In quantum field theory, plane-wave solutions to the Dirac equation, are standard basis solutions to the Dirac equation describing the propagation of Dirac spinors. These are spinors that describe all known fundamental particles that are fermions, with the possible exception of neutrinos. They are a certain combination of two Weyl spinors, that transforms "spinorially" under the action of the Lorentz group. For each momentum value p {\displaystyle p} , the four basis plane-wave solutions are commonly labeled as u s ( p ) {\displaystyle u^{s}(p)} and v s ( p ) {\displaystyle v^{s}(p)} , where s {\displaystyle s} is a spin index, taking two possible values for the spin up and spin down states. Dirac spinors are important and interesting in numerous ways. Foremost, they are important as they do describe all of the known fundamental particle fermions in nature; this includes the electron and the quarks. Algebraically they behave, in a certain sense, as the "square root" of a vector. This is not readily apparent from direct examination, but it has slowly become clear over the last 60 years that spinorial representations are fundamental to geometry. For example, effectively all Riemannian manifolds can have spinors and spin connections built upon them, via the Clifford algebra. The Dirac spinor is specific to that of Minkowski spacetime and Lorentz transformations; the general case is quite similar. This article is devoted to the Dirac spinor in the Dirac representation. This corresponds to a specific representation of the gamma matrices, and is best suited for demonstrating the positive and negative energy solutions of the Dirac equation. There are other representations, most notably the chiral representation, which is better suited for demonstrating the chiral symmetry of the solutions to the Dirac equation. The chiral spinors may be written as linear combinations of the Dirac spinors presented below; thus, nothing is lost or gained, other than a change in perspective with regard to the discrete symmetries of the solutions. The remainder of this article focuses primarily on the algebra of the plane-wave solutions. The manner in which the Dirac spinor transforms under the action of the Lorentz group is discussed in the article on Dirac spinors.
Definition The Dirac spinor u ( p ) {\displaystyle u\left(\mathbf {p} \right)} in the plane-wave ansatz
ψ ( x ) = u ( p ) e − i p ⋅ x {\displaystyle \psi (x)=u\left(\mathbf {p} \right)\;e^{-ip\cdot x}}
of the free Dirac equation for a spinor with mass m {\displaystyle m} ,
( i ℏ γ μ ∂ μ − m c ) ψ ( x ) = 0 {\displaystyle \left(i\hbar \gamma ^{\mu }\partial _{\mu }-mc\right)\psi (x)=0}
which, in natural units becomes
( i γ μ ∂ μ − m ) ψ ( x ) = 0 {\displaystyle \left(i\gamma ^{\mu }\partial _{\mu }-m\right)\psi (x)=0}
and with Feynman slash notation may be written
( i ∂ / − m ) ψ ( x ) = 0 {\displaystyle \left(i\partial \!\!\!/-m\right)\psi (x)=0}
An explanation of terms appearing in the ansatz is given below.
The Dirac field is ψ ( x ) {\displaystyle \psi (x)} , a relativistic spin-1/2 field, or concretely a function on Minkowski space R 1 , 3 {\displaystyle \mathbb {R} ^{1,3}} valued in C 4 {\displaystyle \mathbb {C} ^{4}} , a four-component complex vector function. The Dirac spinor related to a plane-wave with wave-vector p {\displaystyle \mathbf {p} } is u ( p ) {\displaystyle u\left(\mathbf {p} \right)} , a C 4 {\displaystyle \mathbb {C} ^{4}} vector which is constant with respect to position in spacetime but dependent on momentum p {\displaystyle \mathbf {p} } . The inner product on Minkowski space for vectors p {\displaystyle p} and x {\displaystyle x} is p ⋅ x ≡ p μ x μ ≡ E p t − p ⋅ x {\displaystyle p\cdot x\equiv p_{\mu }x^{\mu }\equiv E_{\mathbf {p} }t-\mathbf {p} \cdot \mathbf {x} } . The four-momentum of a plane wave is p μ = ( ± m 2 + p 2 , p ) := ( ± E p , p ) {\textstyle p^{\mu }=\left(\pm {\sqrt {m^{2}+\mathbf {p} ^{2}}},\,\mathbf {p} \right):=\left(\pm E_{\mathbf {p} },\mathbf {p} \right)} where p {\displaystyle \mathbf {p} } is arbitrary, In a given inertial frame of reference, the coordinates are x μ {\displaystyle x^{\mu }} . These coordinates parametrize Minkowski space. In this article, when x μ {\displaystyle x^{\mu }} appears in an argument, the index is sometimes omitted. The Dirac spinor for the positive-frequency solution can be written as
u ( p ) = [ ϕ σ ⋅ p E p + m ϕ ] , {\displaystyle u\left(\mathbf {p} \right)={\begin{bmatrix}\phi \\{\frac {{\boldsymbol {\sigma }}\cdot \mathbf {p} }{E_{\mathbf {p} }+m}}\phi \end{bmatrix}}\,,}
where
ϕ {\displaystyle \phi } is an arbitrary two-spinor, concretely a C 2 {\displaystyle \mathbb {C} ^{2}} vector.
