sign in sign up
<<
Home , 3D rotation group

PHY–396 K. Solutions for homework set #6.

Problem 1(a):

h ˆi ˆji 1 ik` jmn h ˆk` ˆmni J , J ≡ 4   J , J = hh by eq. (1) ii 1 ik` jmn  km ˆ`n kn ˆ`m `m ˆkn `n ˆkm = 4   −ig J + ig J + ig J − g J hh by antisymmetry of the ’s ii = ik`jmn × −igkmJˆ`n   = iJˆ`n × −gkmik`jmn = +δkmik`jmn = δijδ`n − δinδ`j = 0 − iJˆji = +iJˆij ≡ +iijkJˆk. (S.1)

h ˆi ˆ ji 1 ik` h ˆk` ˆ0ji J , K ≡ 2  J , J = hh by eq. (1) ii 1 ik`  k0 ˆ`j kj ˆ`0 `0 ˆkj `j ˆk0 = 2  −ig J + ig J + ig J − ig J 1 ik`  kj ˆ`0 `j ˆk0 = 2  0 − iδ J + 0 + iδ J 1 ik`  kj ˆ ` `j ˆ k ≡ 2  +iδ K − iδ K 1 ij` ˆ ` 1 ikj ˆ j = 2  K − 2  K = ijkKˆ k, (S.2)

h i h i Kˆ i, Kˆ j ≡ Jˆ0i, Jˆ0j = hh by eq. (1) ii = −ig00Jˆij + ig0jJˆi0 + igi0Jˆ0j − igijJˆ00 = −iJˆij + 0 + 0 + 0, ≡ −iijkJˆk. (S.3)

Problem 1(b):

h ˆ i ˆji 1 jk` h ˆ i ˆk`i 1 jk`  ik ˆ ` i` ˆ k V , J ≡ 2  V , J = 2  ig V − ig V 1 jk`  ik ˆ ` i` ˆ k i ji` ˆ ` i jki ˆ k = 2  −iδ V + δ V = − 2  V + 2  V = iijkVˆ k, (S.4)

1 h ˆ 0 ˆji 1 jk` h ˆ 0 ˆk`i 1 jk`  0k ˆ ` 0` ˆ k V , J = 2  V , J = 2  i\g V − i\g V = 0, (S.5) h i h i Vˆ i, Kˆ j = Vˆ i, Jˆ0j = i\gi0Vˆ j − igijVˆ 0 = +iδijVˆ 0, (S.6) h i h i Vˆ 0, Kˆ j = Vˆ 0, Jˆ0j = ig00Vˆ j − i\g0jVˆ 0 = +iVˆ j. (S.7)

Note that the Hamiltonian of a relativistic theory is a member of a 4-vector multiplet Pˆµ = (H,ˆ Pˆ) where Pˆ is the net momentum operator. Applying the above equations to the Pˆµ vectors, we obtain h i Pˆi, Jˆj = iijkPˆk, h i H,ˆ Jˆj = 0, h i (S.8) Pˆi, Kˆ j = +iδijH,ˆ h i H,ˆ Kˆ j = +iPˆj.

In particular, the Hamiltonian Hˆ commutes with the three angular momenta Jˆj but it does not commute with the three generators Kˆ k of the Lorentz boosts.

Problem 1(c): In the ordinary , it is often said that generators of continuous symmetries must commute with the Hamiltonian operator. However, this is true only for the symmetries that act in a time independent manner — for example, rotating the 3D space by the same at all times t. But when the transformation rules of a depend on time, the Hamiltonian must change to account for this time dependence.

In a Lorentz boost, the transform x → x0 obviously depends on time, which changes the way the transformed quantum fields such as Φˆ 0(x, t) depend on t. Consequently, the Hamiltonian Hˆ of the theory must change so that the new Heisenberg equations would match the new time dependence. In terms of the generators, this means that the boost generators Kˆ i should not commute with the Hamiltonian.

