The Dirac equation for a free particle is given by:

\[(i \gamma^\mu \partial_\mu - m) \psi = 0\]

where \(\psi\) is the Dirac spinor, \(\gamma^\mu\) are the gamma matrices, \(\partial_\mu\) denotes the four-gradient, and \(m\) is the mass of the particle. The conjugate of the Dirac spinor \(\psi\) is denoted by \(\bar{\psi} = \psi^\dagger \gamma^0\).

Step-by-Step Derivation

1. Start with the Dirac Equation:

   \[(i \gamma^\mu \partial_\mu - m) \psi = 0\]

2. Multiply from the left by \(\bar{\psi}\):

   \[\bar{\psi} (i \gamma^\mu \partial_\mu - m) \psi = 0\]

   This expands to:

   \[i \bar{\psi} \gamma^\mu \partial_\mu \psi - m \bar{\psi} \psi = 0\]

3. Take the Dirac conjugate of the Dirac equation:

   The Dirac conjugate of the Dirac equation is:

   \[\psi^{\dagger} ( -i \overleftarrow{\partial_\mu} \gamma^{\mu\;\dagger} - m) = 0\]

   Here, \(\overleftarrow{\partial_\mu}\) indicates that the derivative acts to the left. Multiplying both sides by \(\gamma^0\) on R.H.S we get

   \[\psi^{\dagger} ( -i \overleftarrow{\partial_t} \gamma^{0\;\dagger} -i \overleftarrow{\partial_k} \gamma^{k\;\dagger} - m)\gamma^0 = 0\]

   and using the property \(\gamma^{k \dagger}=-\gamma^{k}\) and \(\gamma^0 \gamma^k=\gamma^k\gamma^0\) and $k\neq0$ we get

   \[\psi^{\dagger} ( -i \overleftarrow{\partial_t} \gamma^{0}\gamma^0 -i \overleftarrow{\partial_k} \gamma^0\gamma^{k} - m\gamma^0) = 0\]

   Now using \(\psi^{\dagger}\overleftarrow{\partial_t} \gamma^0=\partial_t\bar{\psi}\) we get

   \[-i \partial_t\bar{\psi} \gamma^0 -i \partial_k\bar{\psi} \gamma^k - m\bar{\psi} = 0\]

   In einstein summation form we get

   \[-i \partial_\mu\bar{\psi} \;\gamma^\mu - m\bar{\psi} = 0\]

   Multiplying by \(\psi\) on R.H.S and cancelling negative sign we get

   \[i \partial_\mu\bar{\psi} \;\gamma^\mu\psi + m\bar{\psi}\psi = 0\]

4. Combine the results:

   From steps 2 and 3, we have:

   \[i \bar{\psi} \gamma^\mu \partial_\mu \psi = m \bar{\psi} \psi\]

   And:

   \[i\partial_\mu \bar{\psi} \gamma^\mu \psi = - m \bar{\psi} \psi\]

5. Add the conjugate equation from the original:

   \[i \bar{\psi} \gamma^\mu \partial_\mu \psi + \partial_\mu \bar{\psi} \gamma^\mu \psi = 0\]

   This simplifies to:

   \[i \partial_\mu (\bar{\psi} \gamma^\mu \psi) = 0\]

6. Define the current density \(j^\mu\):

   The current density \(j^\mu\) is defined as:

   \[j^\mu = \bar{\psi} \gamma^\mu \psi\]

   Hence, the above equation becomes:

   \[\partial_\mu j^\mu = 0\]

Conclusion

The current density conservation equation derived from the Dirac equation is:

\[\partial_\mu j^\mu = 0\]

This equation expresses the conservation of probability current in relativistic quantum mechanics. The four-current \(j^\mu = \bar{\psi} \gamma^\mu \psi\) is conserved, indicating that the total probability (or charge) is conserved in the system.

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Plane Wave Solution of Dirac Equation

Sure, let’s start with the plane wave given by

\[\psi(x^\mu) = \begin{pmatrix} u_A \\ u_B \end{pmatrix} e^{-i p_\mu x^\mu}\]

where \(\begin{pmatrix} u_A \\ u_B \end{pmatrix}\) is a spinor, and \(p_\mu x^\mu = p^0 x^0 - \mathbf{p} \cdot \mathbf{x}\).

and use Dirac equation to solve for the spinors \(u_A\), \(u_B\) and Energy \(E\). Here is step by step solution:

1. Substitute the Plane Wave Solution:

   Substitute \(\psi(x)\) into the Dirac equation:

   \[(i \gamma^\mu \partial_\mu - m) \psi(x) = 0\]

   Applying the derivative \(\partial_\mu\):

   \[\color{red}{\partial_\mu \psi(x) = -ip_\mu \begin{pmatrix} u_A \\ u_B \end{pmatrix} e^{-i p_\mu x^\mu}}\]

   Thus, the Dirac equation becomes:

   \[\left[i \gamma^\mu (-ip_\mu) - m\right] \begin{pmatrix} u_A \\ u_B \end{pmatrix} e^{-i p_\mu x^\mu} = 0\]

   Simplifying, we get:

   \[(\gamma^\mu p_\mu - m) \begin{pmatrix} u_A \\ u_B \end{pmatrix} = 0\]

   which is a Dirac equation in momentum space.

