Dirac spinor

 ( Quantum Mechanics ) 

In quantum field theory, the Dirac spinor is the spinor that describes all known fundamental particles that are , with the possible exception of . It appears in the Plane wave solution to the Dirac equation, and is a certain combination of two , specifically, a bispinor that transforms "spinorially" under the action of the Lorentz group.

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 . 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 built upon them, via the Clifford algebra. See section 1.8. 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 bispinor, 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 regards to the discrete symmetries of the solutions.

The remainder of this article is laid out in a pedagogical fashion, using notations and conventions specific to the standard presentation of the Dirac spinor in textbooks on quantum field theory. It 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 .

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Definition The Dirac spinor is the bispinor $u\backslash left(\backslash vec\{p\}\backslash right)$ in the plane wave ansatz $$\backslash psi(x)\; =\; u\backslash left(\backslash vec\{p\}\backslash right)\backslash ;\; e^\{-i\; p\; \backslash cdot\; x\}$$ of the free Dirac equation for a spinor with mass $m$, $$\backslash left(i\backslash hbar\backslash gamma^\backslash mu\; \backslash partial\_\backslash mu\; -\; mc\backslash right)\backslash psi(x)\; =\; 0$$ which, in natural units becomes $$\backslash left(i\backslash gamma^\backslash mu\; \backslash partial\_\backslash mu\; -\; m\backslash right)\backslash psi(x)\; =\; 0$$ and with Feynman slash notation may be written $$\backslash left(i\backslash partial\backslash !\backslash !\backslash !/\; -\; m\backslash right)\backslash psi(x)\; =\; 0$$

An explanation of terms appearing in the ansatz is given below.

• The Dirac field is $\backslash psi(x)$, a relativistic spin-1/2 field, or concretely a function on Minkowski space $\backslash mathbb\{R\}^\{1,3\}$ valued in $\backslash mathbb\{C\}^4$, a four-component complex vector function.
• The Dirac spinor related to a plane-wave with wave-vector $\backslash vec\{p\}$ is $u\backslash left(\backslash vec\{p\}\backslash right)$, a $\backslash mathbb\{C\}^4$ vector which is constant with respect to position in spacetime but dependent on momentum $\backslash vec\{p\}$.
• The inner product on Minkowski space for vectors $p$ and $x$ is $p\; \backslash cdot\; x\; \backslash equiv\; p\_\backslash mu\; x^\backslash mu\; \backslash equiv\; E\_\backslash vec\{p\}\; t\; -\; \backslash vec\{p\}\; \backslash cdot\; \backslash vec\{x\}$.
• The four-momentum of a plane wave is $p^\backslash mu\; =\; \backslash left(\backslash pm\backslash sqrt\{m^2\; +\; \backslash vec\{p\}^2\},\backslash ,\; \backslash vec\{p\}\backslash right)\; :=\; \backslash left(\backslash pm\; E\_\backslash vec\{p\},\; \backslash vec\{p\}\backslash right)$ where $\backslash vec\{p\}$ is arbitrary,
• In a given inertial frame of reference, the coordinates are $x^\backslash mu$. These coordinates parametrize Minkowski space. In this article, when $x^\backslash mu$ appears in an argument, the index is sometimes omitted.

The Dirac spinor for the positive-frequency solution can be written as $$$$

 u\left(\vec{p}\right) = \begin{bmatrix}
   \phi \\ \frac{\vec{\sigma} \cdot \vec{p}}{E_\vec{p} + m} \phi
 \end{bmatrix} \,,
     

where

• $\backslash phi$ is an arbitrary two-spinor, concretely a $\backslash mathbb\{C\}^2$ vector.
• $\backslash vec\{\backslash sigma\}$ is the Pauli vector,
• $E\_\backslash vec\{p\}$ is the positive square root $E\_\backslash vec\{p\}\; =\; +\; \backslash sqrt\{m^2\; +\; \backslash vec\{p\}^2\}$. For this article, the $\backslash vec\{p\}$ subscript is sometimes omitted and the energy simply written $E$.

In natural units, when is added to or when is added to $\{p\backslash !\backslash !\backslash !/\}$, means in ordinary units; when is added to , means in ordinary units. When m is added to $\backslash partial\_\backslash mu$ or to $\backslash nabla$ it means $\backslash frac\{mc\}\{\backslash hbar\}$ (which is called the inverse reduced Compton wavelength) in ordinary units.