2 Note that this non-commutativity is not caused by the Lorentz boosts affecting the time 0 0 µ itself, t = L µx 6= t. Even in non-relativistic theories — where the time is absolute — the generators of symmetries which affect the other variables in a time-dependent matter do not commute with the Hamiltonian Hˆ .

Indeed, consider a Galilean transform from one non-relativistic moving frame into an- other, x0 = x+vt but t0 = t. This is a good symmetry of non-relativistic particles interacting with each other but not subject to any external potential,

X 1 X Hˆ = ˆp2 + 1 V (xˆ − xˆ ). (S.9) 2M a 2 a b a a6=b

A unitary operator Gˆ realizing a Galilean symmetry acts on coordinate and momentum operators as

† † GˆxˆaGˆ = xˆa + vt, GˆˆpaGˆ = ˆpa + Mv, (S.10)

and it also transforms the Hamiltonian into

ˆ ˆ ˆ† ˆ ˆ 1 2 GHG = H + v · Ptot + 2 Mtotv . (S.11)

In terms of the Galilean boost generators Kˆ G,

 Gˆ = exp −iv · Kˆ G , (S.12) hence under an infinitesimal boost v = ~, various operators transform according to

i h ˆ i i Ob → Ob + δOb, δOb = −i KG, Ob . (S.13)

Hence, the commutation relations with the boost generators follow from the infinitesimal boosts, for example

h i ˆ j i ij δxˆa = ~t =⇒ xˆa, KG = iδ × t, h i ˆ j i ij δpˆa = ~M =⇒ pˆa, KG = iδ × M, (S.14) ˆ ˆ h ˆ ˆ j i ˆi δH = ~ · Ptot =⇒ H, KG = iPtot .

In particular, the Hamiltonian does NOT commute with the Galilean boost generators.

3 Problem 2(a): 1 ˆµν Consider a linear combination 2 NµνJ of Lorentz generators with some generic coefficients iφ ˆµν Nµν = −Nνµ. The Lorentz symmetries L(N, φ) = exp( 2 NµνJ ) generated by this combi- nation preserve the momentum pµ of the particle state |pi if and only if

iφ ˆµν ˆα iφ ˆµν ˆµ exp 2 NµνJ P exp − 2 NµνJ |pi = P |pi . (S.15)

For an infinitesimal φ, this condition becomes

h iφ ˆµν ˆαi 2 NµνJ , P |pi = 0. (S.16)

Applying eqs. (13) to the left hand side of this formula gives us

h iφ ˆµν ˆαi iφ  αµ ˆν αν ˆµ 2 NµνJ , P |pi = 2 Nµν −ig P + g P |pi αν = φN Pˆν |pi (S.17) αν = φN pν |pi ,

1 ˆµν µ so the condition for the generator 2 NµνJ to preserve particle’s momentum p is simply

αν N × pν = 0. (S.18)

In 3D terms, N ij = ijkak and N 0k = −N k0 = bk for some 3-vectors a and b, the generator in question is

1 ˆµν ˆ ˆ 2 NµνJ = a · J + b · K, (S.19) and the condition (S.18) becomes

a × p − bE = 0 and b · p = 0. (S.20)

Actually, the second condition here is redundant, so the general solution is p any a, b = a × . (S.21) E In terms of eq. (S.19), these solutions mean

(a × p) · Kˆ a   1 N Jˆµν = a · Jˆ + = · EJˆ + p × Kˆ for any a. (S.22) 2 µν E E In other words, the Lorentz symmetries preserving the momentum pµ have 3 generators,

4 namely the components of the 3-vector

Rˆ = E Jˆ + p × Kˆ . (4)

Quod erat demonstrandum.

