Dr Phill’s Science Made Simple

Quantum Theory

Paul Adrienne Maurice Dirac

Paul Dirac was an eccentric genius. When asked if he played a musical instrument he replied, "I do not know; I have never tried." He manipulated equations to solve problems. If an equation didn’t work he discarded it. Wolfgang Pauli characterized Dirac’s form of theorizing as acrobatic. He discovered the Dirac Equation that described relativistic fermions such as the electron in 1928 when he was 25 years old.

Notation

The three dimensional differential operator is:

\[\nabla = \left(\frac{\partial}{\partial x}, \frac{\partial}{\partial y},\frac{\partial}{\partial z}\right)\]

The metric tensor for Minkowski flat spacetime is:

\[g_{\mu\nu} = g^{\mu\nu} = \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & -1 \end{bmatrix}\]

The covariant four-dimensional differential operator is:

\[\partial_\mu = \left(\frac{1}{c}\frac{\partial}{\partial t}, -\frac{\partial}{\partial x}, -\frac{\partial}{\partial y}, -\frac{\partial}{\partial z}\right)\]

The contravariant form is:

\[\partial^\mu = g^{\mu\nu}\partial_\nu = \left(\frac{1}{c}\frac{\partial}{\partial t}, \frac{\partial}{\partial x}, \frac{\partial}{\partial y}, \frac{\partial}{\partial z}\right)\]

The second order form is:

\[\partial^\mu\partial_\nu = \left(\frac{1}{c^2}\frac{\partial^2}{\partial t^2}, -\frac{\partial^2}{\partial x^2}, -\frac{\partial^2}{\partial y^2}, -\frac{\partial^2}{\partial z^2}\right)\]

Dirac’s Requirements

Dirac’s requirements were for an equation that described a spin-1/2 fermion. Neither the Schrödinger equation nor the Klein-Gordon equation can do this.

It must be a first-order linear differential equation in both space and time.
It must have plane-wave solutions that satisfy the Planck-Einstein and de Broglie relations.
It should reproduce the relativistic energy-momentum relation \(E^2 = p^2c^2 + m^2c^4\).

The Planck-Einstein relation related energy to frequency \(E = h\nu\).

The de Broglie relation relates wavelength to energy \(\lambda E = hc\).

The solutions must take the form:

\[\psi(x) = u(\vec{p})e^{i(-Et+\vec{p}\cdot\vec{x})/\hbar}\]

The Dirac field \(\psi(x)\) is a relativistic spin-1/2 field describing any fermion.
The Dirac spinor \(u(\vec{p})\) is constant with respect to position in spacetime but dependent on momentum \(\vec{p}\).
The inner product on Minkowski space for vectors \(\vec{p}\) and \(\vec{x}\) is \(\vec{p}\cdot\vec{x} = p_\mu x^\mu\).

Dirac Equation

Paul Dirac had the idea of taking the square root of the Energy-Momentum equation. Start by rearranging the terms.

\[E^2 - \vec{p}^2c^2 - m^2c^4 = 0\]

Rewrite it using the 4-momentum \(p_\mu = (E/c, -p_x, -p_y, -p_z)\), \(p^\mu = (E/c, p_x, p_y, p_z)\) and dividing by \(c^2\).

\[p^\mu p_\mu - m^2c^2 = 0\]

He tried to factorise this.

\[p^{\mu}p_{\mu} - m^2c^2 = (\beta^\kappa p_\kappa + mc)(\gamma^\lambda p_\lambda - mc)\]

Expand the left hand side ofEquation (9).

\[p^{\mu}p_{\mu} - m^2c^2 = p_0^2 - \vec{p}\cdot\vec{p} = p_0^2 - p_1^2 - p_2^2 - p_3^2 - m^2c^2\]

Expand the right hand side of Equation (9).

\[p^{\mu}p_{\mu} - m^2c^2 = \beta^\kappa\gamma^\lambda p_\kappa p_\lambda - m^2c^2 + mc\gamma^\lambda p_\lambda - mc\beta^\kappa p_\kappa\]

