# Dirac spinor

In quantum field theory, the Dirac spinor is the bispinor in the plane-wave solution

${\displaystyle \psi =\omega _{\vec {p}}\;e^{-ipx}\;}$

of the free Dirac equation,

${\displaystyle \left(i\gamma ^{\mu }\partial _{\mu }-m\right)\psi =0\;,}$

where (in the units ${\displaystyle \scriptstyle c\,=\,\hbar \,=\,1}$)

${\displaystyle \scriptstyle \psi }$ is a relativistic spin-1/2 field,
${\displaystyle \scriptstyle \omega _{\vec {p}}}$ is the Dirac spinor related to a plane-wave with wave-vector ${\displaystyle \scriptstyle {\vec {p}}}$,
${\displaystyle \scriptstyle px\;\equiv \;p_{\mu }x^{\mu }\;\equiv \;Et-{\vec {p}}\cdot {\vec {x}}}$,
${\displaystyle \scriptstyle p^{\mu }\;=\;\left\{\pm {\sqrt {m^{2}+{\vec {p}}^{2}}},\,{\vec {p}}\right\}}$ is the four-wave-vector of the plane wave, where ${\displaystyle \scriptstyle {\vec {p}}}$ is arbitrary,
${\displaystyle \scriptstyle x^{\mu }}$ are the four-coordinates in a given inertial frame of reference.

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

${\displaystyle \omega _{\vec {p}}={\begin{bmatrix}\phi \\{\frac {{\vec {\sigma }}\cdot {\vec {p}}}{E_{\vec {p}}+m}}\phi \end{bmatrix}}\;,}$

where

${\displaystyle \scriptstyle \phi }$ is an arbitrary two-spinor,
${\displaystyle \scriptstyle {\vec {\sigma }}}$ are the Pauli matrices,
${\displaystyle \scriptstyle E_{\vec {p}}}$ is the positive square root ${\displaystyle \scriptstyle E_{\vec {p}}\;=\;+{\sqrt {m^{2}+{\vec {p}}^{2}}}}$

## Derivation from Dirac equation

The Dirac equation has the form

${\displaystyle \left(-i{\vec {\alpha }}\cdot {\vec {\nabla }}+\beta m\right)\psi =i{\frac {\partial \psi }{\partial t}}\,}$

In order to derive the form of the four-spinor ${\displaystyle \scriptstyle \omega }$  we have to first note the value of the matrices α and β:

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

These two 4×4 matrices are related to the Dirac gamma matrices. Note that 0 and I are 2×2 matrices here.

The next step is to look for solutions of the form

${\displaystyle \psi =\omega e^{-ip\cdot x}}$ ,

while at the same time splitting ω into two two-spinors:

${\displaystyle \omega ={\begin{bmatrix}\phi \\\chi \end{bmatrix}}\,}$ .

### Results

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

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

This matrix equation is really two coupled equations:

{\displaystyle {\begin{aligned}\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{aligned}}}

Solve the 2nd equation for ${\displaystyle \scriptstyle \chi \,}$  and one obtains

${\displaystyle \omega ={\begin{bmatrix}\phi \\{\frac {{\vec {\sigma }}\cdot {\vec {p}}}{E+m}}\phi \end{bmatrix}}\,}$ .

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

Alternatively, solve the 1st equation for ${\displaystyle \phi \,}$  and one finds

${\displaystyle \omega ={\begin{bmatrix}-{\frac {{\vec {\sigma }}\cdot {\vec {p}}}{-E+m}}\chi \\\chi \end{bmatrix}}\,}$ .

In this case one needs to enforce that ${\displaystyle E=-{\sqrt {p^{2}+m^{2}}}}$  for this solution to be valid in a frame where the particle has ${\displaystyle {\vec {p}}={\vec {0}}}$ . This can be shown analogously to the previous case.

This solution is useful for showing the relation between anti-particle and particle.

## Details

### Two-spinors

The most convenient definitions for the two-spinors are:

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

and

${\displaystyle \chi ^{1}={\begin{bmatrix}0\\1\end{bmatrix}}\quad \quad \chi ^{2}={\begin{bmatrix}1\\0\end{bmatrix}}\,}$

### Pauli matrices

The Pauli matrices are

${\displaystyle \sigma _{1}={\begin{bmatrix}0&1\\1&0\end{bmatrix}}\quad \quad \sigma _{2}={\begin{bmatrix}0&-i\\i&0\end{bmatrix}}\quad \quad \sigma _{3}={\begin{bmatrix}1&0\\0&-1\end{bmatrix}}}$

Using these, one can calculate:

${\displaystyle {\vec {\sigma }}\cdot {\vec {p}}=\sigma _{1}p_{1}+\sigma _{2}p_{2}+\sigma _{3}p_{3}={\begin{bmatrix}p_{3}&p_{1}-ip_{2}\\p_{1}+ip_{2}&-p_{3}\end{bmatrix}}}$

