Notebook Entry
Reading notes for Modern Particle Physics, Mark Thomson (Chapter 4)
Chapter 4 The Dirac equation
The Klein-Gordon Equation
The naïve thought of the K-G equation:
the energy-mass equation:
\[E^2 = p^2+m^2\]To change this equation into wave equation, let $E \to \mathrm i \frac{\partial }{\partial t}$, $ p \to -\mathrm i \nabla$, hence
\[\left(\frac{\partial ^2}{\partial t^2}-\nabla^2+m^2\right)\psi = 0\]i.e.,
\[(\partial^\mu\partial_\mu+m^2)\psi =0\]In order to obtain the probability density and probability current of this wave function, consider
\[\psi^*\frac{\partial^2}{\partial t^2}\psi = \psi^*(\nabla^2-m^2)\psi\\ \psi\frac{\partial^2}{\partial t^2}\psi^* = \psi(\nabla^2-m^2)\psi^*\]Take the subtraction of these 2 equations above, we get
\[\frac{\partial}{\partial t}\mathrm i(\psi^*\frac{\partial \psi}{\partial t}-\psi\frac{\partial \psi^*}{\partial t})-\nabla \cdot \mathrm i(\psi^*\nabla\psi -\psi\nabla\psi^*)=0\]For freedom wave function, we have
\[\psi = N\mathrm e^{\mathrm i (\vec p \cdot \vec x-Et)}\] \[\rho =\mathrm i(\psi^*\frac{\partial \psi}{\partial t}-\psi\frac{\partial \psi^*}{\partial t})=2|N|^2E\\ \vec j = -\mathrm i(\psi^*\nabla\psi -\psi\nabla\psi^*)= 2|N|^2\vec p\]Obviously when E<0, the probability density comes to be below zero, which sounds ridiculous for physicists.
The Dirac Equation
To limit the wave function to that gives the positive probability density, Dirac tried to constrain the equation into 1-partial-derivation, i.e.,
\[\mathrm i \frac{\partial}{\partial t}\psi = (\vec \alpha \cdot \hat {\vec p} +\beta m)\psi = (-\mathrm i\vec \alpha\cdot\nabla+\beta m )\psi\]Apparently this function needs to accommodate with the energy-mass equation, i.e.
\[\left(\mathrm i \frac{\partial}{\partial t}\right)\cdot\left(\mathrm i \frac{\partial}{\partial t}\right)\psi = (-\mathrm i\vec \alpha\cdot\nabla+\beta m )\cdot (-\mathrm i\vec \alpha\cdot\nabla+\beta m )\psi= (p^2+m^2)\psi\]Hence,
\[\alpha_x^2= \alpha_y^2=\alpha_z^2=\beta^2 = 1\\ \alpha_i\beta+\beta\alpha_i=0,i = x, y, z\\ \alpha_i\alpha_j+\alpha_j\alpha_i = 0, i, j = x, y ,z, i \neq j\]To satisfy the algebra above, we have
\[\alpha_i = \begin{bmatrix} 0 & \sigma_i \\ \sigma_i & 0 \end{bmatrix}, \quad \beta = \begin{bmatrix} I & 0 \\ 0 & -I \end{bmatrix}\]where $\sigma_i$ is one of Pauli matrices.
Probability density and probability current
Similar to the calculation of the K-G equation, we have
\[\psi^\dagger\left(\mathrm i \frac{\partial}{\partial t}\right)\psi = \psi^\dagger(-\mathrm i\vec \alpha\cdot\nabla+\beta m )\psi\\ \left(-\mathrm i \frac{\partial}{\partial t}\right)\psi^\dagger\psi = (\mathrm i\vec \alpha\cdot\nabla+\beta m )\psi^\dagger\psi\]Take the subtraction of these 2 equations above, we gain
\[\frac{\partial}{\partial t}(\psi^\dagger\psi) +\nabla\cdot(\psi^\dagger\vec\alpha\psi)=0\]i.e.,
\[\rho = \psi^\dagger\psi,\vec j = \psi^\dagger\vec \alpha\psi\]Spin and the Dirac Equation
(TO BE CONTINUED)
Covariant form of the Dirac Equation
The Dirac equation:
\[\mathrm i \frac{\partial}{\partial t}\psi = (-\mathrm i \vec \alpha\cdot\nabla+\beta m)\psi\]Multiply this by $\beta$, we have
\[(\mathrm i\gamma^\mu\partial_\mu-m)\psi= 0\]where
\[\gamma_0 =\beta, \gamma_i = \beta\alpha_i\]We regard this as the covariant form of the Dirac equation, albeit with the constant factor gamma. One step more, we get the transformer rule for the Dirac spinor.
