There are two possible ways to interpret this: one, you could envision that the particles are fundamental. The photons are particles, and if you pack enough of them, that gives rise to a classical field; two, you could imagine that it’s the field that is fundamental, and when you quantize it, it’s the ripples that give rise to particles.
So why quantum field theory?
i) Because all particles of the same type are indistinguishable. This is an amazing fact. For instance, a proton created in a far, far way galaxy, billions of years ago would be exactly the same as a proton freshly created in a collision on earth. It would mean that the universe is filled with a proton field, and when one of its ripple is converted into a particle, it is always the same particle, regardless of where and when it was created.
ii) Because QM + SR implies that the number of particles in a reaction is not conserved. Basically, this means that if you take the Schroedinger’s equation, which describes a single particle, and apply Special Relativity, you won’t go far with this scheme. You get two versions: the Klein-Gordon equation and the Dirac equation, but keeping the interpretation of the wave function to a typically single particle leads you to deep troubles. You’ll find that you get negative probabilities, of which no one can make sense, and there are infinite towers of energies, which are unbounded and not quite clear what to make of them.
What is Quantum Field Theory?
In Quantum Mechanics, you figure out the classical degrees of freedom. You then promote them to operators that act on a Hilbert space. This is also true in QFT. Here, the classical degrees of freedom are the fields, and we do the same, promote them to operators. What that means is that we have a field that is quantized, and the basic object is an operator valued function acting on the Hilbert space. Every point in space will have an operator. This immediately tells us one of the major problems in QFT. There is an infinite number of points in space. So we are dealing with an infinite number of degrees of freedom. And so infinity will come back to bite us many, many times in understanding QFT. All the richness of QFT will be in making sense of those infinities. Basically, QFT is the language in which the laws of nature are written.
Units and Scales
[ c ] = LT^{-1}
[ ℏ ] = L^{2}M^{2}T^{-1}
[ G ] = L^{3}M^{-1}T^{-2}
Setting ℏ =c=1, this means we can express everything in terms of mass, or equivalently, energy (eV).
For example, λ = ℏ/mc ⇒ [ L ]= [ M^{-1}]
For the electron, M_{e} = .5 x 10^{6} eV ⇔ λ_{e} = 2 x 10^{-12} m.
At the Planck scale, G = ℏc/M_{p}^{2} ⇒ M_{p}^{-2} , where M_{p} = 10^{18} GeV.
From Classical Physics to Quantum Mechanics
QM is formulated directly from Classical Physics according to the following recipe: construct a Lagrangian (L) from an action principle (S),
(1) S = ∫ dt L(x(t),v(t));
where x(t) is position and v(t)= ∂x/∂t is the corresponding velocity.
This yields the equation of motion, also known as the Euler-Lagrange equations,
d(∂L/∂v)/dt = ∂L/∂q, (Newton's Second Law of Motion)
Define the Hamiltonian through a Legenre transformation,
(2) H = pv – L ;
where p is the canonical momentum defined as p = ∂L/∂v
The equation of motion then becomes,
(3) ∂H/∂p = v ; ∂H/∂q = -∂L/∂q = –p
The advantage is that the Hamiltonian equations are first order, while the Euler-Lagrange equations are second order.
Define the Poisson bracket,
(4) {A,B} = (∂A/∂x)(∂B/∂p) – (∂A/∂p)∂B/∂x,
where A and B are any two functions.
In particular, if B = H, then dA/dt = {A,H}
The equations of motion are further simplified as,
dx/dt = {x, H} = ∂H/∂p ; dp/dt = {p, H} = –∂H/∂p
For the position (A = x) and the momentum (B = p), the Poisson bracket becomes,
(5) {x,p} = 1
Then replace that Poisson bracket with the commutation relation
[x,p] = iℏ, (equation 35 in The Essential Quantum Mechanics (EQM) )
which yields the Heisenberg Uncertainty Principle.
And voilà, we have QM. Note: there are other ramifications in QM to keep in mind, for instance, a state "|a>" is differentiated from the observable "A", and probability is the square of the amplitude, etc. For our purposes, we should remember that we have another process of going from Classical Physics to Quantum Mechanics, which is embodied in the transformation:
(6) {x,p} = 1 → [x,p] = iℏ
Classical Field Theory
A field is a physical quantity defined at every point in space and time. In classical particle mechanics, there are a finite number of degrees of freedom. For instance, the coordinate q_{a}(t), where a is a label that might indicate direction. In field theory, we have an infinite number of degrees of freedom. We denote it by ϕ_{a}(x,t), where a and x are considered as labels. Important Note: x is no longer a dynamical operator but just a label.
