PHY–396 K. Problem set #5. Due October 13, 2011. 1. First, an ...

PHY–396 K. Problem set #5. Due October 13, 2011.
1. First, an exercise in Dirac matrices γ µ . Please do not assume any specific form of these
4 × 4 matrices, just use the anti-commutation relations
γ µ γ ν + γ ν γ µ = 2g µν . (1)
In class, I have defined the spin matrices
S µν = −S νµ def
= i 4 [γµ , γ ν ] (2)
and showed that
[
S µν , γ λ] = ig νλ γ µ − ig µλ γ ν . (3)
(a) Show that the spin matrices S µν have commutation relations of the Lorentz generators,
[
S κλ , S µν] = ig λµ S κν − ig λν S κµ − ig κµ S λν + ig κν S λµ . (4)
A continuous Lorentz transform obtains from integrating infinite sequences of infinitesimal
transforms X ′µ = X µ + ɛ Θ µ νX ν for antisymmetric Θ µν = −Θ νµ (when both indices are
down or both up); the finite transform is X ′µ = L µ νX ν where
L = exp(Θ), i. e., L µ ν = δ µ ν + Θ µ ν + 1 2 Θµ λ Θλ ν + 1 6 Θµ κΘ κ λ Θλ ν + · · · . (5)
(b) Let L be a Lorentz transform of the form (5), and let M D
(L) = exp ( − i 2 Θ αβS αβ) .
Show that MD −1 (L)γµ M D
(L) = L µ νγ ν .
Hint: use Hadamard Lemma e A Be −A = B+[A, B]+
2 1[A, [A, B]]+ 1 6
[A, [A, [A, B]]]+· · ·.
Next, a little more algebra:
(c) Calculate {γ ρ , γ λ γ µ γ ν }, [γ ρ , γ κ γ λ γ µ γ ν ] and [S ρσ , γ λ γ µ γ ν ].
(d) Show that γ α γ α = 4, γ α γ ν γ α = −2γ ν , γ α γ µ γ ν γ α = 4g µν , and γ α γ λ γ µ γ ν γ α =
−2γ ν γ µ γ λ .
Hint: use γ α γ ν = 2g να − γ ν γ α repeatedly.
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2. Now consider the γ 5 def = iγ 0 γ 1 γ 2 γ 3 matrix.
(a) Show that γ 5 anticommutes with each of the γ µ matrices — γ 5 γ µ = −γ µ γ 5 — and
commutes with all the spin matrices S µν .
(b) Show that γ 5 is hermitian and that (γ 5 ) 2 = 1.
(c) Show that γ 5 = (i/24)ɛ κλµν γ κ γ λ γ µ γ ν and γ [κ γ λ γ µ γ ν] = −24iɛ κλµν γ 5 .
(d) Show that γ [λ γ µ γ ν] = −6iɛ κλµν γ κ γ 5 .
(e) Show that any 4 × 4 matrix Γ is a unique linear combination of the following 16
matrices: 1, γ µ , 1 2 γ[µ γ ν] , γ 5 γ µ , and γ 5 .
∗ My conventions here are: ɛ 0123 = −1, ɛ 0123 = +1, γ [µ γ ν] = γ µ γ ν − γ ν γ µ ,
γ [λ γ µ γ ν] = γ λ γ µ γ ν − γ λ γ ν γ µ + γ µ γ ν γ λ − γ µ γ λ γ ν + γ ν γ λ γ µ − γ ν γ µ γ λ , etc.
Now consider Dirac matrices in spacetime dimensions d ≠ 4. Such matrices always satisfy
the Clifford algebra (1), but their sizes depend on d.
Let Γ = i n γ 0 γ 1 · · · γ d−1 be the generalization of the γ 5 to d dimensions; the pre-factor
i n = ±i or ±1 is chosen such that Γ = Γ † and Γ 2 = +1.
(f) For even d, Γ anticommutes with all the γ µ . Prove this, and use this fact to show that
there are 2 d independent products of the γ µ matrices, and consequently the matrices
should be 2 d/2 × 2 d/2 .
(g) For odd d, Γ commutes with all the Γ µ — prove this. Consequently, one can set Γ = +1
or Γ = −1; the two choices lead to in-equivalent sets of the γ µ .
Classify the independent products of the γ µ for odd d and show that their net number
is 2 d−1 ; consequently, the matrices should be 2 (d−1)/2 × 2 (d−1)/2 .
