February 2016
PX408: Relativistic Quantum Mechanics
Tim Gershon
(T.J.Gershon@warwick.ac.uk)
More selected worked answers
Handout 3: Solving the Dirac Equation
Q1 Show that Σz commutes with β.
−→
Σz β =
Q3 Show explicitly that
σz 0
0 σz
2
1
2Σ
−→
1 0
0 −1
=
σz
0
0 −σz
= βΣz
(1)
= 43 .
2
3
1
1
1 2
Σ = (Σ.Σ) =
Σx + Σ2y + Σ2z = ,
2
4
4
4
(2)
since Σ2x = Σ2y = Σ2z = 1.
Q4 Show that Σ ... does not commute with (α.p + βm).
−→ The βm bit does commute, as we have just seen, so we only need to look at the α.p part. Let’s
look at the x-component of Σ:
0
σx σ.p
σx 0
0 σ.p
(3)
=
Σx α.p =
σx σ.p
0
0 σx
σ.p 0
while
α.pΣx =
0
σ.pσx
σ.pσx
0
(4)
Now, σx σ.p = σx (σx px + σy py + σz pz ) = (σx px − σy py − σz pz )σx , so Σx does not commute with the
Hamiltonian, and by obvious extension neither does Σ.
Q9 Show that ψR and ψL are eigenstates of the chirality operator γ 5 , and give their eigenvalues.
Q10 Show that ψ = ψR + ψL .
Q11 Find expressions for PR ψR , PR ψL , PL ψR and PL ψL .
Q12 Using the standard convention, give matrix forms for PR and PL .
Q13 Using matrix representation, show that (PR )2 = PR , (PL )2 = PL and PR PL = PL PR = 0.
−→
γ 5 ψR,L = γ 5
1
1 5
1 ± γ5 ψ =
γ ± 1 = ±ψR,L ,
2
2
(5)
where we have used (γ 5 )2 = 1. Hence ψR,L are eigenvectors of the helicity operator γ 5 with eigenvectors ±1 (+1 for ψR and −1 for ψL ).
ψR + ψL =
1
1
1 + γ5 ψ +
1 − γ5 ψ = ψ
2
2
PR ψR =
=
=
PR ψL =
=
1
2
1
4
1
2
1
2
1
4
(6)
1
1 + γ5 ψ
2
1 + 2γ 5 + (γ 5 )2 ψ
1 + γ5
(7)
(8)
1 + γ 5 ψ = ψR
(9)
1
1 − γ5 ψ
2
5 2
1 − (γ ) ψ = 0
1 + γ5
(10)
(11)
Similarly PL ψR = 0 and PL ψL = ψL .
PR =
PL =
1
1
1 1 1
0 1
1 0
=
+
1 + γ5 =
1 0
0 1
2
2
2 1 1
1
1
1
0 1
1 0
1 −1
5
=
−
1−γ
=
1 0
0 1
2
2
2 −1 1
(PR )2 =
(PL )2 =
PR PL =
1
4
1
4
1
4
(13)
1 1 1
= PR
2 1 1
1
1 −1
1 −1
1 −1
=
= PL
−1 1
−1 1
2 −1 1
1 0 0
1 1
1 −1
=
= 0 = PL PR
1 1
−1 1
2 0 0
1 1
1 1
(12)
1 1
1 1
=
Q16 By giving the non-relativistic expression for E, find the limit of
σ.p
E−m
(14)
(15)
(16)
as |p| → 0.
−→ Non-relativistic: E = m + |p|2 /2m so E − m = |p|2 /2m. So the denominator of
zero faster than the numerator, and the limit goes to infinity.
σ.p
E−m
goes to
Q17 What are the values of ū(p, s)u(p, s) and v̄(p, s)v(p, s)?
Q18 What are the values of ū(p, s)γ 5 u(p, s) and v̄(p, s)γ 5 v(p, s)?
−→
1/2
u(p, s) = (E + m)
† 0
−→ ū(p, s) = u(p, s) γ
1/2
= (E + m)
φs
σ.p
s
(E+m) φ
†
s †
s † σ .p
(φ ) − (φ )
(E + m)
(17)
(18)
so that
†
s † s
s † σ .pσ.p s
ū(p, s)u(p, s) = (E + m) (φ ) φ − (φ )
φ .
(E + m)2
(19)
The definitions of the Pauli matrices are such that σi† = σi (i.e. they are Hermitian), and hence
p2
2m
s † s
= (E + m)
ū(p, s)u(p, s) = (E + m)(φ ) φ 1 −
= 2m ,
(20)
2
(E + m)
E+m
using (φs )† φs = 1. Similarly
s † s
v̄(p, s)v(p, s) = (E + m)(χ ) χ
p2
−1
(E + m)2
= −(E + m)
2m
= −2m .
E+m
†
0 1
s †
s † σ .p
ū(p, s)γ u(p, s) = (E + m) (φ ) − (φ )
1 0
(E + m)
= σ.p (φs )† φs − (φs )† φs = 0 .