Problem 2(b): Consider a massive particle moving at a slower-than-light speed β in z direction, so the energy is E = γm (where γ = 1/p1 − β2) while the 3-momentum is p = (0, 0, βγm). Consequently, the three components of the Rˆ vector this energy-momentum are

Rˆx = γmJˆx − βγmKˆ y, Rˆy = γmJˆy + βγmKˆ x, Rˆz = γmJˆz. (S.23)

These 3 operators generate the little G(p) of the particle’s 4–momentum pµ = (E, p). To see that this little group happens to be isomorphic to the 3D group SO(3), we need to find 3 linear combinations J˜x,y,z of the operators (S.23) which obey the angular- momentum commutation relations h i J˜i, J˜j = iijkJ˜k. (S.24)

My choice of the (properly normalized) generators J˜x,y,z is spelled out in eqs. (5). Let’s see that they indeed obey the commutation relations (S.24): h i h i h i Jˆz, Jex = γ Jˆz, Jˆx − βγ Jˆz, Kˆ y = γ × iJˆy − βγ × (−iKˆ x)

= iJey, h i h i h i Jˆz, Jey = γ Jˆz, Jˆy + βγ Jˆz, Kˆ x = γ × (−iJˆx) − βγ × (+iKˆ y) (S.25)

= −iJex, h x yi 2 h ˆx ˆyi 2 h ˆ y ˆyi 2 h ˆ ˆ x i 2 2 h ˆ y ˆ xi Je , Je = γ J , J − βγ K , J + βγ Jx, K , − β γ K , K = γ2 × iJˆz − 0 + 0 − β2γ2 × iJˆz = iJˆz × γ2(1 − β2) = 1 = iJˆz.

Quod erat demonstrandum.

5 Problem 2(c): For a massless particle, we cannot rescale the little group generators as in eqs. (5) since the 1/m factor for the J˜x,y generators would be infinite. Instead, the best we can do is to use the 1/E factor for all three rescaled generators, hence eqs. (6) and (7). Consequently, instead of eqs. (S.25) for the of the rescaled generators, we get

h i h i h i Jˆz, Iˆx = Jˆz, Jˆx − Jˆz, Kˆ y = iJˆy − (−iKˆ x) = iIˆy, h i h i h i Jˆz, Iˆy = Jˆz, Jˆy + Jˆz, Kˆ x = (−iJˆx) − (+iKˆ y) = −iIˆx, (S.26) h i h i h i h i h i Iˆx, Iˆy = Jˆx, Jˆy − Kˆ y, Jˆy + Jˆx, Kˆ x − Kˆ y, Kˆ x = iJˆz − 0 + 0 − iJˆz = 0, precisely as in eq. (8).

The commutation relations (8) are different from the angular-momentum commutation relations, and they cannot be brought to the form (S.24) by any finite rescaling of the generators. Consequently, the little group of a light-like momentum pµ = (E, 0, 0,E) is NOT isomorphic to the SO(3).

Instead, the commutation relations (8) are similar to the commutation relations between the z component of the and the x and y components of the linear momentum,

h i h i h i Jˆz, Pˆx = +iPˆy, Jˆz, Pˆy = −iPˆx, Pˆx, Pˆy = 0. (S.27)

The Jˆz operator generates around the z axis, i.e., within the xy plane, while the Pˆx and Pˆi operators generate translations in that plane. Together, they generate the group ISO(2) of — rotations and translations — in 2 space dimensions.

Thus, we see that the little group G(p) of a light-like momentum of a massless particle is isomorphic to ISO(2).

6 Problem 2(d): Finally, consider a particle moving faster-than-light in the z direction, so its momentum pµ = (p0, 0, 0, p3) has p3 > p0. Such a tachyon must have negative mass2, so let us denote

0 3 2 2 µ p p Mi = −m = −pµp > 0, γi = , β = (S.28) Mi p0 where the subscript i stands for ‘imaginary’ and β > 1 is the faster-than-light speed. In these notations, let us rescale the little group generators (14) according to

x +1 y x y K˜ = Rˆ = βγiKˆ + γiJˆ , Mi y −1 x y x K˜ = Rˆ = βγiKˆ − γiJˆ , (S.29) Mi 1 J˜z = Rˆz = Jˆz, the helicity. γiMi

