Title: ETH_Sem5 / AdvQC / 06_Dirac_Theory_Electron

Q: What is the problem of the Klein-Gordon equation?
A: If one derives the continuity equation
\[\partial_\mu j^\mu = \dot{\rho}+\dop\bm{j}=0\]
using the Klein-Gordon equation, one gets for the zeroth component of the 4-current \(j^\mu(c\rho,\bm{j})\):
\[\rho = \frac{\imag\,\hbar}{2m_ec^2}\left[\Psi^\star\pdv{}{t}\Psi-\Psi\pdv{}{t}\Psi^\star\right]\]
Since the Klein-Gordon equation is a second-order differential equation in \(t\), two integration constants arise which allow one to choose the initial values of \(\Psi\) and \(\partial\Psi/\partial t\) independently of one another in such a way that \(\rho\) may take positive or negative values.
Consequently, \(\rho\) is not positive definite and, hence, does not represent a probability density distribution, which must take strictly positive values for any coordinates \(t\) and \(x_k\).
Hence, we shall reject the Klein-Gordon equation as a fundamental quantum mechanical equation!

Q: Write down the Dirac equation! What is the reasoning for the used ansatz?
A: The correct relativistic equation for the electron requires:
1. treat all variables on equal footing (Lorentz transformations mix time and space coordinates!)
2. must not have second derivatives w.r.t. time (negative Klein-Gordon density!)
Hence, the ansatz involving only first derivatives is proposed (for a freely moving electron):
\[\begin{aligned} \mathrm{i} \hbar \frac{\partial}{\partial t} \Psi &=\left[\frac{\hbar c}{\mathrm{i}} \alpha^k \partial_k+\beta m_e c^2\right] \Psi \\ &= \left[c\bmalpha\cdot\pop + \beta m_ec^2\right]\Psi \equiv H^D \Psi \end{aligned} \]
Upon multiplication with the operator identity \(\imag\hbar\partial/\partial t = \hbar c/\imag\, \alpha^k\partial_k + \beta m_ec^2\) one gets:
\[\begin{aligned} & \pb{\alpha^i}{\alpha^j} = 2\delta_{ij} \quad;\\ & \pb{\alpha^i}{\beta} = 0 \quad;\\ & \beta = \beta^{-1} \\ & \Rightarrow \quad \text{tr}[\alpha^i] = \text{tr}[\beta] = 0 \end{aligned}\]
Hence, \(\alpha^i\) and \(\beta\) feature an equal number of positive and negative eigenvalues and therefore an even dimension. And they also cannot be numbers but are of matrix form!
\[\begin{aligned} & \text{Or shorthand with}\\ & \gamma^\mu = (\gamma^0,\gamma^i) = (\beta,\beta\alpha^i)\,: \\ & {\left[-\imag\gamma^\mu\partial_\mu+\frac{m_ec}{\hbar}\right]\Psi = 0} \end{aligned}\]

Q: What is the Standard representation?
A: The smallest possible dimension for the \(\alpha^i\) and \(\beta\) parameters in the Dirac equation would be two, but this yields only 3 linearly independent matrices, not 4, as requested.
The next even dimension is four, and we find the following set of \(4\times4\) matrices (called the standard representation):
\[\alpha^i=\left(\begin{array}{cc} 0 & \sigma_i \\ \sigma_i & 0\end{array}\right) \quad \text { and } \quad \beta=\left(\begin{array}{cc}\bm{1}_2 & 0 \\ 0 & -\bm{1}_2\end{array}\right)\]
This particular choice is neither unique nor is the dimension 4 the only dimension that fulfills the constraints (higher even dimensions are also possible)!
Consequences:
In the standard representation, the quantum mechanical state \(\bm{\Psi}\) is a vector of four functions, called a 4-spinor.
Note the connection to the spin operator
\[\hat{S}=\left(\hat{s}_x, \hat{s}_y, \hat{s}_z\right)=\frac{\hbar}{2}\left(\hat{\sigma}_x, \hat{\sigma}_y, \hat{\sigma}_z\right)\]
for a spin 1/2.

Q: Dirac equation in split-notation
A: The Dirac equation can be cast into the split notation
\[\begin{aligned} c\bmsigma\cdot\pop\Psi^S + m_ec^2\Psi^L &= \imag\hbar\pdv{}{t} \Psi^L \\ c\bmsigma\cdot\pop\Psi^L - m_ec^2\Psi^S &= \imag\hbar\pdv{}{t} \Psi^S \end{aligned}\]
with
\[\begin{array}{ccccc} \Psi^L = \begin{pmatrix}\Psi_1 \\ \Psi_2\end{pmatrix} & ; & \Psi^S = \begin{pmatrix}\Psi_3 \\ \Psi_4\end{pmatrix} & ; & \bm{\sigma} = (\sigma_x,\sigma_y,\sigma_z) \\ \sigma_x = \begin{pmatrix}0&1 \\ 1&0\end{pmatrix} & ; & \sigma_y = \begin{pmatrix}0&-\imag \\ \imag&0\end{pmatrix} & ; & \sigma_z = \begin{pmatrix}1&0 \\ 0&-1\end{pmatrix} \end{array}\]

