Lancaster-Blundell-Quantum-Field-Theory-for-the-Gifted-Amateur-Solutions.pdf

Selected Solutions Quantum Field Theory for the Gifted Amateur ∗ Nischal Dwivedi† Sandeep Joshi‡

∗

Authors: Tom Lancaster and Stephen J. Blundell [email protected] ‡ [email protected] †

1 The following is an amateur attempt at solving some of the exercises problems of the book Quantum Field Theory for the Gifted Amateur by Tom Lancaster and Stephen J. Blundell. The current version contains the solutions to first twelve chapters of the book. Updated versions shall follow. We would like to thank Dr.Sudhir R. Jain for his companionship and guidance. Suggestions for improvement are welcome. N.D. S.J.

Chapter 1 √

2

2

1.1 : T1 = Time taken by light to go from A to O = av1+x , √2 b +(d−x)2 T2 = Time taken by the light to go from O to C = , v2 √2 √ 2 2 2 b +(d−x) ∴ Total Time T = av1+x + . v2

dT x (d − x) = √ − p 2 , dx v1 a2 + x2 v2 b − (d − x)2 According to Fermat‘s principle of least action, dT = 0, dx ⇒ Using vi =

c ni ,

v1

√

(d − x) x − p 2 = 0, 2 2 v2 b − (d − x)2 a +x

and relations: √

x (d − x) = sin θ1 ; p 2 = sin θ2 , b − (d − x)2 a2 + x2

we get, n1 sin θ1 = n2 sin θ2 . 2

3

1.2 : δH 1 = lim δf (z) →0  Z

=

Z

G(x, y)[f (y) + δ(y − z)]dy −



Z

G(x, y)f (y)dy ,

G(x, y)δ(y − z)dy,

=G(x, z). δI[f 3 ] 1 = lim δf (x1 ) →0 

Z +1 −1

Z +1

=3 −1

(

=

∴

3

[f (x) + δ(x − x1 )] dx −

Z +1

3

[f (x)] dx , −1

[f (x)]2 δ(x − x1 )dx,

3[f (x1 )]2 0

for − 1 < x1 < +1 . otherwise

 δ3[f (x1 )]2 ] 1 = lim 3([f (x1 )] + δ(x1 − x0 ))2 − 3[f (x1 )]2 , →0  δf (x0 ) =6[f (x1 )]δ(x1 − x0 ), (

6[f (x1 )]δ(x1 − x0 ) δ 2 I[f 3 ] ⇒ = δf (x0 )δf (x1 ) 0

for − 1 < x1 < +1 . otherwise 2

δJ[f ] 1 ∂ = lim [f (y) + δ(y − x)] dy δf (x) →0  ∂y Z   1 ∂f , = lim dy 2 δ 0 (y − x) →0  ∂y Z



−

∂f dy ∂y 

Z

Integrating by parts, we get: δJ[f ] ∂2f = −2 2 . δf (x) ∂x

1.3 :



δG[f ] 1 = lim δf (x) →0 

 Z



g(y, f + δ(y − x))dy − g(y, f ) ,

Taylor Expanding g(y, f + δ(y − x)) upto first order, we get:

2 !

,

4

1 δG[f ] = lim δf (x) →0 

∂g g(y, f ) +  δ(y − x) dy − ∂f

Z 



g(y, f )dy ,

δG[f ] ∂g(x, f ) = . δf (x) ∂f

∴

1.4 :



Z

δφ(x) 1 = lim (φ(x) + δ(x − y) − φ(x)) , δφ(y) →0  δφ(x) = δ(x − y). δφ(y)

∴ ˙ 1 δ φ(t) = lim δφ(t0 ) →0  ∴



d dφ (φ(t) + δ(t − t0 )) − , dt dt 

˙ δ φ(t) d = δ(t − t0 ). δφ(t0 ) dt

1.5 : τ = d3 x(∇ψ)2 , 2 τ = (−2∇2 ψ(y)). 2   Z ρ ∂ψ 2 = , d3 x 2 ∂t ρ ∂ 2 ψ(y) = (−2 . 2 ∂t2 Z

V δV δψ(y) T δT δψ(y) Using L = T − V and

∂L ∂ψ

= 0, ∇2 ψ =

1 ∂2ψ , τ /ρ ∂t2

which is wave equation with velocity of the wave as τ /ρ.

1.6 : δZ0 [J] 1 1 = lim {exp − δJ(z1 ) →0  2 

Z

4



4

d xd y[J(x) + δ(x − z1 )∆(x − y)[J(y) + δ(y − z) 1 −exp − 2 

Z

4

4



d xd y[J(x)∆(x − y)J(y) },

5 Considering the first part Z

4

4

d xd y[J(x) + δ(x − z1 )∆(x − y)[J(y) + δ(y − z) = Z

+

d4 yJ(y)∆(z1 − y) +

Z

Z

d4 xJ(x)∆(x − z1 ) + O(2 ),

= A + B + O(2 ), 1

∴ Z0 [J] = e 2 A,  1 δZ0 [J] 1  1 (A+B) ∴ e2 − e2A . = lim δJ(z1 ) →0  1

Expanding e− 2 B = 1 − 12 B we get, δZ0 [J] B = −Z0 [J] , δJ(z1 ) 2 = −Z0 [J]

Z

d4 xd4 yJ(x)δ(x − y)J(y)

d4 yJ(y)∆(z1 − y).

