Gamma matrices

From Wikipedia, the free encyclopedia

The gamma matrices, also known as Dirac matrices, were developed by P.A.M. Dirac in order to serve as coefficients of the Dirac equation. The Dirac matrices form a four-vector which is represented as:

$\gamma^\mu = \left(\gamma^0, \gamma^1, \gamma^2, \gamma^3 \right) \,$

$\gamma_\mu = \left(\gamma^0, -\gamma^1, -\gamma^2, -\gamma^3 \right). \,$

And one possible representation of the four contravariant gamma matrices is

$\gamma^0 = \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & -1 \end{pmatrix}, \gamma^1 \!=\! \begin{pmatrix} 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \\ 0 & -1 & 0 & 0 \\ -1 & 0 & 0 & 0 \end{pmatrix}$

$\gamma^2 \!=\! \begin{pmatrix} 0 & 0 & 0 & -i \\ 0 & 0 & i & 0 \\ 0 & i & 0 & 0 \\ -i & 0 & 0 & 0 \end{pmatrix}, \gamma^3 \!=\! \begin{pmatrix} 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & -1 \\ -1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \end{pmatrix}.$

1 Mathematical structure
2 Identities
3 Physical structure
4 Expressing the Dirac equation
5 The Fifth Gamma Matrix, γ5
6 Other representations
7 Euclidean Dirac matrices
- 7.1 Chiral representation
- 7.2 Non-relativistic representation
8 See also
9 References

[edit] Mathematical structure

The gamma matrices form a Clifford algebra whose defining property is the anticommutation relation:

$\displaystyle\{ \gamma^\mu, \gamma^\nu \} = \gamma^\mu \gamma^\nu + \gamma^\nu \gamma^\mu = 2 \eta^{\mu \nu} I$

where ,

$\eta \,$ is the Minkowski metric with signature (+ − − −), and

$I \,$ is the 4x4 identity matrix.

This defining property is considered to be more fundamental than the numerical values used in the gamma matrices, so other sign conventions for the metric necessitate a change in the definitions of the gamma matrices.

Covariant gamma matrices are defined by

$\displaystyle \gamma_\mu = \eta_{\mu \nu} \gamma^\nu$ ,

and Einstein notation is assumed.

[edit] Identities

The following identities follow from the fundamental anticommutation relation, so they hold in any basis.

[edit] Trace

Num	Identity
1	trace of any product of an odd number of $γ μ$ is zero
2	$\operatorname{tr} (\gamma^\mu\gamma^\nu) = 4\eta^{\mu\nu}$
3	$\operatorname{tr}(\gamma^\mu\gamma^\nu\gamma^\rho\gamma^\sigma)=4(\eta^{\mu\nu}\eta^{\rho\sigma}-\eta^{\mu\rho}\eta^{\nu\sigma}+\eta^{\mu\sigma}\eta^{\nu\rho})$
4	$\operatorname{tr}(\gamma^5)=\operatorname{tr} (\gamma^\mu\gamma^\nu\gamma^5) = 0$
5	$\operatorname{tr} (\gamma^\mu\gamma^\nu\gamma^\rho\gamma^\sigma\gamma^5) = -4i\epsilon^{\mu\nu\rho\sigma}$

Proving the above involves use of three main properties of the Trace operator:

tr(A + B) = tr(A) + tr(B)
tr(rA) = r tr(A)
tr(ABC) = tr(CAB) = tr(BCA).

Proofs of 1 & 2

To show

$\operatorname{tr} (\gamma^\mu\gamma^\nu) = 4\eta^{\mu\nu}$

Begin with,

$\operatorname{tr} (\gamma^\mu\gamma^\nu) \,$	$= \frac{1}{2} \left(\operatorname{tr} (\gamma^\mu\gamma^\nu) + \operatorname{tr} (\gamma^\nu\gamma^\mu) \right) \,$
	$= \frac{1}{2} \operatorname{tr} (\gamma^\mu\gamma^\nu + \gamma^\nu\gamma^\mu) = \frac{1}{2} \operatorname{tr} \left( \{\gamma^\mu, \gamma^\nu\} \right) \,$
	$= \frac{1}{2} 2 \eta^{\mu \nu} \operatorname{tr} (I) = 4 \eta^{\mu \nu} \,$