2. Writing the Dirac Equation in Matrix Form:

   The Dirac equation becomes:

   \[\left( \gamma^0 p^0 - \gamma^i p^i - m \right) \begin{pmatrix} u_A \\ u_B \end{pmatrix} = 0\]

   Expanding the terms, we have:

   \[\color{red}{ \left[ \begin{pmatrix} I & 0 \\ 0 & -I \end{pmatrix} p^0 - \begin{pmatrix} 0 & \sigma^i \\ -\sigma^i & 0 \end{pmatrix} p^i - m \right] \begin{pmatrix} u_A \\ u_B \end{pmatrix} = 0 }\]

   where the gamma matrices are expanded with the form given by:

   \[\gamma^0 = \begin{pmatrix} I & 0 \\ 0 & -I \end{pmatrix}, \quad \gamma^i = \begin{pmatrix} 0 & \sigma^i \\ -\sigma^i & 0 \end{pmatrix}\]

   where \(I\) is the \(2 \times 2\) identity matrix, and \(\sigma^i\) (with \(i = 1, 2, 3\)) are the Pauli matrices:

   \[\sigma^1 = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}, \quad \sigma^2 = \begin{pmatrix} 0 & -i \\ i & 0 \end{pmatrix}, \quad \sigma^3 = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}\]

3. Substitute the Pauli spin matrices and Simplify:

   On simplifying the above matrix we get:

   \[\begin{pmatrix} I(p^0-m) & -\sigma^i p^i \\ \sigma^i p^i & - I(p^0+m) \end{pmatrix} \begin{pmatrix} u_A \\ u_B \end{pmatrix} = 0\]

   Or,

   \[\begin{pmatrix} I(E-m) & -\sigma^i p^i \\ \sigma^i p^i & - I(E+m) \end{pmatrix} \begin{pmatrix} u_A \\ u_B \end{pmatrix} = 0\]

   Or,

   \[\begin{pmatrix} E-m & 0 & -p_z & -(p_x-ip_y) \\ 0 & E-m & -(p_x+ip_y) & p_z\\ p_z & (p_x-ip_y)&-(E+m)&0\\ (p_x+ip_y) & -p_z &0 &-(E+m) \end{pmatrix} \begin{pmatrix} u_A \\ u_B \end{pmatrix} = 0\]

Here, the spinor \(\begin{pmatrix} u_A \\ u_B \end{pmatrix}\) can take four possible form; two for positive energy \(E=\sqrt{p^2+m^2}\) and two for negative energy \(E=-\sqrt{p^2+m^2}\) as:

1. For Positive energy:

   \[u_1=N_1\begin{pmatrix} 1\\0 \\ A_1\\B_1 \end{pmatrix}\qquad u_2=N_2\begin{pmatrix} 0\\1 \\ A_2\\B_2 \end{pmatrix}\]

   where, \((A_1,\;B_1)\) and \((A_2,\;B_2)\) are \((\frac{p_z}{E+m} , \frac{p_x+i p_y}{E+m})\) and \((\frac{p_x-i p_y}{E+m} , \frac{-p_z}{E+m})\) respectively. \(N_i\) are normalization constant.

2. For Negative energy:

   \[u_3=N_3\begin{pmatrix} A_3\\B_3 \\ 1\\0 \end{pmatrix}\qquad u_4=N_4\begin{pmatrix} A_4\\B_4 \\ 0\\1 \end{pmatrix}\]

   where, \((A_3,\;B_3)\) and \((A_4,\;B_4)\) are \((\frac{p_z}{E-m} , \frac{p_x+i p_y}{E-m})\) and \((\frac{p_x-i p_y}{E-m} , \frac{-p_z}{E-m})\) respectively.

Therefore the eigenfunctions corresponding to positive energy states are:

\[\psi_1(x^\mu)=u_1e^{-ip_\mu x^\mu}\qquad \psi_2(x^\mu)=u_2e^{-ip_\mu x^\mu}\]

and for negative energy state the eigenfunctions are:

\[\psi_3(x^\mu)=u_3e^{ip_\mu x^\mu}\qquad \psi_4(x^\mu)=u_4e^{ip_\mu x^\mu}\]

Interpretation of negative energy states

To be continued …