User:EverettYou/Perturbation Theory

This article is to be merged with Perturbation theory (quantum mechanics).

Hamiltonian and Force Operator
From the differential geometric point of view, a parameterized Hamiltonian is considered as a function defined on the parameter manifold that maps each particular set of parameters $$x^\mu = (x^1,x^2,\cdots)$$ to an Hermitian operator $$H(x^\mu)$$ that acts on the Hilbert space. The parameters here can be external field, interaction strength, or driving parameters in the quantum phase transition. Let $$E_n(x^\mu)$$ and $$|n(x^\mu)\rangle$$ be the nth eigenenergy and eigenstate of $$H(x^\mu)$$ respectively. In the language of deferential geometry, the states $$|n(x^\mu)\rangle$$ form a vector bundle over the parameter manifold, on which derivatives of these states can be defined. The perturbation theory is to answer the following question: given $$E_n(x^\mu_0)$$ and $$|n(x^\mu_0)\rangle$$ at a reference point $$x^\mu_0$$, how to estimate the $$E_n(x^\mu)$$ and $$|n(x^\mu)\rangle$$ at $$x^\mu$$ close to that reference point.

Without loss of generality, the coordinate system can be shifted, such that the reference point $$x^\mu_0 = 0$$ is set to be the origin. The following linearized model Hamiltonian is frequently used
 * $$H(x^\mu)= H(0)+x^\mu F_\mu.$$

If the parameters $$x^\mu$$ are considered as generalized coordinates, then $$F_\mu$$ should be identified as the generalized force operators related to those coordinates. Different indices $$\mu$$ label the different forces along different directions in the parameter manifold. For example, if $$x^\mu$$ denotes the external magnetic field in the $$\mu$$-direction, then $$F_\mu$$ should be the magnetization in the same direction.

Perturbation as Power Series Expansion
The validity of the perturbation theory lies on the adiabatic assumption, which assumes the eigenenergies and eigenstates of the Hamiltonian are smooth functions of parameters such that their values in the vicinity region can be calculated in power series (like Taylor expansion) of the parameters:
 * $$E_n(x^\mu)= E_n + x^\mu\partial_\mu E_n + \frac{1}{2!}x^\mu x^\nu\partial_\mu\partial_\nu E_n+\cdots,$$
 * $$| n(x^\mu)\rangle= | n\rangle + x^\mu|\partial_\mu n\rangle + \frac{1}{2!}x^\mu x^\nu|\partial_\mu\partial_\nu n\rangle+\cdots.$$

Here $$\partial_\mu$$ denotes the derivative with respect to $$x^\mu$$. When applying to the state $$|\partial_\mu n\rangle$$, it should be understood as the Lie derivative if the vector bundle is equipped with non-vanishing connection. All the terms on the right-hand-side of the series are evaluated at $$x^\mu=0$$, e.g. $$E_n\equiv E_n(0)$$ and $$|n\rangle\equiv |n(0)\rangle$$. This convention will be adopted in though out this section, that all functions without the parameter dependence explicitly stated are assumed to be evaluated at the origin. The power series may converge slowly or even not converging when the energy levels are close to each other. The adiabatic assumption breaks down when there is energy level degeneracy, and hence the perturbation theory is not applicable in that case.

Hellmann-Feynman Theorems
The above power series expansion can be readily evaluated if there is a systematic approach to calculate the derivates to any order. Using the chain rule, the derivatives can be broken down to the single derivative on either the energy or the sate. The Hellmann-Feynman theorems are used to calculated these single derivatives. The first Hellmann-Feynman theorem gives the derivative of the energy,
 * $$\partial_\mu E_n=\langle n|\partial_\mu H | n\rangle$$

The second Hellmann-Feynman theorem gives the derivative of the state (resolved by the complete basis with $$m \neq n$$),
 * $$\langle m|\partial_\mu n\rangle=\frac{\langle m|\partial_\mu H | n\rangle}{E_n-E_m}$$
 * $$\langle\partial_\mu m| n\rangle=\frac{\langle m|\partial_\mu H | n\rangle}{E_m-E_n}$$

The theorems can be simply derived by applying the differential operator $$\partial_\mu$$ to both sides of the Schrödinger equation $$H|n\rangle=E_n|n\rangle$$, which reads
 * $$\partial_\mu H|n\rangle + H|\partial_\mu n\rangle=\partial_\mu E_n|n\rangle+E_n|\partial_\mu n\rangle$$

Then overlap with the state $$\langle m|$$ from left and make use of the Schrödinger equation again $$\langle m|H=\langle m|E_m$$,
 * $$\langle m|\partial_\mu H|n\rangle + E_m\langle m|\partial_\mu n\rangle=\partial_\mu E_n\langle m|n\rangle+E_n\langle m|\partial_\mu n\rangle$$

Given that the states from a set of orthonormal basis $$\langle m|n \rangle = \delta_{mn}$$, by rearranging the terms in the above equation, it is straight forward to prove both the Hellmann-Feynman theorems. With the differential rules given by the Hellmann-Feynman theorems, the perturbative correction to the energies and states can be calculated systematically.

