Quantum Field Theory Set 1: solutions Exercise 1

Quantum Field Theory

Set 1: solutions

Exercise 1

The system of natural units is a choice of units of measure commonly employed in QFT in order to simplify the notation. It corresponds to measuring velocities in units of speed of light c and quantities with dimensions of an action in units of ~ . Using this trick, one can reduce every physical unit to the only remaining one, the energy, usually measured in eV (or GeV = 10 ⁹ eV). In the following we use the fact that c ∼ L/T and ~ ∼ M L ² /T .

1. M ∼ M c ² c ² ∼ eV

c ² ∼ eV 2. L ∼ L ~ c

~ c ∼ L ~ L/T

M L ² /T c ∼ (eV) ⁻¹ ~ c ∼ (eV) ⁻¹ 3. T ∼ T c

c ∼ L

c ∼ (eV) ⁻¹ ~ ∼ (eV) ⁻¹ 4. v ∼ L

T ∼ (eV) ⁰ 5. F ∼ M L

T ² ∼ (eV) ² 1

~ c ∼ (eV) ² 6. Coulomb Force ∼ e ²

L ² ⇒ e ² ∼ F L ² ∼ (eV) ⁰ ~ c ∼ (eV) ⁰ 7. E ∼ F _e

e ∼ (eV) ²

8. Lorentz Force ∼ evB ⇒ B ∼ F M

ev ∼ (eV) ² ~ ^−3/2 c ^−5/2

Notice that one could infer the dimensionality of time and length considering for example an incoming wave with energy E. The associated frequency (inverse of time) is defined by ν = E/h and the wavelength by λ = c/ν. Indeed highly energetic waves correspond to X-rays while, decreasing the energy (and correspondingly the frequency), one reaches UV-rays, IR-rays radio-waves and so on.

Exercise 2

Let us perform calculations in the system of natural units, without factors of c and ~ . Of course once we need to compare the obtained results to numerical data we must convert them into the usual units like cm, g, s, ... . The relation is unique as one can see immediately:

~ = 0.66 · 10 ⁻¹⁵ eV sec = 1 = ⇒ 1 sec = 1, 51 · 10 ¹⁵ eV ⁻¹ .

Vice versa, given a time T expressed in eV ⁻¹ one can obtain the equivalent in standard units (t _e and t _s ∈ R):

T = t e eV ⁻¹ = t s sec = ⇒ t s = 0.66 · 10 ⁻¹⁵ t e . In the same way:

~ c = 197 · 10 ⁻⁷ eV cm = 1 = ⇒ 1 cm = 5.08 · 10 ⁴ eV ⁻¹ .

Vice versa, given a length L expressed in eV ⁻¹ one can obtain the equivalent in standard units (l e and l c ∈ R):

L = l _e eV ⁻¹ = l _c cm = ⇒ l _c = 197 · 10 ⁻⁷ l _e .

The electric charge:

1 esu ² = 1 erg · cm = 10 ⁻⁷ J · cm = 10 ⁻⁷ · 6.25 · 10 ¹⁸ eV · 5.08 · 10 ⁴ eV ⁻¹ = 3.18 · 10 ¹⁶

Exercise 3

Let us consider a particle of mass M and energy ˜ E moving along the ˆ z direction. We define this frame as the laboratory frame and denote all quantities measured in this frame with a ’∼’. The four momentum of the particle is defined by

P ˜ _µ = ( ˜ E, 0, 0, p ˜ _z ), and the energy and the spatial momentum are related by the constraint

P ˜ ² ≡ P ˜ _µ P ˜ ^µ = M ² = ˜ E ² − p ˜ ² _z .

We recall that the square of a four momentum is a Lorentz invariant, which means that its value doesn’t change regardless of the reference frame in which it is computed. In particular, let us call P µ the four momentum of the particle in its center of mass frame. By definition the particle is at rest in this frame, hence its spatial momentum is zero. Thus the above constraint tells us that

P ² ≡ P µ P ^µ = M ² ⇒ P µ = (M, 0, 0, 0),

and we conclude that the energy of the particle in the center of mass is given by the mass M , which is expected since a particle at rest doesn’t have kinetic energy. Finally let us compute the velocity of the particle in the laboratory frame:

v ≡ p ˜ z

E ˜ =

p E ˜ ² − M ² E ˜ =

s 1 − M ²

E ˜ ²

The laboratory frame and the center of mass frame are connected by a Lorentz boost along the ˆ z direction with velocity v. Let us recall how the quantities in the laboratory frame are related to those of the center of mass. Let us start with time and space coordinates (c = 1):