Example: E_{r}(x,t) and B_{r}(x,t) in E & M, where r = 1,2,3. So we have six fields. But later it was found out using Maxwell’s equations that there are 4 fields, A_{μ}(x,t), where μ = 0,1,2,3, and
E_{r} = ∂A_{r}/∂t − ∂A_{0}/∂x^{r}
B_{r} = ½ ϵ_{rst}∂A_{t}/∂x^{s}
The dynamics of the fields are governed by the Lagrangian, which is a function of ϕ(x,t) , ∂ϕ(x,t)/∂t and ∇ϕ(x,t). We define,
(7) L(t) = ∫dx^{3}ℒ(ϕ_{a},∂_{μ}ϕ_{a})
⇒ S = ∫dtL(t) = ∫dx^{4} ℒ(ϕ_{a},∂_{μ}ϕ_{a})
We determine the equations of motion by the principle of least action, that is,
(8) δS = 0, keeping the end points fixed.
(9) δS = ∫dx^{4} {(∂ℒ/∂ϕ_{a})δϕ_{a} + (∂ℒ/∂(∂_{μ}ϕ_{a}))δ(∂_{μ}ϕ_{a})}
= ∫dx^{4}[{∂ℒ/∂ϕ_{a} − ∂_{μ}(∂ℒ/∂(∂_{μ}ϕ_{a}))}δϕ_{a}
(10) For δS = 0
(11) ∂_{μ}(∂ℒ/∂(∂_{μ}ϕ_{a})) − ∂ℒ/∂ϕ_{a} = 0
These are called the Euler-Lagrange equations for fields.
Example: the Klein-Gordon equations
(12) ℒ = ½ η_{μν}∂_{μ}ϕ∂_{ν}ϕ − ½ m^{2}ϕ^{2}
where, η_{μν} is the Minkowski metric.
Substituting equation (12) into equation (11),
i) 1st term: ∂ℒ/∂(∂_{μ}ϕ_{a}) = ∂_{μ}ϕ ⇒ ∂_{μ}(∂ℒ/∂(∂_{μ}ϕ_{a})) = ∂_{μ}∂_{μ}ϕ
ii) 2nd term: ∂ℒ ∂ϕ_{a} = − m^{2}ϕ
This gives,
(13) ∂_{μ}∂_{μ}ϕ + m^{2}ϕ = 0
Note: This is the Klein-Gordon Equation for fields. So far the term m has not been determined and is considered at this moment as a constant into the equation.
Hamiltonian Formalism For Fields
In analogy to p = ∂L/∂v and v = ∂q/∂v (see equ. 1 and 2), we define the conjugate momentum as,
(14) π(x) = ∂ℒ/(∂ϕ/∂t)
The Hamiltonian density,
(15) H(π,ϕ)= π(x)(∂ϕ(x)/∂t) − ℒ(ϕ,∂ϕ/∂t,∇ϕ)
Example:
Consider ℒ = ½ (∂ϕ/∂t)^{2} − ½ (∇ϕ) ^{2} − V(ϕ).
⇒ π(x) = ∂ℒ/(∂ϕ/∂t) = ∂ϕ/∂t
(16) H = ½ π^{2} + ½ (∇ϕ)^{2} + V(ϕ)
And the Hamiltonian, H = ∫ d^{3}x H .
Canonical Quantization of the Fields
Recall in QM, canonical quantization tells us to take the coordinates q_{a} and the momenta p_{a} and promote them to operators (ℏ=1).
(17) [q_{a},p_{b}] = iδ_{ab} ⇒ [q,p] = (2π)^{3}δ(q-p) (in 3-D)
In field theory, we do the same for ϕ(x) and π(x).
(18) [ϕ_{a}(x),π_{b}(y)] = iδ_{ab} δ^{(3)}(x−y)
In the Schroedinger picture, ϕ depends on space but not on time. The time-dependence resides in the states, which evolve according to the Schroedinger equation.