Proof of 3

$\operatorname{tr} (\gamma^\mu \gamma^\nu \gamma^\rho \gamma^\sigma) \,$	$= \operatorname{tr} \left(\gamma^\mu \gamma^\nu (2\eta^{\rho \sigma} - \gamma^\sigma \gamma^\rho ) \right) \,$
	$= 2 \eta^{\rho \sigma} \operatorname{tr} \left(\gamma^\mu \gamma^\nu \right) - \operatorname{tr} \left( \gamma^\mu \gamma^\nu \gamma^\sigma \gamma^\rho \right) \quad \quad (1) \,$

For the term on the right, we'll continue the pattern of swapping $\gamma^\sigma \,$ with its neighbor to the left,

$\operatorname{tr} \left( \gamma^\mu \gamma^\nu \gamma^\sigma \gamma^\rho \right) \,$	$= \operatorname{tr} \left(\gamma^\mu (2 \eta^{\nu \sigma} - \gamma^\sigma \gamma^\nu ) \gamma^\rho \right) \,$
	$= 2 \eta^{\nu \sigma} \operatorname{tr} \left(\gamma^\mu \gamma^\rho \right) - \operatorname{tr} \left(\gamma^\mu \gamma^\sigma \gamma^\nu \gamma^\rho \right) \quad \quad (2) \,$

Again, for the term on the right swap $\gamma^\sigma \,$ with its neighbor to the left,

$\operatorname{tr} \left( \gamma^\mu \gamma^\sigma \gamma^\nu \gamma^\rho \right) \,$	$= \operatorname{tr} \left( (2 \eta^{\mu \sigma} - \gamma^\sigma \gamma^\mu ) \gamma^\nu \gamma^\rho \right) \,$
	$= 2 \eta^{\mu \sigma} \operatorname{tr} \left( \gamma^\nu \gamma^\rho \right) - \operatorname{tr} \left( \gamma^\sigma \gamma^\mu \gamma^\nu \gamma^\rho \right)\quad \quad (3) \,$

Eq (3) is the term on the right of eq (2), and eq (2) is the term on the right of eq (1). We'll also use identity number 2 to simplify terms like so:

$2 \eta^{\rho \sigma} \operatorname{tr} \left(\gamma^\mu \gamma^\nu \right) = 2 \eta^{\rho \sigma} (4 \eta^{\mu \nu}) = 8 \eta^{\rho \sigma} \eta^{\mu \nu} .\,$

So finally Eq (1), when you plug all this in information in gives

$\operatorname{tr} (\gamma^\mu \gamma^\nu \gamma^\rho \gamma^\sigma) = 8 \eta^{\rho \sigma} \eta^{\mu \nu} - 8 \eta^{\nu \sigma} \eta^{\mu \rho} + 8 \eta^{\mu \sigma} \eta^{\nu \rho} \,$

$- \ \operatorname{tr} \left( \gamma^\sigma \gamma^\mu \gamma^\nu \gamma^\rho \right) \quad \quad \quad \quad \quad \quad \quad (4) \,$

The terms inside the trace can be cycled, so

$\operatorname{tr} \left( \gamma^\sigma \gamma^\mu \gamma^\nu \gamma^\rho \right) = \operatorname{tr} (\gamma^\mu \gamma^\nu \gamma^\rho \gamma^\sigma). \,$

So really (4) is

$2 \ \operatorname{tr} (\gamma^\mu \gamma^\nu \gamma^\rho \gamma^\sigma) = 8 \eta^{\rho \sigma} \eta^{\mu \nu} - 8 \eta^{\nu \sigma} \eta^{\mu \rho} + 8 \eta^{\mu \sigma} \eta^{\nu \rho} \,$

$\operatorname{tr} (\gamma^\mu \gamma^\nu \gamma^\rho \gamma^\sigma) = 4 \left( \eta^{\rho \sigma} \eta^{\mu \nu} - \eta^{\nu \sigma} \eta^{\mu \rho} + \eta^{\mu \sigma} \eta^{\nu \rho} \right) \,$

Proofs of 4 & 5

[edit] Other

Num	Identity
1	$\displaystyle\gamma^\mu\gamma_\mu=4$
2	$\displaystyle\gamma^\mu\gamma^\nu\gamma_\mu=-2\gamma^\nu$
3	$\displaystyle\gamma^\mu\gamma^\nu\gamma^\rho\gamma_\mu=4\eta^{\nu\rho}$
4	$\displaystyle\gamma^\mu\gamma^\nu\gamma^\rho\gamma^\sigma\gamma_\mu=-2\gamma^\sigma\gamma^\rho\gamma^\nu$

