Quartic interaction

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In quantum field theory, a quartic interaction is a theory about a scalar field φ which contains an interaction term $φ 4$ , and is considered by many teachers and students to be the simplest example of interacting fields. This theory consists of subtracting a $\frac{\lambda}{4\!} \phi^4$ term from the Klein–Gordon Lagrangian, where λ is a dimensionless coupling constant.

This article uses the + − − − sign convention

1 The Lagrangian
2 Canonical quantization
3 Schwinger-Dyson equations
4 Path integrals
5 Renormalization
6 Spontaneous symmetry breaking
7 See also

[edit] The Lagrangian

For a real scalar field the Lagrangian takes on the following form.

$\mathcal{L}=\frac{1}{2}\partial^\mu \phi \partial_\mu \phi -\frac{m^2}{2}\phi^2 -\frac{\lambda}{4!}\phi^4$

This Lagrangian has a global Z₂ symmetry mapping φ to −φ.

For a complex scalar field the Lagrangian is,

$\mathcal{L}=\partial^\mu \phi^* \partial_\mu \phi -m^2 \phi^* \phi -\frac{\lambda}{4}(\phi^* \phi)^2$

With n real scalar fields, we can have a φ⁴ model with a global SO(N) symmetry

$\mathcal{L}=\frac{1}{2}\partial^\mu \phi_a \partial_\mu \phi_a - \frac{m^2}{2}\phi_a \phi_a -\frac{\lambda}{8}(\phi_a \phi_a)^2.$

With a background in Group Theory one could see that the model with one complex scalar is equivalent to the model with two real scalars with an SO(2) symmetry. See 1/N expansion.

In all of the models above, the coupling constant λ has to be nonnegative to preserve stability. This means that one is only concerned with relevant and marginal coupling constants. This is merely an experimental constraint as experimentalists do not have a means to probe energies where irrelevant coupling constants are significant.

In 4 dimensions, φ⁴ only exists as an interacting theory as an effective field theory. This is because of the Landau pole. Otherwise, renormalization will render the theory trivial.

If Λ is the cutoff scale of the theory, the renormalized mass would usually be of the order of Λ. m² is said to have a quadratic divergence. See also supersymmetry.

[edit] Canonical quantization

Main article: canonical quantization

This approach is not manifestly covariant.

Let the conjugate field to φ be π. Both fields are Hermitian. Start with the Schrödinger picture. Then, at an equal time, t,

$[\phi(\vec{x}),\phi(\vec{y})]=[\pi(\vec{x}),\pi(\vec{y})]=0$

and

$[\phi(\vec{x}),\pi(\vec{y})]=i \delta(\vec{x}-\vec{y}).$

See canonical commutation relations.

The Hamiltonian is

$H=\int d^3x \left[{1\over 2}\pi^2+{1\over 2}(\nabla \phi)^2+{m^2\over 2}\phi^2+{\lambda \over 4!}\phi^4\right].$

This is only partially true - see the Wick ordering correction later on in this article.

Use Fourier transforms on the fields to get to momentum space.

$\tilde{\phi}(\vec{k})=\int d^3x e^{-i\vec{k}\cdot\vec{x}}\phi(\vec{x})$

$\tilde{\pi}(\vec{k})=\int d^3x e^{-i\vec{k}\cdot\vec{x}}\pi(\vec{x}).$

The quantity $\sqrt{k^2+m^2}$ is called the energy, E, for reasons that will become apparent.

Define an operator a as follows:

$a(\vec{k})=\left(E\tilde{\phi}(\vec{k})+i\tilde{\pi}(\vec{k})\right).$

Then, its adjoint,

$a^\dagger(\vec{k})=\left(E\tilde{\phi}(\vec{k})-i\tilde{\pi}(\vec{k})\right).$

These operators satisfy the commutation relations

$[a(\vec{k}_1),a(\vec{k}_2)]=[a^\dagger(\vec{k}_1),a^\dagger(\vec{k}_2)]=0$

$[a(\vec{k}_1),a^\dagger(\vec{k}_2)]=(2\pi)^3 2E \delta(\vec{k}_1-\vec{k}_2).$

These have the structure of creation and annihilation operators. The occupancy number

$n^\dagger(\vec{k})=a^\dagger(\vec{k})a(\vec{k})$

and the total number of particles (in the interaction picture),

$N=\int {d^3k \over (2\pi)^3}{1\over 2E}n(\vec{k}),$

which is always a nonnegative integer (well, only in the free field case, otherwise, it would be infinite).

a always reduces N by 1 wheareas $a^\dagger$ raises it by 1.