(19) id|ψ>/dt = H|ψ>
The question arises: what are the eigenvalues of the Hamiltonian. So, we want to know the spectrum of H. Recall that in QM, this can be solved for the harmonic oscillator and the hydrogen atom. When we try it on the helium atom, we run into trouble as we are dealing with six degrees of freedom. In QFT, we are dealing with an infinite number of degrees of freedom – at least one for every point in space. So the task becomes horrible. Thankfully, there is one class of fields in which we can do something about, and these are called free field theories. If you write the theory with "good" coordinates, the fields all decouple from each other, and each of these degrees of freedom evolves independently. In these free field theories, the Lagrangian are quadratic in the fields, and the equations of motion are linear. The simplest free field theory is the Klein-Gordon equation (K-G) for a real scalar, where m has dimension of mass.
∂_{μ}∂_{μ}ϕ + m^{2}ϕ = 0 , equation (13)
Solving for ϕ, we do a Fourier transformation,
(20) Φ(x,t) = ∫ d^{3}p(2π)^{-3} e ^{ip∙x} Φ(p,t)
Substituting (20) into the K-G equation,
(21)⇒ [∂^{2}/∂t^{2} + (p^{2} +m^{2})]Φ(p,t) = 0
This is the equation for a harmonic oscillator with frequency,
(22) ω_{p} = (p^{2}+ m^{2})^{½ }
The solutions to the classical K-G equations are superpositions of simple harmonic oscillators (SHO). To quantize Φ(x,t), we need to quantize SHO. (see Harmonic Oscillators, Vacuum Energy, Pauli Exclusion Principle)
In the Schroedinger picture,
(23) H = ½ p^{2} + ½ ω^{2}q^{2} , [q,p] = i
To find the spectrum of H, we define
(24) a = i(2ω)^{½}p + (ω/2)^{½}q
(25) a^{†} = −i(2ω)^{½}p + (ω/2)^{½}q
(26) ⇒ [a, a^{†}] = 1
(27) ⇒ H = ½ ω(aa^{†} + a^{†}a)
(28) or H = ω(a^{†}a + ½ )
(29) ⇒ [H,a^{†}] = ωa^{†}
(30) and, [H,a] = −ωa
The importance of the last result means that if we have an eigenstate of E, we can construct the other states.
(31) That is,
if H|E> = E|E>, ⇒ Ha^{†}|E> = (E + ω)a^{†}|E> .
(32) Similarly, Ha|E> = (E − ω)a|E> .
We get these towers of energies, E, E + ω, E + 2ω… from above, and E − ω, E − 2ω going down below. If we want a spectrum that is bounded below, then this process must stop. This implies the existence of a ground state, |0> , such that a|0> = 0. (This is also true for <0|a^{†} = 0)
(33) ⇒ H|0> = ω(a^{†}a + ½ )|0> = ½ ω|0>, which is the ground state energy.
For all the excited states,
(34) |n> = (a^{†})^{n}|0>, and H|n> = (n + ½ )ω|0>
Free Field Theory
From the Klein-Gordon equation, consider the following,
(35) H = ½∫d^{3}x[π^{2} + (∇ϕ)^{2} + m^{2}ϕ^{2}], (see equ. 16)
(36) ω_{p} = + (p^{2}+ m^{2})^{½ } and [ϕ(x),π(y)] = iδ^{(3)}(x−y)
We define the creation operators a_{p}^{†} and annihilation operators a_{p} from the following,
(37) Φ(x)=∫d^{3}p(2π)^{-3}(2ω_{p})^{-½} [a_{p}e^{ip.x} + a_{p}^{†}e^{-ip.x}]
(38) π(x)= ∫d^{3}p(2π)^{-3}(-i)(ω_{p}/2)^{½}[a_{p}e^{ip.x} - a_{p}^{†}e^{-ip.x}]
Note that both Φ(x) and π(x) are hermitian operators, with Φ^{†} = Φ and π^{†} = π.