Chapter 2 2.1 :

r

a ˆ=

mω ι x ˆ+ pˆ , 2~ mω

aˆ† =

r

mω ∴ [ˆ a, a ˆ] = 2~







ι mω x ˆ− pˆ . 2~ mω 



We use the relation [ˆ x, pˆ] = i~; i i [ˆ x, pˆ] + [ˆ p, x ˆ] = 0. mω mω 

mω i i ∴ [aˆ† , aˆ† ] = − [ˆ x, pˆ] − [ˆ p, x ˆ] = 0. 2~ mω mω 



mω i i ∴ [ˆ a, aˆ† ] = − [ˆ x, pˆ] + [ˆ p, x ˆ] = 1. 2~ mω mω 



2 ˆ = pˆ + 1 mω 2 x H ˆ2 . 2m 2

iω ˆ = 1 (ˆ p + imωˆ x2 )(ˆ p − imωˆ x2 ) − [ˆ x, pˆ], ∴H 2m 2 ~ω ˆ = ~ω aˆ† a ∴H ˆ+ , 2   1 ˆ = ~ω aˆ† a ∴H ˆ+ . 2

2.2 : We know from equations 2.12, 2.13, 2.28 and 2.29 that: x ˆ=

~ (ˆ a + aˆ† ), 2mω 6

7 ~ (ˆ a − aˆ† ), 2mω √ a ˆ |ni = n |n − 1i , √ aˆ† |ni = n + 1 |n + 1i . pˆ = −i

ˆ = pˆ2 + mω2 xˆ + λˆ ˆ = The Hamiltonian is H x4 which is of the form H 2m 2 0 4 0 ˆ ˆ H0 + H where H = λˆ x is the perturbation. The perturbation in the state |ni is given by, En0 = hn| Hˆ 0 |ni . Expressing H 0 in terms of a ˆ and aˆ† , we get: H0 = λ



~ 2mω

2

(aˆ† + a ˆ)4 .

We expand (aˆ† + a ˆ)4 . Only the terms with equal number of a ˆ and aˆ† will contribute. Using relations of 2.28 and 2.29: hn| a ˆa ˆaˆ† aˆ† |ni = (n + 1)(n + 2), hn| a ˆaˆ† a ˆaˆ† |ni = (n + 1)2 , hn| aˆ† aˆ† a ˆa ˆ |ni = n(n − 1), hn| aˆ† a ˆa ˆaˆ† |ni = (n + 1)n, hn| aˆ† a ˆaˆ† a ˆ |ni = n2 , ˆ |ni = n(n + 1), hn| a ˆaˆ† aˆ† a ∴ T otal = 6n2 + 6n + 3, En0

∴ ∴ En = E0 +

En0



~ =λ 2mω

2

(6n2 + 6n + 3),

1 3λ ~ω + = n+ 2 4 





~ mω

2

(2n2 + 2n + 1).

2.3 : By substituting 2.62 in equation 2.46, we get: 1 xˆj = √ N



~ m

1/2 X k

1 [ˆ ak eikja + aˆ† −k eikja ], (2ωk )1/2

Since we are summing over all k values, we can write −k → k. 1 ∴ xˆj = √ N



~ m

1/2 X k

1 [ˆ ak eikja + aˆ† k e−ikja ]. (2ωk )1/2

8

2.4 : a ˆ |0i = 0,

(1)

∴ hx| a ˆ |ni = 0. r

a ˆ= aˆ† =

(1a)

mω ι x ˆ+ pˆ , 2~ mω

r





(2)

mω ι x ˆ− pˆ . 2~ mω 



(3)

From (1a) and (2), we have: ι hx| x ˆ+ pˆ |0i = 0, mω 



∴ hx| x ˆ |0i +

ι hx| pˆ |0i = 0. mω

d Using hx| x ˆ |0i = x hx | 0i and hx| pˆ |0i = −i dx hx | 0i ,

~ d x+ hx | 0i = 0, mω dx





d hx | 0i mω =− xdx, hx | 0i ~ Integrating,

2 /2~

hx | 0i = Ae−mωx

,

Using the normalization condition, Z ∞

| hx | 0i |2 dx = 1,

−∞



∴A= 

=⇒ hx | 0i =

mω π~

mω π~

1/4

1/4

, 2 /2~

e−mωx

.

Chapter 3 3.1 :

1 X i(¯p.¯x−¯q.¯y) 1 X i(¯p.¯x−¯q.¯y) e [ˆ ap¯, a ˆ†q¯] = e δp,q , V p¯q¯ V p¯q¯ =

1 X i¯p(¯x−¯y) e , V p¯ = δ 3 (¯ x − y¯).

For Fermions

n

o

a ˆp¯, a ˆ†q¯ = δp,q ,

∴

o 1 X i(¯p.¯x−¯q.¯y) 1 X i(¯p.¯x−¯q.¯y) n e a ˆp¯, a ˆ†q¯ = e δp,q , V p¯q¯ V p¯q¯

=

1 X i¯p(¯x−¯y) e , V p¯ = δ 3 (¯ x − y¯).

3.2 : (a) We can prove this by mathematical induction: For n=1, [ˆ a, aˆ† ] = 1 holds. Lets assume it to be true for n = k, then [ˆ a, (aˆ† )k ] = k(aˆ† )k−1 . For n=k+1, [ˆ a, (aˆ† )k+1 ] = [ˆ a, aˆ† (aˆ† )k ]. Using the relation: [A, BC] = [A, B]C + B[A, C]; [ˆ a, aˆ† (aˆ† )k ] = [ˆ a, aˆ† ](aˆ† )k + aˆ† [ˆ a, (aˆ† )k ], k ∴ = (k + 1)aˆ† , So, true for n=k+1. Hence proved. (b) (ˆ a† )m |0i =

√

m! |mi. 9

10 For m = n and √ by taking adjoint, we get: h0| a ˆn = hn| n!, √ ∴ h0| a ˆn (aˆ† )m |0i = n!m! hn | mi = n!δn,m . √ (c) aˆ† |ni = n + 1 |n + 1i, √ √ ∴ hm| aˆ† |ni = n + 1 hm | n + 1i = n + 1δm,n+1 . √ (d) a ˆ |ni = n |n − 1i, √ √ ∴ hm| a ˆ |ni = n hm | n − 1i = nδm,n−1 .

