# Dirac spinor

Jump to navigation Jump to search

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

$\psi =\omega _{\vec {p}}\;e^{-ipx}\;$ of the free Dirac equation,

$\left(i\gamma ^{\mu }\partial _{\mu }-m\right)\psi =0\;,$ where (in the units $c\,=\,\hbar \,=\,1$ )

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

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

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

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

The Dirac equation has the form

$\left(-i{\vec {\alpha }}\cdot {\vec {\nabla }}+\beta m\right)\psi =i{\frac {\partial \psi }{\partial t}}\,$ In order to derive the form of the four-spinor $\omega$ we have to first note the value of the matrices α and β:

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

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

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

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

### 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:

{\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 $\chi \,$ and one obtains

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

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

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

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

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

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

$\chi ^{1}={\begin{bmatrix}0\\1\end{bmatrix}}\quad \quad \chi ^{2}={\begin{bmatrix}1\\0\end{bmatrix}}\,$ ### 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 can calculate:

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

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

$u\left({\vec {p}},s\right)={\sqrt {E+m}}{\begin{bmatrix}\phi ^{(s)}\\{\frac {{\vec {\sigma }}\cdot {\vec {p}}}{E+m}}\phi ^{(s)}\end{bmatrix}}\,$ where s = 1 or 2 (spin "up" or "down")

Explicitly,

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

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

$v({\vec {p}},s)={\sqrt {E+m}}{\begin{bmatrix}{\frac {{\vec {\sigma }}\cdot {\vec {p}}}{E+m}}\chi ^{(s)}\\\chi ^{(s)}\end{bmatrix}}$ Here we choose the $\chi$ solutions. Explicitly,

$v\left({\vec {p}},1\right)={\sqrt {E+m}}{\begin{bmatrix}{\frac {p_{3}}{E+m}}\\{\frac {p_{1}+ip_{2}}{E+m}}\\1\\0\\\end{bmatrix}}\quad \mathrm {and} \quad v({\vec {p}},2)={\sqrt {E+m}}{\begin{bmatrix}{\frac {p_{1}-ip_{2}}{E+m}}\\{\frac {-p_{3}}{E+m}}\\0\\1\end{bmatrix}}$ Note that these solutions are readily obtained by substituting the ansatz $\psi =ve^{+ipx}$ into the Dirac equation.

## Completeness relations

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

{\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

${p\!\!\!/}=\gamma ^{\mu }p_{\mu }\,$ (see Feynman slash notation)
${\bar {u}}=u^{\dagger }\gamma ^{0}\,$ ## 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 $\gamma ^{\mu }$ where $\mu$ 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 $\gamma ^{\mu }$ by $i$ .

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

{\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:

$P_{(a,b,c)}={\tfrac {1}{2}}\left(1+\sigma _{(a,b,c)}\right)$ Now we must choose a charge, +1 (positron) or −1 (electron). Following the conventions of Peskin & Schroeder, the operator for charge is $Q\,=\,-\gamma ^{0}$ , 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 $Q$ is also a square root of unity. Furthermore, $Q$ commutes with $\sigma _{(a,b,c)}$ . 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 $Q$ into a projection operator for charge = −1, we have

$P_{-Q}={\frac {1}{2}}\left(1-Q\right)={\frac {1}{2}}\left(1+\gamma ^{0}\right)$ The projection operator for the spinor we seek is therefore the product of the two projection operators we've found:

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

$P_{(0,0,1)}={\frac {1}{2}}\left(1+i\gamma _{1}\gamma _{2}\right)$ and so our desired projection operator is

$P={\frac {1}{2}}\left(1+i\gamma ^{1}\gamma ^{2}\right)\cdot {\frac {1}{2}}\left(1+\gamma ^{0}\right)={\frac {1}{4}}\left(1+\gamma ^{0}+i\gamma ^{1}\gamma ^{2}+i\gamma ^{0}\gamma ^{1}\gamma ^{2}\right)$ The 4×4 gamma matrices used in the Weyl representation are

{\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 $\sigma ^{i}$ are the usual 2×2 Pauli matrices. Substituting these in for P gives

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

$\left|e^{-},\,+{\frac {1}{2}}\right\rangle ={\begin{bmatrix}1\\0\\1\\0\end{bmatrix}}$ More generally, for electrons and positrons with spin oriented in the (a, b, c) direction, the projection operator is

${\frac {1}{4}}{\begin{bmatrix}1+c&a-ib&\pm (1+c)&\pm (a-ib)\\a+ib&1-c&\pm (a+ib)&\pm (1-c)\\\pm (1+c)&\pm (a-ib)&1+c&a-ib\\\pm (a+ib)&\pm (1-c)&a+ib&1-c\end{bmatrix}}$ 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 $a^{2}+b^{2}+c^{2}\,=\,1$ 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.