Proofs of 1 & 2

To show

$\displaystyle\gamma^\mu\gamma_\mu=4$

one begins with the standard commutation relation

$\displaystyle\{ \gamma^\mu, \gamma^\nu \} = \gamma^\mu \gamma^\nu + \gamma^\nu \gamma^\mu = 2 \eta^{\mu \nu} I.$

One can make this situation look similar by using the metric $η$ :

$\gamma^\mu \gamma_\mu \,$	$= \gamma^\mu \eta_{\mu \nu} \gamma^\nu = \eta_{\mu \nu} \gamma^\mu \gamma^\nu = \frac{1}{2} \eta_{\mu \nu} \{\gamma^\mu, \gamma^\nu \} \,$
	$= \frac{1}{2} \eta_{\mu \nu} \left(2 \eta^{\mu \nu} I \right) = \eta_{\mu \nu} \eta^{\mu \nu} I = 4 I. \,$

To show

$\gamma^\mu\gamma^\nu\gamma_\mu=-2\gamma^\nu . \,$

We again will use the standard commutation relation. So start:

$\gamma^\mu \gamma^\nu \gamma_\mu \,$	$= \gamma^\mu \left(2 \eta_\mu^\nu - \gamma_\mu \gamma^\nu \right) \,$
	$= 2 \gamma^\mu \eta_\mu^\nu - \gamma^\mu \gamma_\mu \gamma^\nu \,$
	$= 2 \gamma^\nu - 4 \gamma^\nu = -2 \gamma^\nu. \,$

Proofs of 3 & 4

To show

$\gamma^\mu \gamma^\nu \gamma^\rho \gamma_\mu = 4 \eta^{\nu\rho} . \,$

Use the anticommutator to shift $γ μ$ to the right

$\gamma^\mu \gamma^\nu \gamma^\rho \gamma_\mu \,$	$= \{\gamma^\mu,\gamma^\nu\} \gamma^\rho \gamma_\mu - \gamma^\nu \gamma^\mu \gamma^\rho \gamma_\mu \,$
	$= 2\ \eta^{\mu\nu} \gamma^\rho \gamma_\mu - \gamma^\nu \{\gamma^\mu,\gamma^\rho\} \gamma_\mu + \gamma^\nu \gamma^\rho \gamma^\mu \gamma_\mu . \,$

Using the relation $γ μ γ μ = 4$ we can contract the last two gammas, and get

$\gamma^\mu \gamma^\nu \gamma^\rho \gamma_\mu \,$	$= 2\ \gamma^\rho \gamma^\nu - \gamma^\nu 2 \eta^{\mu\rho} \gamma_\mu + 4\ \gamma^\nu \gamma^\rho \,$
	$= 2\ \gamma^\rho \gamma^\nu - 2\ \gamma^\nu \gamma^\rho + 4\ \gamma^\nu \gamma^\rho \,$
	$= 2\ (\gamma^\rho \gamma^\nu + \gamma^\nu \gamma^\rho) \,$
	$= 2\ \{\gamma^\rho, \gamma^\nu\}. \,$

Finally using the anticommutator identity, we get

$\gamma^\mu \gamma^\nu \gamma^\rho \gamma_\mu = 4\ \eta^{\rho \nu}. \,$

[edit] Feynman slash notation

The gamma matrices act like a vector, and their contraction with a vector acts like a scalar. So, it is common to write some of identities using Feynman slash notation, defined by

$a\!\!\!/ := \gamma^\mu a_\mu.$

For example, here are similar identities to the ones above, but involving slash notation:

$\operatorname{tr}(a\!\!\!/b\!\!\!/) = 4 a \cdot b$

$\operatorname{tr}(a\!\!\!/b\!\!\!/c\!\!\!/d\!\!\!/) = 4 \left[(a\cdot b)(c \cdot d) - (a \cdot c)(b \cdot d) + (a \cdot d)(b \cdot c) \right]$

$\operatorname{tr}(\gamma_5 a\!\!\!/b\!\!\!/c\!\!\!/d\!\!\!/) = 4 i \epsilon_{\mu \nu \lambda \sigma} a^\mu b^\nu c^\lambda d^\sigma$

$\gamma_\mu a\!\!\!/ \gamma^\mu = -2 a\!\!\!/$

$a\!\!\!/b\!\!\!/ = a \cdot b - 2i a_\mu S^{\mu\nu} b_\nu$

$a\!\!\!/a\!\!\!/ = a^2$

where

$\epsilon_{\mu \nu \lambda \sigma} \,$ is the Levi-Civita symbol and $S^{\mu\nu} = \frac{i}{4} [\gamma^\mu, \gamma^\nu]$ .