3.3 : The given Hamiltonian is essentially sum of three independent 1-D Harmonic oscillators. So it can be written as: ˆ = H

3 X pˆ2 i=0

3 X 1 1 + mω 2 xˆi 2 = ~ω aˆi † aˆi + . 2m 2 2 i=0





The angular momentum(say L3 ) can be written as: L3 = xˆ1 pˆ2 − xˆ2 pˆ1 , Using 2.12 and 2.13, 



L3 = −i~ aˆ1 † aˆ2 − aˆ2 † aˆ1 . Doing similarly for L1 , L2 , we get a general formula: Li = −i~ijk aˆj † aˆk . Now, [b0 , b†1 ]

1 = [a3 , a1 + ia2 ] − √ 2 



= 0,

[b0 , b†−1 ] = 0, 1 [b1 , b†−1 ] = − [[a1 , a†1 ] − [a2 , a†2 ]] = 0, 2 And [b0 , b†0 ] = [a3 , a†3 ] = 1, 1 1 [b1 , b†1 ] = [[a1 , a†1 ] + [a2 , a†2 ]] = .2 = 1. 2 2 Similarly [b−1 , b†−1 ] = 1, =⇒ [bi , b†j ] = δi,j .

11

3.3 : Consider the case N=3, for which 1 ψ(r1 , r2 , r3 ) = √ [ψ1 (r1 )ψ2 (r2 )ψ3 (r3 )−ψ1 (r1 )ψ3 (r2 )ψ2 (r3 )+ψ3 (r1 )ψ1 (r2 )ψ2 (r3 ) 3! − ψ3 (r1 )ψ2 (r2 )ψ1 (r3 ) + ψ2 (r1 )ψ3 (r2 )ψ1 (r3 ) − ψ2 (r1 )ψ1 (r2 )ψ3 (r3 )], which can be written in form of Slater determinant as ψ (r ) ψ (r ) 2 1 1 1 1 ψ(r1 , r2 , r3 ) = √ ψ1 (r2 ) ψ2 (r2 ) 3 ψ (r ) ψ (r ) 1 3 2 3



ψ3 (r1 ) ψ3 (r2 ) . ψ3 (r3 )

This can be generalised for arbitrary N as eq.(3.47).

Chapter 4 4.1 : From the relations given in the problem, we can write: ψ(x)ψ † (y) = ζψ † (y)ψ(x) + δ(x − y), ψ(x)ψ(y) = ζψ(y)ψ(x). Consider Vwrong = ψ(x)ψ † (x)ψ(y)ψ † (y). ∴ Vwrong = ζψ † (x)ψ † (y)ψ(x)ψ(y) + δ(x − y)ψ † (x)ψ † (y), = ζ 2 ψ † (x)ψ † (y)ψ(y)ψ(x) + δ(x − y)ψ † (x)ψ † (y). Now, the term ζ 2 is same for bosons and fermions (i.e. 1). Hence, the it yields the same result for boson and fermions.

4.2 : From equations 4.7 and 4.8: X ˆ = 1 ψ †ˆ(x)ψ(y) e−i(¯q.¯x−¯p.¯y) a ˆ†q¯a ˆp¯, V p¯,¯q

Thus, X ˆ >= 1 e−i(¯q.¯x−¯p.¯y) < a ˆ†q¯a ˆp¯ > . ρ(x − y) =< ψ †ˆ(x)ψ(y) V p¯,¯q

4.3 :(a) If t=0, then the matrix 4.68 reduces to a diagonal matrix with eigen values U, 0, 0, U . (b) By taking the result obtained by diagonalising the matrix 4.68, we get E = 0, U, U2 ± 12 (U 2 + 16t2 )1/2 For t/u  1, U 1 2 2 1/2 = U + 4t2 + ... 2 + 2 (U + 16t ) U U 1 2 + 16t2 )1/2 = − 4t2 + ... − (U 2 2 U 12

Chapter 5 5.1 :

∂L dL ∂L ∂L x˙ i + x ¨i . = + dt ∂t ∂xi ∂ x˙ i

Writing

∂L d ∂L ∂L = and = pi =⇒ ∂xi dt ∂ x˙ i ∂ x˙ i dL ∂L dpi = + x˙ i + pi x ¨i , dt ∂t dt ∴

Since H = pi x˙ i − L ;

dL ∂L d = + (pi x˙ i ). dt ∂t dt ∂L dH =− . ∂t dt

5.3 : [A, B]† = (AB)† − (BA)† = B † A† − A† B † = [B † , A† ] = [B, A] = −[A, B].

5.4 : γ −1 ∴L=

v2 =1− 2 +O 2c

v4 c4

!

,

mv 2 −mc2 = −mc2 + + O(v 4 ). γ 2 p=

∂L = mv, ∂v

mv 2 ∴ H = px˙ − L = mv − −mc + + O(v 4 ) 2 2

2

13

!

= mc2 +

mv 2 . 2

14

5.5 : For a straight world line ∆x2 = 0, ds1 =cdt. Z b

s1 = a

ds1 = c(tb − ta ).

