Dirac spinor
dirac spin or, dirac spinorsIn quantum field theory, the Dirac spinor is the bispinor in the planewave solution ψ = ω p → e − i p x }\;e^\;}
of the free Dirac equation,
i γ μ ∂ μ − m ψ = 0 , \partial _m\right\psi =0\;,}where in the units c = ℏ = 1
ψ is a relativistic spin1/2 field, ω p → }} is the Dirac spinor related to a planewave with wavevector p → }} , p x ≡ p μ x μ ≡ E t − p → ⋅ x → x^\;\equiv \;Et}\cdot }} , p μ = \;=\;\left\+}^}},\,}\right\}} is the fourwavevector of the plane wave, where p → }} is arbitrary, x μ } are the fourcoordinates in a given inertial frame of referenceThe Dirac spinor for the positivefrequency solution can be written as
ω p → = [ ϕ σ → ⋅ p → E p → + m ϕ ] , }=\phi \\}\cdot }}}+m}}\phi \end}\;,}where
ϕ is an arbitrary twospinor, σ → }} are the Pauli matrices, E p → }} is the positive square root E p → = + m 2 + p → 2 }\;=\;++}^}}}Contents
 1 Derivation from Dirac equation
 11 Results
 2 Details
 21 Twospinors
 22 Pauli matrices
 3 Fourspinors
 31 For particles
 32 For antiparticles
 4 Completeness relations
 5 Dirac spinors and the Dirac algebra
 51 Conventions
 52 Construction of Dirac spinor with a given spin direction and charge
 6 See also
 7 References
Derivation from Dirac equation
The Dirac equation has the form
− i α → ⋅ ∇ → + β m ψ = i ∂ ψ ∂ t }\cdot }+\beta m\right\psi =i}\,}In order to derive the form of the fourspinor ω we have to first note the value of the matrices α and β:
α → = [ 0 σ → σ → 0 ] β = [ I 0 0 − I ] }=\mathbf &}\\}&\mathbf \end}\quad \quad \beta =\mathbf &\mathbf \\\mathbf &\mathbf \end}}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
ψ = ω e − i p ⋅ x } ,while at the same time splitting ω into two twospinors:
ω = [ ϕ χ ] \phi \\\chi \end}\,}Results
Using all of the above information to plug into the Dirac equation results in
E [ ϕ χ ] = [ m I σ → ⋅ p → σ → ⋅ p → − m I ] [ ϕ χ ] \phi \\\chi \end}=m\mathbf &}\cdot }\\}\cdot }&m\mathbf \end}\phi \\\chi \end}}This matrix equation is really two coupled equations:
E − m ϕ = σ → ⋅ p → χ E + m χ = σ → ⋅ p → ϕ \leftEm\right\phi &=\left}\cdot }\right\chi \\\leftE+m\right\chi &=\left}\cdot }\right\phi \end}}Solve the 2nd equation for χ and one obtains
ω = [ ϕ σ → ⋅ p → E + m ϕ ] \phi \\}\cdot }}}\phi \end}\,}Alternatively, solve the 1st equation for ϕ and one finds
ω = [ − σ → ⋅ p → − E + m χ χ ] }\cdot }}}\chi \\\chi \end}\,}This solution is useful for showing the relation between antiparticle and particle
Details
Twospinors
The most convenient definitions for the twospinors are:
ϕ 1 = [ 1 0 ] ϕ 2 = [ 0 1 ] =1\\0\end}\quad \quad \phi ^=0\\1\end}\,}and
χ 1 = [ 0 1 ] χ 2 = [ 1 0 ] =0\\1\end}\quad \quad \chi ^=1\\0\end}\,}Pauli matrices
The Pauli matrices are
σ 1 = [ 0 1 1 0 ] σ 2 = [ 0 − i i 0 ] σ 3 = [ 1 0 0 − 1 ] =0&1\\1&0\end}\quad \quad \sigma _=0&i\\i&0\end}\quad \quad \sigma _=1&0\\0&1\end}}Using these, one can calculate:
σ → ⋅ p → = σ 1 p 1 + σ 2 p 2 + σ 3 p 3 = [ p 3 p 1 − i p 2 p 1 + i p 2 − p 3 ] }\cdot }=\sigma _p_+\sigma _p_+\sigma _p_=p_&p_ip_\\p_+ip_&p_\end}}Fourspinors
For particles
Particles are defined as having positive energy The normalization for the fourspinor ω is chosen so that the total probability is invariant under Lorentz transformation The total probability is:
P = ∫ V ω † ω d V \omega ^\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 ω † ω = 2 E \omega \;=\;2E\,} Hence the spinors, denoted as u are:
u p → , s = E + m [ ϕ s σ → ⋅ p → E + m ϕ s ] },s\right=}\phi ^\\}\cdot }}}\phi ^\end}\,}where s = 1 or 2 spin "up" or "down"
Explicitly,
u p → , 1 = E + m [ 1 0 p 3 E + m p 1 + i p 2 E + m ] a n d u p → , 2 = E + m [ 0 1 p 1 − i p 2 E + m − p 3 E + m ] },1\right=}1\\0\\}}\\+ip_}}\end}\quad \mathrm \quad u\left},2\right=}0\\1\\ip_}}\\}}\end}}For antiparticles