Correction of Energy
To the second order
 * $$E_n(x^\mu)=\langle n|H|n\rangle +\langle n|\partial_\mu H|n\rangle x^\mu+\sum _{m\neq n} \frac{\langle n|\partial_\nu H|m\rangle \langle m|\partial_\mu H|n\rangle}{E_n-E_m}x^\mu x^\nu+\cdots$$

Correction of State
To the second order $$
 * n(x^\mu)\rangle =|n\rangle +\sum _{m\neq n} \frac{\langle m|\partial_\mu H|n\rangle }{E_n-E_m}|m\rangle x^\mu$$
 * $$+\left(\sum _{m\neq n} \sum _{l\neq n} \frac{\langle m|\partial_\mu H|l\rangle \langle l|\partial_\nu H|n\rangle }{(E_n-E_m)(E_n-E_l)}|m\rangle -\sum _{m\neq n} \frac{\langle m|\partial_\mu H|n\rangle \langle n|\partial_\nu H|n\rangle }{(E_n-E_m)^2}|m\rangle -\frac{1}{2}\sum _{m\neq n} \frac{\langle n|\partial_\mu H|m\rangle \langle m|\partial_\nu H|n\rangle }{(E_n-E_m)^2}|n\rangle \right)x^\mu x^\nu+\cdots$$

Effective Hamiltonian
Let $$H(0)$$ be the Hamiltonian completely restricted either in the low-energy subspace $$\mathcal{H}_L$$ or in the high-energy subspace $$\mathcal{H}_H$$, such that there is no matrix element in $$H(0)$$ connecting the low- and the high-energy subspaces, i.e. $$\langle m|H(0)|l\rangle=0$$ if $$ m\in \mathcal{H}_L, l\in\mathcal{H}_H$$. Let $$F_\mu=\partial_\mu H$$ be the coupling terms connecting the subspaces. Then when the high energy degrees of freedoms are integrated out, the effective Hamiltonian in the low energy subspace reads
 * $$H_{m n}^{\text{eff}}\left(x^{\mu }\right)=\langle m|H|n\rangle +\langle m|\partial _{\mu }H|n\rangle x^{\mu }+\frac{1}{2!}\sum _{l\in\mathcal{H}_H} \left(\frac{\langle m|\partial _{\mu }H|l\rangle \langle l|\partial _{\nu }H|n\rangle }{E_m-E_l}+\frac{\langle m|\partial _{\nu }H|l\rangle \langle l|\partial _{\mu }H|n\rangle }{E_n-E_l}\right)x^{\mu }x^{\nu }+\cdots$$

Here $$m,n\in\mathcal{H}_L$$ are restricted in the low energy subspace. The above result can be derived by power series expansion of $$\langle m|H(x^\mu)|n \rangle$$.

Quantum Field Theory
For finite temperature case, we calculate the correction to the free energy instead of the ground state energy.

Kernel, Free Energy and Propagator
In quantum field theory, instead of Hamiltonian, an action is used to describe a model. In general, the action of a free field $$\phi$$ takes the bilinear form,
 * $$ S = \phi^\dagger\cdot K\cdot\phi $$

which involves the convolution with the action kernel $$K$$. The action kernel can be $$K=-i\omega+H$$ for diffusive dynamics, or $$K=\nu^2+\Omega^2$$ for wave dynamics. In any case, the free energy is given by
 * $$F=\eta\mathrm{Tr}\ln K$$

where $$\eta = +1$$ for bosonic field and $$\eta = -1$$ for fermionic field. The perturbation expansion of $$F$$ can be calculated in Taylor series term by term as long as we know how to take partial derivatives of the action kernel. Because the free energy involves the logarithm of $$K$$, so after first order derivative, $$K^{-1}$$ will appear in the formula. Therefore the propagator
 * $$G=-K^{-1}$$

is introduced to denote the inverse of the action kernel.

The free energy and the propagator in the quantum field theory take the role of the energy and the state in quantum mechanics respectively. So the goal of perturbation theory at the field theory level is to calculate the free energy and the propagator in power series, given the action kernel of the model.

The Derivative Formula
Each term in the Taylor expansion involves evaluation of derivatives. The derivative formula for the propagator reads
 * $$\partial_\mu G = G \cdot \partial_\mu K \cdot G$$

This can be derived by starting from the identity $$G\cdot K = -1$$, and take partial derivative on both sides
 * $$\partial_\mu G \cdot K + G\cdot \partial_\mu K = 0$$

then the derivative formula is easily obtained by rearranging the terms.

