Chapter 20 The Dirac Equation
20.1 The Free-Particle Dirac Equation
Exercise 20.1.1 Derive the continuity equation
∂ P ∂ t + ∇ ⋅ j = 0 \dfrac{\partial P}{\partial t}+\nabla\cdot\mathbf{j}=0 ∂ t ∂ P + ∇ ⋅ j = 0 where P = ψ † ψ P=\psi^{\dagger}\psi P = ψ † ψ j = c ψ † α ψ \mathbf{j}=c\psi^{\dagger}\mathbf{\alpha}\psi j = c ψ † α ψ
Solution.
Start from the Dirac equation
i ℏ ∂ ψ ∂ t = ( c α ⋅ P + β m c 2 ) ψ ∂ ψ ∂ t = − c α ⋅ ∇ ψ − i ℏ β m c 2 ψ \begin{aligned}
\mathrm{i}\hbar\dfrac{\partial\psi}{\partial t}&=\left(c\mathbf{\alpha}\cdot\mathbf{P}+\beta mc^{2}\right)\psi\\
\dfrac{\partial\psi}{\partial t}&=-c\mathbf{\alpha}\cdot\nabla\psi-\dfrac{\mathrm{i}}{\hbar}\beta mc^{2}\psi
\end{aligned} i ℏ ∂ t ∂ ψ ∂ t ∂ ψ = ( c α ⋅ P + β m c 2 ) ψ = − c α ⋅ ∇ ψ − ℏ i β m c 2 ψ where P = − i ℏ ∇ \mathbf{P}=-\mathrm{i}\hbar\nabla P = − i ℏ∇ † \dagger †
∂ ψ † ∂ t = − c ∇ ψ † ⋅ α † + i ℏ ψ † β † m c 2 \dfrac{\partial\psi^{\dagger}}{\partial t}=-c\nabla\psi^{\dagger}\cdot\mathbf{\alpha}^{\dagger}+\dfrac{\mathrm{i}}{\hbar}\psi^{\dagger}\beta^{\dagger} mc^{2} ∂ t ∂ ψ † = − c ∇ ψ † ⋅ α † + ℏ i ψ † β † m c 2 Since
α † = α β † = β \begin{aligned}
\mathbf{\alpha}^{\dagger}&=\mathbf{\alpha}\\
\beta^{\dagger}&=\beta
\end{aligned} α † β † = α = β We get
∂ ψ † ∂ t = − c ∇ ψ † ⋅ α + i ℏ ψ † β m c 2 \dfrac{\partial\psi^{\dagger}}{\partial t}=-c\nabla\psi^{\dagger}\cdot\mathbf{\alpha}+\dfrac{\mathrm{i}}{\hbar}\psi^{\dagger}\beta mc^{2} ∂ t ∂ ψ † = − c ∇ ψ † ⋅ α + ℏ i ψ † β m c 2 Thus
∂ P ∂ t = ∂ ∂ t ( ψ † ψ ) = ∂ ψ † ∂ t ψ + ψ † ∂ ψ ∂ t = − c ∇ ψ † ⋅ α ψ + i ℏ ψ † β m c 2 ψ − ψ † c α ⋅ ∇ ψ − i ℏ ψ † β m c 2 ψ = − c ( ∇ ψ † ⋅ α ψ + ψ † α ⋅ ∇ ψ ) = − ∇ ⋅ ( c ψ † α ψ ) = − ∇ ⋅ j \begin{aligned}
\dfrac{\partial P}{\partial t}=\dfrac{\partial}{\partial t}(\psi^{\dagger}\psi)&=\dfrac{\partial\psi^{\dagger}}{\partial t}\psi+\psi^{\dagger}\dfrac{\partial \psi}{\partial t}\\
&=-c\nabla\psi^{\dagger}\cdot\mathbf{\alpha}\psi+\cancel{\dfrac{\mathrm{i}}{\hbar}\psi^{\dagger}\beta mc^{2}\psi}-\psi^{\dagger}c\mathbf{\alpha}\cdot\nabla\psi-\cancel{\dfrac{\mathrm{i}}{\hbar}\psi^{\dagger}\beta mc^{2}\psi}\\
&=-c\left(\nabla\psi^{\dagger}\cdot\mathbf{\alpha}\psi+\psi^{\dagger}\mathbf{\alpha}\cdot\nabla\psi\right)\\
&=-\nabla\cdot(c\psi^{\dagger}\mathbf{\alpha}\psi)=-\nabla\cdot\mathbf{j}
\end{aligned} ∂ t ∂ P = ∂ t ∂ ( ψ † ψ ) = ∂ t ∂ ψ † ψ + ψ † ∂ t ∂ ψ = − c ∇ ψ † ⋅ α ψ + ℏ i ψ † β m c 2 ψ − ψ † c α ⋅ ∇ ψ − ℏ i ψ † β m c 2 ψ = − c ( ∇ ψ † ⋅ α ψ + ψ † α ⋅ ∇ ψ ) = − ∇ ⋅ ( c ψ † α ψ ) = − ∇ ⋅ j Finally, we get the continuity equation
∂ P ∂ t + ∇ ⋅ j = 0 \dfrac{\partial P}{\partial t}+\nabla\cdot\mathbf{j}=0 ∂ t ∂ P + ∇ ⋅ j = 0 ■ ~\tag*{$\blacksquare$} ■ 20.2 Electromagnetic Interaction of the Dirac Paritlce
Exercise 20.2.1 Derive Eq. (20.2.16).