This is where Dirac acrobatics come in to play. He had to make the equation work by applying conditions on \(\beta\) and \(\gamma\). He needed to eliminate the terms on the right hand side that have linear \(p\) terms. This can be done by setting \(\beta = \gamma\).

\[p^{\mu}p_{\mu} - m^2c^2 = \gamma^\kappa\gamma^\lambda p_\kappa p_\lambda - m^2c^2\]

Expand the right hand side of Equation (12) for \(\kappa = \lambda\).

\[(\gamma^\lambda)^2 p_\lambda^2 - m^2c^2\]

This is the right hand side of Equation (10) if \((\gamma^0)^2 = 1\) and \((\gamma^\lambda)^2 = -1\).

This leaves the cross terms where \(\kappa \ne \lambda\) and \(p_\kappa p_\lambda = p_\lambda p_\kappa\).

\[(\gamma^\kappa\gamma^\lambda + \gamma^\lambda\gamma^\kappa)p_\kappa p_\lambda\]

Dirac realised that the \(\gamma\) terms had to anti-commute for \(\kappa \ne \lambda\).

\[\{\gamma^\kappa, \gamma^\lambda\} = \gamma^\kappa\gamma^\lambda + \gamma^\lambda\gamma^\kappa = 0\]

Dirac then realised that the \(\gamma\) terms had to be 4x4 Hermitian matrices.

\[\{\gamma^\mu, \gamma^\nu\} = 2g^{\mu\nu}\]

The \(\gamma^\mu\) matrices, or Dirac matrices, are 4x4 matrices that are defined by the 2x2 zero matrix \(0\), the 2x2 identity matrix \(I\), and the 2x2 Pauli matrices \(\sigma_\mu\).

\[\sigma_0 = I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \quad \sigma_1 = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix} \quad \sigma_2 = \begin{bmatrix} 0 & -i \\ i & 0 \end{bmatrix} \quad \sigma_3 = \begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix}\]

\[\gamma^0 = \begin{bmatrix} I & 0 \\ 0 & -I \end{bmatrix} \quad \gamma^\mu = \begin{bmatrix} 0 & \sigma_\mu \\ -\sigma_\mu & 0 \end{bmatrix}\]

The Hermitian conjugates are:

\[\gamma^{0\dagger} = \gamma^{0}\quad \gamma^{1\dagger} = -\gamma^{1}\quad \gamma^{2\dagger} = -\gamma^{2}\quad \gamma^{3\dagger} = -\gamma^{3}\]

Now we have the solution to Equation (9).

\[p^{\mu}_{\mu} - m^2c^2 = (\gamma^{\mu}p_{\mu} + mc)(\gamma^{\mu}p_{\mu} - mc) = 0\]

The Dirac equation is one of the two factors. The right hand factor is usually used.

\[\gamma^{\mu}p_{\mu} - mc = 0\]

Using the standard substitution \(p_\mu \rightarrow i\hbar\partial_\mu\). Where \(\hbar\) is the reduced Planck constant, \(m\) is the particle’s mass.

The Dirac equation is:

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

It has a four-component wave function \(\psi\). This is not a 4-vector as it has no spatial or time components. It is actually a bispinor consisting of the two 2x1 Weyl spinors.

\[\psi = \begin{bmatrix} \psi_R \\ \psi_L \end{bmatrix}\]

The right-handed spinor represents the spin-up and spin-down components of an electron, or any other fermion. The left-handed spinor represents the spin-up and spin-down components of the antiparticle. This predicted the existence of the positron which was discovered in 1932.

The Dirac equation explains electron spin as a form of angular momentum S. The electron’s angular momentum L is not conserved, the combined angular momentum \(L + S\) is conserved.

Dirac Spinor

The plane-wave ansatz is an example of Dirac acrobatics that is an intelligent guess at a solution of the Dirac equation.

The Dirac equation is:

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

Expand the first term.

\[\left(\frac{i\hbar}{c}\gamma^0\frac{\partial}{\partial t} + i\hbar\gamma^\mu\frac{\partial}{\partial x_\mu} - mc\right)\psi = 0\]

Using the requirements, look for a solution.

\[\psi = u(\vec{p})e^{i(-Et + \vec{p}\cdot\vec{x})/\hbar}\]

Where \(u(\vec{p})\) are four Dirac spinors to be determined.