Consequently, the commutation relations of the rescaled operators become

h z xi h z xi h z yi Jˆ , K˜ = βγi Jˆ , Kˆ + γi Jˆ , Jˆ y i x = βγi(+iKˆ ) + γ (−iJˆ ) = +iK˜ y, (S.30)

h z yi h z yi h z xi Jˆ , K˜ = βγi Jˆ , Kˆ − γi Jˆ , Jˆ y i x = βγi(−iKˆ ) + γ (+iJˆ ) = −iK˜ x, (S.31)

h ˜ x ˜ yi 2 2 h ˆ x ˆ yi 2 h ˆy ˆ yi 2 h ˆ x ˆxi 2 h ˆy ˆxi K , K = β γi K , K + βγi J , K − βγi K , J − γi J , J 2 2 ˆz 2 2 2 ˆz = β γi × (−iJ ) + βγi × 0 − βγi × 0 − γi × (−iJ ) 2 2 ˆz = −i(β − 1)γi × J = −iJˆz, (S.32)

where the last equality follows from the kinematic relation

2 2 2 2 2 2 2 p3 − p0 p0 p3 − p0 (β − 1) × γi = 2 × 2 = 2 = 1 (S.33) p0 Mi Mi

7 Altogether, the generators (S.29) obey the commutation relations

h i h i h i Jˆz, K˜ x = +iK˜ y, Jˆz, K˜ y = −iK˜ x, K˜ x, K˜ y = −iJˆz, (S.34) which are exactly similar to the SO+(2, 1) commutation relations

h i h i h i Jˆz, Kˆ x = +iKˆ y, Jˆz, Kˆ y = −iKˆ x, Kˆ x, Kˆ y = −iJˆz. (S.35)

Therefore, the little group G(p) of a tachyonic momentum is isomorphic to the continuous SO+(2, 1) in 2 + 1 dimensions.

Problem 3(a): µ The quantum state |p, λi has definite momentum p , thus Pˆµ |p, λi = pµ |p, λi and likewise

1 ˆαβ ˆγ 1 γ ˆαβ 2 µαβγJ P |p, λi = 2 µαβγp J |p, λi . (S.36)

For simplicity, let’s assume the particle moves in the z direction and spell out the operator

ˆ def 1 γ ˆαβ Qµ = 2 µαβγp J (S.37) in components:

ˆ 1 3 ˆij 3 ˆ12 z z Q0 = 2 0ij3p × J = p × J = p × J , ˆ 1 3 ˆ02 1 3 ˆ20 1 0 ˆjk Q1 = 2 1023p × J + 2 1203p × J + 2 1jk0p × J = −p3 × Jˆ02 − p0 × Jˆ23 = −pz × Kˆ y − p0 × Jˆx, (S.38) ˆ 1 3 ˆ01 1 3 ˆ10 1 0 ˆjk Q2 = 2 2013p × J + 2 2103p × J + 2 2jk0p × J = +p3 × Jˆ01 − p0 × Jˆ31 = +pz × Kˆ x − p0 × Jˆy, ˆ 1 0 ˆij 0 ˆ12 0 ˆz Q3 = 2 3ij0p × J = −p × J = −p J .

For a massless particle with pz = p0 = E, these components simplify to

z x y z Qˆ0 = +E × Jˆ , Qˆx = −E × Iˆ , Qˆy = −E × Iˆ , Qˆz = −E × Jˆ , (S.39) where the operators Iˆx and Iˆy are as in eq. (7).