For ∆x2 6= 0, p

ds2 = c2 dt2 − dx2 = cdt/γ, Z b

ds2 =

s2 = a

γ ≥1,

Since, hence,

c (tb − ta ). γ

s2 ≤s1 .

2

5.6 : L = − mc γ + qA · v − qV. ⇒

∂L ∂v

= γmv + qA.

∂L =(∇L)i = (∇(qA · v) − q∇V )i , ∂xi =q((A · ∇)v + (v · ∇)A + v × ∇ × A + A × ∇ × v − ∇V )i , =q((v × B − ∇V )i . Now, Euler-Lagrange equation gives d ∂L ∂L − =0, dt ∂v ∂x d (γmv + qA) =q(v × B − ∇V ), dt   d ∂A (γmv) =q −∇V − + qv × B, dt ∂t d ⇒ =q(E + v × B). dt 



2

5.7 : L = − mc γ + qA · v − qV. ⇒p=

∂L ∂v

= γmv + qA.

15 For v  c,



v2 2c2

γ ≈ 1+



,

Hence, p = mv + qA. H =v · p − L, 1 =v · (mv + qA) − (−mc2 + mv 2 + qA · v − qV ), 2 1 =mc2 + mv 2 + qV, 2 1 2 =mc + (p − qA)2 + qV. 2m

5.9 : First consider ∂µ F µν = J µ , For µ = 0, ∂0 F 00 + ∂i F 0i = ∇ · E = 0. For µ = i, ∂0 F i0 + ∂j F ij = E˙ i + ∂j (−ijk B k ) = 0, ⇒ (∇ × B)i = E˙ i .

(∵ F ij = −ijk B k )

Also we have ∂λ Fµν + ∂µ Fνλ + ∂ν Fλµ = 0. For λ = 0, ∂0 Fµν + ∂µ Fν0 + ∂ν F0µ = 0. For µ = i and ν = j, we have ∂0 Fij + ∂i Fj0 + ∂j F0i =0, ∂0 (−ijk B k ) − ∂i Ej + ∂j Ei =0, − B˙ k − (∂ E − ∂ E ) =0, ijk

ijk

−

i

j

j

i

ijk B˙ k + ijk (∂i Ej − ∂j Ei ) =0, B˙ k + (∇ × E)k =0, or, (∇ × E) +

∂B =0. ∂t

For λ = k, µ = i and ν = j , we have ∂k Fij + ∂i Fjk + ∂j Fki =0, lmn ∂l (ijk Fmn ) =0, ijk ∂l (lmn Fmn ) =0, ijk ∂l B l =0, ∇ · B =0.

16

5.10 : ∂α F αβ = J β , 1 ∂β ∂α F αβ = (∂β ∂α F αβ + ∂α ∂β F βα ), 2 1 = (∂β ∂α F αβ − ∂β ∂α F αβ ), 2 =0. Hence ∂β J β = 0. Since J µ = (ρ, J), we have ∂µ J µ = which is equation of continuity.

∂ρ + ∇ · J = 0, ∂t

(∵ Fαβ = −Fβα )

Chapter 6 6.1 : The Euler-Lagrange equation is ∂L ∂L − ∂µ = 0. ∂φ ∂(∂µ φ) For L = 21 (∂µ φ)2 − 12 m2 φ2 , Euler-Lagrange equation gives −m2 φ − ∂µ (∂ µ φ) = 0 ⇒ (∂ 2 + m2 )φ = 0. Conjugate Momentum ∂L ∂ 1 ˙2 1 ˙ π= = (φ − (∇φ)2 ) − m2 φ2 = φ. ˙ ˙ 2 ∂φ ∂φ 2 



Hamiltonian density H = π φ˙ − L = φ˙ 2 −



1 1 1 1 ˙2 (φ − (∇φ)2 ) − m2 φ2 , = (φ˙ 2 + (∇φ)2 ) + m2 φ2 . 2 2 2 2 

17

Chapter 7 7.1 : For the given Lagrangian ∞ X ∂L = −m2 φ − λn (2n + 2)φ2n+1 , ∂φ n=1

∂L = ∂ µ φ. ∂(∂µ φ)

Therefore the Euler-Lagrange equation gives ∞ X ∂L ∂L − ∂µ = 0 ⇒ −m2 φ − λn (2n + 2)φ2n+1 − ∂µ (∂ µ φ) = 0, ∂φ ∂(∂µ φ) n=1

(∂ 2 + m2 )φ +

∞ X

λn (2n + 2)φ2n+1 = 0.

n=1

7.2 :

∂L = ∂ µ φ. ∂(∂µ φ)

∂L = −m2 φ + J, ∂φ

Euler-Lagrange equation:

(∂ 2 + m2 )φ = J.

7.3 : For the φ1 field ∂L = −m2 φ1 − 4g(φ21 + φ2 )φ1 , ∂φ1 Euler-Lagrange equation:

∂L = ∂ µ φ1 . ∂(∂µ φ1 )

(∂ 2 + m2 )φ1 + 4g(φ21 + φ2 )φ1 = 0.

Similarly for the φ2 field we get (∂ 2 + m2 )φ2 + 4g(φ21 + φ2 )φ2 = 0.