5
φs
σ.p
s
(E+m) φ
(21)
(22)
(23)
Similarly v̄(p, s)γ 5 v(p, s) = 0.
Q19 Find the solution to ...
−→
cos θ −i sin θ
i sin θ − cos θ
a
b
=
a
b
(24)
i.e.
a(cos θ − 1) − ib sin θ = 0
(25)
ia sin θ − b(cos θ + 1) = 0
b2
−→ a(cos θ − 1) − (cos θ + 1) = 0
a
(26)
(27)
It’s is convenient to write this in terms of 2θ , using the trigonometric identities cos 2θ = cos2 θ − sin2 θ
and sin 2θ = 2 sin θ cos θ. A little bit of rearranging then gives the following expression:
−2a2 sin2
θ
θ
− 2b2 cos2
2
2
= 0,
(28)
which can be solved together with the normalisation condition |a|2 + |b|2 = 1 (which originates from
(φ0+ )† φ0+ = φ†+ φ+ = 1) to obtain a = cos( 2θ ), b = i sin( 2θ ). (Note that we can multiply a and b
simultaneously by any phase (e.g. i, or −1) and still get an acceptable solution.)
Q20 Find a similar description for the transformation of the negative helicity state φ− .
Q21 Give the values of (φ0+ )† φ0− and (φ0− )† φ0+ .
−→ For the negative helicity state we have
σ.p0 0
φ = −φ0− ,
|p0 | −
(29)
which for p along the z-axis becomes
cos θ −i sin θ
a
a
=−
i sin θ − cos θ
b
b
with
φ0−
=
a
b
.
(30)
Following the same steps as before, we would now obtain
−2a2 cos2
θ
θ
− 2b2 sin2
2
2
= 0,
i.e. we get the same solutions as before except with a ←→ b: a = i sin( 2θ ), b = cos( 2θ ).
With these results, it is easy to see that
θ
θ
θ
θ
0 † 0
sin
− i sin
cos
= 0 = (φ0− )† φ0+ ,
(φ+ ) φ− = i cos
2
2
2
2
(31)
(32)
i.e. this transformation preserves the orthonormality of the helicity eigenstates.
Q24 Show that the Dirac probability density is invariant under rotations.
Q25 Show that the Dirac (3-)current transforms as a vector under rotations.
−→
The Dirac probability density is ρDirac = ψ̄γ 0 ψ = ψ † ψ. Under transformations, this becomes
ρ0Dirac = (ψ 0 )† ψ 0 = ψ † e−iΣ
† .n̂ θ/2
eiΣ.n̂ θ/2 ψ = ρDirac ,
(33)
where we have used the fact that the Σ matrices are Hermitian (they inherit this feature from the σ
matrices).
The Dirac current is j Dirac = ψ̄γψ. Under transformations, this becomes
† .n̂ θ/2
j 0Dirac = (ψ 0 )† γ 0 γψ 0 = ψ † e−iΣ
γ 0 γeiΣ.n̂ θ/2 ψ ,
(34)
†
so we have to consider the transformation properties of e−iΣ .n̂ θ/2 γ 0 γeiΣ.n̂ θ/2 = e−iΣ.n̂ θ/2 αeiΣ.n̂ θ/2 .
This looks, very familiar: recall that we started by looking for the transformation S such that
S −1 γ µ S = Λµν γ ν , where Λµν is a Lorentz transformation. We have not got that far yet, but we have
discovered a transformation such that S −1 γS gives a rotation, and it is precisely that which appears
above. In other words, if we write R(θ, n̂) to denote a rotation of angle θ about a direction n̂, we
can say e−iΣ.n̂ θ/2 αeiΣ.n̂ θ/2 = R(θ, n̂)α. Hence the Dirac current transforms like a 3-vector.
q
0
0 +m
Q28 Show that cosh(θ/2) = E2m
and sinh(θ/2) = √ px 0
.
2m(E +m)
2
2
hint: cosh 2θ = cosh θ + sinh θ.
−→
cosh θ = γ = E 0 /m = (1 − β 2 )−1 (where β and γ are the Lorentz boost factors), by the definition
of rapidity. Also, by trigonometric identify, cosh θ = cosh2 2θ + sinh2 2θ = 2 cosh2 2θ − 1. Hence
q
q
q
q
q
2 θ
θ
E 0 +m
θ
E 0 −m
E 02 −m2
cosh 2θ = 1+cosh
=
.
We
then
obtain
sinh
=
cosh
−
1
=
=
2
2m
2
2
2m
2m(E 0 +m) =
√
p0x
,
2m(E 0 +m)
where the momentum is aligned with the x-axis.
Q31 Verify by expanding in power series that eαx θ/2 = cosh(θ/2) + αx sinh(θ/2).
Q32 Find similar expressions for eαy θ/2 and eαz θ/2
−→ eαx θ/2 = 1 + (αx θ/2) + (αx θ/2)2 /2 + . . . Since αx2 = 1, the even terms give 1 + (θ/2)2 /2 + . . . while
the odd terms give αx (θ/2+(θ/2)3 /(3!)+. . ., so the series can be identified as cosh(θ/2)+αx sinh(θ/2).