The adjoint spinor and the covariant current
For $\vec j^\mu =(\rho, \vec j)$,
\[\vec j = \psi^\dagger\vec \alpha\psi= \psi^\dagger\beta\beta\alpha\psi=\psi^\dagger\gamma^0\gamma^\mu\psi=\bar \psi\gamma^\mu\psi\]where $\bar\psi = \psi^\dagger\gamma^0$, called adjoint spinor. This operation is also suitable for probability density.
Solutions to the Dirac equation
Scenario #1: particle with zero momentum
Dirac Eq.:
\[\mathrm i \frac{\partial}{\partial t}\psi = (-\mathrm i \alpha\cdot\nabla+\beta m)\psi\]With $\psi = u\mathrm e^{-\mathrm i Et}$, we have
\[Eu\mathrm e^{-\mathrm i Et}= \begin{bmatrix} I & 0 \\ 0 & -I \end{bmatrix} m u\mathrm e^{-\mathrm i Et}\]i.e.,
\[\begin{bmatrix} -m & 0 \\ 0 & m \end{bmatrix} \begin{bmatrix} U_1\\ U_2 \end{bmatrix} = E \begin{bmatrix} U_1\\ U_2 \end{bmatrix}\]For eigen value $E_+= m$ (the energy-mass equation), we have
\[u_1= \begin{bmatrix} 1\\ 0\\ 0\\ 0 \end{bmatrix}, \quad u_2= \begin{bmatrix} 0\\ 1\\ 0\\ 0 \end{bmatrix}\]For eigen value $E_-=-m$ (the energy-mass equation), we have
\[u_3= \begin{bmatrix} 0\\ 0\\ 1\\ 0 \end{bmatrix}, \quad u_4= \begin{bmatrix} 0\\ 0\\ 0\\ 1 \end{bmatrix}\]Thus
\[u = A_1u_1+A_2u_2+A_3u_3+A_4u_4\]Scenario #2: particle with momentum $p$ in direction z.
Dirac Eq.:
\[\mathrm i \frac{\partial}{\partial t}\psi = (-\mathrm i \alpha\cdot\nabla+\beta m)\psi\]With $\psi = u\mathrm e^{\mathrm i (\vec p\cdot \vec x-Et)}$, we have
\[\begin{bmatrix} m & \sigma\cdot p \\ \sigma\cdot p & -m \end{bmatrix} \begin{bmatrix} U_1\\ U_2 \end{bmatrix} = E \begin{bmatrix} U_1\\ U_2 \end{bmatrix}\]Useful equation: \((\sigma\cdot \vec A)(\sigma \cdot \vec B)=\sigma_iA_i\sigma_jb_j=\sigma_i\sigma_ja_ib_j= a_ib_i+\mathrm i \epsilon_{ijk}\sigma_ka_ib_j = \vec A \cdot \vec B + \sigma \cdot(\vec A\times\vec B)\)
Solving the eigen value equation, we have
\[\det \begin{bmatrix} m-E & \sigma\cdot p \\ \sigma\cdot p & -m-E \end{bmatrix} = E^2-m^2-p^2=0\]Thus $E_1= \sqrt{p^2+m^2}, E_2 = -\sqrt{p^2+m^2}$, $U_1 = \frac{1}{E-m}\sigma\cdot pU_2, U_2 = \frac{1}{E+m}\sigma\cdot p U_1$.
Therefore for eigen value $E_1= \sqrt{p^2+m^2}$,
\[u_1= \begin{bmatrix} 1\\ 0\\ \frac{1}{E_1+m}p_z\\ \frac{1}{E_1+m}(p_x-\mathrm i p_y) \end{bmatrix}, \quad u_2= \begin{bmatrix} 0\\ 1\\ \frac{1}{E_1+m}(p_x+\mathrm i p_y)\\ -\frac{1}{E_1+m}p_z \end{bmatrix}\]Therefore for eigen value $E_2 = -\sqrt{p^2+m^2}$,
\[u_3= \begin{bmatrix} \frac{1}{E_2-m}p_z\\ \frac{1}{E_2-m}(p_x-\mathrm i p_y)\\ 1\\ 0\\ \end{bmatrix}, \quad u_4= \begin{bmatrix} \frac{1}{E_2-m}(p_x+\mathrm i p_y)\\ -\frac{1}{E_2-m}p_z\\ 0\\ 1\\ \end{bmatrix}\] \[u = A_1u_1+A_2u_2+A_3u_3+A_4u_4\]Antiparticles
For physics reason, we have to get proper explanation for these negative energy.