In terms of the creation and annihilation operators,

$H=\int {d^3k\over (2\pi)^3}{1\over 2E}E\left(a^\dagger(\vec{k})a(\vec{k})+(2\pi)^3 E\delta(\vec{0})\right)+$

$+{\lambda\over 4!}\iiiint {d^3k_1\over (2\pi)^3 2E_1}{d^3k_2\over (2\pi)^3 2E_2}{d^3k_3\over (2\pi)^3 2E_3}{d^3k_4\over (2\pi)^3 2E_4} (2\pi)^3 \delta(\vec{k}_1+\vec{k}_2+\vec{k}_3+\vec{k}_4)\left(a^\dagger(\vec{k}_1)+a(\vec{k}_1)\right)\left(a^\dagger(\vec{k}_2)+a(\vec{k}_2)\right)\left(a^\dagger(\vec{k}_3)+a(\vec{k}_3)\right)\left(a^\dagger(\vec{k}_4)+a(\vec{k}_4)\right).$

The first integral diverges badly. In particular, the Dirac delta term is called the zero point term. However, in the absence of general relativity, it does not matter what the total energy is, and since that term is infinite but constant, the Hamiltonian can simply be redefined by dropping it. The second term also has a huge divergence problem which can be resolved by replacing each polynomial in a and $a^\dagger$ with the Wick ordered polynomial where the creation operators always lie to the left of the annihilation operators. This definitely changes the Hamiltonian dynamics, but since quantization always has factor ordering ambiguities, this is perfectly alright.

The new and improved Hamiltonian,

$:H:=\int d^3x \left[{1\over 2}:\pi^2:+{1\over 2}:(\nabla \phi)^2:+{m^2\over 2}:\phi^2:+{\lambda \over 4!}:\phi^4:\right]$

has a state |0> with N|0>=0 called the bare vacuum which is an eigenstate of :H: with zero energy. It is not the lowest energy, though, because of the last term.

The quadratic part is called the free Hamiltonian and the remainder the interaction Hamiltonian. Looking at the free Hamiltonian, it can be seen that each particle with a momentum of $\vec{k}$ has the energy $\sqrt{k^2+m^2}$ .

Use the Dyson series to undertake perturbation theory. This leads directly to Feynman diagrams.

While canonical quantization may look nice, Haag's theorem show the results are incorrect.

[edit] Schwinger-Dyson equations

Main article: Schwinger-Dyson equation

Using the source field method, we get the following Schwinger-Dyson equation for the generating functional:

$i\partial_\mu \partial^\mu \frac{\delta}{\delta J(x)}Z[J]+im^2\frac{\delta}{\delta J(x)}Z[J]-\frac{i\lambda}{3!}\frac{\delta^3}{\delta J(x)^3}Z[J]+J(x)Z[J]=0.$

Note that since

$\frac{\delta^3}{\delta J(x)^3}$

is not well-defined because

$\frac{\delta^3}{\delta J(x_1)\delta J(x_2) \delta J(x_3)}Z[J]$

is a distribution in

x₁, x₂ and x₃,

this equation needs to be regularized.