8 Besides a definite momentum, the state |p, λi has a definite helicity λ, Jˆz |p, λi = λ |p, λi. Moreover, it is annihilated by other two generators of the little group of its momentum, Iˆx |p, λi = Iˆy |p, λi = 0. Therefore, when we apply the operators (S.39) to this state, we obtain

Qˆ0 |p, λi = +Eλ |p, λi ,

Qˆx |p, λi = 0, (S.40) Qˆy |p, λi = 0,

Qˆz |p, λi = −Eλ |p, λi ,

Comparing the right hand sides here to the particle momentum (with a lower index) pα = (+E, 0, 0, −E) we see that

Qˆα |p, λi = λpα |p, λi . (S.41)

In other words,

1 ˆαβ ˆγ ˆ 2 µαβγJ P |p, λi = λPµ |p, λi . (10)

What if a massless particle moves in another direction? In 3-vector notations, eqs. (S.38) become

Qˆ0 = Eβ~ · Jˆ, Qˆ = EJˆ + β~ × Kˆ  = Rˆ = E ˆI. (S.42)

Up the overall factor E, the components of Qˆ are the generators (6) of the particle’s mo- mentum’s little group. The definite-helicity state |p, λi is annihilated by the two generators ˆ ˆ ~ ˆ I⊥p and is an eigenstate of the third generator Ikp. Indeed, for a massless particle β · J is the helicity operator, hence

Qˆ0 |p, λi = Eλ |p, λi , β~ · Qˆ |p, λi = Eβ~ · Jˆ |p, λi = Eλ |p, λi , (S.43)

while

β~ × Qˆ |p, λi = 0. (S.44)

9 Consequently,   Qˆ |p, λi = β~ β~ · Qˆ |p, λi − β~ × β~ × Qˆ |p, λi = β~ (Eλ |p, λi) = pλ |p, λi , (S.45)

and combining this formula with Qˆ0 |p, λi = Eλ |p, λi we obtain

Qˆµ |p, λi = 2λpµ |p, λi . (S.46)

This proves eq. (10) for a massless particle moving in any direction.

Problem 3(b): µ 0µ µ ν Consider a continuous Lorentz transform x → x = L νx acting on the |p, λi state of a massless particle. The operators on both sides of both sides of eq. (10) transform as Lorentz vectors,

ˆ ˆ ˆ† µ ˆ ˆ  1 ˆαβ ˆγ ˆ† µ  1 ˆαβ ˆγ D(L)PνD (L) = Lν Pµ , D(L) 2 ναβγJ P D (L) = Lν 2 µαβγJ P . (S.47)

Consequently, the transformed state

Dˆ(L) |p, λi = |Lp, ??i (S.48) satisfies the same eq. (10) as the original state |p, λi. Indeed,

µ  1 ˆαβ ˆγ ˆ  1 ˆαβ ˆγ ˆ† ˆ Lν 2 µαβγJ P |Lp, ??i = D(L) 2 ναβγJ P D (L) × D |p, λi ˆ  1 ˆαβ ˆγ = D(L) × 2 ναβγJ P |p, λi hh by eq. (10) ii (S.49) = Dˆ(L) × λPˆν |p, λi † = λ × Dˆ(L)PˆνDˆ (L) × Dˆ |p, λi µ ˆ = λ × Lν Pµ |Lp, ??i , and hence

 1 ˆαβ ˆγ ˆ 2 µαβγJ P |Lp, ??i = λPµ |Lp, ??i . (S.50)

Note that this equation for the transformed state |Lp, ??i has exactly the same helicity eigenvalue λ as the original eq. (10). In other words, the transformed state has the same

10 helicity as the original state, and since the momentum and the helicity completely determine the quantum state of a particle up to an overall phase, it follows that

Dˆ(L) |p, λi = |Lp, same λi × ei phase. (11)

For massless particles, the continuous Lorentz transforms preserve helicity!