Put the solution into the Dirac equation and multiply by \(c\).

\[(\gamma_0E - c\hbar\gamma^\mu p_\mu - mc^2)u = 0\]

Multiply by \(\gamma^0\) on the left.

\[(IE - c\gamma^0\gamma^\mu p_\mu - \gamma^0 mc^2)u = 0\]

Rewrite as a 2x2 matrix of 2x2 matrices and spinors.

\[\begin{bmatrix} I(E - mc^2) & -c\sigma^\mu p_mu \\ -c\sigma^\mu p_mu & I(E + mc^2) \end{bmatrix} \begin{bmatrix} u_0 \\ u_1 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}\]

Where.

\[\sigma^\mu p_mu = \begin{bmatrix} p_z & p_x - ip_y \\ p_x + ip_y & -p_z \end{bmatrix}\]

For equation Equation (29) to have a solution, its determinant must be zero.

\[E^2 - m^2c^4 - p^2 = 0\]

This is the relativistic energy momentum equation as required.

Expand equation Equation (29).

\[(E - mc^2)u_1 = c \begin{bmatrix} p_z & p_x - ip_y \\ p_x + ip_y & -p_z \end{bmatrix} u_2\]

\[c \begin{bmatrix} p_z & p_x - ip_y \\ p_x + ip_y & -p_z \end{bmatrix} u_1 = (E + mc^2)u_2\]

In equation Equation (32), \(u_2\) has two possible values for spin-up and spin-down. We can solve for \(u_1\) for both of them.

\[u_2 = \begin{bmatrix} 1 \\ 0 \end{bmatrix} \quad u_1 = \begin{bmatrix} \frac{cp_z}{E - mc^2} \\ \frac{c(px + ip_y)}{E - mc^2} \end{bmatrix}\]

\[u_2 = \begin{bmatrix} 0 \\ 1 \end{bmatrix} \quad u_1 = \begin{bmatrix} \frac{c(px - ip_y)}{E - mc^2} \\ \frac{-cp_z}{E - mc^2} \end{bmatrix}\]

Likewise, in equation ({@eq:sol3}), \(u_1\) has two possible values for spin-up and spin-down. We can solve for \(u_2\) for both of them.

\[u_1 = \begin{bmatrix} 1 \\ 0 \end{bmatrix} \quad u_2 = \begin{bmatrix} \frac{cp_z}{E + mc^2} \\ \frac{c(px + ip_y)}{E + mc^2} \end{bmatrix}\]

\[u_1 = \begin{bmatrix} 0 \\ 1 \end{bmatrix} \quad u_2 = \begin{bmatrix} \frac{c(px - ip_y)}{E + mc^2} \\ \frac{-cp_z}{E + mc^2} \end{bmatrix}\]

There is a problem with the last two solutions in that the energy is negative. The Feynman-Stückelberg interpretaion of this is that the solution describes either a negative energy particle travelling backwards in time or a positive energy antipartcle travelling forward in time.

Conserved Probability Current

This was what made Dirac seek an equation that was first order in the time derivative.

Start with the Dirac equation reorganised.

\[i\gamma^\mu\partial_\mu\psi - \frac{mc}{\hbar}\psi = 0\]

Expand the first term.

\[i\gamma^0\frac{1}{c}\frac{\partial\psi}{\partial t} - i\gamma^1\frac{\partial\psi}{\partial x} - i\gamma^2\frac{\partial\psi}{\partial y} - i\gamma^3\frac{\partial\psi}{\partial z} - \frac{mc}{\hbar}\psi = 0\]

Take the Hermitian conjugate of Equation (39).

\[-i\frac{1}{c}\frac{\partial\psi^\dagger}{\partial t}\gamma^0 - i\frac{\partial\psi^\dagger}{\partial x}\gamma^1 - i\frac{\partial\psi^\dagger}{\partial y}\gamma^2 - i\frac{\partial\psi^\dagger}{\partial z}\gamma^3 - \frac{mc}{\hbar}\psi^\dagger = 0\]