˜ t = γ(v)(t + vx) x ˜ = γ(v)(x + vt) One can write a similar expression for the energy and the spatial momentum

E ˜ = γ(v)(E + vp _z ) p ˜ _z = γ(v)(p _z + vE) Let us check in our simple case that the above formulas are correct:

E ˜ = γ(v)(E + vp _z ) = γ(v)M = 1

√ 1 − v ² M = 1 q

1 − 1 + ^M _˜

E

M = ˜ E

˜

p z = γ(v)(p z + vE) = γ(v)vM = ˜ E s

1 − M ² E ˜ ² = p

E ˜ ² − M ²

The particle decays into two particles of given mass m 1 and m 2 . We now show that the energies and the modulus of the momenta in the center of mass are completely fixed by the kinematics of the problem. In order to do this, let us introduce the four momenta of the particles in the center of mass:

P _µ = (M, 0, 0, 0) , P _1µ = (E ₁ , ~ p ₁ ) , P _2µ = (E ₂ , ~ p ₂ ) and we impose the conservation of the total four momentum:

P _µ = P _1µ + P _2µ ⇒ M = E ₁ + E ₂

0 = ~ p ₁ + ~ p ₂ ,

In addition we have the usual relations between masses, energies, and momenta:

m ² _i = E _i ² + |~ p i | ² , i = 1, 2.

One can try to solve the above set of equations and extract the values of E ₁ and E ₂ . Although this is in principle possible is not the shortest way. A nice trick consists in taking the right product of four momenta. Indeed from the conservation rule one can deduce:

P µ − P 1µ = P 2µ , and taking the square of the left and right hand side one gets

(P − P 1 ) ² = P ₂ ² ⇒ P ² + P ₁ ² − 2P P 1 = P ₂ ² ⇒ M ² + m ² ₁ − 2(M E 1 + 0) = m ² ₂ . Exchanging 1 ↔ 2 one finally gets

E ₁ = M ² + m ² ₁ − m ² ₂

2M E ₂ = M ² + m ² ₂ − m ² ₁

2M .

In addition one can notice that in the center of mass the two particles are emitted back to back and therefore the modulus of their spatial momenta are equal:

|~ p ₁ | = |~ p ₂ | = q

E ₁ ² − m ² ₁ = q

E ₂ ² − m ² ₂ = M 2

s

1 − m ₁ + m ₂

M 1 + m ₁ + m ₂

M 1 − m ₁ − m ₂

M 1 + m ₁ − m ₂ M

The last equality of the above formula can be proved with a bit of manipulation.

We now want to compute the expression of the energy of particle 1 in the laboratory frame as a function of the scattering angle. Let us call ˜ θ ₁ the angle between ~ p ˜ ₁ and ˆ z:

~ ˜

p 1 · ~ p ˜ = ˜ p 1 p ˜ cos ˜ θ 1

Again we can consider the square of the conservation law, this time computed in the laboratory frame:

( ˜ P − P ˜ 1 ) ² = ˜ P ₂ ² ⇒ P ˜ ² + ˜ P ₁ ² − 2 ˜ P P ˜ 1 = ˜ P ₂ ² ⇒ M ² + m ² ₁ − 2( ˜ E E ˜ 1 − p ˜ 1 p ˜ cos ˜ θ 1 ) = m ² ₂ which can also be written as

M ² + m ² ₁ − m ² ₂ − 2 ˜ E E ˜ ₁ = ˜ p ₁ p ˜ cos ˜ θ ₁ = q

( ˜ E − M ² )( ˜ E ₁ ² − m ² ₁ ) cos ˜ θ ₁ Squaring the above equation we get a quadratic equation in ˜ E 1 that we can invert.

Let us now consider the simpler case of m 1 = m 2 = m 6= 0. The expression for E 1 , E 2 , |p| simplifies:

E ₁ = E ₂ = M

2 |p| = 1

2

p M ² − 4m ²

We want to compute the maximum energy that one of the two particles can have in the laboratory frame. This clearly happens when the particle is emitted in the center of mass in the ˆ z direction while it will have the minimal energy when is emitted in the −ˆ z direction. We can understand this looking at the relation between ˜ E 1 and E 1 , p 1 , which can be rewritten as:

E ˜ = γ(v)(E + vp z ) = γ(v)(E + vp cos θ) = γ(v) M

2 + v 2

p M ² − 4m ² cos θ

where θ is the emission angle of the particle 1 in the center of mass (while particle 2 is emitted at θ + π). Hence E ˜ max = M

2 γ(v) 1 + v r

1 − 4m ² M ²

!

