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Master Dissertation

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2.3 Spinor Representations of the Lorentz Group<br />

We want to extend the spinor representation to four dimensions. 1 Doing<br />

this we add the unit matrix<br />

<br />

1 0<br />

σ0 = , (2.6)<br />

0 1<br />

to the usual Pauli matrices:<br />

σ1 =<br />

0 1<br />

1 0<br />

<br />

, σ2 =<br />

0 −i<br />

i 0<br />

<br />

1 0<br />

, σ3 =<br />

0 −1<br />

<br />

. (2.7)<br />

These form a basis of H(2), the vector space over R of all complex<br />

Hermitian 2 × 2 matrices. Actually, given a 2 × 2 matrix H ∈ H(2) we get<br />

by direct calculation that H can be represented by a vector x ∈ R4 in the<br />

following way:<br />

H := σ(x) = x µ σµ = 1<br />

3<br />

tr(Hσµ)σµ<br />

(2.8)<br />

2<br />

The map x ↦→ σ(x) defined in this way is indeed an isomorphism of R 4<br />

onto H(2).<br />

Fixme: should this be written out?<br />

Further we can define another isomorphism by<br />

µ=0<br />

x ↦→ σ ′ (x) = x 0 σ0 − x k σk =<br />

3<br />

xµσµ.<br />

Now consider the transform of σ(x) with a complex 2 × 2 matrix A:<br />

σ(x) ↦→ Aσ(y)A ∗ .<br />

Clearly σ(y) := Aσ(x)A ∗ ∈ H(2). And from equation (2.8) we see that its<br />

components in the basis of Pauli Matrices are given by<br />

y ν = 1<br />

2 tr(σ(y)σν) = 1<br />

2 tr(Aσ(x)A∗σν) = 1<br />

2<br />

µ=0<br />

3<br />

µ=0<br />

tr(AσµA ∗ σν)x µ .<br />

In this way we get a linear map ΛA : x → y of R 4 into it self. Note that<br />

Hence if det A = 1, then<br />

σ(x) = x µ <br />

x0 + x3 x1 − ix2 σµ =<br />

x 1 + ix 2 x 0 − x 3<br />

<br />

.<br />

1 The two-component spinor formalism is treated in [8] chapter 3.2-3.3.<br />

17

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