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Derivation from Dirac equation The Dirac equation has the form $$\backslash left(-i\; \backslash vec\{\backslash alpha\}\; \backslash cdot\; \backslash vec\{\backslash nabla\}\; +\; \backslash beta\; m\; \backslash right)\; \backslash psi\; =\; i\; \backslash frac\{\backslash partial\; \backslash psi\}\{\backslash partial\; t\}$$

In order to derive an expression for the four-spinor , the matrices and 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 $$$$

 \vec\alpha = \begin{bmatrix}
   \mathbf{0} & \vec{\sigma} \\
   \vec{\sigma} & \mathbf{0}
 \end{bmatrix} \quad \quad
 \beta = \begin{bmatrix}
   \mathbf{I} & \mathbf{0} \\
   \mathbf{0} & -\mathbf{I}
 \end{bmatrix}
     

These two 4×4 matrices are related to the gamma matrices. Note that and are 2×2 matrices here.

The next step is to look for solutions of the form $$\backslash psi\; =\; \backslash omega\; e^\{-i\; p\; \backslash cdot\; x\}\; =\; \backslash omega\; e^\{\; -i\; \backslash left(E\; t\; -\; \backslash vec\{p\}\; \backslash cdot\; \backslash vec\{x\}\backslash right)\; \},$$ while at the same time splitting into two two-spinors: $$\backslash omega\; =\; \backslash begin\{bmatrix\}\; \backslash phi\; \backslash \backslash \; \backslash chi\; \backslash end\{bmatrix\}\; \backslash ,.$$

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Results Using all of the above information to plug into the Dirac equation results in $$$$

 E \begin{bmatrix}
   \phi \\
   \chi
 \end{bmatrix} =
 \begin{bmatrix}
   m \mathbf{I} & \vec{\sigma}\cdot\vec{p} \\
   \vec{\sigma}\cdot\vec{p} & -m \mathbf{I}
 \end{bmatrix}\begin{bmatrix}
   \phi \\
   \chi
 \end{bmatrix}.
     

This matrix equation is really two coupled equations: $$\backslash begin\{align\}$$

 \left(E - m \right) \phi &= \left(\vec{\sigma} \cdot \vec{p} \right) \chi \\
 \left(E + m \right) \chi &= \left(\vec{\sigma} \cdot \vec{p} \right) \phi
     

\end{align}

Solve the 2nd equation for and one obtains $$\backslash omega\; =\; \backslash begin\{bmatrix\}$$

   \phi \\
   \frac{\vec{\sigma} \cdot \vec{p}}{E + m} \phi
     

\end{bmatrix} .

Note that this solution needs to have $E\; =\; +\backslash sqrt\{\backslash vec\; p^2\; +\; m^2\}$ in order for the solution to be valid in a frame where the particle has $\backslash vec\; p\; =\; \backslash vec\; 0$.

To derive the sign of the energy in this case, we consider the potentially problematic term $\backslash frac\{\backslash vec\backslash sigma\backslash cdot\; \backslash vec\{p\}\}\{E\; +\; m\}\; \backslash phi$.

• If $E\; =\; +\backslash sqrt\{p^2\; +\; m^2\}$, clearly $\backslash frac\{\backslash vec\backslash sigma\backslash cdot\backslash vec\; p\}\{E\; +\; m\}\; \backslash rightarrow\; 0$ as $\backslash vec\; p\; \backslash rightarrow\; \backslash vec\; 0$.
• On the other hand, let $E\; =\; -\backslash sqrt\{p^2\; +\; m^2\}$, $\backslash vec\; p\; =\; p\backslash hat\{n\}$ with $\backslash hat\; n$ a unit vector, and let $p\; \backslash rightarrow\; 0$.

$$\backslash begin\{align\}$$

    E = -m\sqrt{1 + \frac{p^2}{m^2}} &\rightarrow -m\left(1 + \frac{1}{2}\frac{p^2}{m^2}\right) \\
 \frac{\vec\sigma\cdot\vec p}{E + m} &\rightarrow p\frac{\vec\sigma\cdot\hat n}{-m - \frac{p^2}{2m} + m} \propto \frac{1}{p} \rightarrow \infty
     

\end{align}

Hence the negative solution clearly has to be omitted, and $E\; =\; +\backslash sqrt\{p^2\; +\; m^2\}$. End derivation.