( See Equ 19 in The Essential Quantum Mechanics (EQM) for the importance of Hermitian operators)
(39) ⇒ [a_{p},a_{q}^{†}] = (2π)^{3}δ^{(3)}(p−q),
Note: for p = q, [a_{p},a_{p}^{†}] = (2π)^{3}δ^{(3)}(0)
(40) From H = ½∫d^{3}x[π^{2} + (∇ϕ)^{2} + m^{2}ϕ^{2}], (equ. 35)
= ½∫d^{3}x d^{3}p d^{3}q(2π)^{-6}
{−(ω_{q}ω_{p}/2)^{-½} [a_{p}e^{ip.x} - a_{p}^{†}e^{-ip.x}][a_{q}e^{iq.x} - a_{q}^{†}e^{-iq.x}]
+i2^{-1}(ω_{p}ω_{q})^{-½}[pa_{p}e^{ip.x} - pa_{p}^{†}e^{-ip.x}][qa_{q}e^{iq.x} - qa_{q}^{†}e^{-iq.x}]
+m^{2}2^{-1}(ω_{p}ω_{q})^{-½}[a_{p}e^{ip.x} + a_{p}^{†}e^{-ip.x}][a_{q}e^{iq.x} + a_{q}^{†}e^{-iq.x}]}
Using the definition of the delta function,
(41) δ(x − y) = (2π)^{-3}∫d^{3}ke^{ik(x-y)},
we integrate over x and q.
(42)H = 1/4∫d^{3}p(2π)^{-3}(ω_{p})^{-1}
[(−ω_{p}^{2} + p^{2} + m^{2})(a_{p}a_{-p} + a_{p}^{†}a_{-p}^{†})
+(ω_{p}^{2} + p^{2} + m^{2})(a_{p}a^{†}_{p} + a_{p}^{†}a_{p})]
But ω_{p}^{2} = p^{2} + m^{2}, therefore the first term in the square bracket vanishes, and we are left with,
(43) H = ½∫ d^{3}p(2π)^{-3}ωp(a_{p}a^{†}_{p} + a_{p}^{†}a_{p})
This is the energy of an infinite number of uncoupled SHO, as expected. Also from the commutation relationship, we get,
(44) H = ∫ d^{3}p(2π)^{-3}ω_{p}(a_{p}a^{†}_{p} + ½(2π)^{3}δ^{(3)}(0))
The last term is problematic as it is infinite.
The Vacuum
Define the vacuum as in the case of the SHO,
a_{p}|0> = 0, for all p.
Its energy is then,
(45) H|0 > = E_{0} |0 > = [∫d^{3}p ω_{p}δ^{(3)}(0)]|0 >
Note that we have infinity coming from two sources:
(i) from the delta function, because space extends to infinity in all directions - these are called infra-red (IF) divergences.
(ii) Consider a box of size L. In the limit that L → ∞,
(2π)^{3}δ^{(3)}(0) = lim ∫_{-L/2}^{+L/2} d^{3}xe^{ip.x} = V
Where V is the volume of the box.
Therefore, E_{0} = ½V ∫d^{3}p(2π)^{-3}ω_{p}
This indicates that we should work with energy density,
(46) E = E_{0}/V = ½ ∫d^{3}p(2π)^{-3}ω_{p}
This energy density is still infinite due to space being infinitesimally small (as p → ∞, x gets smaller and smaller, that is, we have bigger and bigger oscillations over smaller and smaller distances). This is the second source of infinity, which is called an ultra-violet (UV) divergence. Here we have assumed that our theory is valid over arbitrarily short distance scales.
Since we are only interested in energy differences, we remove the infinities and redefine,
(47) H = ∫d^{3}p(2π)^{-3}ω_{p}a_{p}a^{†}_{p}
(48) ⇒ H|0> = 0
Interacting Picture
i) Starting with the Schroedinger picture:
(49) id|ψ(t) >_{s}/dt = H_{s}|ψ(t) >_{s}, (equ. 19)
States are time-dependent and operators O_{s} are time-independent.
ii) In the Heisenberg picture,
(50) O_{H}(t) = e^{iHt} O_{s} e^{-iHt} and |ψ>_{H} = e^{iHt}|ψ>_{s}
States are time-independent and operators O_{s} are time-dependent.
The interaction picture is a hybrid of the two. We write the Hamiltonian as,
(51) H = H_{0} + H_{int}
Time-dependence of operators is governed by H_{0}. And time-dependence of states is governed by H_{int}.