7.4 : (∂µ φ)2 = ∂ µ φ∂µ φ = φ˙ 2 − (∇φ)2 , Therefore

L = 21 (φ˙ 2 − (∇φ)2 ) − 12 m2 φ2 . 18

19

π=

∂L ∂ φ˙

=

∂ ∂ φ˙

h

1 ˙2 2 (φ

H = π φ˙ − L = φ˙ 2 − Πµ =

∂L ∂(∂µ φ)

Π0 =

∂L ∂ φ˙

=

i

− (∇φ)2 ) − 12 m2 φ2 = φ˙ . h

∂ ∂(∂µ φ)

˙ = π = φ.

1 ˙2 2 (φ

h

i

− (∇φ)2 ) − 21 m2 φ2 , = 12 (φ˙ 2 + (∇φ)2 ) + 21 m2 φ2 .

1 2 2 (∂µφ)

i

− 21 m2 φ2 = ∂ µ φ.

Chapter 8 ˆ (t2 , t1 ) = exp[−iH(t ˆ 2 − t1 )], 8.1 : U ˆ (t1 , t1 ) = 1. (1) U ˆ (t3 , t2 )U ˆ (t2 , t1 ) = exp[−iH(t ˆ 3 − t2 )]exp[−iH(t ˆ 2 − t1 )] = exp[−iH(t ˆ 3 − t1 )] = U ˆ (t3 , t1 ). (2) U ˆ (t2 , t1 ) = i d (exp[−iH(t ˆ 2 − t1 )]) = H ˆU ˆ (t2 , t1 ). (3) i dtd2 U dt2 ˆ (t1 , t2 ) = exp[−iH(t ˆ 1 − t2 )] = exp[iH(t ˆ 2 − t1 )] = U ˆ −1 (t2 , t1 ). (4) U ˆ † (t2 , t1 )U ˆ (t2 , t1 ) = exp[iH(t ˆ 2 − t1 )]exp[−iH(t ˆ 2 − t1 )] = 1. (5) U

8.2 : Heisenberg’s equation of motion dˆ a†k (t) 1 † ˆ = [ˆ a (t), H]. dt i~ k ˆ = P Ek a ˆk , we have For H ˆ†k a k X dˆ a†k (t) 1 † X = (ˆ ak En a ˆ†n a ˆn − En a ˆ†n a ˆn a ˆ†k ), dt i~ n n X 1 X = ( En a ˆ†n a ˆ†k a ˆn − En a ˆ†n a ˆn a ˆ†k ), i~ n n X 1 X = ( En a ˆ†n (ˆ an a ˆ†k − δnk ) − En a ˆ†n a ˆn a ˆ†k )), i~ n n 1 † = − Ek a ˆk . i~

Integrating both sides, we get i log(ˆ a†k (t)/ˆ a†k (0)) = Ek t, ~ a ˆ†k (t) =ˆ a†k (0)eiEk t/~ . 20

21 Similarly, for the operator a ˆk we get a ˆk (t) = a ˆk (0)e−iEk t/~ . ˆ = Xlm a 8.3 : The Heisenberg’s equation of motion for operator X ˆ†l a ˆm is ˆ dX 1 ˆ = [Xlm a ˆ†l a ˆm , H], dt i~ ˆ = P Ek a where H ˆ†k a ˆk . k ˆ X X dX 1 = Xlm (ˆ a†l a ˆm Ek a ˆ†k a ˆk − Ek a ˆ†k a ˆk a ˆ†l a ˆm ). dt i~ k k a ˆ†l a ˆm a ˆ†k a ˆk =ˆ a†l (δmk + a ˆ†k a ˆm )ˆ ak , =ˆ a†l a ˆm + a ˆ†k (ˆ ak a ˆ†l − δlk )ˆ am , =ˆ a†l a ˆm + a ˆ†k a ˆk a ˆ†l a ˆm − a ˆ†l a ˆm , =ˆ a†k a ˆk a ˆ†l a ˆm , ˆ dX = 0, dt ˆ = constant. or, X ⇒

ˆ H is 8.4 : The Heisenberg’s equation of motion of the spin operator S i

ˆH dS ˆ ˆ H , H]. = [S dt

x dSˆH x x =(SˆH H − H SˆH ), dt y ˆx x ˆy =ω(SˆH SH − SˆH SH ), z x ˆy z =iω SˆH , (∵ [SˆH , SH ] = iSˆH ) x dSˆH z ⇒ =ω SˆH . dt

i

Similarly , z dSˆH y ˆz z ˆy x =ω(SˆH SH − SˆH SH ) = −iω SˆH , dt z dSˆH x ⇒ = − ω SˆH . dt

i

z ˆy x ([SˆH , SH ] = −iSˆH ),

Chapter 9 9.1 : The translation operator is ˆ (a) = e−iˆp.a , U

Hence, pˆ =

ˆ (a − 1i ∂ U ∂a

. a=0

9.2 : cosh φ1 sinh φ1  sinh φ1 cosh φ1  D(φ1 ) =   0 0 0 0 

0 0  . 0 1



1 0 0 0

0 1  = −i  0 0

1 ∂D(φ1 ) K = i ∂φ1 φ1 =0

1



0 0 1 0

0 0 0 0



0 0  . 0 0

Similarly, we get K 1 and K 2 .

9.3 : For infinitesimal boost by v j along the xj axis, we have x00 = x0 + vi xi , 0j

j

j 0

x =x +v x .

(1) (2)

Also, we have for Lorentz transformations x00 =Λ0ν xν = Λ00 x0 + Λ0i xi , x

0j

=Λjν xν

=

Λj0 x0

+

Λjk xk .

