0
$\begingroup$

The Hamiltonian of the Dirac equation in with an electromagnetic field in SI units is $$H=\gamma^0\left[ mc^2+c\gamma^k\left( p_k-\frac{q}{c}A_k \right) \right]+qA^0$$

(from https://en.wikipedia.org/wiki/Dirac_equation#Identification_of_observables)

My question is simple: How does it look like in CGS unis?

$\endgroup$
  • $\begingroup$ You have to be careful when studying this topic as to which units are being used. In his preface to the third edition of Classical Electrodynamics, Jackson states that the first 10 chapters are in SI units but Gaussian (cgs) units are retained for the later chapters "since such units seem more suited to relativity and relativistic electrodynamics then SI" $\endgroup$ – Farcher Aug 28 '17 at 8:23
0
$\begingroup$

I believe you are mistaken - that expression is already in CGS units. The giveaway is the form of the mechanical momentum $\vec P = (\vec p - \frac{q}{c}\vec A$) ; in SI units, there would be no factor of $c$ there, and you would have

$$H = \gamma^0 \left[mc^2 + c\gamma^k\left(p_k - qA_k\right)\right] + q\phi$$


In any system of units, the electric field is always defined as "force per unit charge." Because the electric field (in the static case) can be written as the gradient of the scalar potential, this tells us that the dimensions of the scalar potential are always energy per unit charge.

Electric charge has different dimensions depending on whether you use SI units or cgs units, and this difference is reflected in different dimensions for the scalar potential. However, the dimensions of $[q\phi]$ are always energy.

The crucial difference between the SI and CGS systems is that in CGS, the dimensions of the electric field and magnetic field are the same. This means that the dimensions of the scalar potential and vector potential are the same - in other words, $$[\phi]_{cgs}=[A]_{cgs} = \frac{\text{Energy}}{[q]_{cgs}}$$

On the other hand, in SI units, $$[E] = [B]\cdot \text{Velocity}$$ so $$[A]_{SI} = \frac{\text{Momentum}}{[q]_{SI}}$$ whereas $$[\phi]_{SI} = \frac{\text{Energy}}{[q]_{SI}}$$ as stated earlier.


The TL;DR summary is that in the SI system, the vector potential has dimensions of momentum per unit SI charge; in the CGS system, the vector potential has dimensions of energy per unit CGS charge. In contrast, the scalar potential always has dimensions of energy divided by the relevant charge. This manifests itself in an extra factor of $c$ in the expression for a particle's mechanical momentum.


Additional edit: When collecting the scalar and vector potentials into a 4-vector, all of the components obviously need to have the same dimensions. Typically one gives the four-vector $\bf{A}$ the same dimensions as the three-vector $\vec A$ and scales the term involving the scalar potential. Therefore, in SI units, $$ {\bf{A}} = \left(\frac{\phi}{c},\vec A\right)$$ so if you write the Hamiltonian in terms of $A^0$ rather than $\phi$, you would get

$$H = \gamma^0 \left[mc^2 + c\gamma^k\left(p_k - qA_k\right)\right] + qcA^0$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.