Since the Schrödinger Pauli equation describes a non-relativistic spin ½ particle. This equation must be an approximation of the Dirac equation in an electromagnetic field. I was trying to derive this but I got stuck at a point. The free particle Dirac can be reduced to the equations \begin{align} σ^{i}(p_{i}+eA_{i}) u_B & = (E-m+eA_0)u_A. \\ \sigma^{i}(p_{i}+eA_{i})u_A & = (E+m+A_{0})u_B \end{align}

I multiplied both sides of the first equation by $(E+m+eA_0)$ to get the Schrödinger Pauli equation. I was not able to eliminate $u_B$ completely from the equation. Can someone help me with the derivation?

  • $\begingroup$ Hi user215742, I've started MathJax for you, but please edit the post to ensure it's done correctly. For reference on the math fonts, search "notation" in help center. $\endgroup$
    – Kyle Kanos
    Mar 25, 2019 at 14:34
  • $\begingroup$ Which components of $A_\mu$ are non-zero here? $\endgroup$ Mar 25, 2019 at 14:40
  • $\begingroup$ @GuestGuestGuest1 none of the components are zero. $\endgroup$ Mar 25, 2019 at 15:12
  • 2
    $\begingroup$ Solve second for $u_B$ and plug it back in first equation. $\endgroup$
    – Sunyam
    Mar 25, 2019 at 20:20
  • $\begingroup$ You can find more information about this by searching for Foldy-Wouthuysen transformation. $\endgroup$ Mar 26, 2019 at 0:54

2 Answers 2


Dirca's equation has is the following form:

$$i\hbar\gamma^{\mu}\partial_{\mu}\psi - mc\psi = 0$$

where $\mu = 0,1,2,3$ and $\gamma^{\mu}$ are the $4\times4$ matrices(in Dirac's representation):

\begin{align} \gamma^0 = \begin{bmatrix} \mathbb{1} & 0 \\ 0 & -\mathbb{1} \end{bmatrix} \gamma^k = \begin{bmatrix} 0 & \sigma_k \\ -\sigma_k & 0 \end{bmatrix} \end{align}

Where $\sigma_k$ are the Pauli Matrices ($k = 1,2,3$):

\begin{align} \sigma_1 = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix} \sigma_2 = \begin{bmatrix} 0 & -i \\ i & 0 \end{bmatrix} \sigma_3 = \begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix} \end{align}.

Using the momentum e energy in quantum mechanics differential operators; $\vec{p} = -i\hbar\vec{\nabla}$, $E = ih\partial_t = i\hbar c\partial_0$ the Dirac's equation become: $$(E\gamma^0 -c\gamma^{k} p_k - mc^2)\psi = 0$$

Breaking the four dimensional spinnor $\psi$ in two components the equation in a matrix for become:

\begin{align} \begin{bmatrix} (E - mc^2)\mathbb{1}& -\sigma_k p_k c\\ \sigma_k p_k c & -(E + mc^2)\mathbb{1} \end{bmatrix} \begin{bmatrix} u\\ v \end{bmatrix} =0 \end{align}

To coupling the EM field the energy and the momentum change

\begin{align} E \rightarrow E - e \Phi \\ \vec{p} \rightarrow \vec{p} - e \vec{A} \end{align}

So the EM weak coupling Dirac's equation in matricial form become:

\begin{align} \begin{bmatrix} (E - e \Phi - mc^2)\mathbb{1}& -\sigma_k (p_k - e A_k) c\\ \sigma_k (p_k - e A_k) c & -(E - e \Phi + mc^2)\mathbb{1} \end{bmatrix} \begin{bmatrix} u\\ v \end{bmatrix} =0 \end{align}

Now writing it in function of $u$ and $v$:

\begin{align} (E - e \Phi - mc^2) u -(\sigma_k (p_k - e A_k) c)v =0 \\ \sigma_k (p_k - e A_k)c u -(E - e \Phi + mc^2)v =0 \end{align}

In the second equation the $u$ and $v$ relation become:

\begin{align} \frac{\sigma_k (p_k - e A_k)c}{E - e \Phi + mc^2}u=v \end{align}

For the non relativistic approach $e \Phi << mc^2$ and $E = mc^2$

So the $u$ and $v$ relation became: \begin{align} \frac{\sigma_k (p_k - e A_k)c}{2mc^2}u=v \end{align}

Plugging this non relativistic relation in the first equation:

\begin{align} (E - e \Phi - mc^2) u -\frac{(\sigma_k (p_k - e A_k)c)^2}{2mc^2}u =0 \end{align}

Renaming $E - mc^2= E_{NR}$ and reorganizing: \begin{align} \left( e \Phi +\frac{(\sigma_k (p_k - e A_k)c)^2}{2mc^2}\right)u =E_{NR}u \end{align}

Focus in the $(\sigma_k (p_k - e A_k))^2$ term we have: \begin{align} (\sigma_k (p_k - e A_k))^2 &= (\sigma_i (p_i - e A_i)(\sigma_j (p_j - e A_j)\\ &= \sigma_i \sigma_j \Pi_i \Pi_j \end{align}

Where $\Pi_i = p_i - e A_i$

From Pauli anti-commutation and commutation relations:

$$\sigma_i \sigma_j = \delta_{ij} + i \varepsilon_{ijk}\sigma_k $$

With that the quadratic term become:

\begin{align} (\sigma_k (p_k - e A_k))^2 &= \Pi_i \Pi_i + i \varepsilon_{ijk}\sigma_k \Pi_i \Pi_j \end{align}