There is a problem with Equation (40) in that the time and spacial terms now have different signs than in Equation (39). This can be fixed by multiplying on the right by \(\gamma^0\) and applying the anti-commute \(\gamma^\mu\gamma^0 = -\gamma^0\gamma^\mu\).

\[-i\frac{1}{c}\frac{\partial\psi^\dagger}{\partial t}\gamma^0\gamma^0 + i\frac{\partial\psi^\dagger}{\partial x}\gamma^0\gamma^1 + i\frac{\partial\psi^\dagger}{\partial y}\gamma^0\gamma^2 + i\frac{\partial\psi^\dagger}{\partial z}\gamma^0\gamma^3 - \frac{mc}{\hbar}\psi^\dagger\gamma^0 = 0\]

Now, define the adjoint \(\bar{\psi} = \psi^\dagger\gamma^0\) and negate.

\[i\frac{1}{c}\frac{\partial\bar{\psi}}{\partial t}\gamma^0 - i\frac{\partial\bar{\psi}}{\partial x}\gamma^1 - i\frac{\partial\bar{\psi}}{\partial y}\gamma^2 - i\frac{\partial\bar{\psi}}{\partial z}\gamma^3 + \frac{mc}{\hbar}\bar{\psi} = 0\]

Rewrite using the summation convention.

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

Multiply Equation (38) on the left by \(\bar{\psi}\) and Equation (43) on the right by \(\psi\).

\[i\bar{\psi}\gamma^\mu\partial_\mu\psi - \frac{mc}{\hbar}\bar{\psi}\psi = 0\]

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

Adding Equation (44) and Equation (45) eliminates the mass term.

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

Now write the differential as a 4-current.

\[j^\mu = \bar{\psi}\gamma^\mu\psi = (\rho, \vec{j})\quad \partial_\mu j^\mu = 0\]

Where \(\rho\) is the probability density and \(\vec{j}\) is the probability current. This is an equation of continuity describing the scattering of one fermion off another.

Dirac Equation Solutions

The Dirac equation is:

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

The solution must take the form:

\[\psi = u(E, \vec{p})e^{i(\vec{p}\cdot\vec{x} - Et)/\hbar}\]

The derivatives can be calculated.

\[\partial_0\psi = \frac{1}{c}\frac{\partial\psi}{\partial t} = -\frac{iE}{\hbar c}\psi\]

\[\partial_1\psi = \frac{\partial\psi}{\partial x} = \frac{ip_x}{\hbar}\]

\[\partial_2\psi = \frac{\partial\psi}{\partial y} = \frac{ip_y}{\hbar}\]

\[\partial_3\psi = \frac{\partial\psi}{\partial z} = \frac{ip_z}{\hbar}\]

Substituting Equation (50) through Equation (53) into Equation (48) gives Dirac equation in terms of momentum that has no derivatives.

\[(\gamma_0\frac{E}{c} - \gamma_1 p_x - \gamma_2 p_y - \gamma_3 p_z - mc)u = (\gamma^\mu p_\mu - mc)u = 0\]

Solution for a Particle at Rest

Consider a particle at rest.

For a particle at rest the spatial derivatives are all zero, just leaving the time derivative for Equation (54).

\[\gamma^0\frac{E}{c}u = mcu\]

Rearrange.

\[E\gamma^0 u = mc^2 u\]

Expand the terms.

\[E \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & -1 \end{bmatrix} \begin{bmatrix} u_1 \\ u_2 \\ u_3 \\ u_4 \end{bmatrix} = mc^2 \begin{bmatrix} u_1 \\ u_2 \\ u_3 \\ u_4 \end{bmatrix}\]

This has two orthogonal solutions where \(E = mc^2\).

\[u_1(m, 0) = \begin{bmatrix} 1 \\ 0 \\ 0 \\ 0 \end{bmatrix}\]

\[u_2(m, 0) = \begin{bmatrix} 0 \\ 1 \\ 0 \\ 0 \end{bmatrix}\]

