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Dirac equation in a central field (Schiff)

  1. Sep 27, 2014 #1

    DrClaude

    User Avatar

    Staff: Mentor

    Not really a homework problem, but I think it fits better in this section.

    1. The problem statement, all variables and given/known data
    I'm having a problem with eq. (53.12) in the book Quantum Mechanics by Schiff. In the context of the Dirac equation, we have
    $$
    \hbar^2 k^2 = (\vec{\sigma}' \cdot \vec{L})^2 + 2\hbar (\vec{\sigma}' \cdot \vec{L}) + \hbar^2 = (\vec{L} + \frac{1}{2} \hbar \vec{\sigma}')^2 + \frac{1}{4} \hbar^2
    $$
    The first equality is fine, it is the second one that I can't reproduce. The point of the equation is to recover ##(\vec{L} + \frac{1}{2} \hbar \vec{\sigma}')^2 = (\vec{L} + \vec{S})^2 = \vec{J}^2##.

    2. Relevant equations
    $$
    \vec{\sigma}' = ( \sigma_x', \sigma_y', \sigma_z')
    $$
    where the ##\sigma_i'## are ##4\times4## matrices related to the Pauli matrices ##\sigma_i## through
    $$
    \sigma_i' \equiv \begin{pmatrix} \sigma_i & 0 \\ 0 & \sigma_i \end{pmatrix}
    $$
    ##\vec{L}## is the orbital angular momentum (actually an operator, but that's not important in the present context).

    3. The attempt at a solution
    I start by looking at the square term on the LHS:
    $$
    \begin{array}{}
    (\vec{\sigma}' \cdot \vec{L})^2 &= ( \sigma_x' L_x + \sigma_y' L_y + \sigma_z' L_z)^2 \\
    &= (\sigma_x')^2 L_x^2 + \sigma_x' \sigma_y' L_x L_y + \sigma_x' \sigma_z' L_x L_z \\
    &\quad + \sigma_y' \sigma_x' L_y L_x + (\sigma_y')^2 L_y^2 + \sigma_y' \sigma_z' L_y L_z \\
    & \quad + \sigma_z' \sigma_x' L_z L_x + \sigma_z' \sigma_y' L_z L_y + (\sigma_z')^2 L_z^2 \\
    &= L_x^2 \mathbf{1} + L_y^2 \mathbf{1} + L_z^2 \mathbf{1} = \vec{L}^2
    \end{array}
    $$
    where I have used the properties of the Pauli matrices, namely ##\sigma_i^2 = \mathbf{1}##, with ##\mathbf{1}## the identity matrix, and ##\sigma_i \sigma_j + \sigma_j \sigma_i = 2\delta_{ij}##.

    Now, for the square term on the RHS:
    $$
    \begin{array}{}
    (\vec{L} + \frac{1}{2} \hbar \vec{\sigma}')^2 &= \vec{L} \cdot \vec{L} + \frac{1}{2} \hbar \vec{L} \cdot \vec{\sigma}' + \frac{1}{2} \hbar \vec{\sigma}' \cdot \vec{L} + \frac{1}{4} \hbar^2 \vec{\sigma}' \cdot \vec{\sigma}' \\
    &= \vec{L}^2 + \hbar \vec{\sigma}' \cdot \vec{L} + \frac{1}{4} \hbar^2 (\sigma_x^2 + \sigma_y^2 + \sigma_z^2) \\
    &= \vec{L}^2 + \hbar \vec{\sigma}' \cdot \vec{L} + \frac{3}{4} \hbar^2
    \end{array}
    $$

    Putting all this together, I can rewrite the original equality as
    $$
    \vec{L}^2 + 2\hbar (\vec{\sigma}' \cdot \vec{L}) + \hbar^2 = \vec{L}^2 + \hbar (\vec{\sigma}' \cdot \vec{L}) + \hbar^2
    $$
    There is a factor 2 missing in the RHS that I can't find.
     
  2. jcsd
  3. Sep 27, 2014 #2

    TSny

    User Avatar
    Homework Helper
    Gold Member

    Consider $$ \sigma_x' \sigma_y' L_x L_y + \sigma_y' \sigma_x' L_y L_x $$ Does this reduce to zero or something else?
     
  4. Sep 28, 2014 #3

    DrClaude

    User Avatar

    Staff: Mentor

    I was so concentrated on the ##\sigma'## that I only saw ##\sigma_x' \sigma_y' + \sigma_y' \sigma_x' = 0##. But of course
    $$
    \begin{array}{}
    \sigma_x' \sigma_y' L_x L_y + \sigma_y' \sigma_x' L_y L_x &= i \sigma_z' L_x L_y - i \sigma_z' L_y L_x \\
    &= i \sigma_z' (L_x L_y - L_y L_x) \\
    &= i \sigma_z' (i \hbar L_z) =-\hbar \sigma_z' L_z
    \end{array}
    $$
    and so on, such that
    \begin{array}{}
    (\vec{\sigma}' \cdot \vec{L})^2 &= ( \sigma_x' L_x + \sigma_y' L_y + \sigma_z' L_z)^2 \\
    &= (\sigma_x')^2 L_x^2 + \sigma_x' \sigma_y' L_x L_y + \sigma_x' \sigma_z' L_x L_z \\
    &\quad + \sigma_y' \sigma_x' L_y L_x + (\sigma_y')^2 L_y^2 + \sigma_y' \sigma_z' L_y L_z \\
    & \quad + \sigma_z' \sigma_x' L_z L_x + \sigma_z' \sigma_y' L_z L_y + (\sigma_z')^2 L_z^2 \\
    &= (L_x^2 + L_y^2 + L_z^2) \mathbf{1} - \hbar ( \sigma_x' L_x + \sigma_y' L_y + \sigma_z' L_z)\\
    &= \vec{L}^2 - \hbar ( \vec{\sigma}' \cdot \vec{L}),
    \end{array}
    leading to the correct equality.

    Thanks a lot for pointing me in the right direction.
     
  5. Sep 28, 2014 #4

    TSny

    User Avatar
    Homework Helper
    Gold Member

    OK. Good.
     
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