3. Let’s go back to d = 3 + 1. Since all the spin matrices S µν commute with the γ 5 , all the
M D
(L) = exp ( −
2 i Θ αβS αβ) matrices are block-diagonal in the eigenbasis of the γ 5 . In the
Weyl convention for the γ matrices,
( )
ML (L) 0
M D (L) =
(6)
0 M R (L)
where all blocks are 2 × 2. This makes the Dirac spinor a reducible representation of the
continuous Lorentz group SO + (3, 1).
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Write down the explicit S µν matrices in the Weyl convention, then use them to show that
the M L (L) block in eq. (6) is precisely the SL(2, C) matrix M from problem 4 of the
last homework while the M R block is the M = ( M †) −1 = σ2 M ∗ σ 2 matrix from the same
problem. In particular, for a pure rotation through angle ϕ around axis n
M L = M R = exp(−
2 i ϕ n · σ). (7)
while for a pure boost of rapidity r in the direction n, M L = exp(−
2 r n · σ) but M R =
exp(+
2 r n · σ); in terms of the boost’s β and γ parameters,
M L = √ γ × √ 1 − β n · σ , M R = √ γ × √ 1 + β n · σ . (8)
4. Finally, consider the plane-wave solutions e −ipx u(p, s) and e +ipx v(p, x) of the Dirac equation.
The 4–component spinors u(p, s) and v(p, s) satisfy
(̸p − m)u(p, s) = 0, (̸p + m)v(p, s) = 0, u † (p, s)u(p, s ′ ) = v † (p, s)v(p, s ′ ) = 2Eδ s,s
′ .
(9)
Let’s writing down explicit formulae for these spinors in the Weyl basis for the γ µ matrices.
(a) Show that for p = 0,
u(p = 0, s) =
( √ )
m ξs
√ m ξs
(10)
where ξ s is a two-component SO(3) spinor encoding the electron’s spin state. The ξ s
are normalized to ξ † sξ s
′ = δ s,s ′.
(b) For other momenta, u(p, s) = M(boost)u(p = 0, s) for the boost that turns (m,⃗0) to
p µ . Use eqs. (8) to show that
u(p, s) =
( √ )
E − p · σ ξ s
√ . (11)
E + p · σ ξ s
(c) Use similar arguments to show that
v(p, s) =
( √ )
+ E − p · σ η s
− √ E + p · σ η s
(12)
where η s are two-component SO(3) spinors normalized to η † sη s
′ = δ s,s ′.
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Physically, the η s should have opposite spins from ξ s — the holes in the Dirac sea have
opposite spins (as well as p µ ) from the missing negative-energy particles. Mathematically,
this requires η sSη † s = −ξ sSξ † s ; we may solve this condition by letting η s = σ 2 ξs ∗ = ±iξ −s .
(d) Check that this is a solution, then show that it leads to v(p, s) = γ 2 u ∗ (p, s).
(e) Show that for ultra-relativistic electrons or positrons of definite helicity λ = ± 1 2 , the
Dirac plane waves become chiral — i.e., dominated by one of the two irreducible
components 2 or ¯2 of the Dirac spinor 2 ⊕ ¯2 while the other component becomes
negligible. (The 2 component is the left-handed Weyl spinor while the ¯2 component
is the right-handed Weyl spinor. I shall discuss them later in class.) Specifically,
u(p, −
2 1) ≈ √ ( )
ξL
2E , u(p, + 1 2
0
) ≈ √ ( )
0
2E ,
ξ R
( )
0
v(p, − 1 2 ) ≈ −√ 2E , v(p, + 1 2
η ) ≈ √ ( )
ηR
2E .
L
0
(13)
Note that for electrons the left/right chirality is same as the helicity, but for positrons
the chirality is opposite from the helicity.
Finally, let’s establish some basis-independent properties of the Dirac spinors u(p, s) and
v(p, s) — although you may use the Weyl basis to verify them.
(f) Show that
ū(p, s)u(p, s ′ ) = +2mδ s,s
′ , ¯v(p, s)v(p, s ′ ) = −2mδ s,s
′ ; (14)
note that the normalization here is different from eq. (9) for the v † u and v † v.
(g) There are only two independent SO(3) spinors, hence ∑ s ξ sξ † s = ∑ s η† sη s = 1 2×2 . Use
this fact to show that
∑
s=1,2
u α (p, s)ū β (p, s) = (̸p+m) αβ
and
∑
s=1,2
v α (p, s)¯v β (p, s) = (̸p−m) αβ . (15)
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