Equating Eq.(1) with Eq.(3), and Eq.(2) with Eq.(4), we have Λ00 =1

Λ0i = vi ,

Λi0 =v i

Λij = δji . 22

(3) (4)

23 Hence, the transformation matrix is given by 1 v1 v2 v3 v 1 1 0 0    Λµν =  2 . v 0 1 0 v3 0 0 1 



For infinitesimal rotations by θj about xj , the rotation matrix is given by 



1 θ3 −θ2  3  i 1 θ1  . Rj = −θ 2 1 θ −θ 1

(5)

The Lorentz transformation under rotation gives x00 = x0 ,

x0i = Rij xj .

(6)

Hence the Lorentz transformation matrix is 



1 0 0 0 0 3 2 1 θ −θ   Λµν =  . 0 −θ 3 1 θ1  0 θ2 −θ1 1 General infinitesimal Lorentz transformation x0µ = Λµν xν = (δνµ + ωνµ )xν , where

0 v1 v2 v3 v 1 3 0 θ −θ2    ωνµ =  2 . v −θ3 0 θ1  v 3 θ2 −θ1 0 



1 0 0 0 0 v1 v2 v3 0 −1 0   0  v 1 0 θ3 −θ2    = gµλ ωνλ =   2 , 0 0 −1 0  v −θ3 0 θ1  0 0 0 −1 v 3 θ2 −θ1 0 

ωµν



0 v1 v2 v3 −v 1 0 −θ3 θ2    = 2 , −v θ3 0 −θ1  −v 3 −θ2 θ1 0 

⇒ ωµν



which is antisymmetric i.e. ω T = −ω.



24 Clearly, v i = ω 0i . Also, ω ij =ijk θk , ilm ω ij =ilm ijk θk , =(δ lj δ mk − δ lk δ mj )θk , =δ lj θm − δ mj θl , ⇒ δlj ilm ω ij =δlj δ lj θm − δlj δ mj θl , ijm ω ij =2θm , 1 or, θm = − mij ω ij . 2

Chapter 10 10.1 : [φ(x), P α ] =[φ(x),

Z

d3 yT 0α ],

Z

d3 yΠ0 (y)∂ α φ(y)],

=[φ(x), Z

d3 y[φ(x), Π0 (y)]∂ α φ(y),

Z

d3 y(iδ (3) (x − y))∂ α φ(y),

= =

=i∂ α φ(x).

10.2 : L = L(φ1 , ..., φN ; ∂µ φ1 , ..., ∂µ φN ; xµ ). !

δL =

X i

=

X  ∂L i

=

∂L ∂L δφi + δ(∂µ φi ) , δφi ∂(∂µ φi ) δφi



δφi +

X  ∂L i

δφi

−

Πµi δ(∂µ φi )

∂µ Πµi



δφi +

, ∂µ (Πµi δφi )



.

Using Euler-Lagrange equation the first term vanishes, therefore δL =

X

=

X

∂µ (Πµi δφi ),

i

∂µ (Πµi Dφi )δλ,

i

!

=∂µ

X

Πµi Dφi

i

25

δλ.

26 Also, δL = (∂µ W µ )δλ. Using these two equations, we have !

δL =∂µ

X

Πµi Dφi

−W

µ

δλ = 0,

i

∂µ J µ =0,

or, where J µ =

µ i Πi Dφi

P

− W µ.

10.3 : L = 21 (∂µ φ)2 − 12 m2 φ2 . Πµ =

∂L ∂(∂µ φ)

= ∂ µ φ.

T µν =Πµ ∂ ν φ − g µν L, µ

ν

=(∂ φ)(∂ φ) − g

µν



1 1 (∂λ φ)2 − m2 φ2 , 2 2 

1 1 = (∂ µ φ)(∂ ν φ) + g µν m2 φ2 . 2 2 1 1 T 00 = φ˙ 2 + m2 φ2 = H. 2 2

∂µ T µν =∂µ



1 µ 1 (∂ φ)(∂ ν φ) + g µν m2 φ2 , 2 2 

1 = ((∂µ ∂ µ φ)(∂ ν φ) + (∂ µ φ)(∂µ ∂ µ φ)) + m2 (∂ ν φ)φ, 2 Using Euler- Lagrange equation: (∂ 2 + m2 )φ = 0, we have  1 µ (∂ φ)(∂µ ∂ ν φ) + m2 (∂ ν φ)φ , 2   1 = (∂ ν φ) ∂ 2 φ + m2 φ , 2 =0.

∂µ T µν =

27 10.4 : L = − 41 Fµν F µν = 21 (E2 − B2 ). ∂L , ∂(∂σ Aρ )   1 ∂ µ ν ν µ − (∂µ Aν − ∂ν Aµ )(∂ A − ∂ A ) , = ∂(∂σ Aρ ) 4 i 1h σ ρ (δµ δν − δνσ δµρ )(∂ µ Aν − ∂ ν Aµ ) + (∂µ Aν − ∂ν Aµ )(δ σµ δ ρν − δ σν δ ρµ , =− 4 = − F σρ .

Πσρ =

Tνµ =Πµσ ∂ν Aσ − δνµ L, ⇒ T µν =Tλµ g λν = Πµσ ∂ ν Aσ − g µν L, 1 = − F µσ ∂ ν Aσ + g µν F αβ Fαβ . 4 T˜µν =T µν + ∂λ X λµν , =T µν + ∂λ (F µλ Aν ), =T µν + F µλ ∂λ Aν , 1 = − F µλ ∂ ν Aλ + g µν F αβ Fαβ + F µλ ∂λ Aν , 4 1 =F µλ (∂λ Aν − ∂ ν Aλ ) + g µν F αβ Fαβ , 4 1 µν αβ µλ ν =F Fλ + g F Fαβ . 4 1 T˜00 =F 0λ Fλ0 + F αβ Fαβ , 4 1 =F 0i Fi0 − (E2 − B2 ), 2 1 2 2 =E − (E − B2 ), 2 1 2 = (E + B2 ). 2 T˜i0 =F iλ Fλ0 , =F ij Fj0 , = − kij Bk (−Ej ), =ijk Ej Bk , =(E × B)i .

