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,

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

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

${\displaystyle \psi }$ is a relativistic spin-1/2 field,

${\displaystyle \omega _{\vec {p}}}$ is the Dirac spinor related to a plane-wave with wave-vector ${\displaystyle {\vec {p}}}$,

${\displaystyle px\;\equiv \;p_{\mu }x^{\mu }\;\equiv \;Et-{\vec {p}}\cdot {\vec {x}}}$,

${\displaystyle p^{\mu }\;=\;\left\{\pm {\sqrt {m^{2}+{\vec {p}}^{2}}},\,{\vec {p}}\right\}}$ is the four-wave-vector of the plane wave, where ${\displaystyle {\vec {p}}}$ is arbitrary,

${\displaystyle x^{\mu }}$ 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

${\displaystyle \phi }$ is an arbitrary two-spinor,

${\displaystyle {\vec {\sigma }}}$ are the Pauli matrices,

${\displaystyle E_{\vec {p}}}$ is the positive square root ${\displaystyle E_{\vec {p}}\;=\;+{\sqrt {m^{2}+{\vec {p}}^{2}}}}$

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 ${\displaystyle {\vec {p}}={\vec {0}}}$.

Derivation of the sign of the energy in this case

We consider the potentially problematic term ${\displaystyle {\frac {{\vec {\sigma }}\cdot {\vec {p}}}{E+m}}\phi }$.

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

${\displaystyle E=-m{\sqrt {1+{\frac {p^{2}}{m^{2}}}}}\rightarrow -m(1+{\frac {1}{2}}{\frac {p^{2}}{m^{2}}})}$

${\displaystyle {\frac {{\vec {\sigma }}\cdot {\vec {p}}}{E+m}}\rightarrow p{\frac {{\vec {\sigma }}\cdot {\hat {n}}}{-m-{\frac {p^{2}}{2m}}+m}}\propto {\frac {1}{p}}\rightarrow \infty }$

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

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,

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

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.

See also

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

References

• Aitchison, I.J.R.; A.J.G. Hey (September 2002). Gauge Theories in Particle Physics (3rd ed.). Institute of Physics Publishing. ISBN 0-7503-0864-8.
• Miller, David (2008). "Relativistic Quantum Mechanics (RQM)" (PDF). pp. 26–37.