(52) |ψ>_{I} = e^{iH0t} |ψ>_{s}
(53) O_{I}(t) = e^{iH0t}O_{s}e^{-iH0t}
Note: H_{I} ≡ (H_{int})_{I} = e^{iH0t}(H_{int})_{s}e^{-iH0t}
From the Schroedinger equation,
id|ψ>_{s}/dt = H_{s}|ψ>_{s} (equ. 49)
Substituting 52 and 53,
⇒ id(e^{-iH0t}|ψ>I)/dt = (H_{0} + H_{int})_{s}e^{-iH0t}|ψ>_{I}
⇒ ie^{-iH0t}d|ψ>_{I}/dt = (H_{int})_{s}e^{-iH0t}|ψ>_{I}
⇒ id|ψ>_{I}/dt = e^{iH0t}(H_{int})_{s}e^{-iH0t}|ψ>_{I}
(54) id|ψ>_{I}/dt = H_{I}(t)|ψ>_{I}
In solving this equation we write,
(55) |ψ(t) >_{I} = U(t,t_{0})|ψ(t_{0}) >_{I}, where U(t,t_{0}) is the time evolution operator.
(56) ⇒ idU/dt = H_{I}(t)U
This leaves us with an equation of operators, instead of an equation of states.
If U and H_{I} were ordinary functions, we could solve this equation as,
(57) U(t,t_{0}) = exp(−i∫_{t0}^{t} H_{I}(t')dt')
However, this is not correct because,
(58) [H_{I}(t), H_{I}(t')] ≠ 0, when t ≠ t'
Claim: the correct solution is given by Dyson’s formula,
(59) U(t,t_{0}) = T exp(−i∫_{t0}^{t} H_{I}(t')dt') , where T is the time-ordering operator defined as follow:
(60) T[O(t_{1})O(t_{2})] = O(t_{1})O(t_{2}), if t_{1} > t_{2}
= O(t_{2})O(t_{1}), if t_{2} > t_{1}
The solution is then,
(61) U(t,t_{0}) = 1 − iexp(∫_{t0}^{t} H_{I}(t')dt'
+ (−i)^{2}/2{ ∫_{t0}^{t}dt'∫_{t0}^{t'}dt"H_{I}(t")H_{I}(t')
+ ∫_{t0}^{t}dt'∫_{t0}^{t'}dt"H_{I}(t')H_{I}(t")} + …)
Scattering
We’ll work with scalar Yukawa theory,
(62) ℒ = ½∂_{μ}ϕ∂^{μ}ϕ + ∂_{μ}Ψ*∂^{μ}Ψ − ½m^{2}ϕ − M^{2}Ψ*Ψ − gΨ*Ψϕ ,
where ϕ is a real field with its particle of mass m, Ψ is a complex field with its particle of mass M, and the last term is for the interaction between the two fields with coupling constant g. We’ll take g << M,m to ensure weak coupling.
(63) H_{int} = gΨ*Ψϕ
In this theory, particles number is not conserved. In particular,
Φ ~ a + a^{†} can create/destroy Φ-particles (mesons).
Ψ ~ b + c^{†} can destroy Ψ^{+}particles and create Ψ^{-} (nucleons).
Ψ^{†} ~ b^{†} + c can create Ψ^{+}particles and destroy Ψ^{-} (nucleons).
At first order perturbation theory, we will have terms as Ψ*Ψϕ ~ c^{†}b^{†}a which will destroy a meson and create a pair of nucleon and an anti-nucleon.
ϕ → Ψ^{+} + Ψ^{−} (decay)
This can be represented by the following diagram:
At second order perturbation theory, this will include terms as, (c^{†}b^{†}a)(cba^{†}) which will create a meson, destroy a pair; then destroy the meson and create a pair.
Ψ^{+} + Ψ^{−} → ϕ → Ψ^{+} + Ψ^{−} (scattering)
To calculate amplitudes for these processes, we need an important assumption: the initial and final states are non-interacting particles, meaning that the initial state |i > at t → −∞ and the final state |f > at t → +∞ are eigenstates of H_{0}.
Some caveats:
i) For instance, an electron and a proton interact and form a hydrogen atom. The final state is an interacting state.
ii) Particles are never alone. They can have a cloud of pairs of particles/anti-particles surrounding them.