Assembling these pieces, the full positive energy solution is conventionally written as $$\backslash psi^\{(+)\}\; =\; u^\{(\backslash phi)\}(\backslash vec\; p)e^\{-i\; p\; \backslash cdot\; x\}\; =\; \backslash textstyle\; \backslash sqrt\{\backslash frac\{E\; +\; m\}\{2m\}\}\; \backslash begin\{bmatrix\}$$

   \phi \\
   \frac{\vec{\sigma} \cdot \vec{p}}{E + m} \phi
     

\end{bmatrix} e^{-i p \cdot x} The above introduces a normalization factor $\backslash sqrt\{\backslash frac\{E+m\}\{2m\}\},$ derived in the next section.

Solving instead the 1st equation for $\backslash phi$ a different set of solutions are found: $$\backslash omega\; =\; \backslash begin\{bmatrix\}\; -\backslash frac\{\backslash vec\{\backslash sigma\}\; \backslash cdot\; \backslash vec\{p\}\}\{-E\; +\; m\}\; \backslash chi\; \backslash \backslash \; \backslash chi\; \backslash end\{bmatrix\}\; \backslash ,.$$

In this case, one needs to enforce that $E\; =\; -\backslash sqrt\{\backslash vec\; p^2\; +\; m^2\}$ for this solution to be valid in a frame where the particle has $\backslash vec\; p\; =\; \backslash vec\; 0$. 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 $$\backslash psi^\{(-)\}\; =\; v^\{(\backslash chi)\}(\backslash vec\; p)\; e^\{i\; p\; \backslash cdot\; x\}\; =\; \backslash textstyle\backslash sqrt\{\backslash frac\{E\; +\; m\}\{2m\}\}\; \backslash begin\{bmatrix\}$$

   \frac{\vec{\sigma} \cdot \vec{p}}{E + m} \chi \\ \chi
     

\end{bmatrix} e^{i p \cdot x}

In further development, the $\backslash psi^\{(+)\}$-type solutions are referred to as the particle solutions, describing a positive-mass spin-1/2 particle carrying positive energy, and the $\backslash psi^\{(-)\}$-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.

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Spin orientation

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Two-spinors In the Dirac representation, the most convenient definitions for the two-spinors are: $$$$

 \phi^1 = \begin{bmatrix} 1 \\ 0 \end{bmatrix} \quad \quad
 \phi^2 = \begin{bmatrix} 0 \\ 1 \end{bmatrix}
     

and $$$$

 \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.

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Pauli matrices The Pauli matrices are $$$$

 \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: $$\backslash vec\{\backslash sigma\}\backslash cdot\backslash vec\{p\}\; =\; \backslash sigma\_1\; p\_1\; +\; \backslash sigma\_2\; p\_2\; +\; \backslash sigma\_3\; p\_3\; =$$

 \begin{bmatrix}
   p_3         & p_1 - i p_2 \\
   p_1 + i p_2 & - p_3
     

\end{bmatrix}

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Orthogonality The Dirac spinors provide a complete and orthogonal set of solutions to the Dirac equation. See Chapter 3.

Claude Itzykson (1980). 9780070320710, McGraw-Hill. ISBN 9780070320710

See Chapter 2. 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-momentum vanishes: $\backslash vec\; p\; =\; \backslash vec\; 0,$ one may define four spinors $$u^\{(1)\}\backslash left(\backslash vec\{0\}\backslash right)\; =$$

 \begin{bmatrix}
   1 \\
   0 \\
   0 \\
   0
 \end{bmatrix}
     

\qquad

 u^{(2)}\left(\vec{0}\right) =
 \begin{bmatrix}
   0 \\
   1 \\
   0 \\
   0
 \end{bmatrix}
     

\qquad

 v^{(1)}\left(\vec{0}\right) =
 \begin{bmatrix}
   0 \\
   0 \\
   1 \\
   0
 \end{bmatrix}
     

\qquad

 v^{(2)}\left(\vec{0}\right) =
 \begin{bmatrix}
   0 \\
   0 \\
   0 \\
   1
 \end{bmatrix}
     

Introducing the Feynman slash notation $$\{p\backslash !\backslash !\backslash !/\}\; =\; \backslash gamma^\backslash mu\; p\_\backslash mu$$

the boosted spinors can be written as $$u^\{(s)\}\backslash left(\backslash vec\{p\}\backslash right)\; =\; \backslash frac\; u^\{(s)\}\backslash left(\backslash vec\{0\}\backslash right)$$

 \begin{bmatrix}
   \phi^{(s)}\\
   \frac {\vec\sigma \cdot \vec p} {E+m} \phi^{(s)}
 \end{bmatrix}
     

and $$$$

 v^{(s)}\left(\vec{p}\right) =
 \frac{-{p\!\!\!/} + m}{\sqrt{2m(E+m)}} v^{(s)}\left(\vec{0}\right)
     

 \begin{bmatrix}
   \frac {\vec\sigma \cdot \vec p} {E+m} \chi^{(s)} \\
   \chi^{(s)}
 \end{bmatrix}
     

