# Spin in Schrodingers Equations

With the usage of Dirac notation we've gotten around a large amount of of inconvenience that would be dealing with spin. But, I was wondering, do we merely do this because it is inconvenient to try and create a differential equation with the corresponding spin or is it actually impossible to write that into the PDEs? If it's merely a matter of convenience, what would Schrodingers equation (or some kind of generalization that allows for you to include spin) look like?

• I suspect either a) you basically would be needing to somehow remap your dimensions of the differential equations to some kind of basis that represents the group of your system and solve it there or b)you would be able to include a new separable term that is very weird to deal with calculus wise – Skyler Apr 25 '15 at 1:56
• Possible duplicates: physics.stackexchange.com/q/10837/2451 and links therein. – Qmechanic Apr 25 '15 at 19:13

The Schrödinger equation is only correct in the non-relativistic limit $v << c$, for particles without spin. The correct equation for spinless (=spin $0$) particles is the Klein-Gordon equation, which reduces in the non-relativistic limit to the Schrödinger equation.

If we want to talk about spin $\frac{1}{2}$, the correct, relativistic equation is the Dirac equation. The Dirac equation reduces in the non-relativistic limit to the Pauli-equation, which is the analogue to the Schrödinger equation for particles with spin $\frac{1}{2}$.

For spin $1$ particles we have the Proca equation, which is known in the massless limit as the famous inhomogeneous Maxwell equation. For massless particles we have no non-relativistic limit and unfortunately I don't know the name, if there is any, for the non-relativistic limit of the Proca equation.

• Note that the Schroedinger equation $i\hbar \dot{\Psi} = \hat{H} \Psi$ for the probability amplitudes $\Psi$ is always valid. The single-particle Shroedinger equation fails in the relativistic limit due to particle production and annihilation. This problem is remedied by using the correct many-body Hamiltonian $\hat{H}$ and wave function(al) $\Psi$ in the Schroedinger equation. – Mark Mitchison Apr 25 '15 at 17:49
• @MarkMitchison r u saying u can use the Schrödinger equation for relativistic stuff if you do this many body method ur describing? – Stan Shunpike Dec 26 '15 at 8:20
• @StanShunpike Yes. The Schroedinger equation is a fundamental postulate of quantum mechanics and holds in both non-relativistic and relativistic settings. However, the degrees of freedom it describes in each of these settings are very different. – Mark Mitchison Dec 29 '15 at 18:00
• @MarkMitchinson do you know some source that compares this method with the Dirac equation? – Stan Shunpike Dec 29 '15 at 20:37
• @StanShunpike "This method" that I am describing is just relativistic quantum field theory. The Dirac equation is only correct on shell, i.e. classically. However, the dynamics of the quantum state of the Dirac field is described exactly by the Schrodinger equation using the appropriate Hamiltonian (from which the Dirac equation can also be derived). – Mark Mitchison Dec 30 '15 at 13:08

The Dirac notation is simply an alternative to vector notation. Certainly there are PDEs describing the quantum state of a lone particle with spin and they are:

1. The Pauli equation (see Wiki page of this name) was historically the first, and here the quantum state is two $\mathcal{L}^2(\mathbb{R}^3, \mathbb{R})$ functions of space and time. The two components encode the spin. Pauli built his equation with a phenomenological approach, looking primarily to the results of the Stern-Gerlach experiment. Importantly, Pauli consciously built the notion of spin into his equation;

2. Paul Dirac formulated the Dirac Equation in 1928; his motivation was to come up with a relativistic equation (unlike the 1926 Schrödinger and 1927 Pauli equations) and his method was to seek a "square root" of the second order operator in the Klein-Gordon equation, which was built to be relativistic but suffered from having negative as well as positive energy solutions. He hoped that a first order equation coming from the "square root" would overcome the negative energy solution problem. He realized he could not build such an equation with co-efficients in commutative fields such as $\mathbb{R}$ and $\mathbb{C}$ and so was forced to the idea of co-efficients from Clifford Algebras, represented by $4\times 4$ matrices (in case you've not met this before, one can represent complex numbers as a field of $2\times 2$ real element matrices, so the idea is not as unnatural as it first seemed to Dirac). So the operator takes a $4\times 1$ column vector as input, hence the question arose as to what the four components meant. It turned out "by chance" that the solutions modelled particles with spin; unlike Pauli, this had not been Dirac's main aim. His equation also had negative energy solutions, and he was forced to his Dirac Sea hypothesis to solve his original problem.

The Dirac equation is mentioned in other answers as PDE describing spin. As you ask "what would Schrodingers equation (or some kind of generalization that allows for you to include spin) look like?", the following may be relevant.

Yes, the Dirac equation adequately describes spin. However, it is actually a system of four partial differential equations for four functions (components of the Dirac spinor). It came as a surprise to me that the Dirac equation is generally equivalent to one PDE for just one function. I don't want to write it out here, but you may look at my article http://akhmeteli.org/wp-content/uploads/2011/08/JMAPAQ528082303_1.pdf (published in the Journal of mathematical physics). The equation is written in a more general and symmetric form in my recent article http://arxiv.org/abs/1502.02351 .