It also has two orthogonal solutions where \(E = -mc^2\).

\[u_3(m, 0) = \begin{bmatrix} 0 \\ 0 \\ 1 \\ 0 \end{bmatrix}\]

\[u_4(m, 0) = \begin{bmatrix} 0 \\ 0 \\ 0 \\ 1 \end{bmatrix}\]

This gives two positive energy spin states.

\[\psi_1 = \begin{bmatrix} 1 \\ 0 \\ 0 \\ 0 \end{bmatrix} e^{-imc^2t/\hbar}\]

\[\psi_2 = \begin{bmatrix} 0 \\ 1 \\ 0 \\ 0 \end{bmatrix} e^{-imc^2t/\hbar}\]

It also gives two negative enegy spin states corresponding to an anti-particle.

\[\psi_3 = \begin{bmatrix} 0 \\ 0 \\ 1 \\ 0 \end{bmatrix} e^{imc^2t/\hbar}\]

\[\psi_4 = \begin{bmatrix} 0 \\ 0 \\ 0 \\ 1 \end{bmatrix} e^{imc^2t/\hbar}\]

Plane Wave Solutions

Write the term in parentheses from Equation (54) in matrix form writing 4x4 matrices as a 2x2 matrix of 2x2 matrices.

\[\begin{bmatrix} I & 0 \\ 0 & -I \end{bmatrix} \frac{E}{c} - \begin{bmatrix} 0 & \sigma \\ -\sigma & 0 \end{bmatrix} \cdot\vec{p} - mc \begin{bmatrix} I & 0 \\ 0 & -I \end{bmatrix}\]

Combine the terms.

\[\begin{bmatrix} (E/c - mc)I & -\sigma\cdot\vec{p} \\ \sigma\cdot\vec{p} & -(E/c + mc)I \end{bmatrix}\]

Substitute into Equation (54).

\[\begin{bmatrix} (E/c - mc)I & -\sigma\cdot\vec{p} \\ \sigma\cdot\vec{p} & -(E/c + mc)I \end{bmatrix} \begin{bmatrix} u_A \\ u_B \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}\]

Multiply the matrix term and rearrange.

\[(\sigma\cdot\vec{p})u_B = (E/c - mc)u_A\]

\[(\sigma\cdot\vec{p})u_A = (E/c + mc)u_B\]

Expand the \(\sigma\) term.

\[\sigma\cdot\vec{p} = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix} p_x + \begin{bmatrix} 0 & -i \\ i & 0 \end{bmatrix} p_y + \begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix} p_z = \begin{bmatrix} p_z & p_x - ip_y \\ p_x + ip_y & -p_z \end{bmatrix}\]

Reorganise Equation (70) and substitute Equation (71).

\[u_A = \frac{\sigma\cdot\vec{p}}{E/c + mc}u_B = \frac{c}{E + mc^2} \begin{bmatrix} p_z & p_x - ip_y \\ p_x + ip_y & -p_z \end{bmatrix} u_B\]

We can make simple arbitrary coices for \(u_A\).

\[u_A = \begin{bmatrix} 1 \\ 0 \end{bmatrix}\]

\[u_A = \begin{bmatrix} 0 \\ 1 \end{bmatrix}\]

We now have solutions.

\[u_1 = N_1 \begin{bmatrix} 1 \\ 0 \\ \frac{cp_z}{E + mc^2} \\ \frac{c(p_x + ip_y)}{E + mc^2} \\ \end{bmatrix}\]

\[u_2 = N_2 \begin{bmatrix} 0 \\ 1 \\ \frac{c(p_x - ip_y)}{E + mc^2} \\ \frac{-cp_z}{E + mc^2} \\ \end{bmatrix}\]

Where \(N_1\) and \(N_2\) are normalisation factors.

Note that in the case \(\vec{p} = 0\) these are the positive energy particle at rest solutions.