The conjugate spinors are defined as $\backslash overline\; \backslash psi\; =\; \backslash psi^\backslash dagger\; \backslash gamma^0$ which may be shown to solve the conjugate Dirac equation $$\backslash overline\; \backslash psi\; (i\{\backslash partial\backslash !\backslash !\backslash !/\}\; +\; m)\; =\; 0$$

with the derivative understood to be acting towards the left. The conjugate spinors are then $$$$

 \overline u^{(s)}\left(\vec{p}\right) =
 \overline u^{(s)}\left(\vec{0}\right) \frac
     

and $$$$

 \overline v^{(s)}\left(\vec{p}\right) =
 \overline v^{(s)}\left(\vec{0}\right) \frac{-{p\!\!\!/} + m}{\sqrt{2m(E+m)}}
     

The normalization chosen here is such that the scalar invariant $\backslash overline\backslash psi\; \backslash psi$ really is invariant in all Lorentz frames. Specifically, this means $$$$

 \begin{align}
    \overline u^{(a)} (p) u^{(b)} (p) &=  \delta_{ab} & \overline u^{(a)} (p) v^{(b)} (p) &= 0 \\
    \overline v^{(a)} (p) v^{(b)} (p) &= -\delta_{ab} & \overline v^{(a)} (p) u^{(b)} (p) &= 0
 \end{align}
     

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Completeness The four rest-frame spinors $u^\{(s)\}\backslash left(\backslash vec\{0\}\backslash right),$ $\backslash ;v^\{(s)\}\backslash left(\backslash vec\{0\}\backslash right)$ indicate that there are four distinct, real, linearly independent solutions to the Dirac equation. That they are indeed solutions can be made clear by observing that, when written in momentum space, the Dirac equation has the form $$(\{p\backslash !\backslash !\backslash !/\}\; -\; m)u^\{(s)\}\backslash left(\backslash vec\{p\}\backslash right)\; =\; 0$$ and $$(\{p\backslash !\backslash !\backslash !/\}\; +\; m)v^\{(s)\}\backslash left(\backslash vec\{p\}\backslash right)\; =\; 0$$

This follows because $$\{p\backslash !\backslash !\backslash !/\}\{p\backslash !\backslash !\backslash !/\}\; =\; p^\backslash mu\; p\_\backslash mu\; =\; m^2$$ which in turn follows from the anti-commutation relations for the gamma matrices: $$\backslash left\backslash \{\backslash gamma^\backslash mu,\; \backslash gamma^\backslash nu\backslash right\backslash \}\; =\; 2\backslash eta^\{\backslash mu\backslash nu\}$$ with $\backslash eta^\{\backslash mu\backslash nu\}$ the metric tensor in flat space (in curved space, the gamma matrices can be viewed as being a kind of vielbein, although this is beyond the scope of the current article). It is perhaps useful to note that the Dirac equation, written in the rest frame, takes the form $$\backslash left(\backslash gamma^0\; -\; 1\backslash right)u^\{(s)\}\backslash left(\backslash vec\{0\}\backslash right)\; =\; 0$$ and $$\backslash left(\backslash gamma^0\; +\; 1\backslash right)v^\{(s)\}\backslash left(\backslash vec\{0\}\backslash right)\; =\; 0$$ so that the rest-frame spinors can correctly be interpreted as solutions to the Dirac equation. There are four equations here, not eight. Although 4-spinors are written as four complex numbers, thus suggesting 8 real variables, only four of them have dynamical independence; the other four have no significance and can always be parameterized away. That is, one could take each of the four vectors $u^\{(s)\}\backslash left(\backslash vec\{0\}\backslash right),$ $\backslash ;v^\{(s)\}\backslash left(\backslash vec\{0\}\backslash right)$ and multiply each by a distinct global phase $e^\{i\backslash eta\}.$ This phase changes nothing; it can be interpreted as a kind of global gauge freedom. This is not to say that "phases don't matter", as of course they do; the Dirac equation must be written in complex form, and the phases couple to electromagnetism. Phases even have a physical significance, as the Aharonov–Bohm effect implies: the Dirac field, coupled to electromagnetism, is a U(1) fiber bundle (the circle bundle), and the Aharonov–Bohm effect demonstrates the holonomy of that bundle. All this has no direct impact on the counting of the number of distinct components of the Dirac field. In any setting, there are only four real, distinct components.