Solution. To compute the π × π \boldsymbol{\pi}\times\boldsymbol{\pi} π × π i i i
( π × π ) i = ε i j k π j π k = 1 2 ε i j k π j π k + 1 2 ε i j k π j π k = 1 2 ε i j k π j π k + 1 2 ε i k j π k π j = 1 2 ε i j k π j π k − 1 2 ε i j k π k π j = 1 2 ε i j k [ π j , π k ] \begin{aligned}
(\boldsymbol{\pi}\times\boldsymbol{\pi})_{i}&=\varepsilon_{ijk}\pi_{j}\pi_{k}\\
&=\dfrac{1}{2}\varepsilon_{ijk}\pi_{j}\pi_{k}+\dfrac{1}{2}\varepsilon_{ijk}\pi_{j}\pi_{k}\\
&=\dfrac{1}{2}\varepsilon_{ijk}\pi_{j}\pi_{k}+\dfrac{1}{2}\varepsilon_{ikj}\pi_{k}\pi_{j}\\
&=\dfrac{1}{2}\varepsilon_{ijk}\pi_{j}\pi_{k}-\dfrac{1}{2}\varepsilon_{ijk}\pi_{k}\pi_{j}\\
&=\dfrac{1}{2}\varepsilon_{ijk}[\pi_{j},\pi_{k}]
\end{aligned} ( π × π ) i = ε ijk π j π k = 2 1 ε ijk π j π k + 2 1 ε ijk π j π k = 2 1 ε ijk π j π k + 2 1 ε ikj π k π j = 2 1 ε ijk π j π k − 2 1 ε ijk π k π j = 2 1 ε ijk [ π j , π k ] Therefore, we need to compute the commutative relation between the mechanical momentum operator components, by definition, it is
[ π j , π k ] = [ P j − q c A j , P k − q c A k ] = [ P j , P k ] − q c [ P j , A k ] − q c [ A j , P k ] + ( q c ) 2 [ A j , A k ] \begin{aligned}
[\pi_{j},\pi_{k}]&=\left[P_{j}-\dfrac{q}{c}A_{j},P_{k}-\dfrac{q}{c}A_{k}\right]\\
&=[P_{j},P_{k}]-\dfrac{q}{c}[P_{j},A_{k}]-\dfrac{q}{c}[A_{j},P_{k}]+\left(\dfrac{q}{c}\right)^{2}[A_{j},A_{k}]
\end{aligned} [ π j , π k ] = [ P j − c q A j , P k − c q A k ] = [ P j , P k ] − c q [ P j , A k ] − c q [ A j , P k ] + ( c q ) 2 [ A j , A k ] Now let's compute the commutators:
[ P j , P k ] ψ = ( P j P k − P k P j ) ψ = P j ( − i ℏ ∂ ∂ x k ) − P k ( − i ℏ ∂ ∂ x j ) = ( − i ℏ ) 2 ∂ ∂ x j ( ∂ ψ ∂ x k ) − ( − i ℏ ) 2 ∂ ∂ x k ( ∂ ψ ∂ x j ) = − ℏ 2 ∂ 2 ψ ∂ x j ∂ x k + ℏ 2 ∂ 2 ψ ∂ x k ∂ x j = 0 \begin{aligned}
[P_{j},P_{k}]\psi&=(P_{j}P_{k}-P_{k}P_{j})\psi=P_{j}\left(-\mathrm{i}\hbar\dfrac{\partial}{\partial x_{k}}\right)-P_{k}\left(-\mathrm{i}\hbar\dfrac{\partial}{\partial x_{j}}\right)\\
