \begin{pmatrix} \psi_\uparrow(\mathbf r,t) \\ \psi_\downarrow(\mathbf r,t)\end{pmatrix}.$$ The Hamiltonian will now also contain terms coupling spins together, but it depends on which forces are present. For example in $$d=3$$ you could have $$\left[-\frac{\hbar^2}{2m}\nabla^2\,\mathbf 1 + V(\mathbf r)\,\mathbf 1 + \alpha \,\mathbf B\cdot\mathbf\sigma\right]\psi(\mathbf r,t) = i\hbar \frac{\partial}{\partial t}\psi(\mathbf r, t).$$ Here $$\mathbf 1$$ is a $$2\times 2$$ identity matrix and $$\mathbf\sigma = (\sigma_x,\sigma_y,\sigma_z)$$ are the Pauli matrices.

Sometimes people write down a Schrödinger equation for only position or only spin degrees of freedom, if they don't couple to other degrees of freedom in the problem at hand.

For example, if the Hamiltonian only has $$\mathbf 1$$ for the spin then the Hamiltonian will be block diagonal and $$\psi_\uparrow$$ and $$\psi_\downarrow$$ will be completely decoupled. You could therefore forget spin, and only keep position in your description. Or if the particle is trapped somewhere and cannot move, then only the spin degrees of freedom are relevant for the description.

But the Schrödinger equation is valid for any quantum system, it just describes how the system evolves in time.

• I think you need to add the word nonrelativistic into your first paragraph
– Joe
Apr 4 at 6:33
• The first equation I wrote is true even for relativistic theories, the different type of Hamiltonian determines the specific theory. The equation Schrödinger wrote was indeed for a specific non-relativistic Hamiltonian. But nowadays we call this more general equation, with arbitrary Hamiltonian, for Schrödinger's equation. Apr 4 at 11:02
• Why should we extend the notion of 'Schrödinger equation' to mean any QM wave equation? That does not make sense. Apr 4 at 22:22
• @my2cts Well, do you have any other name for it? It appears to me a simple accident of history that the only common term we have for the most general equation is "the Schrödinger equation."
– HTNW
Apr 5 at 6:55
• In research many different quantum systems are used: QFT's, spin models (Ising, Heisenberg model etc), fermions/bosons on lattices, anyonic models etc etc. All of these come with a Hamiltonian. In the research literature when people say "Schrödinger equation" they either mean $H|\psi\rangle = E|\psi\rangle$ or the time-dependent one above. The essence of Schrödinger's equation is that time evolution is unitary and governed by a Hamiltonian, not just the specific non-relativistic Hamiltonians undergrads see for the first time. What else are these equations called in research literature? Apr 5 at 11:18

The correct wave equation for an electron is the famous Dirac equation (in fact it is correct for all spin $$\frac{1}{2}$$ particles).

The non-relativistic limit of the Dirac equation is called Pauli equation. The derivation can be found here. In this equation $$|\psi \rangle$$ is a two-component spinor wavefunction.

Pauli equation (general)

$$\left[{\frac {1}{2m}}({\boldsymbol {\sigma }}\cdot (\mathbf {\hat {p}} -q\mathbf {A} ))^{2}+q\phi \right]|\psi \rangle =i\hbar {\frac {\partial }{\partial t}}|\psi \rangle }$$

If you substitute $$\mathbf {A}=0$$ above you get back the Schrödinger equation. In fact, a relativistic wave equation for any spin should under some limit give back the Schrödinger equation be it Klein–Gordon equation or Proca action or Rarita–Schwinger equation.

• But here $|\psi\rangle$ represents the combined spinor wave function? Does it also hold for the spinor only without wave function information? Apr 3 at 16:55
• @ChrisYang in this case the Hamiltonian doesn't depend on the position of space. So the "real space solution" also doesn't depend on the position. The time dependence can be absorbed into the "spin space" part. So we can directly neglect the "real space part" and consider the 2 spinor. Apr 3 at 17:00
• The Pauli equation is not the nonrelativistic limit of the Dirac equation. That’s the usual Schrödinger equation. The Pauli equation is the semi relativistic approximation because it keeps $O(1/c)$ terms (in front of $A$ and $B$) which vanish in the limit $c\rightarrow \infty$. Apr 4 at 13:15
• @JakobElias the standard terminology refers to it as the nonrelativistic limit. For reference check 2.3 of RELATIVISTIC QUANTUM MECHANICS by W Greiner or 2.4 of Relativistic Quantum Mechanics by Armin Wachter. Apr 4 at 14:04
• @JakobElias I don't find it misleading. In classical electromagnetism we really don't take $c\to \infty$. Electromagnetism always comes relativistic irrespective of whether you do non relativistic classical or quantum mechanics. Apr 4 at 14:16

If you want to describe non-relativistic, interacting spins you need to extend the Schrödinger equation with the Pauli interaction, $$H_{spin} = -\gamma {\bf B} \cdot {\bf S}$$. This is the simplest of a class of hamiltonians known as spin hamiltonians. For non-translating spins, as in your case, you only need the Pauli term.