With an appropriate choice of the gamma matrices, it is possible to write the Dirac equation in a purely real form, having only real solutions: this is the Majorana equation. However, it has only two linearly independent solutions. These solutions do not couple to electromagnetism; they describe a massive, electrically neutral spin-1/2 particle. Apparently, coupling to electromagnetism doubles the number of solutions. But of course, this makes sense: coupling to electromagnetism requires taking a real field, and making it complex. With some effort, the Dirac equation can be interpreted as the "complexified" Majorana equation. This is most easily demonstrated in a generic geometrical setting, outside the scope of this article.

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Energy eigenstate projection matrices It is conventional to define a pair of projection matrices $\backslash Lambda\_\{+\}$ and $\backslash Lambda\_\{-\}$, that project out the positive and negative energy eigenstates. Given a fixed Lorentz coordinate frame (i.e. a fixed momentum), these are $$\backslash begin\{align\}$$

 \Lambda_{+}(p) = \sum_{s=1,2}{u^{(s)}_p \otimes \bar{u}^{(s)}_p} &= \frac
 \begin{bmatrix}
   \phi\\
   \frac{\vec{\sigma} \cdot \vec{p}}{E + m} \phi
 \end{bmatrix}
  e^{-ip\cdot x}
     

is carried to its charge conjugate $$\backslash psi^\{(+)\}\_c$$

 \begin{bmatrix}
   i\sigma_2 \frac{\vec{\sigma}^* \cdot \vec{p}}{E + m} \phi^*\\
   -i\sigma_2 \phi^*
 \end{bmatrix}
  e^{ip\cdot x}
     

Note the stray complex conjugates. These can be consolidated with the identity $$\backslash sigma\_2\; \backslash left(\backslash vec\backslash sigma^*\; \backslash cdot\; \backslash vec\; k\backslash right)\; \backslash sigma\_2\; =\; -\; \backslash vec\backslash sigma\backslash cdot\backslash vec\; k$$ to obtain $$\backslash psi^\{(+)\}\_c$$

 \begin{bmatrix}
  \frac{\vec{\sigma} \cdot \vec{p}}{E + m} \chi \\
   \chi
 \end{bmatrix}
  e^{ip\cdot x}
     

with the 2-spinor being $$\backslash chi\; =\; -i\backslash sigma\_2\; \backslash phi^*$$ As this has precisely the form of the negative energy solution, it becomes clear that charge conjugation exchanges the particle and anti-particle solutions. Note that not only is the energy reversed, but the momentum is reversed as well. Spin-up is transmuted to spin-down. It can be shown that the parity is also flipped. Charge conjugation is very much a pairing of Dirac spinor to its "exact opposite".

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Chiral basis In the chiral (or Weyl) representation of $\backslash gamma^\backslash mu$, the solution space for the Dirac equation can be parameterized by two‐component complex spinors $\backslash xi\_s$ and $\backslash eta\_s$. The general Dirac spinor solutions in this representation are often written as

Michael Edward Peskin (2025). 9780367320560, CRC Press, Taylor & Francis Group. ISBN 9780367320560

Matthew Dean Schwartz (2025). 9781107034730, Cambridge university press. ISBN 9781107034730

$u\_s(p)\; =\; \backslash begin\{pmatrix\}\backslash sqrt\{p\; \backslash cdot\; \backslash sigma\}\backslash ,\backslash xi\_s\backslash \backslash \; \backslash sqrt\{p\; \backslash cdot\; \backslash bar\backslash sigma\}\backslash ,\backslash xi\_s\backslash end\{pmatrix\},\; \backslash quad\; \backslash quad\; v\_s(p)\; =\; \backslash begin\{pmatrix\}\backslash sqrt\{p\; \backslash cdot\; \backslash sigma\}\backslash ,\backslash eta\_s\backslash \backslash \; -\backslash sqrt\{p\; \backslash cdot\; \backslash bar\backslash sigma\}\backslash ,\backslash eta\_s\backslash end\{pmatrix\},$

where $\backslash sigma^\backslash mu\; =\; (I\_2,\; \backslash sigma^i),~\; \backslash bar\backslash sigma^\backslash mu\; =\; (I\_2,\; -\backslash sigma^i)$ are Pauli 4-vectors and $\backslash sqrt\{\backslash cdot\}$ denotes the Hermitian matrix square-root. In many practical calculations, it is convenient to choose $\backslash mathbf\{p\}$ along the $z$ axis. With this choice, the contractions read as