[edit] Physical structure

The 4-tuple $\displaystyle\gamma^\mu=(\gamma^0,\gamma^1,\gamma^2,\gamma^3)$ has the same numerical values regardless of reference frame, so it is not a vector but a scalar quantity with the trivial transformation $\gamma^\mu\to\gamma^\mu$ . On the other hand, if $\displaystyle\lambda$ is the spinor representation of an arbitrary Lorentz transformation $\displaystyle\Lambda$ , then we have the identity

$\displaystyle\gamma^\mu=\Lambda^\mu{}_\nu\lambda\gamma^\nu\lambda^{-1}$

and so $γ μ$ also transforms appropriately as an object with one contravariant 4-vector index and one covariant and one contravariant Dirac spinor index.

Given the above transformation properties of $γ μ$ , if $ψ$ is a Dirac spinor then the product $γ μ ψ$ transforms as if it were the product of a contravariant 4-vector with a Dirac spinor. In expressions involving spinors, then, it is often appropriate to treat $γ μ$ as if it were simply a vector.

There remains a key difference between $γ μ$ and any nonzero 4-vector: $γ μ$ does not point in any direction. More precisely, the only way to make a true vector from $γ μ$ is to contract its spinor indices, leaving a vector of traces

$\displaystyle tr(\gamma^\mu)= (0, 0, 0, 0)$

This property of the gamma matrices is essential for them to serve as coefficients in the Dirac equation.

[edit] Expressing the Dirac equation

In natural units, the Dirac equation may be written as

$(i \gamma^\mu \partial_\mu - m) \psi = 0$

where ψ is a Dirac spinor. Here, if $γ μ$ were an ordinary 4-vector, then it would pick out a preferred direction in spacetime, and the Dirac equation would not be Lorentz invariant.

Switching to Feynman notation, the Dirac equation is

$(i \not\!\;\partial - m) \psi = 0.$

Applying $-(i \not\!\;\partial + m)$ to both sides yields

$(\not\!\;\partial^2 + m^2) \psi = (\partial^2 + m^2) \psi = 0,$

which is the Klein-Gordon equation. Thus, as the notation suggests, the Dirac particle has mass m.

[edit] The Fifth Gamma Matrix, $γ 5$

It is useful to define the product of the four gamma matrices as follows:

$\gamma^5 := i\gamma^0\gamma^1\gamma^2\gamma^3 = \begin{pmatrix} 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \end{pmatrix}$ (in the Dirac basis).

Although $γ 5$ uses the letter gamma, it is not one of the gamma matrices. The number 5 is a relic of old notation in which $γ 0$ was called " $γ 4$ ".

This matrix is useful in discussions of quantum mechanical chirality. For example, a Dirac field can be projected onto its left-handed and right-handed components by:

$\psi_L= \frac{1-\gamma^5}{2}\psi, \qquad\psi_R= \frac{1+\gamma^5}{2}\psi$ .

Some properties are:

(γ 5) + = γ 5

, it is hermitic.

(γ 5) 2 = I 4

, so its eigenvalues are ±1.

{

γ 5

γ i

}

= 0

, anticommutes with the gamma matrices.

[edit] Other representations

The matrices are also sometimes written using the 2x2 identity matrix, I, and

$\gamma^i = \begin{pmatrix} 0 & \sigma^i \\ -\sigma^i & 0 \end{pmatrix}$

where i runs from 1 to 3 and the σⁱ are Pauli matrices.