Repeat the process exchanging \(u_A\) and \(u_B\) gives the other two solutions.

\[u_3 = N_3 \begin{bmatrix} \frac{cp_z}{E - mc^2} \\ \frac{c(p_x + ip_y)}{E - mc^2} \\ 1 \\ 0 \\ \end{bmatrix}\]

\[u_4 = N_4 \begin{bmatrix} \frac{c(p_x - ip_y)}{E - mc^2} \\ \frac{-cp_z}{E - mc^2} \\ 0 \\ 1 \\ \end{bmatrix}\]

These are negative energy solutions.

Negative Energy Solution Interpretation

The negative energy solutions have been associated with the electron’s anti-particle the positron. Dirac interpreted it as a sea of anti-particles where all of the energy states are full. Holes in the negative energy sea were positive energy positrons.

The Feynman-Stückelberg Interpretation describes a negative energy particle travelling backward in time. This is equivalent to a positive energy anti-particle travelling forward in time.

Anti-Particle Spinors

We want to redefine the negative energy solutions to reflect the real positive energy of the actual particle.

\[E = \left|\sqrt{(\left|\vec{p}\right|^2 - m^2c^2)}\right|\]

Start with the negative energy solution Equation (77) and Equation (78).

We can define the antiparticle solution by changing the sign of the energy and momentum following the Feynman-Stückelburg interpretation to ensure that energy is always positive.

\[v_1(E, \vec{p})e^{-i(\vec{p}\cdot\vec{x} - Et)/\hbar} = u_4(-E, -\vec{p})e^{i(\vec{p}\cdot\vec{x} - Et)/\hbar}\]

\[v_2(E, \vec{p})e^{-i(\vec{p}\cdot\vec{x} - Et)/\hbar} = u_3(-E, -\vec{p})e^{i(\vec{p}\cdot\vec{x} - Et)/\hbar}\]

Note that the exponent is negated.

The Dirac in momentum Equation (54) becomes.

\[(-\gamma_0\frac{E}{c} + \gamma_1 p_x + \gamma_2 p_y + \gamma_3 p_z - mc)v = -(\gamma^\mu p_\mu + mc)v = 0\]

Repeat the process for creating \(u_3\) and \(u_4\).

\[v_1 = N'_1 \begin{bmatrix} \frac{c(p_x - ip_y)}{E + mc^2} \\ \frac{-cp_z}{E + mc^2} \\ 0 \\ 1 \\ \end{bmatrix}\]

\[v_2 = N'_2 \begin{bmatrix} \frac{cp_z}{E + mc^2} \\ \frac{c(p_x + ip_y)}{E + mc^2} \\ 1 \\ 0 \\ \end{bmatrix}\]

Likewise there is also \(v_3\) and \(v_4\).

The solution needs to be a four component spinor. We ca use \({u_1, u_2, u_3, u_4}\) or \({v_1, v_2, v_3, v_4}\). A better coise is to use only positive energy components \({u_1, u_2, v_1, v_2}\).

Wave Function Normalisation

Revisit the component \(u_1\) defined in Equation (75). The solution is:

\[\psi = u_1e^{i(\vec{p}\cdot\vec{r} - ET)/\hbar}\]

Calculate the probabilty density.

\[\rho = \psi^\dagger\psi = (\psi^*)^T\psi = u^\dagger_1 u_1\]

\[u^\dagger_1 u_1 = \left|N_1\right|^2\left(1 + \frac{c^2p_z^2}{(E + mc^2)^2} + \frac{c^2(p_x^2 + p_y^2)}{(E + mc^2)^2}\right) = \frac{\left|N_1\right|^2}{(E + mc^2)^2}\left((E + mc^2)^2 + c^2\left|\vec{p}\right|^2\right)\]

Use the momentum from Equation (7).

\[u^\dagger_1 u_1 = \frac{\left|N_1\right|^2}{(E + mc^2)^2}\left((E + mc^2)^2 + E^2 - m^2c^4\right) = \frac{\left|N_1\right|^2}{(E + mc^2)^2}2E(E + mc^2) = \frac{2E\left|N_1\right|^2}{E + mc^2}\]

For normalisation \(\rho = 2E\).

\[N_1 = \sqrt{(E + mc^2)}\]

The process can repeated for the remaining N values.