Q: What are the Dirac density distribution and the Dirac current density for a single particle?
A: Multiplying the Dirac equation by the complex conjugate of \(\Psi\)
\[\mathrm{i} \hbar \Psi^{\dagger} \frac{\partial}{\partial t} \Psi=\frac{\hbar c}{\mathrm{i}} \Psi^{\dagger} \alpha^k \partial_k \Psi+m_e c^2 \Psi^{\dagger} \beta \Psi\]
whose complex conjugate reads
\[-\mathrm{i} \hbar\left(\frac{\partial}{\partial t} \Psi^{\dagger}\right) \Psi=-\frac{\hbar c}{\mathrm{i}}\left(\partial_k \Psi^{\dagger}\right) \alpha^{k, \dagger} \Psi+m_e c^2 \Psi^{\dagger} \beta^{\dagger} \Psi\]
and by subtracting the latter equation from the first one, and using \(\alpha^{k, \dagger} = \alpha^{k}\) as \(\hat{H}^D\) has to be hermitian, one obtains
\[\begin{aligned} \frac{\partial}{\partial t}\underbrace{\left(\Psi^{\dagger} \Psi\right)}_{=\rho} &=-c\left[\Psi^{\dagger} \alpha^k \partial_k \Psi+\left(\partial_k \Psi^{\dagger}\right) \alpha^{k, \dagger} \Psi\right]+\frac{m_e c^2}{\mathrm{i} \hbar}\underbrace{\left[\Psi^{\dagger} \beta \Psi-\Psi^{\dagger} \beta^{\dagger} \Psi\right]}_{=0} \\ &= -\partial_k\,\underbrace{c\left[\Psi^\dagger\alpha^k\Psi\right]}_{=j^k} \end{aligned}\]
This means:
\[\begin{aligned} & \rho=\Psi^\dagger\Psi \quad;\quad \bm{j} = c\Psi^\dagger\bmalpha\Psi \\ & \thus j^\mu = (j^0,j^k) = (c\rho,j^k) \\ & \thus \partial_\mu j^\mu=0 \end{aligned} \]
NB: Comparing to the classical current density, the relationship \(c\bm{\alpha} = \dot{\rop}\) is implied.

Q: Schematize how the Dirac equation is solved for a free electron! Which energies are obtained for the free Dirac electron?
A: 1. Ansatz of plane waves which is promoted to 4-spinor by 4-vector \(u(\mom)\):
\[\begin{aligned} \Psi(x) &=u(\bm{p}) \exp \left[-\mathrm{i} \frac{p \cdot x}{\hbar}\right]\ ; \\ p\cdot x &= Et-\mom\cdot\pos \end{aligned}\]
2. Determine \(u(\mom)\) by inserting the ansatz into the Dirac equation which leads to
\[\left[c \bm{\alpha} \cdot \bm{p}+\beta m_e c^2\right] u(\bm{p})=E u(\bm{p})\]
leading to the relations
\[\begin{aligned} u^L &=\left[\frac{c \bm{\sigma} \cdot \bm{p}}{E-m_e c^2}\right] u^S \\ u^S &=\left[\frac{c \bm{\sigma} \cdot \bm{p}}{E+m_e c^2}\right] u^L \end{aligned}\]
3. Insertion of the upper into the lower equation yields:
\[{E = \pm\sqrt{c^2\mom^2+m_e^2c^4}}\]
4. Choose natural ansätze for linearly independent 2-vectors yielding \(u(\mom)\).
\[u_1^{L/S}=\left(\begin{array}{l}1 \\ 0\end{array}\right) \quad \text { and } \quad u_2^{L/S}=\left(\begin{array}{c}0 \\ 1\end{array}\right)\]
\[\begin{array}{cccc} & u_1^{(\cdot)}(\mom)/\mathcal{N} & u_2^{(\cdot)}(\mom)/\mathcal{N} \\ (+) & \left(\begin{array}{c}1 \\ 0 \\ \frac{c p_z}{E+m_e c^2} \\ \frac{c\left(p_x+\mathrm{i} p_y\right)}{E+m_e c^2}\end{array}\right) & \left(\begin{array}{c}0 \\ 1 \\ \frac{c\left(p_x-\mathrm{i} p_y\right)}{E+m_e c^2} \\ \frac{-c p_z}{E+m_e c^2}\end{array}\right) \\ (-) & \left(\begin{array}{c}\frac{c p_z}{E-m_e c^2}\\ \frac{c\left(p_x+\mathrm{i} p_y\right)}{E-m_e c^2} \\ 1 \\ 0\end{array}\right) & \left(\begin{array}{c}\frac{c\left(p_x-\mathrm{i} p_y\right)}{E-m_e c^2} \\ \frac{-c p_z}{E-m_e c^2} \\ 0 \\ 1\end{array}\right) \end{array}\]