The last term in $(\sigma_k (p_k - e A_k))^2 u$ is:

\begin{align} i \varepsilon_{ijk}\sigma_k \Pi_i \Pi_j u &= i \varepsilon_{ijk}\sigma_k\left[(-i\hbar\partial_i - e A_i)(-i\hbar \partial_j - e A_j)\right] \\ &= i \varepsilon_{ijk}\sigma_k\left[-\hbar^2 \partial_i \partial_j + e^2 A_i A_j + i\hbar e(\partial_i A_j + A_i \partial_j)\right]u \end{align}

The first two terms are symmetrical so they become:

\begin{align} \varepsilon_{ijk} \partial_i \partial_j &= \frac{1}{2}\varepsilon_{ijk}(\partial_i \partial_j + \partial_j \partial_i) \\ &= \frac{1}{2}(\varepsilon_{ijk}\partial_i \partial_j +\varepsilon_{ijk}\partial_j \partial_i) \\ &= \frac{1}{2}(\varepsilon_{ijk}\partial_i \partial_j -\varepsilon_{jik}\partial_j \partial_i)\\ & = 0 \end{align}

The same thing are value for the $A_i A_j$ so the term become:

\begin{align} i \varepsilon_{ijk}\sigma_k \Pi_i \Pi_j u &= -\hbar e \varepsilon_{ijk}\sigma_k\left[ \partial_i( A_j u) + A_i \partial_j (u)\right] \\ &= -\hbar e \varepsilon_{ijk}\sigma_k\left[ \partial_i( A_j)u+ A_j \partial_i (u) + A_i \partial_j (u)\right] \end{align}

Where were applied the product rule in the first derivative. Now looking in the last two terms we can show that they vanish each other:

\begin{align} \varepsilon_{ijk}\left[ A_j \partial_i + A_i \partial_j \right] &= \varepsilon_{ijk}A_j \partial_i + \varepsilon_{ijk}A_i \partial_j \\ &= \varepsilon_{ijk}A_j \partial_i - \varepsilon_{jik}A_i \partial_j \\ &= 0 \end{align}

The only term that does not vanish is the $k$-th component os the curl of $\vec{A}$

\begin{align} i \varepsilon_{ijk}\sigma_k \Pi_i \Pi_j &= -\hbar e \varepsilon_{ijk}\sigma_k \partial_i( A_j) \\ &= -\hbar e \sigma_k \varepsilon_{kij}\partial_i A_j \\ &= -\hbar e \vec{\sigma}\cdot \vec{B} \end{align}

With that we can write:

\begin{align} (\sigma_k (p_k - e A_k))^2 &= \frac{(p_k - e A_k)^2}{2m} - \frac{\hbar e}{2m}\vec{\sigma}\cdot \vec{B} \end{align}

Now finally putting all together the non relativistic Dirac's equation is: \begin{align} \left( e \Phi + \frac{(p_k - e A_k)^2}{2m} - \frac{\hbar e}{2m}\vec{\sigma}\cdot \vec{B}\right)u =E_{NR}u \end{align}

  • $\begingroup$ Isn't the energy $E=i\hbar \partial_t$ instead of $E=i h \partial_t$ ? $\endgroup$
    – Schiele
    Apr 29, 2023 at 13:18

Let's start with the Dirac equation as you've written (but I choose to use lower indexes only, since they denote 3-vectors here, not 4-vectors): $$\begin{align} \sigma_i(p_i+eA_i)u_B & = (E-m+eA_0)u_A \\ \sigma_i(p_i+eA_i)u_A & = (E+m+eA_0)u_B \end{align} \tag{1}$$

As @Sunyam suggested in his comment you can solve the second equation of (1) for $u_B$ $$u_B=\frac{1}{E+m+eA_0}\sigma_j(p_j+eA_j)u_A$$

and then you plug that into the first equation of (1): $$\frac{1}{E+m+eA_0}\sigma_i(p_i+eA_i)\sigma_j(p_j+eA_j)u_A = (E-m+eA_0)u_A \tag{2}$$

Since you want to derive the Pauli-Schrödinger equation you restrict to the non-relativistic situation. Therefore the total energy $E$ is positive and very close to the rest energy $m$, in other words $$E=m+E_{NR} \quad \text{with } E_{NR} \ll m \tag{3}$$ where $E_{NR}$ denotes the non-relativistic energy. Inserting (3) into (2) you get $$\frac{1}{2m+E_{NR}+eA_0}\sigma_i(p_i+eA_i)\sigma_j(p_j+eA_j)u_A = (E_{NR}+eA_0)u_A \tag{4}$$

This equation is still exact. But in the non-relativistic situation, the non-relativistic energy $E_{NR}$ and the electric energy $eA_0$ both are much smaller than the rest energy $m$.

Therefore you can neglect these in the denominator on the left-hand-side of (4) and get $$\frac{1}{2m}\sigma_i(p_i+eA_i)\sigma_j(p_j+eA_j)u_A \approx (E_{NR}+eA_0)u_A. \tag{5}$$

Now you're nearly finished. Using $\sigma_i\sigma_j=\delta_{ij}+i\epsilon_{ijk}\sigma_k$, $\vec{p}=-i\vec{\nabla}$, and doing some simple but tedious math you can rewrite equation (5) to $$\frac{1}{2m}\left((p_i+eA_i)(p_i+eA_i)+e\sigma_i\epsilon_{ijk}\nabla_j A_k\right)u_A \approx (E_{NR}+eA_0)u_A. \tag{6}$$

Here you recognize $\epsilon_{ijk}\nabla_j A_k$ as the magnetic field $B_i$ and have the Pauli-Schrödinger equation $$\frac{1}{2m}\left((p_i+eA_i)(p_i+eA_i)+e\sigma_i B_i\right)u_A \approx (E_{NR}+eA_0)u_A. \tag{7}$$


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