## Four-spinors

### For particles

Particles are defined as having positive energy. The normalization for the four-spinor ω is chosen so that the total probability is invariant under Lorentz transformation. The total probability is:

${\displaystyle P=\int _{V}\omega ^{\dagger }\omega dV}$

where ${\displaystyle V}$  is the volume of integration. Under Lorentz transformation, the volume scales as the inverse of Lorentz factor: ${\displaystyle (E/m)^{-1}}$ . This implies that the probability density must be normalized proportional to ${\displaystyle E}$  so the total probability is Lorentz invariant. The usual convention is to choose ${\displaystyle \omega ^{\dagger }\omega \;=\;2E\,}$ . Hence the spinors, denoted as u are:

${\displaystyle u\left({\vec {p}},s\right)={\sqrt {E+m}}{\begin{bmatrix}\phi ^{(s)}\\{\frac {{\vec {\sigma }}\cdot {\vec {p}}}{E+m}}\phi ^{(s)}\end{bmatrix}}\,}$

where s = 1 or 2 (spin "up" or "down")

Explicitly,

${\displaystyle u\left({\vec {p}},1\right)={\sqrt {E+m}}{\begin{bmatrix}1\\0\\{\frac {p_{3}}{E+m}}\\{\frac {p_{1}+ip_{2}}{E+m}}\end{bmatrix}}\quad \mathrm {and} \quad u\left({\vec {p}},2\right)={\sqrt {E+m}}{\begin{bmatrix}0\\1\\{\frac {p_{1}-ip_{2}}{E+m}}\\{\frac {-p_{3}}{E+m}}\end{bmatrix}}}$

### For anti-particles

Anti-particles having positive energy ${\displaystyle \scriptstyle E}$  are defined as particles having negative energy and propagating backward in time. Hence changing the sign of ${\displaystyle \scriptstyle E}$  and ${\displaystyle \scriptstyle {\vec {p}}}$  in the four-spinor for particles will give the four-spinor for anti-particles:

${\displaystyle v({\vec {p}},s)={\sqrt {E+m}}{\begin{bmatrix}{\frac {{\vec {\sigma }}\cdot {\vec {p}}}{E+m}}\chi ^{(s)}\\\chi ^{(s)}\end{bmatrix}}}$

Here we choose the ${\displaystyle \scriptstyle \chi }$  solutions. Explicitly,

${\displaystyle v\left({\vec {p}},1\right)={\sqrt {E+m}}{\begin{bmatrix}{\frac {p_{3}}{E+m}}\\{\frac {p_{1}+ip_{2}}{E+m}}\\1\\0\\\end{bmatrix}}\quad \mathrm {and} \quad v({\vec {p}},2)={\sqrt {E+m}}{\begin{bmatrix}{\frac {p_{1}-ip_{2}}{E+m}}\\{\frac {-p_{3}}{E+m}}\\0\\1\end{bmatrix}}}$

Note that these solutions are readily obtained by substituting the ansatz ${\displaystyle \psi =ve^{+ipx}}$  into the Dirac equation.

## Completeness relations

The completeness relations for the four-spinors u and v are

{\displaystyle {\begin{aligned}\sum _{s=1,2}{u_{p}^{(s)}{\bar {u}}_{p}^{(s)}}&={p\!\!\!/}+m\\\sum _{s=1,2}{v_{p}^{(s)}{\bar {v}}_{p}^{(s)}}&={p\!\!\!/}-m\end{aligned}}}

where

${\displaystyle {p\!\!\!/}=\gamma ^{\mu }p_{\mu }\,}$       (see Feynman slash notation)
${\displaystyle {\bar {u}}=u^{\dagger }\gamma ^{0}\,}$

## Dirac spinors and the Dirac algebra

The Dirac matrices are a set of four 4×4 matrices that are used as spin and charge operators.

### Conventions

There are several choices of signature and representation that are in common use in the physics literature. The Dirac matrices are typically written as ${\displaystyle \scriptstyle \gamma ^{\mu }}$  where ${\displaystyle \scriptstyle \mu }$  runs from 0 to 3. In this notation, 0 corresponds to time, and 1 through 3 correspond to x, y, and z.

The + − − − signature is sometimes called the west coast metric, while the − + + + is the east coast metric. At this time the + − − − signature is in more common use, and our example will use this signature. To switch from one example to the other, multiply all ${\displaystyle \scriptstyle \gamma ^{\mu }}$  by ${\displaystyle \scriptstyle i}$ .

After choosing the signature, there are many ways of constructing a representation in the 4×4 matrices, and many are in common use. In order to make this example as general as possible we will not specify a representation until the final step. At that time we will substitute in the "chiral" or "Weyl" representation.