Problem 4(a): The Lorentz generators Jˆ and Kˆ obey commutation relations (2). Consequently, for two

components of Jˆ+ or two components of Jˆ− we have

h ˆi ˆj i 1 h ˆi ˆji i h ˆ i ˆji i h ˆi ˆ ji 1 h ˆ i ˆ ji J±, J± = 4 J , J ± 4 K , J ± 4 J , K − 4 K , K = 1 iijkJˆk ± i iijkKˆ k ± i iijkKˆ k − 1 (−i)ijkJˆk 4 4 4 4 (S.51) ijk  i ˆk 1 ˆ k ijk  1 ˆk i ˆ k =  2 J ∓ 2 K = i × 2 J ± 2 K ijk ˆk = i J± ,

while for one component of Jˆ+ and one component of Jˆ0 (or the other way around) we get

h ˆi ˆj i 1 h ˆi ˆji i h ˆ i ˆji i h ˆi ˆ ji 1 h ˆ i ˆ ji J±, J∓ = 4 J , J ± 4 K , J ∓ 4 J , K + 4 K , K = 1 iijkJˆk ± i iijkKˆ k ∓ i iijkKˆ k + 1 (−i)ijkJˆk 4 4 4 4 (S.52) i ijk ˆk ˆ k ˆ k ˆk = 4  J ± K ∓ K − J = 0.

Problem 4(b):

Among the 3 σ, the σ1 and the σ3 matrices are real while the σ2 is imaginary.

At the same time, the σ1 and the σ3 anticommute with the σ2 while the σ2 commutes with

11 itself. Consequently,

∗ σ2(σ1,3) σ2 = +σ2σ1,3σ2 = −σ1,3σ2σ2 = −σ1,3 , (S.53)

while

∗ σ2(σ2) σ2 = −σ2σ2σ2 = −σ2 , (S.54) thus for all 3 Pauli matrices

∗ σ2σ σ2 = −σ. (S.55)

Now let c be any complex 3–vector while c∗ is its complex conjugate. Without connection to any Lorentz symmetries, let

M = exp(−ic · σ) and M = exp(−ic∗ · σ), (S.56) then

∗ ∗ M = σ2M σ2 and M = σ2M σ2. (S.57)

To prove this relation, note that eq. (S.55) implies

∗ ∗ ∗ ∗ σ2(−ic · σ) σ2 = +ic · (σ2σ σ2) = −ic · σ, (S.58) hence

 2∗ ∗ 2 ∗ ∗ ∗ 2 σ2 (−ic·σ) σ2 = σ2(−ic·σ) σ2 ×σ2(−ic·σ) σ2 = (−ic ·σ)×(−ic ·σ) = (−ic ·σ) , (S.59) and likewise for any integer power n,

 n∗ ∗ n σ2 (−ic · σ) σ2 = (−ic · σ) . (S.60)

12 Finally,

∞ !∗ ∞ X 1 X 1 ∗ σ M ∗σ = σ (−ic · σ)n σ = σ (−ic · σ)n σ 2 2 2 n! 2 n! 2 2 n=0 n=0 (S.61) ∞ X 1 = (−ic∗ · σ)n = M, n! n=0

∗ and likewise σ2M σ2 = M.

Now that we have verified eqs. (S.57) for the matrices (S.56), let’s apply this formula to the 2 and the 2 representations of the Lorentz group. By inspection of eqs. (33) and (34) of my notes on Lorentz representations, we see that for a space rotation R through angle φ around axis n, the M(R) and the M(R) matrices have form (S.56) for a real 3–vector 1 c = 2 φn; likewise, for a boost B of r in direction n, the M(B) and the M(B) r also have form (S.56) but for an imaginary 3–vector c = i 2 n. Consequently, for both pure rotations and pure boosts, the M(L) and the M(L) matrices are related via eqs. (14).

To verify eqs. (14) for all continuous Lorentz symmetries L, we note that any such symmetry is a product of a pure rotation and a pure boost, L = RB. Consequently, for any faithful representation of the Lorentz group such as 2 or 2, we have similar decompositions of the matrices representing L = RB, thus

M2(L) = M2(R) × M2(B) and M2(L) = M2(R) × M2. (S.62)

And since we have already verified eqs. (14) for the rotations and the boosts, it follows that

∗ ∗ ∗ ∗ σ2 M(L) σ2 = σ2 M(R) × M(B) σ2 = σ2 M(R) σ2 × σ2 M(B) σ2 (S.63) = M(R) × M(B) = M(L).