The derivative of free energy to any order can be calculated by recursively using the derivative formula for the propagator. For example,
 * $$\partial_\mu F= -\eta\mathrm{Tr}\, G\cdot\partial_\mu K$$
 * $$\partial_\mu\partial_\nu F= -\eta\mathrm{Tr}(G\cdot\partial_\mu \partial_\nu K+ G\cdot\partial_\mu K\cdot G\cdot\partial_\nu K)$$

The expression grows complicated with the order quickly. For single variable, if the derivative of the Kernel is cut-off at the second order, i.e. $$\partial^2 K = 0$$, then the derivative of free energy has the following simple form
 * $$\partial^n F= -\eta\,(n-1)!\,\mathrm{Tr}(G\cdot\partial K)^{n}$$

Free Energy Correction
Then the correction of free energy for the single parameter perturbation can be obtained by plugging the derivative formula into the Taylor series,
 * $$F=F_0-\eta\sum_{n=1}^\infty \frac{1}{n}\mathrm{Tr}(G_0\cdot\partial K)^n x^n$$

$$F_0$$ and $$G_0$$ are respectively the free energy and the propagator evaluated at the origin $$x = 0$$ of the perturbation parameter. The series may be formally expressed as a logarithm
 * $$F=F_0+\eta\mathrm{Tr}\ln(1-(G_0\cdot\partial K) x)$$

Dynson's Equation and Self-Energy Correction
Notice the relation between the free energy and the propagator $$F=\eta \mathrm{Tr}\ln(-G^{-1})$$. The above free energy correction implies
 * $$\eta\mathrm{Tr} \ln(-G^{-1})=\eta\mathrm{Tr} \ln(-G_0^{-1})+\eta\mathrm{Tr}\ln(1-(G_0\cdot\partial K)x)=\eta\mathrm{Tr} \ln(-G_0^{-1}+x \partial K))$$

which is the Dynson's equation of the propagator,
 * $$G^{-1} = G_0^{-1}-x\partial K$$

One immediately identifies $$\Sigma_0=\partial K$$ to be the bare self-energy, which becomes obvious from the fact that the Dynson's equation can be also obtained from the series expansion of $$K$$ as $$K = K_0 + x\partial K$$, with $$K = -G^{-1}$$ and $$K_0= -G_0^{-1}$$, while $$K$$ carries the energy (action) dimension.

The Dynson's equation may be formally solved as
 * $$G=(G_0^{-1}-\Sigma_0)^{-1}=(1-G_0\cdot\Sigma_0)^{-1}G_0$$

however the difficulty lies in the evaluation of the matrix inverse. A way to circumvent the inversion is to use perturbative expansion, such that
 * $$G=G_0+G_0\cdot\Sigma_0\cdot G_0+G_0\cdot\Sigma_0\cdot G_0\cdot\Sigma_0\cdot G_0+\cdots=G_0\cdot\sum_{n=1}^{\infty}(\Sigma_0\cdot G_0)^n$$

This leads to the definition of the self-energy, which follows form
 * $$G=G_0+G_0\cdot\Sigma\cdot G_0$$

meaning that the normal propagator is dressed from the bare propagator by the self-energy correction.

Effective Hamiltonian under Perturbation
To obtain an effective Hamiltonian under perturbation, we can calculate the effective propagator by series expansion, then restore the Hamiltonian by .....
 * $$\Sigma=\Sigma_0 + \Sigma_0\cdot G_0\cdot\Sigma_0 + \Sigma_0\cdot G_0\cdot\Sigma_0\cdot G_0\cdot\Sigma_0 + \ldots =\Sigma_0\cdot\sum_{n=1}^{\infty}(G_0\cdot\Sigma_0)^n.$$

The bare self-energy $$\Sigma_0$$ stands for the perturbation (connecting low and high energy sectors), and $$G_0$$ contains the dynamics of both energy sectors independently. The resulting self-energy can then be projected to the low-energy subspace to produce the effective Hamiltonian.

The normal self-energy obtained by the above formula is a function of frequency $$\Sigma(\omega)$$. To obtain the effective correction near energy $$\omega_0$$, the trick is to evaluate the following residue
 * $$H_\text{eff}(\omega_0) = \mathrm{Res}_{\omega\rightarrow\omega_0} \frac{\Sigma(\omega)}{\omega-\omega_0}.$$

This is a way to regularize $$\Sigma(\omega)$$ around $$\omega=\omega_0$$ by taking the 0th order term in its Laurent series.

However, when the low-energy Hamiltonian is not trivially zero, we need to set $$\omega$$ to the low-energy Hamiltonian $$H_0$$ itself, and the effective Hamiltonian reads
 * $$H_\text{eff}=H_0 + \Sigma(H_0).$$