Similarly, eαy θ/2 = cosh(θ/2) + αy sinh(θ/2), eαz θ/2 = cosh(θ/2) + αz sinh(θ/2).
Q41 Find the effect of the parity operation on (i) the time-like part and (ii) the space-like part of
the Klein-Gordon current i (φ∗ ∂ µ φ − φ∂ µ φ∗ ).
Q42 Find the effect of the parity operation on (i) the time-like part and (ii) the space-like part of
the Dirac current ψ̄γ µ ψ.
−→ In both cases the time-like part is unchanged, while the space-like part is inverted (sign flipped),
as one would expect for a current.
Q44 How does the Klein-Gordon current transform under charge conjugation?
−→ φC = φ∗ so i (φ∗ ∂ µ φ − φ∂ µ φ∗ ) has its sign flipped. (Note that the current is conserved under
the product of charge conjugation and parity operations.)
Q48 Insert χ = iσ2 φ∗ into (i∂0 + iσ.∇) φ = mχ. Using complex conjugation (noting the properties of the Pauli matrices) and anticommutation relations, show that (i∂0 − iσ.∇) χ = mφ is
obtained.
−→
(i∂0 + iσ.∇) φ = imσ2 φ∗
− (i∂0 + iσ ∗ .∇) φ∗ = −imσ2∗ φ
(i∂0 + iσ ∗ .∇) φ∗ = −imσ2 φ
since σ2∗ = −σ2 . Now left multiply by σ2 , and use σ22 = 1:
(iσ2 ∂0 + iσ2 σ ∗ .∇) φ∗ = −imφ
(i∂0 − iσ.∇) σ2 φ∗ = −imφ
(i∂0 − iσ.∇) χ = mφ
where in the second line we have used σ2 σ ∗ .∇ = −σ.∇σ2 since σ1,3 are real and anticommute with
σ2 , whereas σ2∗ = −σ2 as noted above.
Q50 Give an expression for φ in terms of the elements of χ.
−→ χ = iσ2 φ∗ so φ = −iσ2 χ∗ .
Handout 4: Implications of Relativistic Quantum Mechanics
Q6 By writing out explicitly ∇.Aψ =
∂
∂x Ax ψ + . . .,
prove ∇.Aψ = ∇.(Aψ) = (∇.A)ψ + A.(∇ψ).
−→
∂
∂
∂
(Ax ψ) +
(Ay ψ) +
(Az ψ)
∂x
∂y
∂z
∂
∂
∂
∂
∂
∂
=
Ax +
Ay +
Az ψ + Ax ψ + Ay ψ + Az ψ
∂x
∂y
∂z
∂x
∂y
∂z
= (∇.A)ψ + A.(∇ψ)
∇.Aψ =
(35)
(36)
(37)
Q9 Show that (σ.X)(σ.Y ) = X.Y + iσ.(X × Y ).
−→
(σ.X)(σ.Y ) = (σx .Xx + σy .Xy + σz .Xz )(σx .Yx + σy .Yy + σz .Yz )
= σx .σx .Xx .Yx + σy .σy .Xy .Yy + σz .σz .Xz .Yz +
(38)
(39)
σx .σy .Xx .Yy + σy .σx .Xy .Yx + σy .σz .Xy .Yz + σz .σy .Xz .Yy + σz .σx .Xz .Yx + σx .σz .Xx .Yz
= X.Y + iσz (Xx .Yy − Xy .Yx ) + iσx (Xy .Yz − Xz .Yy ) + iσy (Xz .Yx − Xx .Yz )
(40)
= X.Y + iσ.(X × Y )
(41)
2
2
1
1
Q12 Using V = − er , En = − α2nm
2 and the results for h r i and h r 2 i, prove the result for ∆Hrel .
−→ We are given
1
1
h i= 2
r
n a
2
h
1
1
i=
.
1
2
r
l + 2 n3 a2
(42)
2
Inserting V = − er and En = − α2nm
2 together with the above results, we find
h
p4
i = En2 − 2En hV i + hV 2 i
4m2
α2 m
α2 m
1
1
= (− 2 )2 − 2(− 2 )(−e2 ) 2 + (−e2 )2
2n
2n
n a
l + 21 n3 a2
(43)
(44)
=
α4 m2 α2 me2
e4
−
+
4n4
n4 a
l + 12 n3 a2
(45)
=
α4 m2 α4 m2
α4 m2
3α4 m2
α4 m2
−
+
=
−
+
4n4
n4
4n4
l + 21 n3
l + 12 n3
(46)
where in the last line we have used a = (mα)−1 and e2 = α.
We wish to calculate
1 p4
1
4
p
=
−
h
i
∆Hrel = −
8m3
2m 4m2
!
α4 m
2n
3
−
= − 4
4n
2
l + 12
(47)
(48)