$p\; \backslash cdot\; \backslash sigma\; \backslash equiv\; p\_\{\backslash mu\}\; \backslash sigma^\{\backslash mu\}\; =$

\begin{pmatrix}
  E - p_z & 0 \\
  0 & E + p_z
\end{pmatrix}, \quad
     

p \cdot \bar \sigma \equiv p_{\mu} \bar\sigma^{\mu} =

\begin{pmatrix}
  E + p_z & 0 \\
  0 & E - p_z
\end{pmatrix}.
     

Since the matrices are diagonal, their square roots are

$\backslash sqrt\{p\_\{\backslash mu\}\; \backslash sigma^\{\backslash mu\}\}\; =$

\begin{pmatrix}
  \sqrt{E - p_z} & 0 \\
  0 & \sqrt{E + p_z }
\end{pmatrix}, \quad
     

\sqrt{p_{\mu} \bar\sigma^{\mu}} =

\begin{pmatrix}
  \sqrt{E + p_z} & 0 \\
  0 & \sqrt{E - p_z}
\end{pmatrix}.
     

The most convenient choice for the two‐component spinors is:

$\backslash xi\_\{+\backslash frac\{1\}\{2\}\}\; =\; \backslash begin\{bmatrix\}\; 1\; \backslash \backslash \; 0\; \backslash end\{bmatrix\}\; =\; \backslash eta\_\{+\backslash frac\{1\}\{2\}\}\; \backslash quad\; \backslash quad$

 \xi_{-\frac{1}{2}} = \begin{bmatrix} 0 \\ 1 \end{bmatrix} = \eta_{-\frac{1}{2}}.
     

Then the four independent solutions take the explicit forms

$u\_\{+\backslash frac\{1\}\{2\}\}(p\_z)\; =\; \backslash begin\{pmatrix\}\backslash sqrt\{E\; -\; p\_z\}\backslash ,\backslash \backslash \; 0\; \backslash \backslash \; \backslash sqrt\{E+p\_z\}\backslash ,\; \backslash \backslash \; 0\; \backslash end\{pmatrix\},\; \backslash quad\; u\_\{-\backslash frac\{1\}\{2\}\}(p\_z)\; =\; \backslash begin\{pmatrix\}0\backslash \backslash \; \backslash sqrt\{E+p\_z\}\backslash ,\backslash \backslash \; 0\; \backslash \backslash \; \backslash sqrt\{E\; -\; p\_z\}\backslash ,\; \backslash end\{pmatrix\},$

$v\_\{+\backslash frac\{1\}\{2\}\}(p\_z)\; =\; \backslash begin\{pmatrix\}\backslash sqrt\{E-p\_z\}\backslash ,\backslash \backslash \; 0\; \backslash \backslash \; -\backslash sqrt\{E+p\_z\}\backslash ,\; \backslash \backslash \; 0\; \backslash end\{pmatrix\},\; \backslash quad\; v\_\{-\backslash frac\{1\}\{2\}\}(p\_z)\; =\; \backslash begin\{pmatrix\}0\backslash \backslash \; \backslash sqrt\{E+p\_z\}\backslash ,\; \backslash \backslash \; 0\; \backslash \backslash \; -\backslash sqrt\{E-p\_z\}\backslash ,\; \backslash end\{pmatrix\}.$

The conventional normalization conditions for these spinors are

$\backslash begin\{align\}$

    \overline u_s (p) u_{s'} (p) &=  2m\delta_{ss'} & \overline u_s (p) v_{s'} (p) &=  0 \\
    \overline v_s (p) v_{s'} (p) &=  -2m\delta_{ss'} & \overline v_s (p) u_{s'} (p) &=  0
 \end{align}
     

while the completeness (spin‐sum) relations are

$\backslash begin\{align\}$

    \textstyle \sum_s \displaystyle u_s (p) \overline u_{s} (p) &= {p\!\!\!/} + m \\
    \textstyle \sum_s \displaystyle v_s (p) \overline v_{s} (p) &= {p\!\!\!/} - m.
 \end{align}
     

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See also

• Dirac equation
• Weyl equation
• Majorana equation
• Helicity basis
• Spin(1,3), the double cover of SO(1,3) by a spin group

• I.J.R. Aitchison (2002). 9780750308649, Institute of Physics Publishing. ISBN 9780750308649

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