Quod erat demonstrandum.

13 Problem 4(c): µ µ Let’s start with reality. The V = V σµ is hermitian if and only if the 4-vector V is real. For any matrix M ∈ SL(2, C), the transform

V → V 0 = MVM † (S.64) preserves hermiticity: if V is hermitian, then so is V 0; indeed

† † † V 0 = MVM † = M † V †M † = MVM † = V 0. (S.65)

µ 0µ µ ν In terms of the 4-vectors, this means that if the V is real than the V = L νV is also real. µ In other words, the 4 × 4 matrix L ν(M) is real.

µ αβ Next, let’s prove that L ν(M) ∈ O(3, 1) — it preserves the Lorentz metric g , or µ αβ 0 0 αβ µ equivalently, for any V , g VαVβ = g VαVβ. In terms of the 2 × 2 matrix V = V σµ, the Lorentz square of the 4-vector becomes the :

αβ µ g VαVβ = det V = Vµσ . (S.66)

Indeed, from the explicit form of the 4 matrices

1 0 ! 0 1 ! 0 −i ! +1 0 ! σ0 = , σ1 = , σ2 = , σ3 = (S.67) 0 1 1 0 +i 0 0 −1 we have ! µ V0 + V3 V1 − iV2 V = V σµ = V1 + iV2 V0 − V3 and hence

2 2 2 2 αβ det(V ) = (V0 + V3)(V0 − V3) − (V1 − iV2)(V1 + iV2) = V0 − V3 − V1 − V2 = g VαVβ . (S.68)

The determinant of a matrix product is the product of the individual matrices’ determi- nants. Hence, for the transform (S.64),

det(V 0) = det(M) × det(V ) × det(M †) = det(V ) × |det(M)|2 . (S.69)

The M matrices of interest to us belong to the SL(2, C) group — they are complex matrices

14 with unit . There are no other restrictions, but det(M) = 1 is enough to assure det(V 0) = det(V ), cf. eq. (S.69). Thanks to the relation (S.66), this means

αβ 0 0 0 αβ g VαVβ = det(V ) = det(V ) = g VαVβ (S.70)

µ — which proves that the matrix L ν(M) is indeed Lorentzian.

Problem 4(d): µ To prove that the Lorentz transform L ν(M) is orthochronous, we need to show that for any 2 0 Vµ in the forward light cone — that is, V > 0 and V0 > 0 — the Vµ is also in the forward 2 light cone. In matrix terms, V > 0 means det(V ) > 0 while V0 > 0 means tr(V ) > 0; together, these two conditions mean that the 2 × 2 V is positive-definite. The transform (S.64) preserves positive definiteness: if for any complex 2-vector ξ 6= 0 we have ξ†V ξ > 0, then

ξ†V 0ξ = ξ†MVM †ξ = (M †ξ)†V (M †ξ) > 0. (S.71)

(Note that M †ξ 6= 0 for any ξ 6= 0 because det(M) 6= 0.) Thus, for any M ∈ SL(2, C) the µ 0µ µ Lorentz transform V → V preserves the forward light cone — in other words, the L ν(M) µ + is orthochronous, L ν(M) ∈ O (3, 1).

The simplest proof that the Lorentz transform (15) is proper — det(L) = +1 — for any SL(2, C) matrix M is topological: The SL(2, C) group — which spans all matrices of the form a b ! M = , complex a, b, c, d, ad − bc = 1 (S.72) c d is connected, so all such matrices are continuously connected to the 12×2 matrix. It is µ µ easy to see that for M = 1, the Lorentz transform L(1) is trivial, L ν = δν , hence all Lorentz transforms of the form (15) are continuously connected to the trivial transform. Consequently, they all must be continuous Lorentz transforms and therefore proper and orthochronous.

