Gamma matrices

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In mathematical physics, the gamma matrices, {γ0, γ1, γ2, γ3}, also known as the Dirac matrices, form a matrix-valued representation of a set of orthogonal basis vectors for contravariant vectors in space time, from which can be constructed a Clifford algebra.

This in turn makes possible the introduction of spinors to represent spatial rotations and Lorentz boosts. Spinors facilitate space-time computations in general, and in particular are fundamental to the Dirac equation for relativistic spin-½ particles.

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}.

Contents

[edit] Mathematical structure

The defining property for the gamma matrices to form a Clifford algebra is the anticommutation relation

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

where \eta^{\mu \nu} \, is the Minkowski metric with signature (+ − − −)

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 = \left\{\gamma^0, -\gamma^1, -\gamma^2, -\gamma^3 \right\},

and Einstein notation is assumed.

[edit] Physical structure

The 4-tuple \displaystyle\gamma^\mu=(\gamma^0,\gamma^1,\gamma^2,\gamma^3) = \gamma^0 e^0 + \gamma^1 e^1 + \gamma^2 e^2 + \gamma^3 e^3 is often loosely described as a 4-vector (where e0 to e3 are the basis vectors of the 4-vector space). But this is misleading. Instead \displaystyle\gamma^\mu is more appropriately seen as a mapping operator, taking in a 4-vector \displaystyle a^\mu and mapping it to the corresponding vector in the Clifford algebra representation.

This is symbolised by the useful Feynman slash notation,

a\!\!\!/ := \gamma^\mu a_\mu.

Slashed quantities like a\!\!\!/ "live" in the multilinear Clifford algebra, with its own set of basis directions — they are immune to changes in the 4-vector basis.

On the other hand, one can define a transformation identity for the mapping operator \displaystyle\gamma^\mu. 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}.

This says essentially that an operator mapping from the old 4-vector basis \displaystyle\{e^0,e^1,e^2,e^3\} to the old Clifford algebra basis \displaystyle\{\gamma^0,\gamma^1,\gamma^2,\gamma^3\} is equivalent to a mapping from the new 4-vector basis \displaystyle\Lambda^\mu{}_\nu\{e^0,e^1,e^2,e^3\} to a correspondingly transformed new Clifford algebra basis \displaystyle\lambda\{\gamma^0,\gamma^1,\gamma^2,\gamma^3\}\lambda^{-1}. Alternatively, in pure index terms, it shows that γμ transforms appropriately for an object with one contravariant 4-vector index and one covariant and one contravariant Dirac spinor index.

Given the above transformation properties of γμ, if ψ is a Dirac spinor then the product γμψ 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 γμ as if it were simply a vector.

There remains a final key difference between γμ and any nonzero 4-vector: γμ does not point in any direction. More precisely, the only way to make a true vector from γμ 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 γμ 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] Identities

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

[edit] Normalisation

Because of the above anti-commutation relation, we can show:

\left( \gamma^0 \right)^\dagger = \gamma^0 \,, and \left( \gamma^0 \right)^2 = I \,

and for the other gamma matrices (for k=1,2,3) we have

\left( \gamma^k \right)^\dagger = -\gamma^k \,, and \left( \gamma^k \right)^2 = -I \,

These results can be generalized by the relation

\left( \gamma^\mu \right)^\dagger = \gamma^0 \gamma^\mu \gamma^0 \,

[edit] Miscellaneous identities

Num Identity
1 \displaystyle\gamma^\mu\gamma_\mu=4 I
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} I
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 I

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 I - \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. \,
Proof of 3

To show

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

Use the anticommutator to shift γμ 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 γμγμ = 4I 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^\nu, \gamma^\rho\}. \,

Finally using the anticommutator identity, we get

\gamma^\mu \gamma^\nu \gamma^\rho \gamma_\mu = 4\ \eta^{\nu \rho} I. \,
Proof of 4
\gamma^\mu \gamma^\nu \gamma^\rho \gamma^\sigma \gamma_\mu \, = (2\eta^{\mu\nu}-\gamma^\nu \gamma^\mu) \gamma^\rho \gamma^\sigma \gamma_\mu \, \quad (anticommutator identity)
= 2\eta^{\mu\nu}\gamma^\rho \gamma^\sigma \gamma_\mu - 4 \gamma^\nu \eta^{\rho \sigma} \, \quad (using identity 3)
= 2\gamma^\rho \gamma^\sigma \gamma^\nu - 4 \gamma^\nu \eta^{\rho \sigma} \, (raising an index)
= 2(2\eta^{\rho \sigma} - \gamma^\sigma \gamma^\rho) \gamma^\nu - 4 \gamma^\nu \eta^{\rho \sigma} \, (anticommutator identity)
=-2\gamma^\sigma\gamma^\rho\gamma^\nu \, (2 terms cancel)

[edit] Trace identities

Num Identity
1 trace of any product of an odd number of γμ 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 four main properties of the Trace operator:

  • tr(A + B) = tr(A) + tr(B)
  • tr(rA) = r tr(A)
  • tr(ABC) = tr(CAB) = tr(BCA)
Proof of 1

To show

\operatorname{tr} ( \mathrm{odd \ num \ of \ } \gamma) = 0 \,

First note that obviously

\operatorname{tr} (\gamma^\mu) = 0. \,

We'll also use two facts about the the fifth gamma matrix \gamma^5 \, (shown later in this article) that says:

\left(\gamma^5 \right)^2 = I, \quad \mathrm{and} \quad \gamma^\mu \gamma^5 = - \gamma^5 \gamma^\mu \,

So lets use these two facts to prove this identity for the first non-trivial case: the trace of three gamma matrices. Step one is to put in a pair of \gamma^5 \,'s next to each of the three original \gamma \,'s, and step two is to swap each of the original \gamma \,'s with its neighbor to the left.

