Path integral quantization and interactions Flashcards
(15 cards)
Path integrals in QM
What’s the purpose and principle of path integrals in QM?
The goal is to find scattering cross sections for which we’ll need n-point functions.
Principle: when calculating the transition amplitude for the cross section, we take all possible paths into account and work with infinitely many integrals (and infinitely many d.o.f.) instead of operators: ⟨x|exp(iHt)|y⟩ = ∫ D[x(t)] exp(iS[x(t)/ħ])
- Transition amplitude: separating the exp w/ small ε instead of t.
- We make the operators functions instead.
- Discretization: breaking up the paths into chunks via intermediate integrals.
- Taking the N —» ∞ limit to get rid of the O(ε²) terms.
Relation to the clasical limit: at ħ —» all the paths that are not the minimal ones that Lagrangian mechanics postulates, vanish when integrating over everything.
Path integral for fields
How to generalize the formalization of path integrals for fields (say instead of operators)?
Here we not only have to discretize time but space as well (e.g., N points in time, M in space), to get infinitely many points and integrate over all possible paths. Ultimately, we’re constructing a spacetime lattice and we take the continuum limit of it.
⟨φ(x,t(N)|φ(x,t0)⟩ = Z = ∫ D[φ] exp(iS[φ]),
where
- ∫ D[φ] = lim(M —» ∞, N —» ∞) Π(k=1,N)Π(i=1,M) dφ(x(i), t(k))
- S[φ] = ∫d^4x [L(φ, ∂()μφ) + iε/2 φ²]: we add a small ε number that converges back to zero
Adding ε is equivalent to but a different approach than taking t —» exp(iα)t slightly imaginary time, like previously in the QM case.
Path integral for fields
How to determine the expectation value of operators and fields?
⟨Ω|T φ(x)φ(y)|Ω⟩ = (1/Z)∫ D[φ]exp(iS[φ])φ(x)φ(y)
- same principle for operators
- comes from: ⟨0|A|0⟩ = lim(t —» ∞)⟨x,t/2|A(t(a))|y,–t/2⟩/⟨x,t/2|y,–t/2⟩, where t(a) «_space;t is some intermediate point where A is evaluated; lim(t —» ∞) results in exponential supression and only the groun state survives if we plug in intermediate states
- special case above: two-point function
- φ on left side: operators evaluated at some points, φ on the right: functions evaluated (so, numbers) —» such an eq. only holds because of the time ordering symbol (LOL)
- ⟨Ω|: interactive vacuum, free states in the far past/future at the beginning/end, interactions only in the middle
Generating functional
What is a generating functional in QFT and what’s its purpose?
It’s a differential functional that contains a source. The goal is to be able to differentiate this with respect to the source and get back the path integral formula.
Z[J] = (1/Z) ∫ Dφ exp[i ∫d^4x (L(φ) + (i/2)εφ² + J(x)φ(x))],
where J(x) is the source coupled to the field φ(x).
- normalization: Z[J = 0] = 1
Generating functional
What’s the generating functional for the free case?
Z0[J] = (1/Z0) ∫ Dφ exp[–i/2 ∫ φ(◻ + m² – iε)φ – J(x)φ(x)]]
- L = (1/2) ∂(μ)φ∂^(μ)φ – (1/2)m²φ²
The statement here is that this formula can be written as: Z0[J] = exp[-i/2 ∫ d^4x d^4y J(x)Δ(F)(x–y)J(y)]. Proof by:
- shifting the integration variable: φ —» φ + φ0 —» Z0[J] = exp(i/2 ∫ φ0 J)
- the source: (◻ + m² – iε)φ0 = J
- solving for φ0: (◻ + m² – iε)Δ(F)(x) = –δ^(4)(x) —» φ0 = –∫ d^4y Δ(F)(x–y)J(y)
- the Feyman propagator: Δ(F)(x) = ∫ d^4p/(2π)^4 exp(–ipx) 1/(p² – m² + iε)
The ◻ gives a p² and the Dirac delta a constant when integrating.
Generating functional
What are some of the n-point functions exactly in the free case? What’s the physical content? What is Wick’s theorem?
1-point: G0(x1) = 0 —» no special value for a field at one point
2-point: G0(x1,x2) = Δ(F)(x1 – x2) —» just the propagator
3-point: G0(x1,x2,x3) = 0 —» odd function over a symmetric integral
4-point: G0(x1,x2,x3,x4) = Δ(F)(x1 – x2)Δ(F)(x3 – x4) + Δ(F)(x1 – x4)Δ(F)(x3 – x2) + Δ(F)(x1 – x3)Δ(F)(x2 – x4) —» all the ways the four points can be connected
Wick’s theorem: any n-point function can be written with the 2-point function in the free case
G0(x1,…,x(2n)) = 1/(2^n) Σ(π) G0[x(π1), x(π2)]…G0[x(π(2n–1)),x(π(2n))]
π: permutations of {x1,…,x(2n)}
Generally, odd functions will vanish and if x1 = x2, we get a loop.
Generating functional
What’s the transition amplitude in the interacting case?
The Lagrangian density contains an interaction term now: L = L0 + L(int), where *L(int) = –(λ/4!)φ⁴(x) *gives the simplest interacting field theory.