Chapter 11 R ˆ 11.2 : φ(x) =

d3 p (2π)3/2

√1

2Ep

(ˆ ap e−ip.x + a ˆ†p eip.x ).

ˆ ˆ 0 (y) =∂0 φ(y), Π Z

= Z

=i

d3 p 1 p (ˆ ap (−iEp )e−ip.y + a ˆ†p (iEp )eip.y ), 3/2 2Ep (2π) d3 p (2π)3/2

s

Ep (−ˆ ap e−ip.y + a ˆ†p eip.y ). 2

i d3 p 3 ˆ ˆ [φ(x), Π(y)] = ˆq ]ei(p.x−iq.y) }, d q{[ˆ ap , a ˆ†q ]e−i(p.x−q.y) − [ˆ a†p , a 2 (2π)3 Z d3 p 3 i d q{δ 3 (p − q)e−i(p.x−q.y) + δ 3 (p − q)ei(p.x−q.y) }, = 2 (2π)3 Z i d3 p −ip.(x-y) = (e + eip.(x-y) ) (∵ x0 = y 0 ), 2 (2π)3 Z

Swapping p with −p in first term, ˆ ˆ [φ(x), Π(y)] =i

Z

d3 p ip.(x-y) e = iδ 3 (x − y). (2π)3

28

Chapter 12 12.1 : H = ∂0 ψ † (x)∂0 ψ(x) + ∇ψ † (x).∇ψ(x) + m2 ψ † (x)ψ(x). Using the definition of ψ and ψ † from Eq.(12.5) in the text, we get, 1 1 d3 p d3 q p p Ep Eq (ˆ a†p a ˆq ei(p−q).x 3/2 3/2 2Ep 2Eq (2π) (2π) −a ˆ† ˆb† ei(p+q).x − ˆbp a ˆq e−i(p+q).x + ˆbpˆb† e−i(p−q).x ),

† ˆ ˆ (∂0 ψ(x)) (∂0 ψ(x)) =

Z

Z

p q

q

d3 q d3 p 1 1 p p p.q(ˆ a†p a ˆq ei(p−q).x 3/2 3/2 2Ep 2Eq (2π) (2π) † ˆ† i(p+q).x −i(p+q).x † −i(p−q).x ˆ ˆ ˆ −a ˆ b e − bp a ˆq e + bp b e ),

ˆ ∇ψˆ† (x).∇ψ(x) =

Z

Z

p q

ˆ ψˆ† (x)ψ(x) =

q

Z

d3 q 1 1 d3 p p p (ˆ a†p a ˆq ei(p−q).x 3/2 3/2 2E 2E (2π) (2π) p q † ˆ† i(p+q).x −i(p+q).x ˆ +a ˆ b e + bp a ˆq e + ˆbpˆb† e−i(p−q).x ). Z

p q

q

R ˆ = d3 xH. ˆ Using Now, H R 3 i(p−q).x d xe = δ (3) (p − q), we get

ˆ = H

Z

d3 q d3 p 1 1 p p {(Ep Eq +p.q+m2 )(ˆ a†p a ˆq ei(Ep −Eq )t 3/2 3/2 2Ep 2Eq (2π) (2π) + ˆbpˆb† e−i(Ep −Eq )t )δ (3) (p − q) + (−Ep Eq − p.q + m2 ) Z

q

(ˆ a†pˆb†q ei(Ep +Eq )t + ˆbp a ˆq e−i(Ep +Eq )t )δ (3) (p + q)},

ˆ = H

Z

d3 p 1 {(Ep2 + p2 + m2 )(ˆ a†p a ˆp + ˆbpˆb†p )+ (2π)3 2Ep ˆ−p e−2iEp t )}, (−E 2 + p2 + m2 )(ˆ a† ˆb† e2iEp t + ˆbp a p −p

p

29

30 Using Ep = p2 + m2 , we get, d3 p Ep (ˆ a†p a ˆp + ˆbpˆb†p ), (2π)3 Z d3 p ˆ = Ep (ˆ a†p a ˆp + ˆb†pˆbp ). N [H] (2π)3 ˆ = H

Z

12.2 : ˆ [ψ(x), ψˆ† (y)] =

Z

d3 p 1 p (2π)3/2 2Ep

d3 q 1 p [ˆ ap , a ˆ†q ]e−i(p.x−q.y) + (2π)3/2 2Eq 

Z



[ˆ ap , ˆbq ]e−i(p.x+q.y) + [ˆb†p , a ˆ†q ]ei(p.x+q.y) + [ˆb†p , ˆbq ]ei(p.x−q.y) , Using commutation relations, we get ˆ [ψ(x), ψˆ† (y)] =

Z

d3 p 1 p 3/2 2Ep (2π)

Z

d3 q 1 p δ (3) (p−q)e−i(p.x−q.y) + 3/2 2Eq (2π) 

δ

ˆ x), ψˆ† (t, y)] = [ψ(t,

(3)

i(p.x−q.y)

(q − p)e


d3 p 1 e−ip.(x−y) − eip.(x−y) . 3 (2π) 2Ep

Z

d3 p d3 q [ˆ ap , a ˆ†q ]e−i(p.x−q.y) , (2π)3/2 (2π)3/2 Z d3 p (3) = δ (p − q)e−i(p.x−q.y) , (2π)3

