# I From Pauli spinors to Dirac spinors

1. Jan 23, 2017

### Milsomonk

Hey guys,
Hope all is well. I am trying to understand the process that takes us from the Pauli equation to the Dirac equation. Whilst I understand the motivation is to have a lorrentz covariant equation I don't really understand A.) how this was done B.) what the physical result is of the new covariant equation, what new info are we getting?.

Lorrentz covariance is a fairly new concept to me so just trying to get a proper sense of what it physically means. Any and all enlightenent is much appreciated :) cheers

2. Jan 23, 2017

### Staff: Mentor

Can you give any references that you have already looked at? In particular, what presentations of this process have you already looked at?

3. Jan 24, 2017

### Milsomonk

Thus far I have been reading some paper's by David Hestenes, http://geocalc.clas.asu.edu/html/GAinQM.html , I have been writing my undergrad thesis on Pauli spinors in Geometric Algebra but it has been suggested that I should expand my project to talk about relativistic quantum mechanics as well. Any suggested reading would be great, I've just picked up Griffths Introduction to Elementary Particles in the hope it can shed some light for me. I do want to understand the general process of making an equation lorrentz covariant.

4. Jan 24, 2017

### Staff: Mentor

Hestenes' approach is quite a bit different from the standard approach. Griffiths will give you something more like the standard approach.

Btw, we recently had an interview with Hestenes on PF; see here (this is the comment thread on the interview, a link to the interview itself is also there):

This is too broad a topic for a PF thread. But the reason we do it is simple: because experiments have shown us that our universe is relativistic, so any theory that claims to be fundamental should be Lorentz covariant. Any theory that is not must be an approximation only, and if we can find a Lorentz covariant version of the same theory, that version should be more fundamental.

5. Jan 25, 2017

### Milsomonk

Yeah It is a little daunting to try to get my head around both approaches, but i'm finding Griffiths very useful and am quite well versed in Gometric Algebra so hopefully I can make the connection.

Yeah sorry, I realise it was quite a general query, but your answer was actually pretty much what I was looking for. Thanks for the help :)

6. Jan 26, 2017

### stevendaryl

Staff Emeritus
I wouldn't say that there is a derivation of the Dirac equation starting with the Pauli equation. But the trick that makes the Pauli equation work can be used as a heuristic for developing a relativistic version that is equivalent to the Dirac equation.

Start with the nonrelativistic formula relating energy and momentum for a particle:

$\frac{p^2}{2m} = E$

This is for a free particle, but to get a particle in an electromagnetic field, you can do the substitutions:
$\vec{p} \Rightarrow \vec{p} - e \vec{A}$
$E \Rightarrow E - e \Phi$

where $\vec{A}$ is the electromagnetic vector potential, and $\Phi$ is the scalar potential, and $e$ is the particle's charge.

To get the Schrodinger equation, you do the following:
1. Replace $\vec{p}$ by the operator $-i \vec{\nabla}$ (I'm using units where $\hbar = c = 1$)
2. Replace $E$ by the operator $i \frac{d}{dt}$
3. Interpret the operator equation as applying to a scalar wave function $\psi(\vec{r}, t)$.
To get the Pauli equation, you do a twist on the above:
1. Instead of 1 above, you replace the $\vec{p}$ by the operator $\vec{p} \cdot \vec{\sigma}$, where $\vec{\sigma}$ is the three Pauli spin matrices.
2. Instead of 3 above, you interpret the operator equation as applying to a two component spinor $U(\vec{r}, t)$.
The use of the spin matrices makes no difference for free particles, since $(\vec{p} \cdot \vec{\sigma})^2 = (\vec{p})^2$. But once you introduce the electromagnetic field, this makes a difference, because you get an extra term, $-e \vec{B} \cdot \vec{\sigma}$ where $\vec{B} = \vec{\nabla} \times \vec{A}$

Now, we can do the same thing to almost get the Dirac equation for the relativistic case.

Instead of starting with the equation:

$\frac{(\vec{p})^2}{2m} = E$

$E^2 - \vec{p}^2 = m^2$

Now introduce the substitution: $\vec{p} \Rightarrow \vec{p} \cdot \vec{\sigma}$ to get:

$E^2 - (\vec{p} \cdot \vec{\sigma})^2 = m^2$

Here comes an unmotivated step: Write the above in a "factored" form.

$(E - \vec{p} \cdot \vec{\sigma})(E + \vec{p} \cdot \vec{\sigma}) = m^2$

Factoring it this way makes no difference for free particles, but will make a difference when you introduce the electromagnetic field. Of course, this is interpreted as applied to a two-component spinor, $U$. But here's the huge benefit of factoring:

Let's define a new two-component spinor, $V$ via the equation:

$V = \frac{1}{m} (E + \vec{p} \cdot \vec{\sigma}) U$

Then the second-order factored equation can be written as a pair of coupled first-order equations:
$(E - \vec{p} \cdot \vec{\sigma}) V = m U$
$(E + \vec{p} \cdot \vec{\sigma}) U = m V$

These two equations can be written as a 4-component matrix equation:
$(\gamma^0 E - \vec{\gamma} \cdot \vec{p}) \Psi = m \Psi$

where $\Psi = \left( \begin{array}\\ U \\ V \end{array} \right)$ and where $\gamma^0 = \left( \begin{array}\\ 0 & 1 \\ 1 & 0 \end{array} \right)$ and $\gamma^j = \left( \begin{array}\\ 0 & \sigma_j \\ -\sigma_j & 0 \end{array} \right)$

(All the "1"s and "0"s really mean the 2x2 unit matrix and 2x2 zero matrix, respectively)

That is not the usual choice for $\gamma^0$, but the result is equivalent to the usual covariant form of the Dirac equation.

This was definitely not the way that Dirac came up with his equation. He didn't start with the second-order Pauli equation, but just tried to guess a first-order equation that produced the right energy-momentum relationship.

7. Jan 26, 2017

### Staff: Mentor

It is if you're using the Weyl basis instead of the Dirac basis; the Weyl basis is better adapted to particles that are relativistic, i.e., their energy/momentum is much greater than their rest mass. Basically the Weyl basis diagonalizes $\gamma^5$ instead of $\gamma^0$.

Great post, btw!

8. Jan 27, 2017

### Milsomonk

Amazing! thanks very much, that deffinately clears up some queries I had from the griffiths book as well. Also if anyone is familiar with the Hestenes approach to the Dirac equation i'd love to here your thought's on it, any advantages or disadvantages to using Geometric Algebra this way? From what I've read about non-relativistic spin in GA it appears to give a bit more clarity of what the imaginary numbers mean but the work required and the results are more or less the same as the traditional approach.

9. Feb 2, 2017

### Elemental

Geometric Algebra definitely promises greater insights into Pauli and Dirac equations than traditional vector algebra because of the intrinsic geometric interpretations allowed by Clifford algebra.

The drawback I find with Hestenes' approach is that he introduces more than the required number of dimensions by using C1,3(R) to represent space-time (space-time algebra, or STA). On the other hand, Baylis offers a simpler approach by using C3 (algebra of physical Space, or APS). With APS, real three-dimensional vectors are expressible in terms of Pauli matrices, but the time dimension emerges automatically in the Clifford algebra - and with the correct signature for special relativity! A good introduction can be found at ArXiv:physics/0406158.

Both STA and APS are geometric algebras, but APS seems to be more straightforward. In STA, the 3+1 dimensions of space time and the (+ - - -) metric are posited, while in APS they appear naturally and the relativistic equations (Dirac, Maxwell's, etc.) find an extraordinarily simple expression.