σ {\displaystyle {\boldsymbol {\sigma }}} is the Pauli vector,
E p {\displaystyle E_{\mathbf {p} }} is the positive square root E p = + m 2 + p 2 {\textstyle E_{\mathbf {p} }=+{\sqrt {m^{2}+\mathbf {p} ^{2}}}} . For this article, the p {\displaystyle \mathbf {p} } subscript is sometimes omitted and the energy simply written E {\displaystyle E} . In natural units, when m2 is added to p2 or when m is added to p / {\displaystyle {p\!\!\!/}} , m means mc in ordinary units; when m is added to E, m means mc2 in ordinary units. When m is added to ∂ μ {\displaystyle \partial _{\mu }} or to ∇ {\displaystyle \nabla } it means m c ℏ {\textstyle {\frac {mc}{\hbar }}} (which is called the inverse reduced Compton wavelength) in ordinary units.
Derivation from Dirac equation The Dirac equation has the form
( − i α ⋅ ∇ + β m ) ψ = i ∂ ψ ∂ t {\displaystyle \left(-i{\boldsymbol {\alpha }}\cdot {\boldsymbol {\nabla }}+\beta m\right)\psi =i{\frac {\partial \psi }{\partial t}}}
In order to derive an expression for the four-spinor ω, the matrices α and β must be given in concrete form. The precise form that they take is representation-dependent. For the entirety of this article, the Dirac representation is used. In this representation, the matrices are
α = [ 0 2 σ σ 0 2 ] β = [ I 2 0 2 0 2 − I 2 ] {\displaystyle {\boldsymbol {\alpha }}={\begin{bmatrix}0_{2}&{\boldsymbol {\sigma }}\\{\boldsymbol {\sigma }}&0_{2}\end{bmatrix}}\quad \quad \beta ={\begin{bmatrix}I_{2}&0_{2}\\0_{2}&-I_{2}\end{bmatrix}}}
These two 4×4 matrices are related to the Dirac gamma matrices. Note that 02 is the 2x2 zero matrix and I2 is the 2x2 identity matrix. The next step is to look for solutions of the form
ψ = ω e − i p ⋅ x = ω e − i ( E t − p ⋅ x ) , {\displaystyle \psi =\omega e^{-ip\cdot x}=\omega e^{-i\left(Et-\mathbf {p} \cdot \mathbf {x} \right)},}
while at the same time splitting ω into two two-spinors:
ω = [ ϕ χ ] . {\displaystyle \omega ={\begin{bmatrix}\phi \\\chi \end{bmatrix}}\,.}
Results Using all of the above information to plug into the Dirac equation results in
E [ ϕ χ ] = [ m I 2 σ ⋅ p σ ⋅ p − m I 2 ] [ ϕ χ ] . {\displaystyle E{\begin{bmatrix}\phi \\\chi \end{bmatrix}}={\begin{bmatrix}mI_{2}&{\boldsymbol {\sigma }}\cdot \mathbf {p} \\{\boldsymbol {\sigma }}\cdot \mathbf {p} &-mI_{2}\end{bmatrix}}{\begin{bmatrix}\phi \\\chi \end{bmatrix}}.}
This matrix equation is really two coupled equations:
( E − m ) ϕ = ( σ ⋅ p ) χ ( E + m ) χ = ( σ ⋅ p ) ϕ {\displaystyle {\begin{aligned}\left(E-m\right)\phi &=\left({\boldsymbol {\sigma }}\cdot \mathbf {p} \right)\chi \\\left(E+m\right)\chi &=\left({\boldsymbol {\sigma }}\cdot \mathbf {p} \right)\phi \end{aligned}}}
Solve the 2nd equation for χ and one obtains