In this example, the bare propagator, D is the Green's function for $-\partial^\mu \partial_\mu-m^2$ and so, the SD set of equation goes as

$\langle\psi|\mathcal{T}\{\phi(x_0)\phi(x_1)\}|\psi\rangle=iD(x_0,x_1)+\frac{\lambda}{3!}\int d^dx_2 D(x_0,x_2)\langle\psi|\mathcal{T}\{\phi(x_1)\phi(x_2)\phi(x_2)\phi(x_2)\}|\psi\rangle$

$\langle\psi|\mathcal{T}\{\phi(x_0)\phi(x_1)\phi(x_2)\phi(x_3)\}|\psi\rangle=iD(x_0,x_1)\langle\psi|\mathcal{T}\{\phi(x_2)\phi(x_3)\}|\psi\rangle+iD(x_0,x_2)\langle\psi|\mathcal{T}\{\phi(x_1)\phi(x_3)\}|\psi\rangle+iD(x_0,x_3)\langle\psi|\mathcal{T}\{\phi(x_1)\phi(x_2)\}|\psi\rangle+\frac{\lambda}{3!}\int d^dx_4D(x_0,x_4)\langle\psi|\mathcal{T}\{\phi(x_1)\phi(x_2)\phi(x_3)\phi(x_4)\phi(x_4)\phi(x_4)\}|\psi\rangle$

etc.

|ψ> can be any state, not just the vacuum state.

There are many techniques to solve the Schwinger-Dyson equations iteratively.

[edit] Path integrals

Main article: path integral formulation

Consider Feynman's approach.

In the source field approach,

$Z[J]=\int \mathcal{D}\phi e^{i\int d^4x \left({1\over 2}\partial^\mu \phi \partial_\mu \phi -{m^2 \over 2}\phi^2-{\lambda\over 4!}\phi^4+J\phi\right)}$

where the spacetime integral is slightly Wick rotated to give the correct pole prescriptions.

The time ordered vacuum expectation values of polynomials in φ is given by

$\langle\Omega|\mathcal{T}\{\hat{\phi}(x_1)\cdots \hat{\phi}(x_n)\}|\Omega\rangle=\frac{\int \mathcal{D}\phi \phi(x_1)\cdots \phi(x_n) e^{i\int d^4x \left({1\over 2}\partial^\mu \phi \partial_\mu \phi -{m^2 \over 2}\phi^2-{\lambda\over 4!}\phi^4\right)}}{\int \mathcal{D}\phi e^{i\int d^4x \left({1\over 2}\partial^\mu \phi \partial_\mu \phi -{m^2 \over 2}\phi^2-{\lambda\over 4!}\phi^4\right)}}.$

Here, the hats distinguish the operators from the classical fields.

Everything could be Wick rotated to get a φ⁴ statistical mechanics theory over a 4-dimensional Euclidean space. Set the inverse temperature β to 1. Alternatively, rescale φ so that different temperatures correspond to different λs.

Then,

$Z[J]=\int \mathcal{D}\phi e^{-\int d^4x \left({1\over 2}(\nabla\phi)^2+{m^2 \over 2}\phi^2+{\lambda\over 4!}\phi^4+J\phi\right)}$

and

$\langle\phi(x_1)\cdots \phi(x_n)\rangle=\frac{\int \mathcal{D}\phi \phi(x_1)\cdots \phi(x_n) e^{-\int d^4x \left({1\over 2}(\nabla\phi)^2+{m^2 \over 2}\phi^2+{\lambda\over 4!}\phi^4\right)}}{\int \mathcal{D}\phi e^{-\int d^4x \left({1\over 2}(\nabla\phi)^2+{m^2 \over 2}\phi^2+{\lambda\over 4!}\phi^4\right)}}.$

The standard trick to evaluate this functional integral is to use the Gaussian integral approximation. This gives the following Feynman rules in momentum space:

To evaluate the n-point correlation function, assign an external leg to each φ and assign it the momentum of the Fourier transform. Sum over all possible bubbleless Feynman diagrams. For each diagram, assign a momentum to each edge such that the momenta flowing into each vertex is zero and integrate over all flows and divide by the symmetry factor.

[edit] Renormalization

Main article: renormalization

Making actual calculations using any of the many methods given above (they are all essentially equivalent) would show divergent integrals. That's why the theory needs renormalization.

[edit] Spontaneous symmetry breaking

Main article: spontaneous symmetry breaking

An interesting feature can occur if m² happens to be negative, but with λ still positive. Below the phase transition, the vacuum superselection sector splits into two, each of which is in the ordered phase, spontaneously breaking the Z₂ global symmetry of the original theory. Also, interesting collective states like domain walls can appear.