Path integral quantization and interactions Flashcards

1
Q

Path integrals in QM

What’s the purpose and principle of path integrals in QM?

A

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])

  1. Transition amplitude: separating the exp w/ small ε instead of t.
  2. We make the operators functions instead.
  3. Discretization: breaking up the paths into chunks via intermediate integrals.
  4. 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.

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2
Q

Path integral for fields

How to generalize the formalization of path integrals for fields (say instead of operators)?

A

Here we have 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.

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3
Q

Path integral for fields

How to determine the expectation value of operators and fields?

A

⟨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) &laquo_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
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4
Q

Generating functional

What is a generating functional in QFT and what’s its purpose?

A

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
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5
Q

Generating functional

What’s the generating functional for the free case?

A

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.

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6
Q

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?

A

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.

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7
Q

Generating functional

What’s the transition amplitude in the interacting case?

A

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.

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8
Q

Generating functional

What the 2-point generating functional in the interacting case? What’s the physical meaning of this?

A

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.

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