ω = [ ϕ σ ⋅ p E + m ϕ ] . {\displaystyle \omega ={\begin{bmatrix}\phi \\{\frac {{\boldsymbol {\sigma }}\cdot \mathbf {p} }{E+m}}\phi \end{bmatrix}}.}
Note that this solution needs to have E = + p 2 + m 2 {\textstyle E=+{\sqrt {\mathbf {p} ^{2}+m^{2}}}} in order for the solution to be valid in a frame where the particle has p = 0 {\displaystyle \mathbf {p} =\mathbf {0} } . To derive the sign of the energy in this case, we consider the potentially problematic term σ ⋅ p E + m ϕ {\textstyle {\frac {{\boldsymbol {\sigma }}\cdot \mathbf {p} }{E+m}}\phi } .
If E = + p 2 + m 2 {\textstyle E=+{\sqrt {p^{2}+m^{2}}}} , clearly σ ⋅ p E + m → 0 {\textstyle {\frac {{\boldsymbol {\sigma }}\cdot \mathbf {p} }{E+m}}\rightarrow 0} as p → 0 {\displaystyle \mathbf {p} \rightarrow \mathbf {0} } . On the other hand, let E = − p 2 + m 2 {\textstyle E=-{\sqrt {p^{2}+m^{2}}}} , p = p n ^ {\displaystyle \mathbf {p} =p{\hat {n}}} with n ^ {\displaystyle {\hat {n}}} a unit vector, and let p → 0 {\displaystyle p\rightarrow 0} .
E = − m 1 + p 2 m 2 → − m ( 1 + 1 2 p 2 m 2 ) σ ⋅ p E + m → p σ ⋅ n ^ − m − p 2 2 m + m ∝ 1 p → ∞ {\displaystyle {\begin{aligned}E=-m{\sqrt {1+{\frac {p^{2}}{m^{2}}}}}&\rightarrow -m\left(1+{\frac {1}{2}}{\frac {p^{2}}{m^{2}}}\right)\\{\frac {{\boldsymbol {\sigma }}\cdot \mathbf {p} }{E+m}}&\rightarrow p{\frac {{\boldsymbol {\sigma }}\cdot {\hat {n}}}{-m-{\frac {p^{2}}{2m}}+m}}\propto {\frac {1}{p}}\rightarrow \infty \end{aligned}}}
Hence the negative solution clearly has to be omitted, and E = + p 2 + m 2 {\textstyle E=+{\sqrt {p^{2}+m^{2}}}} . End derivation. Assembling these pieces, the full positive energy solution is conventionally written as
ψ ( + ) = u ( ϕ ) ( p ) e − i p ⋅ x = E + m 2 m [ ϕ σ ⋅ p E + m ϕ ] e − i p ⋅ x {\displaystyle \psi ^{(+)}=u^{(\phi )}(\mathbf {p} )e^{-ip\cdot x}=\textstyle {\sqrt {\frac {E+m}{2m}}}{\begin{bmatrix}\phi \\{\frac {{\boldsymbol {\sigma }}\cdot \mathbf {p} }{E+m}}\phi \end{bmatrix}}e^{-ip\cdot x}}
The above introduces a normalization factor E + m 2 m , {\textstyle {\sqrt {\frac {E+m}{2m}}},} derived in the next section. Solving instead the 1st equation for ϕ {\displaystyle \phi } a different set of solutions are found:
ω = [ − σ ⋅ p − E + m χ χ ] . {\displaystyle \omega ={\begin{bmatrix}-{\frac {{\boldsymbol {\sigma }}\cdot \mathbf {p} }{-E+m}}\chi \\\chi \end{bmatrix}}\,.}
In this case, one needs to enforce that E = − p 2 + m 2 {\textstyle E=-{\sqrt {\mathbf {p} ^{2}+m^{2}}}} for this solution to be valid in a frame where the particle has p = 0 {\displaystyle \mathbf {p} =\mathbf {0} } . The proof follows analogously to the previous case. This is the so-called negative energy solution. It can sometimes become confusing to carry around an explicitly negative energy, and so it is conventional to flip the sign on both the energy and the momentum, and to write this as