&=(-\mathrm{i}\hbar)^{2}\dfrac{\partial}{\partial x_{j}}\left(\dfrac{\partial\psi}{\partial x_{k}}\right)-(-\mathrm{i}\hbar)^{2}\dfrac{\partial}{\partial x_{k}}\left(\dfrac{\partial\psi}{\partial x_{j}}\right)\\
&=-\hbar^{2}\dfrac{\partial^{2}\psi}{\partial x_{j}\partial x_{k}}+\hbar^{2}\dfrac{\partial^{2}\psi}{\partial x_{k}\partial x_{j}}=0
\end{aligned} [ P j , P k ] ψ = ( P j P k − P k P j ) ψ = P j ( − i ℏ ∂ x k ∂ ) − P k ( − i ℏ ∂ x j ∂ ) = ( − i ℏ ) 2 ∂ x j ∂ ( ∂ x k ∂ ψ ) − ( − i ℏ ) 2 ∂ x k ∂ ( ∂ x j ∂ ψ ) = − ℏ 2 ∂ x j ∂ x k ∂ 2 ψ + ℏ 2 ∂ x k ∂ x j ∂ 2 ψ = 0 [ P j , A k ] ψ = ( P j A k − A k P j ) ψ = P j ( A k ψ ) − A k ( P j ψ ) = − i ℏ ∂ ∂ x j ( A k ψ ) − A k ( − i ℏ ∂ ψ ∂ x j ) = − i ℏ ( ∂ A k ∂ x j ψ + A k ∂ ψ ∂ x j ) − A k ( − i ℏ ∂ ψ ∂ x j ) = − i ℏ ∂ A k ∂ x j ψ \begin{aligned}
[P_{j},A_{k}]\psi&=(P_{j}A_{k}-A_{k}P_{j})\psi=P_{j}(A_{k}\psi)-A_{k}(P_{j}\psi)\\
&=-\mathrm{i}\hbar\dfrac{\partial}{\partial x_{j}}(A_{k}\psi)-A_{k}\left(-\mathrm{i}\hbar\dfrac{\partial\psi}{\partial x_{j}}\right)\\
&=-\mathrm{i}\hbar\left(\dfrac{\partial A_{k}}{\partial x_{j}}\psi+A_{k}\dfrac{\partial\psi}{\partial x_{j}}\right)-A_{k}\left(-\mathrm{i}\hbar\dfrac{\partial\psi}{\partial x_{j}}\right)\\
&=-\mathrm{i}\hbar\dfrac{\partial A_{k}}{\partial x_{j}}\psi
\end{aligned} [ P j , A k ] ψ = ( P j A k − A k P j ) ψ = P j ( A k ψ ) − A k ( P j ψ ) = − i ℏ ∂ x j ∂ ( A k ψ ) − A k ( − i ℏ ∂ x j ∂ ψ ) = − i ℏ ( ∂ x j ∂ A k ψ + A k ∂ x j ∂ ψ ) − A k ( − i ℏ ∂ x j ∂ ψ ) = − i ℏ ∂ x j ∂ A k ψ [ A j , P k ] ψ = ( A j P k − P k A j ) ψ = A j ( P k ψ ) − P k ( A j ψ ) = A j ( − i ℏ ∂ ψ ∂ x k ) − ( − i ℏ ) ∂ ∂ x k ( A j ψ ) = A j ( − i ℏ ∂ ψ ∂ x k ) + i ℏ ∂ A j ∂ x k ψ + i ℏ A j ∂ ψ ∂ x k = i ℏ ∂ A j ∂ x k ψ \begin{aligned}
[A_{j},P_{k}]\psi&=(A_{j}P_{k}-P_{k}A_{j})\psi=A_{j}(P_{k}\psi)-P_{k}(A_{j}\psi)\\
&=A_{j}\left(-\mathrm{i}\hbar\dfrac{\partial \psi}{\partial x_{k}}\right)-(-\mathrm{i}\hbar)\dfrac{\partial}{\partial x_{k}}(A_{j}\psi)\\