\operatorname{tr} (\gamma^\mu \gamma^\nu \gamma^\rho) \, = \operatorname{tr} \left( \gamma^5 \gamma^5 \gamma^\mu \gamma^5 \gamma^5 \gamma^\nu \gamma^5 \gamma^5 \gamma^\rho \right) \,
= -\operatorname{tr} \left(\gamma^5 \gamma^\mu \gamma^5 \gamma^5 \gamma^\nu \gamma^5 \gamma^5 \gamma^\rho \gamma^5 \right) \,
= -\operatorname{tr} \left(\gamma^5 \gamma^\mu \gamma^\nu \gamma^\rho \gamma^5 \right) = -\operatorname{tr} \left( \gamma^5 \gamma^5 \gamma^\mu \gamma^\nu \gamma^\rho \right) \,
= -\operatorname{tr} \left(\gamma^\mu \gamma^\nu \gamma^\rho \right) \,

Bring the term on the right over to the left-hand side to finally see

2 \operatorname{tr} \left(\gamma^\mu \gamma^\nu \gamma^\rho \right) = 0 \,

A similar pattern can be used to prove this is true for the general case of any odd number of gamma matrices.

Proof of 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 (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} \,

or

\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

To prove

\operatorname{tr}(\gamma^5)= 0 ,

insert two factors of \gamma^0 \, into the trace, anticommute one with \gamma^5 \, (picking up a minus sign), cycle it back around using the cyclic property of the trace, and conclude that \operatorname{tr}(\gamma^5) is minus itself:

\operatorname{tr}(\gamma^5) = \operatorname{tr} (\gamma^0 \gamma^0 \gamma^5)=-\operatorname{tr} (\gamma^0 \gamma^5 \gamma^0)=-\operatorname{tr} (\gamma^0 \gamma^0 \gamma^5)=-\operatorname{tr}(\gamma^5) \,

So,

\operatorname{tr}(\gamma^5)= 0 \, .

This also works for

\operatorname{tr} (\gamma^\mu\gamma^\nu\gamma^5) = 0 \, .

Simply add two factors of \gamma^\alpha \, , with \alpha \, different from \mu \, and \nu \,. Anticommute three times instead of once, picking up three minus signs, and cycle using the cyclic property of the trace.

So,

\operatorname{tr} (\gamma^\mu\gamma^\nu\gamma^5) = 0 \, .

For a proof of identity 5, the same trick still works unless (\mu \nu \rho \sigma) \, is some permutation of (0123), so that all 4 gammas appear. The anticommutation rules imply that interchanging two of the indices changes the sign of the trace, so \operatorname{tr} (\gamma^\mu\gamma^\nu\gamma^\rho\gamma^\sigma\gamma^5) \, must be proportional to \epsilon^{\mu\nu\rho\sigma} \,. The proportionality constant is -4i \,, as can be checked by plugging in (\mu \nu \rho \sigma)=(0123) \,, writing out \gamma^5 \,, and remembering that the trace of the identity is 4.

[edit] Feynman slash notation

The contraction of the mapping operator γμ with a vector aμ maps the vector out of the 4-vector representation. So, it is common to write identities using the Feynman slash notation, defined by

a\!\!\!/ := \gamma^\mu a_\mu.

Here are some similar identities to the ones above, but involving slash notation:

a\!\!\!/b\!\!\!/ = a \cdot b - 2i a_\mu S^{\mu\nu} b_\nu
a\!\!\!/a\!\!\!/ = a^2
\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 \rho \sigma} a^\mu b^\nu c^\rho d^\sigma
\gamma_\mu a\!\!\!/ \gamma^\mu = -2 a\!\!\!/
\gamma_\mu a\!\!\!/ b\!\!\!/ \gamma^\mu = 4 a \cdot b \,
\gamma_\mu a\!\!\!/ b\!\!\!/ c\!\!\!/ \gamma^\mu = -2 c\!\!\!/ b\!\!\!/ a\!\!\!/ \,
where
\epsilon_{\mu \nu \rho \sigma} \, is the Levi-Civita symbol and S^{\mu\nu} = \frac{i}{4} [\gamma^\mu, \gamma^\nu].

[edit] The Fifth Gamma Matrix, γ5

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 γ5 uses the letter gamma, it is not one of the gamma matrices. The number 5 is a relic of old notation in which γ0 was called "γ4".

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:

  • It is hermitic:
(\gamma^5)^\dagger = \gamma^5 \,,
  • Its eigenvalues are ±1, because:
(\gamma^5)^2 = I \,
  • It anticommutes with the four gamma matrices:
\left\{ \gamma^5,\gamma^\mu \right\} =\gamma^5 \gamma^\mu + \gamma^\mu \gamma^5 = 0 \,,

[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 γi remains the same but γ0 is different, and so γ5 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 ψL and ψR 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. In terms of the Pauli matrices, it can be written as

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

[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,

γ5 = γ1γ2γ3γ4 = γ5 +

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 0-471-88741-2. 
  • 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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