Q: What is the Dirac velocity operator?
A: From the Heisenberg equation of motion
\[\dot{x}^k=\frac{\mathrm{i}}{\hbar}\left[H^D, x^k\right]\]
and the relation
\[\begin{aligned}\left[H^D, x^k\right] &=\left[c \bm{\alpha} \cdot \bm{p}, x^k\right] \\ &= \left[c \alpha^j p_j, x^k\right] \\ &=c \alpha^j \underbrace{\left[p_j, x^k\right]}_{-\mathrm{i} \hbar \delta^k{ }_j}\\ &=-\mathrm{i} \hbar c \alpha^k \end{aligned}\]
we get
\[{\dot{x}^k=c \alpha^k \means \dpos = c\bmalpha}\]
Therefore, the current density can be rewritten as:
\[\bm{j} = \Psi^\dagger\dot{\rop}\Psi\]

Q: Write down the Dirac equation including minimal coupling with electromagnetic fields!
A: Using minimal coupling, which only considers coupling between fields involving only the charge distribution and not higher multipole moments of the charge distribution, we must substitute:
\[\imag\hbar\partial_\mu \quad\rightarrow\quad \imag\hbar\partial_\mu-\frac{q_e}{c}A_\mu \]
which is equivalent to
\[\begin{aligned} \imag\hbar\pdv{}{t} \quad&\rightarrow\quad \imag\hbar\pdv{}{t}-q_e\phi \\ \imag\hbar\pdv{}{x^i} \quad&\rightarrow\quad \imag\hbar\pdv{}{x^i}-\frac{q_e}{c}A_i \end{aligned}\]
Insertion into the Dirac equation leads to:
\[\left[-\gamma^\mu\left(\imag\hbar\partial_\mu-\frac{q_e}{c}A_\mu\right)+m_ec\right]\Psi = 0\]
and can be rewritten as
\[ \imag\hbar\pdv{}{t}\Psi = \left[c\bmalpha\cdot\left(\mom-\frac{q_e}{c}\vecfield\right)+\beta m_ec^2+q_e\phi\right]\Psi\]
Attention: The sign in front of \(\bm{A}\) had to be changed, as it contains the contravariant components of the vector potential \(A^i=-A_i\).

Q: What is the kinematic momentum?
A: The connection of the canonical momentum operator \(\bm{p}\) with the effect of external vector potentials \(\bm{A}\) on a moving electron with charge \(q_e=-e\) is often written as
\[{\mom \quad\rightarrow\quad \mom-\frac{q_e}{c}\vecfield = \mom+\frac{e}{c}\vecfield \equiv \bm{\pi}}\]
which denotes the mechanical-momentum operator \(\bm{\pi}\) (also called kinematic or linear momentum).
The Dirac equation for an electron in external electromagnetic fields thus reduces to:
\[\imag\hbar\pdv{t}\Psi = \left[c\bmalpha\cdot\bm{\pi}+\beta m_ec^2-e\phi\right]\Psi\]

Q: Write down the electromagnetic interaction operator for the Dirac equation! Compare it to the classical potential term!
A: The Dirac equation for a fermion in external electromagnetic fields can be rewritten as:
\[{\imag\hbar\pdv{t}\Psi = \Big[c\bmalpha\cdot\mom+\beta m_ec^2 + \underbrace{q_e\phi-q_e\bmalpha\cdot\vecfield}_{=\hat{V}}\Big]\Psi}\]
Comparing it to the classical expression for the interaction energy
\[V = q\phi -\frac{q}{c}\dot{\pos}\cdot\bm{A}\]
implies again by the correspondence principle that the expression for the Dirac velocity operator reads
\[\dot{\pos} \xrightarrow{\mathrm{c.p.}} c\bm{\alpha}\]