(64) Definition: t± → ±∞ < f|U(t,t_{0})|i > ≡ < f|S|i >, where S is a unitary operator, called the S(cattering)-Matrix. And U(t,t_{0}) was found in equation 61.
Meson Decay
(65) |i > = 2E_{p}a_{p}^{†}|0 >, ( a meson with momentum p)
(66) |f > = 4E_{q1}E_{q2}b_{q1}^{†}c_{q2} ^{†}|0 > (two nucleons with momenta q_{1} and q_{2})
⇒ < f|S|i > = − ig< f|∫dx^{4} Ψ^{†}(x)Ψ(x)ϕ(x) |i > (using equ. 63 and 64)
Using the definition of the field operator(from equ. 37),
Φ(x) = ∫d^{3}k(2π)^{-3}(2Ek)^{-½}[a_{k}e^{-ik.x} + a_{k}^{†}e^{ik.x}]
And using equation (65), we get
(67)< f|S|i > = −ig< f|∫dx^{4} Ψ^{†}(x)Ψ(x)∫d^{3}k(2π)^{-3}(2Ek)^{-½}[a_{k}e^{-ik.x} + a_{k}^{†}e^{ik.x}] 2E_{p}a_{p}^{†}|0 >,
The second term from the square bracket will give a term that contains < f|a_{k}^{†}a_{p}^{†}|0 >. When equation (66) is substituted, the a_{k}^{†} term will hit the < 0| on the left and give zero. What's left will be,
< f|S|i > = − ig < f| ∫dx^{4}Ψ^{†}(x)Ψ(x)∫d^{3}k(2π)^{-3}
x (2E_{p}/2E_{k})^{-½}a_{k}a_{p}^{†}e ^{ik.x}|0>
Then we then pass the a_{k} through a_{p}^{†} to kill the vacuum |0 > but pick up a delta function, which when integrated is non-zero for k = p, making the term in the square-root equal to 1.
(68) < f|S|i > = − ig < f|∫dx^{4}Ψ^{†}(x)Ψ(x)e^{ip.x}|0 >
Using equation 37 for the field operators Ψ^{†}(x)Ψ(x), taking into consideration that Ψ produces "b" and "c" particles instead of "a" particles
(69) < f|S|i > = −ig< f|∫dx^{4}∫d^{3}k_{1}d^{3}k_{2}(2π)^{-6}(4E_{k1}E_{k2})^{-½}
x b_{k1}^{†}c_{k2}^{†}e^{i(k1+k2-p).x}|0>
Substituting for < f| = (4E_{q1}E_{q2})^{½} < 0|c_{q1}b_{q2} from equation (69), we do as previously, passing these two annihilator operators through b_{k1}^{†}c_{k2}^{†} to kill the vacuum |0 > and pick up a delta function which when integrated is non-zero for k_{1} = q_{1}, and k_{2} = q_{2} making the term in the square-root equal to 1.
(70) < f|S|i > = − ig < 0|∫dx^{4}e^{i(q1+q2- p).x}|0 >
= − ig(2π)^{4}δ^{(4)}(q_{1}+ q_{2}− p)
The delta function expresses the conservation of momentum p = q_{1}+ q_{2}
Normal Ordering
In free field theory we define a normal ordered string of operators ϕ(x1)…. ϕ(xn), we write it as :ϕ(x1)…. ϕ(xn): to be the usual product of operators with all the annihilation operators moved to the right. For the Hamiltonian, this would be :H:
Wick’s Theorem
(71) Consider the operator; O = T exp(−i∫H_{I}(x)dx^{4}),
We’ll need to compute terms like < f|T{ H_{I}(x_{1}) H_{I}(x_{2})… H_{I}(x_{n})} |i > . When we do these calculations, we need to move the annihilator operators to the right, each time we get a delta function through the commutation relations, in order to make many terms vanish, that is, the annihilators will destroy the vacuum state. In other words we want to rewrite the time-ordered operators as normal-ordered operators, where all the annihilation operators are to the right.