&=A_{j}\left(-\mathrm{i}\hbar\dfrac{\partial \psi}{\partial x_{k}}\right)+\mathrm{i}\hbar\dfrac{\partial A_{j}}{\partial x_{k}}\psi+\mathrm{i}\hbar A_{j}\dfrac{\partial\psi}{\partial x_{k}}\\
&=\mathrm{i}\hbar\dfrac{\partial A_{j}}{\partial x_{k}}\psi
\end{aligned} [ A j , P k ] ψ = ( A j P k − P k A j ) ψ = A j ( P k ψ ) − P k ( A j ψ ) = A j ( − i ℏ ∂ x k ∂ ψ ) − ( − i ℏ ) ∂ x k ∂ ( A j ψ ) = A j ( − i ℏ ∂ x k ∂ ψ ) + i ℏ ∂ x k ∂ A j ψ + i ℏ A j ∂ x k ∂ ψ = i ℏ ∂ x k ∂ A j ψ [ A j , A k ] ψ = ( A j A k − A k A j ) ψ = ( A j A k − A j A k ) ψ = 0 [A_{j},A_{k}]\psi=(A_{j}A_{k}-A_{k}A_{j})\psi=(A_{j}A_{k}-A_{j}A_{k})\psi=0 [ A j , A k ] ψ = ( A j A k − A k A j ) ψ = ( A j A k − A j A k ) ψ = 0 Therefore,
( π × π ) i = 1 2 ε i j k [ π j , π k ] = − q c ( − i ℏ ∂ A k ∂ x j + i ℏ ∂ A j ∂ x k ) = i q ℏ c ( ∂ A k ∂ x j − ∂ A j ∂ x k ) = i q ℏ c ε i j k ∂ A k ∂ x j = i q ℏ c B i \begin{aligned}
(\boldsymbol{\pi}\times\boldsymbol{\pi})_{i}&=\dfrac{1}{2}\varepsilon_{ijk}[\pi_{j},\pi_{k}]\\
&=-\dfrac{q}{c}\left(-\mathrm{i}\hbar\dfrac{\partial A_{k}}{\partial x_{j}}+\mathrm{i}\hbar\dfrac{\partial A_{j}}{\partial x_{k}}\right)\\
&=\dfrac{\mathrm{i}q\hbar}{c}\left(\dfrac{\partial A_{k}}{\partial x_{j}}-\dfrac{\partial A_{j}}{\partial x_{k}}\right)\\
&=\dfrac{\mathrm{i}q\hbar}{c}\varepsilon_{ijk}\dfrac{\partial A_{k}}{\partial x_{j}}\\
&=\dfrac{\mathrm{i}q\hbar}{c}B_{i}
\end{aligned} ( π × π ) i = 2 1 ε ijk [ π j , π k ] = − c q ( − i ℏ ∂ x j ∂ A k + i ℏ ∂ x k ∂ A j ) = c i q ℏ ( ∂ x j ∂ A k − ∂ x k ∂ A j ) = c i q ℏ ε ijk ∂ x j ∂ A k = c i q ℏ B i Thus we get Eq. (20.2.16):
π × π = i q ℏ c B \boldsymbol{\pi}\times\boldsymbol{\pi}=\dfrac{\mathrm{i}q\hbar}{c}\mathbf{B} π × π = c i q ℏ B ■ ~\tag*{$\blacksquare$} ■ Exercise 20.2.2 Solve for the exact levels of the Dirac particle in a uniform magnetic field B = B 0 k \mathbf{B}=B_{0}\mathbf{k} B = B 0 k A = ( B 0 / 2 ) ( − y i + x j ) \mathbf{A}=(B_{0}/2)(-y\mathbf{i}+x\mathbf{j}) A = ( B 0 /2 ) ( − y i + x j ) χ \chi χ
20.3 More on Relativistic Quantum Mechanics 