ψ ( − ) = v ( χ ) ( p ) e i p ⋅ x = E + m 2 m [ σ ⋅ p E + m χ χ ] e i p ⋅ x {\displaystyle \psi ^{(-)}=v^{(\chi )}(\mathbf {p} )e^{ip\cdot x}=\textstyle {\sqrt {\frac {E+m}{2m}}}{\begin{bmatrix}{\frac {{\boldsymbol {\sigma }}\cdot \mathbf {p} }{E+m}}\chi \\\chi \end{bmatrix}}e^{ip\cdot x}}
In further development, the ψ ( + ) {\displaystyle \psi ^{(+)}} -type solutions are referred to as the particle solutions, describing a positive-mass spin-1/2 particle carrying positive energy, and the ψ ( − ) {\displaystyle \psi ^{(-)}} -type solutions are referred to as the antiparticle solutions, again describing a positive-mass spin-1/2 particle, again carrying positive energy. In the laboratory frame, both are considered to have positive mass and positive energy, although they are still very much dual to each other, with the flipped sign on the antiparticle plane-wave suggesting that it is "travelling backwards in time". The interpretation of "backwards-time" is a bit subjective and imprecise, amounting to hand-waving when one's only evidence are these solutions. It does gain stronger evidence when considering the quantized Dirac field. A more precise meaning for these two sets of solutions being "opposite to each other" is given in the section on charge conjugation, below.
Spin orientation
Two-spinors In the Dirac representation, the most convenient definitions for the two-spinors are:
ϕ 1 = [ 1 0 ] ϕ 2 = [ 0 1 ] {\displaystyle \phi ^{1}={\begin{bmatrix}1\\0\end{bmatrix}}\quad \quad \phi ^{2}={\begin{bmatrix}0\\1\end{bmatrix}}}
and
χ 1 = [ 0 1 ] χ 2 = [ 1 0 ] {\displaystyle \chi ^{1}={\begin{bmatrix}0\\1\end{bmatrix}}\quad \quad \chi ^{2}={\begin{bmatrix}1\\0\end{bmatrix}}}
since these form an orthonormal basis with respect to a (complex) inner product.
Pauli matrices The Pauli matrices are
σ 1 = [ 0 1 1 0 ] σ 2 = [ 0 − i i 0 ] σ 3 = [ 1 0 0 − 1 ] {\displaystyle \sigma _{1}={\begin{bmatrix}0&1\\1&0\end{bmatrix}}\quad \quad \sigma _{2}={\begin{bmatrix}0&-i\\i&0\end{bmatrix}}\quad \quad \sigma _{3}={\begin{bmatrix}1&0\\0&-1\end{bmatrix}}}
Using these, one obtains what is sometimes called the Pauli vector:
σ ⋅ p = σ 1 p 1 + σ 2 p 2 + σ 3 p 3 = [ p 3 p 1 − i p 2 p 1 + i p 2 − p 3 ] {\displaystyle {\boldsymbol {\sigma }}\cdot \mathbf {p} =\sigma _{1}p_{1}+\sigma _{2}p_{2}+\sigma _{3}p_{3}={\begin{bmatrix}p_{3}&p_{1}-ip_{2}\\p_{1}+ip_{2}&-p_{3}\end{bmatrix}}}
Orthogonality The Dirac spinors provide a complete and orthogonal set of solutions to the Dirac equation. This is most easily demonstrated by writing the spinors in the rest frame, where this becomes obvious, and then boosting to an arbitrary Lorentz coordinate frame. In the rest frame, where the three-moment