Z[J] = (1/Z) ∫ Dφ exp[i ∫d⁴x (L0(x) + L(int)(x) + J(x)φ(x))] = [1 – i(λ/4!) ∫dz (6 xloopx + xxxx)] exp(xx/2)
- this is leading order
- the vaccum terms (looploop) cancelled: Z[J] doesn’t contain vacuum diagrams at any order of the perturbation theory with proper normalization
- xloopx: propagation of a physical particle
- xxxx: relevant for the scattering of two physical particle
φ⁴ is relevant for the Higgs sector. Also, φ cannot be on an odd power because the Hamiltonian wouldn’t be bounded from below then.
Generating functional
What the 2-point generating functional in the interacting case? What’s the physical meaning of this?
G(x1,x2) = (1/i²) δ²Z[J]/δJ(x1)J(x2)|J=0 = i ∫d⁴p/(2π)⁴ exp[ip(x1 – x2)] 1/(p² – m(r)² + iε)
- m(r)² = m² + (iλ/2)Δ(F)(0), where m(r) is the renormalized mass
- m(r) ≠ m, where m is the bare mass appearing in L
- m(r) is measured in experiments as an observable quantity
- physical meaning: the interaction changes the mass of the particle
The quantum field as a medium around the particle responds to the particle and contributes to its mass.
Generating functional
What is the mass correction exactly? How does this tie into experimental things?
δm = (iλ/2) ∫ d⁴p/(2π)⁴ 1/(p² – m² + iε)
- the poles are shifted in the imaginary direction where we don’t cross them
- Wick rotation: p4 = ip0 —» p² = p0² -p(vec)² = –p4² –p(vec)² = –p(E)²
- Euclidean measure: p(E) = (p(vec), p4) —» dp0 = –idp4(Λ)
- δm = ∫ d⁴p(E)/(2π)⁴ 1/(–p(E)² – m²) = O(Λ²) + O(m²log(Λ²/m²) + O(m²) = ∞ —» quadratic and logarithmic divergences
- Λ: cutoff —» cutoff regularization
m(r)² = m² + δm² = –∞ + ∞ = finite value
- m(r) is defined by experimental input
- m0(Λ) can tuned —» higher order n-point functions (scattering) can be calculated
If the iε term wasn’t there, there would be poles we can’t integrate over.
Generating functional
What are the contributing diagrams for the four-point function?
G(x1,x2,x3,x4) = (1/i⁴) δ⁴Z[J]/δJ(x1)δJ(x2)δJ(x3)δJ(x4)|(J=0) = 3 (——) – iλ x – 3iλ (—loop— —)
Generating functional
What are the Feyman rules? What types of diagrams are there based on these?
- vertex: x —» –iλ
- propagator: x—y —» –iΔ(F)(x – y)
- symmery factors: S/4!
Diagram types: connected, disconnected, vacuum diagrams
Connected diagrams
What is the generating functional for connected diagrams?
W[J] = logZ[J]
- this can be thought of as free energy, similar to thermodynamics
- G(x1,…,x(n)) = (1/i^n) δZ[J]/δJ(x1)…δJ(x(n))|(J=0)
- this is relevant for scattering
Fermion fields
How is the path integral defined for fermionic fields?
The Lagrangian: L = ψ‾(i∂(slash) – m + iε)ψ —» the action: S = S[ψ,ψ‾]
Different sources for ψ and ψ‾ needed: Z[η, η‾] ~ ∫ DψDψ‾ exp{iS[ψ,ψ‾] + i ∫ d⁴x (ψ‾η + η‾ψ)}
- same method as for scalar fields: (i∂(slash) – m – iε)S(F)(x) = δ⁴(x), where S(F)(x) = (i∂(slash) + m – iε)Δ(F)(x)
- S(F)(x – y) = ∫ d⁴p/(2π)⁴ (p(slash) + m)exp[–ip(x – y)]/(p² – m² +iε)
The transition amplitude: ⟨Ω|T[ψ(x1)…ψ(x(n))]|Ω⟩ = (1/Z) ∫ DψDψ‾ exp(iS[ψ,ψ‾]) ψ(x1)…ψ‾(y1)…
- here ψ and ψ‾ are Grassmann-valued fields
Fermion fields
How do Grassmann numbers work?
These are “numbers” that anticommute.
- {C(i), C(j)} = 0, C(i)² = 0 by definition
- two-point function: f(C1,C2) = a0 + a1 C1 + a2 C2 + a3 C1 C2 (finite Taylor expansion!)
- differentiation: {∂/∂C1, C2} = 0
- integration: Berezin integral, {dC(i), dC(j)} = 0
- Grassmann vectors with independent Grassmann numbers in it exist as well
- ∫ dC‾dC exp[–C‾(k)A(kl)C(l)] ~ detA, where A is a bosonic n×n matrix
Integration and differentiation are the same thing here. Integral of any prdinary number is zero and ∫ dC1 C1 = 1 by definition.
Fermion fields
How do the generating functional and the n-point fcuntions look like for fermionic fields?
Z[η,‾η] = (1/Z0) ∫ DψD‾ψ exp[i ∫ d⁴x (‾ψ(x)(i∂(slash) - m + iε)ψ(x) + ‾η(x)ψ(x) + ‾ψ(x)η(x))] = exp[-i ∫ d⁴xd⁴y ‾η(x)S(F)(x – y) η(x)] = exp( x–>–x)
From this, the n+m point functions:
G0(x1,…,x(n),y1,…,y(m)) = 1/[i^n(-i)^m] δ^(n+m) Z0/[δ‾η(y1)…δ‾η(y(m))δη(x1)…δη(x(n))]|(η=‾η = 0)