15 An alternative proof is more involved. It involves decomposition of M into a product of a unitary U and a Hermitian H, 2 Lemmas showing that L(U) is a pure 3D rotation (without reflections) while L(H) is a pure Lorentz boost, and the group law from part (e) which tells us that

L(M = UH) = L(U) × L(H) = rotation × boost, (S.73)

which makes it a continuous Lorentz transform. But this is a rather long proof, so I leave it as optional exercise for the interested students.

Problem 4(e):

Let L1 = L(M1), L2 = L(M2) and L12 = L(M2M1) be Lorentz transforms constructed according to eq. (15) for some SL(2, C) matrices M1 and M2 and their product M2M1. We want to prove that L12 obtains from a product of consecutive Lorentz transforms, first L2 µ and then L1, so let’s consider how all these transforms act on some 4–vector V . On one hand,

ν ν † ν ν † † L12V σν = (M2M1) × (V σν) × (M2M1) = M2M1 × (V σ ) × M1 M2 . (S.74)

On the other hand,

ν ν  † ν † † L2L1V σν = M2 × (L1V ) σν × M2 = M2 × M1 × (V σν) × M1 × M2 ν ν † † also = M2M1 × (V σ ) × M1 M2 .

Thus we see that ν ν L12V σν = L2L1V σν (S.75) and therefore ν ν L12V = L2L1V . (S.76)

µ Moreover, this holds true for any 4–vector V , hence the Lorentz transforms L12 = L(M2M1) and L2L1 = L(M2)L(M1) must be equal to each other, quod erat demonstrandum.

16 Problem 4(f): For any equivalent to an angular momentum or its analytic continuation, the product of two doublets comprises a triplet and a singlet, 2 ⊗ 2 = 3 ⊕ 1, or in (j) notations, 1 1 ( 2 ) ⊗ ( 2 ) = (1) ⊕ (0). Furthermore, the triplet 3 = (1) is symmetric with respect to permutations of the two doublets while the singlet 1 = (0) is antisymmetric.

For two separate and independent types of angular momenta J+ and J− we combine the

j+ quantum numbers independently from the j− and the j− quantum numbers independently

from the j+. For two bi-, this gives us

1 1 1 1 ( 2 , 2 ) ⊗ ( 2 , 2 ) = (1, 1) ⊕ (1, 0) ⊕ (0, 1) ⊕ (0, 0). (S.77)

Furthermore, the symmetric part of this product should be either symmetric with respect to both the j+ and the j− indices or antisymmetric with respect to both indices, thus

 1 1 1 1  ( 2 , 2 ) ⊗ ( 2 , 2 ) sym = (1, 1) ⊕ (0, 0). (S.78)

Likewise, the antisymmetric part is either symmetric with respect to the j+ but antisym-

metric with respect to the j− or the other way around, thus

 1 1 1 1  ( 2 , 2 ) ⊗ ( 2 , 2 ) antisym = (1, 0) ⊕ (0, 1). (S.79)

+ 1 1 From the SO (3, 1) point of view, the bi- ( 2 , 2 ) is the Lorentz vector. A general 2-index Lorentz tensor transforms like a product of two such vectors, so from the SL(2, C) point of view it’s a product of two bi-spinors, which decomposes to irreducible multiplets according to eq. (S.77).

The Lorentz symmetry respects splitting of a general 2-index tensor into a symmetric tensor T µν = +T νµ and an asymmetric tensor F µν = −F νµ. The symmetric tensor corre- sponds to a symmetrized square of a bi-spinor, which decomposes into irreducible multiplets µ according to eq. (S.78). The singlet (0, 0) component is the Lorentz-invariant trace T µ while the (1, 1) irreducible multiplet is the traceless part of the symmetric tensor.

17 Likewise, the antisymmetric Lorentz tensor F µν = −F νµ decomposes according to eq. (S.79). Here, the irreducible components (1, 0) and (0, 1) are complex but conjugate to each other; individually, they describe antisymmetric tensors subject to complex duality 1 κλµν κλ conditions 2  Fµν = ±iF , or in 3D terms, E = ±iB.

18