ˆ ˆ † (y)] = [Ψ(x), Ψ

Z

Z

ˆ x), Ψ ˆ † (t, y)] = δ (3) (x − y). [Ψ(t,

12.3 :For two scalar fields, µ JN =(∂ µ φ1 )Dφ1 + (∂ µ φ2 )Dφ2 ,

Z

QN =

d

0 xJN

Z

=

d3 x(φ˙ 1 Dφ1 + φ˙ 2 Dφ2 ),

Z

d3 y [φ˙ 1 (y), φ1 (x)]Dφ1 (y) + [φ˙ 2 (y), φ1 (x)]Dφ2 (y) ,

Z

d3 y(−iδ (3) (x − y))dφ1 (y),

[QN , φ1 (x)] = =

3




= − iDφ1 (x).



,

31 Similarly, [QN , φ2 (x)] = −iDφ2 (x). Now ˆ =[Q ˆ N , ψ] ˆ N , √1 (φ1 + iφ2 )], [Q 2
1 ˆ ˆ N , φˆ2 ] , = √ [QN , φˆ1 ] + i[Q 2
1 = √ − iDφˆ1 + i. − iDφˆ2 , 2   1 = − iD √ (φˆ1 + iφˆ2 ) , 2 ˆ ˆ = − iDψ = −i.iψˆ = ψ.



12.4 : L = 2i ∂0 ρ − ρ∂0 θ −

1 1 2 2m 4ρ (∇ρ)



+ ρ(∇θ)2 − g2 ρ2 .

As θ → θ + α δθ = 1, δα ∂L Π0θ = = −ρ, ∂(∂0 θ) 0 JN =Π0θ Dθ = −ρ,

Dθ =

∴ QN c = −

Z

d3 xρ(x),

Using [QN , θ] = −iDθ Z

[−

ˆ x)] = − i, d3 y ρˆ(t, y), θ(t, ˆ x)] =i. ˆ (t), θ(t, [N

12.5 : L = iΨ(x)∂0 Ψ(x) −

1 † 2m ∇Ψ (x).∇Ψ(x)

− V (x)Ψ† (x)Ψ(x).

The Euler-Lagrange equations for Ψ give ∂L ∂L − ∂0 =0, ∂Ψ ∂(∂0 Ψ) 1 2 † ∇ Ψ − V (x)Ψ† − i∂0 Ψ† =0, 2m ∂Ψ† 1 2 † ⇒i = ∇ Ψ − V (x)Ψ† , ∂t 2m

32 Taking transpose ∂Ψ 1 2 =− ∇ Ψ + V (x)Ψ, ∂t 2m which is Schrodinger equation. i

For V (x) = 0 i

1 2 ∂Ψ =− ∇ Ψ. ∂t 2m

Since Ψ∼

X

ap e−ip.x ,

p

we get, i

X

ap (−iEp )e−ip.x = −

p

1 X ap (ip.ip)e−ip.x , 2m p

⇒ Ep =

12.6 : For L = iΨ(x)∂0 Ψ(x) −

p2 . 2m

1 † 2m ∇Ψ (x).∇Ψ(x)

− V (x)Ψ† (x)Ψ(x),

we have, from Eq.(12.25) Π0Ψ = iΨ† ,

Π0Ψ† = 0,

Also

∂L 1 =− (∇Ψ† )i , ∂(∂i Ψ) 2m ∂L 1 ΠiΨ† = =− (∇Ψ)i , † ∂(∂i Ψ ) 2m For an infinitesimal change in phase ΠiΨ =

Ψ → Ψ + iΨδα, Ψ† → Ψ† − iΨ† δα,

DΨ = iΨ, DΨ† = −iΨ† .

Noether currents are given by 0 JN =Π0Ψ DΨ + Π0Ψ† DΨ† ,

=iΨ† .iΨ = −Ψ† Ψ, = − ρ(x) = Probability density. JN =ΠΨ DΨ + ΠΨ† DΨ† , 1 1 =− ∇Ψ† .iΨ + − ∇Ψ. − iΨ† , 2m 2m  i  =− Ψ∇Ψ† − Ψ† ∇Ψ = Current density. 2m

33

ˆ (α) = eiQˆ N α , 12.7 : U ˆ U ˆ iQˆ N α , ˆ † ψ(x) ˆ =e−iQˆ N α ψe U

ˆ N α + 1 (−iQ ˆ N α)2 + ... ψˆ 1 + iQ ˆ N α + 1 (iQ ˆ N α)2 + ... , = 1 − iQ 2! 2! 2 2 α ˆN − ˆ 2 − iαQ ˆ N ψˆ + α2 Q ˆ N ψˆQ ˆN − α Q ˆ 2 ψˆ + ..., =ψˆ + iαψˆQ ψˆQ N 2! 2! N 2 ˆ Q ˆ
+ ..., ˆ Q ˆ N ]Q ˆN + Q ˆ N [Q ˆ N , ψ] ˆ N ] − α [ψ, =ψˆ + iα[ψ, 2! ˆ = ψ (prob. 12.3(c)) ˆ N , ψ] Using [Q 2
ˆ U ˆ † ψ(x) ˆ =ψˆ − αDψˆ − α − ψˆQ ˆN + Q ˆ N ψˆ + ..., U 2! 2 α ˆ + ..., ˆ N , ψ] =ψˆ − iαψˆ − [Q 2! α2 ˆ =(1 − iα − + ...)ψ, 2! ˆ =e−iα ψ.