= M 2

E ˜ M 1 +

s

1 − 4m ²

M ² 1 − M ² E ˜ ²

!

E ˜ _min = E ˜ 2 1 −

s

1 − 4m ²

M ² 1 − M ² E ˜ ²

!

Finally let us consider the special case in which the initial particle decays into two massless particles. We want to

compute what is the minimum angle that the two particles can form in the laboratory frame. First of all we can

compute the value of the scattering angle of particle 1 in the laboratory frame ˜ θ ₁ as a function of the scattering angle in the center of mass θ. This is easily obtained considering the relation

cos ˜ θ ₁ = p ˜ _1z

| ~ p ˜ 1 | = p ˜ _1z

E ˜ ₁ = γ(v)(p _1z + vE ₁ ) γ(v)(E 1 + vp 1z ) =

M

2 cos θ + v ^M ₂

M

2 + v ^M ₂ cos θ = cos θ + v 1 + v cos θ

where in the second equality we made use of the masslessness of the particle: ˜ E ₁ ² − | ~ p ˜ 1 | ² = 0. Similarly:

cos ˜ θ ₂ = p ˜ 2z

E ˜ ₂ = γ(v)(p 2z + vE 2 )

γ(v)(E ₂ + vp _2z ) = cos(θ + π) + v

1 + v cos(θ + π) = v − cos θ 1 − v cos θ

We are interested in computing (˜ θ 1 − θ ˜ 2 ) min . One can use trigonometric functions to express tan (˜ θ 1 − θ ˜ 2 ) as a function of θ only and then minimize, however we can overpass this step noting that the configuration in which (˜ θ ₁ − θ ˜ ₂ ) is minimal must correspond to some maximally symmetric configuration ˜ θ ₁ = − θ ˜ ₂ . This can happen only when θ = 0, π/2, π since the configuration must be symmetric under reflection with respect to the ˆ z axis.

Substituting in the above formulas we get that for θ = 0, π then ˜ θ ₁ − θ ˜ ₂ = π therefore this configuration corresponds to the maximum angular difference. Hence for θ = π/2 we get the minimum angular difference and

cos ˜ θ 1 = cos ˜ θ 2 = v = s

1 − M ² E ˜ ² .

Exercise 4

Let us define the four momenta of the particles in the center of mass frame:

P _aµ = (E _a , 0, 0, p _a ) P _bµ = (E _b , 0, 0, −p _a ) P cµ = (E c , ~ p c ) P dµ = (E d , ~ p d ) P _i ² = E _i ² − |~ p i | ² = m ² _i i = a, b, c, d.

The conservation of total momentum reads,

P cµ + P dµ = P aµ + P bµ ≡ P µ ,

where we have defined the total initial momentum P _µ = (E _a + E _b , 0, 0, 0). Notice that the form of P is similar to that of the decaying particle in the previous exercise, apart from the fact that the mass M is replaced by the total energy in the center of mass √

s ≡ E _a + E _b . Indeed we can interpret this two body scattering in two steps: first a fusion of particles a and b into a new state with four momentum equal to the sum of the four momenta (this in particular implies that in the center of mass the new ”particle” is at rest and has a ”mass” equal to the sum of the energies); then we have a decay of this state into the particles c + d. Clearly in the second step we can make use of the formulae derived in the previous exercise with the substitution M → √

s:

E 1 = s + m ² ₁ − m ² ₂ 2 √

s E 2 = s + m ² ₂ − m ² ₁ 2 √

s

|~ p c | = |~ p d | =

√ s 2

s

1 − m 1 + m 2

√ s 1 + m 1 + m 2

√ s 1 − m 1 − m 2

√ s 1 + m 1 − m 2

√ s

Hence the energies and the modulus of the momenta of the final particles in the center of mass are completely determined by the total energy.