Example: For a real scalar field, from equ. 37, we write,
(72) Φ(x) = Φ^{+} (x) + Φ^{−}(x), where
(73) Φ^{+}(x)) = ∫d^{3}p(2π)^{-3} (2E_{p})^{-1}a_{p}e^{-ip.x}, (1st term)
(74) Φ^{−}(x) = ∫d^{3}p(2π)^{-3}(2E_{p})^{-1}a_{p}^{+}e^{ip.x}, (2nd term)
(75) Recall (equ. 60) that
TΦ(x)Φ(y) = Φ(x)Φ(y) , when x0 > y0
= {Φ^{+}(x) + Φ^{-}(x)} {Φ^{+}(y) + Φ^{-}(y)}
= Φ^{+}(x)Φ^{+}(y) + Φ^{+}(x)Φ^{-}(y) + Φ^{-}(x)Φ^{+}(y) + Φ^{-}(x)Φ^{-}(y)
= Φ^{+}(x)Φ^{+}(y) + Φ^{+}(x)Φ^{-}(y) + Φ^{-}(x)Φ^{+}(y) + Φ^{-}(x)Φ^{-}(y)
Note that Φ^{+}(x)Φ^{-}(y) = Φ^{-}(y)Φ^{+}(x) + [Φ^{+}(x),Φ^{-}(y)]. Substituting in the above,
(76) TΦ(x)Φ(y) = Φ^{+}(x)Φ^{+}(y) + Φ^{-}(y)Φ^{+}(x)
+ [Φ^{+}(x),Φ^{-}(y)] + Φ^{-}(x)Φ^{+}(y) + Φ^{-}(x)Φ^{-}(y)
= :Φ(x)Φ(y): + [Φ^{+}(x),Φ^{-}(y)]
But we already know what the commutation will do when acting on the vacuum.
< 0|Φ(x)Φ(y)|0 >
= ∫d^{3}pd^{3}p'(2π)^{-6}(4E_{p} E_{p}')^{-½}< 0|a_{p}a_{p}'^{†} |0 > e^{-i(p.x- p'.y)}
(77) < 0|Φ(x)Φ(y)|0 > = ∫d^{3}p(2π)^{-3}(2E_{p})^{-1}e^{-ip.(x-y)} (see appendix)
(78)< 0|Φ(x)Φ(y)|0 > ≡ D(x – y), called the propagator
Combining equations (76) and (78),
For x_{0} > y_{0}, TΦ(x)Φ(y) = :Φ(x)Φ(y): + D(x – y)
And for y_{0} > x_{0}, TΦ(x)Φ(y) = :Φ(x)Φ(y): + D(y – x)
In general,
(79) TΦ(x)Φ(y) = :Φ(x)Φ(y): + Δ_{F}(x – y), where
(80) Δ_{F}(x – y) = ∫d^{4}k(2π)^{-4} (ie^{-ip.(x-y)})/(k^{2} – m^{2} +ie), called the Feynman Propagator
We define a contraction of a pair of field in a string of n operators as,
|^{-----------}|
Φ(x_{1})...Φ(x_{i})... Φ(x_{j})... Φ(x_{n})= {Φ(x_{i})Φ(x_{j})}= Δ_{F}(x_{i} – x_{j})
For a complex field:
{Ψ(x)Ψ^{†}(y)} = Δ_{F}(y – x)
{Ψ(x)Ψ(y)} = 0
{Ψ†(x)Ψ^{†}(y)} = 0
(81) Define: Φ(x_{1}) = Φ_{1} and Φ(x_{2}) = Φ_{2}, etc.
