# Dirac Equation and commutation relations

1. Apr 24, 2015

### CAF123

1. The problem statement, all variables and given/known data
Consider the Dirac Hamiltonian $\hat H = c \alpha_i \hat p_i + \beta mc^2$ . The operator $\hat J$ is defined as $\hat J_i = \hat L_i + (\hbar/2) \Sigma_i$, where $\hat L_i = (r \times p)_i$ and $\Sigma_i = \begin{pmatrix} \sigma_i & 0 \\0 & \sigma_i \end{pmatrix}$.

a) Show that $[\hat H, \hat L] = -i\hbar c (\alpha \times \hat p)$
b) Evaluate $[\hat H, \Sigma], [\hat H, J]$ and thus $[\hat H, \hat J^2]$.
c)Evalaute $[\hat H, \Sigma \cdot \Sigma]$

2. Relevant equations
$[\hat x, \hat p] = i\hbar \delta_{ij}$

It is also given that $\alpha_j = \begin{pmatrix} -\sigma_j & 0 \\ 0 & \sigma_j \end{pmatrix}$ and $\beta = \begin{pmatrix} 0 & I \\ I & 0 \end{pmatrix},$ I the 2x2 identity.

3. The attempt at a solution
a) is fine, for b) I am getting zero for the first commutator there but I believe that to be wrong. Here is what I have:
$$[\hat H, \Sigma] = [c \alpha_j p_j, \Sigma_i] + mc^2[\beta, \Sigma_i] = c[\alpha_j p_j, \Sigma_i],$$ the second commutation relation there vanishes by expanding out the matrices. Then I also argued this was zero because the $\alpha_j'$s commute with the $p_j$'s since they are constant complex matrices. So then the whole expression is zero, again by expanding, but I am not sure what I did wrong. My notes say this shouldn't vanish. Thanks!

2. Apr 24, 2015

### stevendaryl

Staff Emeritus
$[ \alpha_j p_j, \Sigma_k]$ is not zero.

3. Apr 24, 2015

### CAF123

Hi stevendaryl,
Thanks, I see my error. Perhaps not related to the topic of the thread, but could you explain why the form of the Dirac Hamiltonian is as it is? My notes just posit the form without saying why.

4. Apr 24, 2015

### stevendaryl

Staff Emeritus
Well, the idea was that he wanted an equation of the form:

$H \psi = E \psi$

and it had to be consistent with Special Relativity, which means that $E^2 = p^2 c^2 + m^2 c^4$

That implies that $H = \sqrt{p^2 c^2 + m^2 c^4}$. But it's really difficult to work with square-roots. Besides that, in SR, energy and momentum are on the same footing, so Dirac reasoned that if the right side is linear in $E$, then the left side should be linear in $p$. So he just guessed that it had the form:

$H = c \vec{p} \cdot \vec{\alpha} + \beta m c^2$

for some constant vector $\vec{\alpha}$ and some constant $\beta$

Since $H^2 \psi = E^2 \psi = (p^2 c^2 + m^2 c^4) \psi$, the constants $\vec{\alpha}$ and $\beta$ had to satisfy certain conditions:

$(c \vec{p} \cdot \vec{\alpha} + \beta m c^2)^2 = c^2 \sum_{i j} (p_i p_j) (\alpha_i \alpha_j) + m c^3 \sum_i p_i (\alpha_i \beta + \beta \alpha_i) + m^2 c^4 \beta^2$

In order for this to equal $c^2 p^2 + m^2 c^4$, it must be that:
1. $\alpha_i \alpha_j + \alpha_j \alpha_i = 0$ if $i \neq j$
2. $\alpha_i \alpha_i = 1$
3. $\alpha_i \beta + \beta \alpha_i = 0$
4. $\beta^2 = 1$
The first two equations are obeyed by the Pauli spin matrices, which are 2x2. In order to accommodate the 3rd and 4th equations, he had to go to 4x4 matrices.