Suppose now that we want to strike a target b at rest with a particle a with energy in the laboratory ˜ E _a in such a way that in the center of mass the energy is √

s = 14 TeV. We need to compute ˜ E _a as a function of √ s. We can use the property that the value of the square of a four momentum is independent of the frame in which it is computed:

P ² = ˜ P ² = ( ˜ P _a + ˜ P _b ) ² ⇒ s = m ² _a + m ² _b + 2m _b E ˜ _a

Substituting √

s = 14 TeV and m a = m b ' 1 GeV (at LHC we collide protons) we get:

E ˜ a = s − m ² _a − m ² _b

2m _b ' 98 × 10 ³ TeV

We see immediately the convenience of having a collider with two beams circulating in opposite direction and equal energy.

Finally let us count the degrees of freedom needed to describe this process completely. We can put ourselves in the center of mass frame and align the colliding particles along the ˆ z: all the other configurations can be obtained from this by a rotation or a boost and therefore don’t contain additional degrees of freedom. Then we can count the parameters in the four momenta written at the beginning of the exercise. We have 11 degrees of freedom. Not all of them are independent because of the relation between energies, momenta and masses. Then the independent ones are 11 − 4 = 7. We still need to impose the conservation of the total energy and total spatial momentum.

This gives 4 constraints, hence 7 − 4 = 3. Since the total energy is known, there is one more constraint, so that the degrees of freedom are 2. A good choice for them are the two angular variables characterizing the direction of

~

p ₁ , i.e. θ and φ.

Exercise 5

(Recall 1 eV = 1.602 · 10 ⁻¹⁹ J, M p = 0.938 GeV/c ² ) Velocity of protons coming from SPS:

M p γc ² = 450 GeV ⇒ 1

1 − β ² = (450 GeV) ²

M _p ² c ⁴ ⇒ β = v u u t

(450)

(0.938)

− 1

(450)

(0.938)

= 0.999998 Protons energy:

E = M p γc ² = 7 TeV Number of protons per beam:

N P = 2.8 · 10 ¹⁴ Total energy of the beam:

E tot = 7 TeV · N P = 1.96 × 10 ¹⁵ TeV = 1.96 × 1.6 × 10 ²⁷⁻¹⁹ J = 314 MJ

One can compare this quantity with the kinetic energy of a running TGV. The velocity of the train can be easily extracted:

1

2 M V _T ² = 314 · 10 ⁶ J ⇒ V T GV = r

2 · 3140

4 · m s ⁻¹ = 39.6 m s ⁻¹ = 143 Km/h

One can also compute the total current circulating inside the ring: in the interval of 1 s a number of bunches N _B = 27 Km/3 · 10 ⁵ −1

passes in a given point of the ring. The total charge passing through that point in a second is given by:

Q = N B · 2800 · 1.6 · 10 ¹¹⁻¹⁹ C = 0.498 C ⇒ I = Q

s = 0.498 A.

Exercise 6

Let’s call ˜ p’s the momenta of the incoming particles in the laboratory frame, p’s the momenta of the incoming particles in the center of mass frame, k’s the momenta of the outgoing particles in the center of mass frame. One has generically

˜

p ^(γ) _µ = ( ˜ E _γ , 0, 0, E ˜ _γ ), p ˜ ^(e) _µ = (m _e , 0, 0, 0), p ^(γ) _µ = (E γ , 0, 0, E γ ), p ^(e) _µ = ( q

E _γ ² + m ² _e , 0, 0, −E γ ),

k ^(γ) _µ = (E _γ ⁰ , 0, E _γ ⁰ sin θ, E _γ ⁰ cos θ), k ^(e) _µ = (E _e ⁰ , 0, k ⁰ _e sin θ ⁰ , k ⁰ _e sin θ ⁰ ), k ⁽⁺⁾ _µ = (E ⁺ , ~ k ⁺ ), k _µ ⁽⁻⁾ = (E ⁻ , ~ k ⁻ ).

Note that the total energy E ^(tot) = E γ + q

E _γ ² + m ² _e is a monotonically increasing function of E γ , so that the kinematical configuration which minimizes E γ also minimizes E ^(tot) .