Theorem: For any collection of fields,
(82) T(Φ_{1}… Φ_{n}) = :Φ_{1}… Φ_{n}: + :all possible contractions:
Example: Take a string of four fields
T(Φ_{1}Φ_{2}Φ_{3}Φ_{4}) = :Φ_{1}Φ_{2}Φ_{3}Φ_{4}:
+{Φ_{1}Φ_{2}}:Φ_{3}Φ_{4}:+ {Φ_{1}Φ_{3}}:Φ_{2}Φ_{4}: + 4 similar terms
+ {Φ_{1}Φ_{2}} {Φ_{3}Φ_{4}} + {Φ_{1}Φ_{3}} {Φ_{2}Φ_{4}} + {Φ_{1}Φ_{4}} {Φ_{2}Φ_{3}}
The second line, we contract one field with another → 6 terms
The third line, we contract one pair of fields with another pair → 3 terms
Nucleon Scattering
Ψ + Ψ → Ψ + Ψ
|i> = (2E_{p1})^{-½}(2E_{p2})^{-½}b_{p1}^{†}b_{p2}^{†}|0> ≡ |p_{1},p_{2} >,
|f> = (2E_{p'1})^{-½}(2E_{p'2})^{-½}b_{p'1}^{†}b_{p'2}^{†}|0> ≡ |p'_{1},p'_{2} >,
We want to compute
⇒ Recall Hint = gΨ*Ψϕ (equ. 63)
Second order is,
⇒ (− ig)^{2}/2 ∫d^{4}x_{1}d^{4}x_{2}T(Ψ^{†}(x_{1})Ψ(x_{1})ϕ(x_{1})Ψ^{†}(x_{2})Ψ(x_{2})ϕ(x_{2}))
There is a term from Wick’s theorem which looks like,
:Ψ^{†}(x_{1})Ψ(x_{1})Ψ^{†}(x_{2})Ψ(x_{2}):{ϕ(x_{1})ϕ(x_{2})}
Leaving the contraction aside for the moment,
⇒ < p'_{1},p'_{2}|:Ψ†(x_{1})Ψ(x_{1}) Ψ†(x_{2})Ψ(x_{2}):| p_{1},p_{2} >
⇒ e^{-ix1.(p1'- p1) + ix2.(p2'- p2)} + e ^{-ix1.(p2'- p1) + ix2.(p2'- p1)} + term (x_{1} ↔ x_{2})
Inserting the contraction term
< f|S|i > = (− ig)^{2}/2 ∫d^{4}x_{1}d^{4}x_{2} [ all exp terms ] ∫d^{4}p(2π)^{-4} (i e^{-ip.(x1- x2)}) /(k^{2} – m^{2} +ie)
⇒ (−ig)^{2}[ ( (p_{1} – p'_{1})^{2} – M^{2} ) ^{-1} + ( (p_{1} – p'_{2})^{2} – M^{2} ) ^{-1} ] (2π)^{4} δ^{(4)} ( p_{1}+ p_{2} – p'_{2} – p'_{2})
Notice again that the delta function expresses the conservation of momentum
Feynman Diagram
This is an alternative to Wick’s theorem. Each diagram corresponds to one of these terms that arise from Wick’s theorem.
General Rules for scalar field (spin zero):
- Draw an external line for each particle in the |i> and |f>
- Draw an arrow on the line to Ψ^{+}, Ψ^{−} particles to denote its charge. Choose incoming (outgoing) arrow for Ψ^{+} (Ψ^{−}) in the |i>, and opposite for |f>.
- Join the lines together with vertices.
- Add a momentum k to each internal line
- To each vertex, there is a factor of (−ig(2π)^{4}δ^{(4)}(Σ_{i}k_{i})
- For each internal line with momentum k, there is a factor of ∫d^{4}k(2π)^{-4} D(k^{2}),
where D = i(k^{2} – m^{2} +ie)^{-1} for ϕ
And D = i(k^{2} – M^{2} +ie)^{-1} for Ψ
Appendix
(A)[a_{p},a_{p'}^{†}] = (2π)^{3}δ^{(3)}(p−p'), (equ. 39)
(B)< 0|a_{p}a_{p'}^{†}|0 > = < 0|a_{p'}^{†}a_{p} + [a_{p},a_{p'}^{†}]|0 >
=< 0|a_{p'}^{†}a_{p} +(2π)^{3}δ^{(3)}(p−p')|0 >
Where we used equ.(A) in the second line.
Now with a_{p}|0 > = 0 and < 0|0 > = 1, we get,
(C)< 0|a_{p}a_{p'}^{†}|0 > = (2π)^{3}δ^{(3)}(p−p')
Rewriting the line above equation (77),
< 0|Φ(x)Φ(y)|0 >
= ∫d^{3}pd^{3}p'(2π)^{-6}(4E_{p} E_{p}')^{-½}< 0|a_{p}a_{p}'^{†} |0 > e^{-i(p.x- p'.y)}
Substituting equ.(C),
= ∫d^{3}pd^{3}p'(2π)^{-6}(4E_{p} E_{p}')^{-½}(2π)^{3}δ^{(3)}(p−p')e^{-i(p.x- p'.y)}
Integrating over p', we get
= ∫d^{3}p(2π)^{-3}(2E_{p})^{-1}e^{-ip.(x-y)},
This is equation (77).