The Dirac sea interpretation
To be concise, Dirac thought the negative energy levels are almost completely occupied, so the particle cannot reach these negative energy levels due to the Pauli’s exclusion principle. When a vacant position with negative energy appears, the positive-energy particle (the normal kind) jump into the hole in the negative Dirac sea, and release energy.
The Feynman-Stückelberg interpretation
(TO BE CONTINUED)
the confusion for propagate backwards in time
Antiparticle spinors
For negative wavefunctions that propagate backwards in time, we can extract the negative sign from E and p, which is
\[v(E, \vec p)\mathrm e^{-\mathrm i (\vec p \cdot \vec x -Et)}= u(-E, -\vec p)\mathrm e ^{\mathrm i ((-\vec p)\cdot \vec x -(-E)t)}\]To satisfy the Dirac function, we get the Lorentz-invariant form of this equation,
\[(\gamma^\mu p_\mu+m)v = 0\] \[\begin{bmatrix} E+m &-\sigma \cdot p\\ \sigma \cdot p & -E+m\\ \end{bmatrix} \begin{bmatrix} V_1\\ V_2\\ \end{bmatrix}= 0\] \[V_1 = \frac{\sigma\cdot p}{E+m}V_2, V_2 = \frac{\sigma\cdot p}{E-m}V_1\]Note that the positive eigen value E corresponds to the “normal” wavefunction with negative function.
So we have
\[v_1 = \begin{bmatrix} \frac{p_x-\mathrm i p_y}{E+m}\\ \frac{-p_z}{E+m}\\ 0\\ 1\\ \end{bmatrix}, \quad v_2 = \begin{bmatrix} \frac{p_z}{E+m}\\ \frac{p_x+\mathrm i p_y}{E+m}\\ 1\\ 0\\ \end{bmatrix}\]for E > 0 and
\[v_3= \begin{bmatrix} 1\\ 0\\ \frac{p_z}{E-m}\\ \frac{p_x+\mathrm i p_y}{E-m} \end{bmatrix}, \quad v_4= \begin{bmatrix} 0\\ 1\\ \frac{p_x-\mathrm i p_y}{E-m}\\ \frac{-p_z}{E-m}\\ \end{bmatrix}\]for E < 0.
With the equivalence of u-function and v-function, we can construct the genal wave function with $(u_1, u_2, v_1, v_2)$.
According to the normalization rule for wave function$\psi^\dagger\psi = 2E$, we have
\[u(v)_{1(2)}^*= \sqrt{E+m}\, u(v)_{1(2)}\]Operators and the antiparticle spinors
Just remember that for antiparticle wavefunction, the physical energy and momentum are given by
\[H^{(v)}= -\mathrm i\partial _t, p^{(v)}=\mathrm i \nabla\]Charge conjugation
(TO BE CONTINUED)
Spin and helicity states
The z-axis ‘spin matrix’
\[S_z = \frac{1}{2} \begin{bmatrix} \sigma_z & 0\\ 0 &-\sigma_z \end{bmatrix}\]apparently not commute with the Dirac Hamiltonian.
To solve this problem, we construct a new concept called helicity, which is
\[h = \frac{S\cdot p}{p} =\frac{1}{2p} \begin{bmatrix} \sigma \cdot p & 0\\ 0 & \sigma \cdot p \end{bmatrix}\]This physical quantity can be commute with the Dirac Hamiltonian, and can help degenerate the eigen vectors for H.
Consider the eigen vector u for h,
\[\frac{1}{2p} \begin{bmatrix} \sigma \cdot p & 0\\ 0 & \sigma \cdot p \end{bmatrix} \begin{bmatrix} U_1\\ U_2 \end{bmatrix} = \lambda \begin{bmatrix} U_1\\ U_2 \end{bmatrix}\] \[\frac{1}{2p}\sigma\cdot p \, U_{1, 2}= \lambda U_{1, 2}\]Apparently $\lambda = \pm \frac{1}{2}$.