[edit] Dirac basis

The gamma matrices we have written so far are appropriate for acting on Dirac spinors written in the Dirac basis; in fact, the Dirac basis is defined by these matrices. To summarize, in the Dirac basis:

$\gamma^0 = \begin{pmatrix} I & 0 \\ 0 & -I \end{pmatrix},\quad \gamma^i = \begin{pmatrix} 0 & \sigma^i \\ -\sigma^i & 0 \end{pmatrix},\quad \gamma^5 = \begin{pmatrix} 0 & I \\ I & 0 \end{pmatrix}.$

[edit] Weyl basis

Another common choice is the Weyl or chiral basis, in which $γ i$ remains the same but $γ 0$ is different, and so $γ 5$ is also different:

$\gamma^0 = \begin{pmatrix} 0 & I \\ I & 0 \end{pmatrix},\quad \gamma^i = \begin{pmatrix} 0 & -\sigma^i \\ \sigma^i & 0 \end{pmatrix},\quad \gamma^5 = \begin{pmatrix} I & 0 \\ 0 & -I \end{pmatrix}.$

The Weyl basis has the advantage that its chiral projections take a simple form:

$\psi_L=\begin{pmatrix} I & 0 \\0 & 0 \end{pmatrix}\psi,\quad \psi_R=\begin{pmatrix} 0 & 0 \\0 & I \end{pmatrix}\psi.$

By a slight abuse of notation we can then identify

$\psi=\begin{pmatrix} \psi_L \\\psi_R \end{pmatrix},$

where now $ψ L$ and $ψ R$ are left-handed and right-handed two-component Weyl spinors.

[edit] Majorana basis

There's also a Majorana basis, in which all of the Dirac matrices are imaginary.

[edit] Euclidean Dirac matrices

In Quantum Field Theory one can wick rotate the time axis to transit from Minkowski space to Euclidean space, this is particularly useful in some renormalization procedures as well as Lattice gauge theory. In Euclidean space, there are two commonly used representation of Dirac Matrices:

[edit] Chiral representation

$\gamma^{1,2,3} = \begin{pmatrix} 0 & -i \sigma^{1,2,3} \\ i \sigma^{1,2,3} & 0 \end{pmatrix}, \quad \gamma^4=\begin{pmatrix} 0 & I \\ I & 0 \end{pmatrix}$

Different from Minkowski space, in Euclidean space,

γ 5 = γ 1 γ 2 γ 3 γ 4 = γ 5 +

So in Chiral basis,

$\gamma^5=\gamma^1 \gamma^2 \gamma^3 \gamma^4 = \begin{pmatrix} I & 0 \\ 0 & -I \end{pmatrix}$

[edit] Non-relativistic representation

$\gamma^{1,2,3} = \begin{pmatrix} 0 & -i \sigma^{1,2,3} \\ i \sigma^{1,2,3} & 0 \end{pmatrix}, \quad \gamma^4=\begin{pmatrix} I & 0 \\ 0 & -I \end{pmatrix}, \quad \gamma^5=\begin{pmatrix} 0 & -I \\ -I & 0 \end{pmatrix}$

[edit] See also

[edit] References

Halzen, Francis; Martin, Alan (1984). Quarks & Leptons: An Introductory Course in Modern Particle Physics. John Wiley & Sons. ISBN 0471887412.
A. Zee, Quantum Field Theory in a Nutshell (2003), Princeton University Press: Princeton, New Jersey. ISBN 0-691-01019-6. See chapter II.1.
M. Peskin, D. Schroeder, An Introduction to Quantum Field Theory (Westview Press, 1995) [ISBN 0-201-50397-2] See chapter 3.2.

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Categories: Quantum field theory | Spinors | Matrices | Clifford algebras

Gamma matrices

From Wikipedia, the free encyclopedia

Contents

[edit] Mathematical structure

[edit] Identities

[edit] Trace

[edit] Other

[edit] Feynman slash notation

[edit] Physical structure

[edit] Expressing the Dirac equation

[edit] The Fifth Gamma Matrix, $γ 5$

[edit] Other representations

[edit] Dirac basis

[edit] Weyl basis

[edit] Majorana basis

[edit] Euclidean Dirac matrices

[edit] Chiral representation

[edit] Non-relativistic representation

[edit] See also

[edit] References

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Gamma matrices

From Wikipedia, the free encyclopedia

Contents

[edit] Mathematical structure

[edit] Identities

[edit] Trace

[edit] Other

[edit] Feynman slash notation

[edit] Physical structure

[edit] Expressing the Dirac equation

[edit] The Fifth Gamma Matrix, γ5

[edit] Other representations

[edit] Dirac basis

[edit] Weyl basis

[edit] Majorana basis

[edit] Euclidean Dirac matrices

[edit] Chiral representation

[edit] Non-relativistic representation

[edit] See also

[edit] References

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Navigation

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[edit] The Fifth Gamma Matrix, $γ 5$