Antiparticles having positive energy E are defined as particles having negative energy and propagating backward in time Hence changing the sign of E and p → }} in the fourspinor for particles will give the fourspinor for antiparticles:
v p → , s = E + m [ σ → ⋅ p → E + m χ s χ s ] },s=}}\cdot }}}\chi ^\\\chi ^\end}}Here we choose the χ solutions Explicitly,
v p → , 1 = E + m [ p 1 − i p 2 E + m − p 3 E + m 0 1 ] a n d v p → , 2 = E + m [ p 3 E + m p 1 + i p 2 E + m 1 0 ] },1=}ip_}}\\}}\\0\\1\end}\quad \mathrm \quad v\left},2\right=}}}\\+ip_}}\\1\\0\\\end}}Note that these solutions are readily obtained by substituting the ansatz ψ = v e + i p x } into the Dirac equation
Completeness relations
The completeness relations for the fourspinors u and v are
∑ s = 1 , 2 u p s u ¯ p s = p / + m ∑ s = 1 , 2 v p s v ¯ p s = p / − m \sum _^}_^}&=+m\\\sum _^}_^}&=m\end}}where
p / = γ μ p μ =\gamma ^p_\,} see Feynman slash notation u ¯ = u † γ 0 }=u^\gamma ^\,}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 γ μ } where μ 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 γ μ } 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
i γ 2 γ 3 , i γ 3 γ 1 , i γ 1 γ 2 = − γ 1 , γ 2 , γ 3 i γ 1 γ 2 γ 3 σ a , b , c = i a γ 2 γ 3 + i b γ 3 γ 1 + i c γ 1 γ 2 i\gamma ^\gamma ^,\;\;i\gamma ^\gamma ^,\;\;i\gamma ^\gamma ^&=\gamma ^,\;\gamma ^,\;\gamma ^i\gamma ^\gamma ^\gamma ^\\\sigma _&=ia\gamma ^\gamma ^+ib\gamma ^\gamma ^+ic\gamma ^\gamma ^\end}}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 subalgebra of the Dirac algebra that has spin oriented in the a, b, c direction:
P a , b , c = 1 2 1 + σ a , b , c =}\left1+\sigma _\right}Now we must choose a charge, +1 positron or −1 electron Following the conventions of Peskin & Schroeder, the operator for charge is Q = − γ 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 σ 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 = 1 2 1 − Q = 1 2 1 + γ 0 =}\left1Q\right=}\left1+\gamma ^\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 \;P_}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 = 1 2 1 + i γ 1 γ 2 =}\left1+i\gamma _\gamma _\right}and so our desired projection operator is
P = 1 2 1 + i γ 1 γ 2 ⋅ 1 2 1 + γ 0 = 1 4 1 + γ 0 + i γ 1 γ 2 + i γ 0 γ 1 γ 2 }\left1+i\gamma ^\gamma ^\right\cdot }\left1+\gamma ^\right=}\left1+\gamma ^+i\gamma ^\gamma ^+i\gamma ^\gamma ^\gamma ^\right}The 4×4 gamma matrices used in the Weyl representation are
γ 0 = [ 0 1 1 0 ] γ k = [ 0 σ k − σ k 0 ] \gamma _&=0&1\\1&0\end}\\\gamma _&=0&\sigma ^\\\sigma ^&0\end}\end}}for k = 1, 2, 3 and where σ i } are the usual 2×2 Pauli matrices Substituting these in for P gives
P = 1 4 [ 1 + σ 3 1 + σ 3 1 + σ 3 1 + σ 3 ] = 1 2 [ 1 0 1 0 0 0 0 0 1 0 1 0 0 0 0 0 ] }1+\sigma ^&1+\sigma ^\\1+\sigma ^&1+\sigma ^\end}=}1&0&1&0\\0&0&0&0\\1&0&1&0\\0&0&0&0\end}}Our answer is any nonzero column of the above matrix The division by two is just a normalization The first and third columns give the same result:
 e − , + 1 2 ⟩ = [ 1 0 1 0 ] ,\,+}\right\rangle =1\\0\\1\\0\end}}More generally, for electrons and positrons with spin oriented in the a, b, c direction, the projection operator is
1 4 [ 1 + c a − i b ± 1 + c ± a − i b a + i b 1 − c ± a + i b ± 1 − c ± 1 + c ± a − i b 1 + c a − i b ± a + i b ± 1 − c a + i b 1 − c ] }1+c&aib&\pm 1+c&\pm aib\\a+ib&1c&\pm a+ib&\pm 1c\\\pm 1+c&\pm aib&1+c&aib\\\pm a+ib&\pm 1c&a+ib&1c\end}}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 +b^+c^\,=\,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
 Spin3,1, the double cover of SO3,1 by a spin group
References
 Aitchison, IJR; AJG Hey September 2002 Gauge Theories in Particle Physics 3rd ed Institute of Physics Publishing ISBN 0750308648
 Miller, David 2008 "Relativistic Quantum Mechanics RQM" PDF pp 26–37
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Dirac spinor
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