Plus the constrains form the Dirac Equation,
\[U_2 =\frac{1}{E+m}\sigma\cdot p U_1\]To simplify these notation, we set $\vec p = p(\sin \theta\cos \phi, \sin \theta \sin \phi, \cos \theta)$,
\[\frac{1}{2} \begin{bmatrix} \cos \theta & \sin \theta \mathrm e ^{-\mathrm i \phi}\\ \sin \theta\mathrm e ^{\mathrm i \phi} &-\cos \theta \end{bmatrix} \begin{bmatrix} a\\ b \end{bmatrix} = \lambda \begin{bmatrix} a\\ b \end{bmatrix}\] \[\frac{b}{a} = \frac{2\lambda-\cos \theta}{\sin\theta}\mathrm e ^{\mathrm i \phi}\]When $\lambda = \frac{1}{2}$, we have
\[\frac{b}{a}= \frac{\sin \frac {\theta}{2}\mathrm e^{\mathrm i \phi}}{\cos \frac{\theta}{2}}\]Thus
\[u_\uparrow = \begin{bmatrix} c\\ s \mathrm e ^{\mathrm i \phi}\\ \frac{p}{E+m} c\\ \frac{p}{E+m} s \mathrm e ^{\mathrm i \phi}\\ \end{bmatrix}, \quad u_\downarrow = \begin{bmatrix} -s\\ c \mathrm e ^{\mathrm i \phi}\\ \frac{p}{E+m} s\\ -\frac{p}{E+m} c \mathrm e ^{\mathrm i \phi}\\ \end{bmatrix}\] \[v_\uparrow = \begin{bmatrix} \frac{p}{E+m} s\\ -\frac{p}{E+m} c \mathrm e ^{\mathrm i \phi}\\ -s\\ c \mathrm e ^{\mathrm i \phi}\\ \end{bmatrix}, \quad v_\downarrow = \begin{bmatrix} \frac{p}{E+m} c\\ \frac{p}{E+m} s \mathrm e ^{\mathrm i \phi}\\ c\\ s \mathrm e ^{\mathrm i \phi}\\ \end{bmatrix}\]where $c = \cos \frac{\theta}{2}, s = \sin\frac{\theta}{2}$.
Intrinsic parity for Dirac fermions
Parity transformation: the spatial inversion through the origin,
\[x, y, z\to x^\prime,y^\prime,z^\prime= -x, -y, -z \text{ , and }t\to t^\prime=t\]And the affect of the parity operator:
\[\psi \to \psi^\prime= \hat P\psi\]The Dirac equation:
\[\mathrm i \gamma^0\frac{\partial}{\partial t}\psi + \mathrm i \gamma^1\frac{\partial}{\partial x}\psi + \mathrm i \gamma^2\frac{\partial}{\partial y}\psi + \mathrm i \gamma^3\frac{\partial}{\partial z}\psi = m\psi\]After the change of coordinates, we have
\[\mathrm i \gamma^0\frac{\partial}{\partial t^\prime}\psi^\prime + \mathrm i \gamma^1\frac{\partial}{\partial x^\prime}\psi^\prime + \mathrm i \gamma^2\frac{\partial}{\partial y^\prime}\psi^\prime + \mathrm i \gamma^3\frac{\partial}{\partial z^\prime}\psi^\prime = m\psi^\prime\]Substitute $\psi$ with $\psi = \hat P\psi^\prime$ and the coordinates in the original equation, we have
\[\mathrm i \gamma^0\hat P\frac{\partial}{\partial t^\prime}\psi^\prime - \mathrm i \gamma^1\hat P\frac{\partial}{\partial x^\prime}\psi^\prime - \mathrm i \gamma^2\hat P\frac{\partial}{\partial y^\prime}\psi^\prime - \mathrm i \gamma^3\hat P\frac{\partial}{\partial z^\prime}\psi^\prime = m\hat P\psi^\prime\]Compared to the equation above, we get
\[\hat P \gamma^0 = \gamma^0 \hat P\\ \hat P\gamma^1 =- \gamma^1\hat P\\ \hat P\gamma^2 = -\gamma^2 \hat P\\ \hat P \gamma^3 = -\gamma^3\hat P\\\]Apparently $\hat P = \pm \gamma^0 = \pm \begin{pmatrix}1&0&0&0\0 &1&0&0\0&0&-1&0\0&0&0&-1\end{pmatrix}$ satisfies these equations above.
Conventionally we choose $+$ for particles and $-$ for antiparticles.