The energies of the outgoing particles satisfy E _γ ⁰ ≥ 0, E _e ⁰ ≥ m e , E ⁺ ≥ m e , and E ⁻ ≥ m e , and the configuration with smallest total energy is the one in which the bounds are minimally satisfied, namely E _γ ⁰ = 0, E _e ⁰ = E ⁺ = E ⁻ = m e . This is the situation in which in the center of mass frame no final-state particle has kinetic energy, i.e. the three massive final-state particles are produced at rest and the final-state photon is soft. Thus, in this threshold configuration

p ^(γ) _µ = (E γ , 0, 0, E γ ), p ^(e) _µ = ( q

E _γ ² + m ² _e , 0, 0, −E γ ), k _µ ^(γ) = (0, 0, 0, 0), k _µ ^(e) = (m e , 0, 0, 0),

k _µ ⁽⁺⁾ = (m e , 0, 0, 0), k ⁽⁻⁾ _µ = (m e , 0, 0, 0).

and

k _µ ^(γ) + k ^(e) _µ + k _µ ⁽⁺⁾ + k ⁽⁻⁾ _µ = (3 m _e , 0, 0, 0) ≡ k ^(tot) _µ .

Now, exploiting the invariance under Lorentz transformations of the square of four-vectors, and using momentum conservation, one can deduce

9 m ² _e = (k ^(tot) ) ² = (p ^(γ) + p ^(e) ) ² = (˜ p ^(γ) + ˜ p ^(e) ) ² = m ² _e + 2 m _e E ˜ _γ

= ⇒ E ˜ _γ

= 4 m _e .

Exercise 7

In the usual cgs units the electron charge is: e cgs = 4 , 8032 · 10 ⁻¹⁰ esu = 1.60217 · 10 ⁻¹⁹ C.

One can, however, use a different convention and define a new elementary charge:

e = √

4πe cgs = 1 .70269 · 10 ⁻⁹ esu

The latter is the commonly used parameter in QFT. With this definition the static potential between two charges at a distance r will be:

e ² cgs

r −→ e ² 4πr .

In quantum electrodynamics (QED), which is the quantum version of classical electrodynamics, one can define an adimensional parameter α using the electric charge, ~ and c. This is given by the ratio:

α cgs = e ² cgs

~ c = (4, 8) ² · 10 ²⁰ esu ²

1.05 · 10 ⁻²⁷ erg · s × 3 · 10 ¹⁰ c/s ' 1 137 If one uses the other convention then ₁₃₇ ¹ = _4π ^e

~ c = _4π ^α . This ratio is the true expansion parameter of QED: we will see later on that a generic quantity computed in QED can be expressed as a power series:

A QED =

∞

X

n=0

a n

α 4π

ⁿ

.

The Bohr radius is the typical length of the Hydrogen atom. It can be obtained combining the constants that characterize the system: since hydrogen is a quantum electron moving in a radial potential one can use e, m e , ~ . Looking at the Schroedinger equation for the fundamental level

− ~ ² 2m e

∂ _r ² + e ² r − E

ψ(r) = 0,

one can deduce that e ² ∼ E · L and _m ^~

e

∼ E · L ² . Therefore in order to obtain a length one can simply consider the ratio:

r B = ~ ²

m _e e ² = 0.53 ˚ A = 0.53 · 10 ⁻⁸ cm,

which is indeed the typical dimension of an atom. Now it’s easy to get the typical energy of the system: the Rydberg constant. Indeed

R = e ² r B

= 27.2 eV.

One should notice that computing the exact energy of the fundamental state of the hydrogen (which requires the solution of the Schroedinger equation in terms of complicated Hypergeometric function) would give E = −R/2.

Moreover the radius of the fundamental orbit is exactly r B . This is a general feature of physics: playing with the constants that describe a system one can build the typical quantities that characterise that system. Usually the exact result will be the typical scale found times some numerical O(1) coefficient.

The Compton wavelength of a particle is the length scale below which the quantum mechanical description of the particle breaks down and quantum field theory becomes crucial. If one tries to localize the particle within its Compton wavelength, its momentum becomes so uncertain that the particle can have enough energy to create an extra particle-antiparticle pair. In the case of the electron, h ∼ ∆p · ∆x ∼ m _e c · λ _e , so

λ _e = h

m _e c = 2.4 · 10 ⁻¹⁰ cm.

Finally one can ask what would be the radius of the electron if its mass energy only came from charge self interaction. Consider indeed the electron as a sphere containing a charge e uniformly distributed. The self interaction energy is ^e _r

e

where the radius has to be determined. If one equates this to the relativistic rest energy one gets:

r e = e ²

m e c ² = 2.8 · 10 ⁻¹³ cm.

Since this is the typical length of an atomic nucleus, this confirms that the considered model is not correct.