### Construction of Dirac spinor with a given spin direction and charge

First we choose a spin direction for our electron or positron. As with the example of the Pauli algebra discussed above, the spin direction is defined by a unit vector in 3 dimensions, (a, b, c). Following the convention of Peskin & Schroeder, the spin operator for spin in the (a, b, c) direction is defined as the dot product of (a, b, c) with the vector

{\displaystyle {\begin{aligned}(i\gamma ^{2}\gamma ^{3},\;\;i\gamma ^{3}\gamma ^{1},\;\;i\gamma ^{1}\gamma ^{2})&=-(\gamma ^{1},\;\gamma ^{2},\;\gamma ^{3})i\gamma ^{1}\gamma ^{2}\gamma ^{3}\\\sigma _{(a,b,c)}&=ia\gamma ^{2}\gamma ^{3}+ib\gamma ^{3}\gamma ^{1}+ic\gamma ^{1}\gamma ^{2}\end{aligned}}}

Note that the above is a root of unity, that is, it squares to 1. Consequently, we can make a projection operator from it that projects out the sub-algebra of the Dirac algebra that has spin oriented in the (a, b, c) direction:

${\displaystyle P_{(a,b,c)}={\tfrac {1}{2}}\left(1+\sigma _{(a,b,c)}\right)}$

Now we must choose a charge, +1 (positron) or −1 (electron). Following the conventions of Peskin & Schroeder, the operator for charge is ${\displaystyle \scriptstyle Q\,=\,-\gamma ^{0}}$ , that is, electron states will take an eigenvalue of −1 with respect to this operator while positron states will take an eigenvalue of +1.

Note that ${\displaystyle \scriptstyle Q}$  is also a square root of unity. Furthermore, ${\displaystyle \scriptstyle Q}$  commutes with ${\displaystyle \scriptstyle \sigma _{(a,b,c)}}$ . They form a complete set of commuting operators for the Dirac algebra. Continuing with our example, we look for a representation of an electron with spin in the (a, b, c) direction. Turning ${\displaystyle \scriptstyle Q}$  into a projection operator for charge = −1, we have

${\displaystyle P_{-Q}={\frac {1}{2}}\left(1-Q\right)={\frac {1}{2}}\left(1+\gamma ^{0}\right)}$

The projection operator for the spinor we seek is therefore the product of the two projection operators we've found:

${\displaystyle P_{(a,b,c)}\;P_{-Q}}$

The above projection operator, when applied to any spinor, will give that part of the spinor that corresponds to the electron state we seek. So we can apply it to a spinor with the value 1 in one of its components, and 0 in the others, which gives a column of the matrix. Continuing the example, we put (a, b, c) = (0, 0, 1) and have

${\displaystyle P_{(0,0,1)}={\frac {1}{2}}\left(1+i\gamma _{1}\gamma _{2}\right)}$

and so our desired projection operator is

${\displaystyle P={\frac {1}{2}}\left(1+i\gamma ^{1}\gamma ^{2}\right)\cdot {\frac {1}{2}}\left(1+\gamma ^{0}\right)={\frac {1}{4}}\left(1+\gamma ^{0}+i\gamma ^{1}\gamma ^{2}+i\gamma ^{0}\gamma ^{1}\gamma ^{2}\right)}$

The 4×4 gamma matrices used in the Weyl representation are

{\displaystyle {\begin{aligned}\gamma _{0}&={\begin{bmatrix}0&1\\1&0\end{bmatrix}}\\\gamma _{k}&={\begin{bmatrix}0&\sigma ^{k}\\-\sigma ^{k}&0\end{bmatrix}}\end{aligned}}}

for k = 1, 2, 3 and where ${\displaystyle \sigma ^{i}}$  are the usual 2×2 Pauli matrices. Substituting these in for P gives

${\displaystyle P={\frac {1}{4}}{\begin{bmatrix}1+\sigma ^{3}&1+\sigma ^{3}\\1+\sigma ^{3}&1+\sigma ^{3}\end{bmatrix}}={\frac {1}{2}}{\begin{bmatrix}1&0&1&0\\0&0&0&0\\1&0&1&0\\0&0&0&0\end{bmatrix}}}$

Our answer is any non-zero column of the above matrix. The division by two is just a normalization. The first and third columns give the same result:

${\displaystyle \left|e^{-},\,+{\frac {1}{2}}\right\rangle ={\begin{bmatrix}1\\0\\1\\0\end{bmatrix}}}$

More generally, for electrons and positrons with spin oriented in the (a, b, c) direction, the projection operator is

${\displaystyle {\frac {1}{4}}{\begin{bmatrix}1+c&a-ib&\pm (1+c)&\pm (a-ib)\\a+ib&1-c&\pm (a+ib)&\pm (1-c)\\\pm (1+c)&\pm (a-ib)&1+c&a-ib\\\pm (a+ib)&\pm (1-c)&a+ib&1-c\end{bmatrix}}}$

where the upper signs are for the electron and the lower signs are for the positron. The corresponding spinor can be taken as any non zero column. Since ${\displaystyle \scriptstyle a^{2}+b^{2}+c^{2}\,=\,1}$  the different columns are multiples of the same spinor. The representation of the resulting spinor in the Dirac basis can be obtained using the rule given in the bispinor article.