Q: What is the kinetic balance condition?
A: To recover the Schrödinger equation, the Dirac equation can be rewritten in split notation:
\[\mathrm{i} \hbar \frac{\partial}{\partial t}\left(\begin{array}{c}\Psi^L \\ \Psi^S\end{array}\right)=c\left(\begin{array}{c}(\bm{\sigma} \cdot \bm{\pi})\Psi^S \\ (\bm{\sigma} \cdot \bm{\pi}) \Psi^L\end{array}\right)-2 m_e c^2\left(\begin{array}{c}0 \\ \Psi^S\end{array}\right)+V\left(\begin{array}{c}\Psi^L \\\Psi^S\end{array}\right)\]
with \(V = q_e\phi\). The origin of the energy scale was shifted by \(-m_ec^2\), as the lowest possible non-relativistic energy of a free particle in Schrödinger quantum mechanics is zero instead of \(+m_ec^2\).
For non-relativistic energies, the energy \(E\rightarrow\imag\hbar\partial/\partial t\) and the potential \(V\) are small compared to the rest energy \(m_ec^2\) so that we can rewrite and approximate the lower part of the above equation to:
\[\left(\imag\hbar\pdv{}{t}+2m_ec^2-V\right)\Psi^S = c(\bm{\sigma}\cdot\bm{\pi})\Psi^L \thus {\Psi^S\approx\frac{\bmsigma\cdot\bm{\pi}}{2m_ec}\Psi^L}\]
This is the so-called kinetic-balance condition. It shows that the lower component of the spinor \(\Psi^S\) is damped by a factor of \(1/c\) compared to \(\Psi^L\) (for small linear momenta), and vanishes in the limit \(c\rightarrow\infty\).

Q: What is the Pauli equation and how does one get to it?
A: Inserting the kinetic balance condition in the upper part of the Dirac notation in split notation and utilizing Dirac's relation
\[\begin{aligned} (\sigma \cdot \bm{X})(\sigma \cdot \bm{Y})=\bm{X} \cdot \bm{Y} 1_2+\mathrm{i} \sigma \cdot(\bm{X} \times \bm{Y})\\ \thus (\bm{\sigma} \cdot \bm{\pi})(\bm{\sigma} \cdot \bm{\pi})=\pi^2-\frac{q_e \hbar}{c} \bm{\sigma} \cdot \bm{B} \end{aligned}\]
yields the Pauli equation
\[{\frac{1}{2m_e}\left(\mom-\frac{q_e}{c}\vecfield\right)^2\Psi^L-\frac{q_e\hbar}{2m_ec}(\bmsigma\cdot\bm{B})\Psi^L+V\Psi^L = \imag\hbar\pdv{}{t}\Psi^L}\]
The non-relativistic limit for infinite speed of light, \(c\rightarrow\infty\) immediately recovers the Schrödinger equation (note, however, the 2-spinor \(\Psi^L\)). Obviously, in the non-relativistic limit all magnetic interactions are eliminated. This also implies that Schrödinger QM is exact if not electromagnetic fields are present.

Q: What is the magnetic moment of the Pauli electron?
A: The Pauli equation can also be rewritten to yield the below equation. The operator quadratic in the vector potential, \(q_e^2 \bm{A}^2/2 m_e c^2\), is often called the diamagnetic term, while the remaining magnetic-potential operators linear in \(\Bfield\) (or \(\vecfield\), respectively) are called paramagnetic terms.
\[\left[\frac{\mom^2}{2 m_e}+\frac{q_e^2 \bm{A}^2}{2 m_e c^2}-\underbrace{\frac{q_e}{2 m_e c}(\bm{l}+2\bm{s})}_{\equiv \bm{\mu}} \cdot \Bfield+V\right] \Psi^L=\mathrm{i} \hbar \frac{\partial}{\partial t} \Psi^L\]
The magnetic moment \(\bm{\mu}\) of the Pauli electron, which interacts with the external magnetic field \(\bm{B}\), is generated by the angular and spin momenta
\[\bm{\mu}=\bm{\mu}_{\text {orbit }}+\bm{\mu}_{\mathrm{spin}}=-\frac{\mu_{\mathrm{B}}}{\hbar}\bm{l}-2 \frac{\mu_{\mathrm{B}}}{\hbar}\bm{s}=-\mu_{\mathrm{B}}\bm{l}-g \mu_{\mathrm{B}}\bm{s}\]
Here, we introduced two important quantities:
1. the Bohr magneton \(\mu_{\mathrm{B}} = e\hbar/2m_ec\) and
2. the gyromagnetic ratio of the electron, \(g = 2\), which is also known as the Land\'e factor and which is already very well (but not exactly, as quantization of fields is neglected \(\rightarrow\) only in QED) reproduced by the Dirac and Pauli theories.

Q: How can one interpret the negative-energy solutions to the Dirac equation?
A: 1. Dirac Sea or Hole Theory
- All (infinite) states of negative energy are occupied except a few of small velocity
- Their distribution is exactly uniform and thus not observable
- Involves existence of new elementary particle: positron
- Problem: implies an infinite negative charge of the universe and an infinite mass without gravitational impact
2. Feynman-Stuckelberg Interpretation
- Negative energy solutions travel backwards in time \(\rightarrow\) ``\(Et\)'' is positive again
- This defines antiparticles \(\rightarrow\) all quantum charges (\(q,s,l,\dots\)) are reversed
- Resolves above stated problems