Perturbation theory (quantum mechanics)

Under some circumstances, perturbation theory is an invalid approach to take. This happens when the system we wish to describe cannot be described by a small perturbation imposed on some simple system. In quantum chromodynamics, for instance, the interaction of quarks with the gluon field cannot be treated perturbatively at low energies because the coupling constant (the expansion parameter) becomes too large, violating the requirement that corrections must be small.

Perturbation theory also fails to describe states that are not generated adiabatically from the "free model", including bound states and various collective phenomena such as solitons. Imagine, for example, that we have a system of free (i.e. non-interacting) particles, to which an attractive interaction is introduced. Depending on the form of the interaction, this may create an entirely new set of eigenstates corresponding to groups of particles bound to one another. An example of this phenomenon may be found in conventional superconductivity, in which the phonon-mediated attraction between conduction electrons leads to the formation of correlated electron pairs known as Cooper pairs. When faced with such systems, one usually turns to other approximation schemes, such as the variational method and the WKB approximation. This is because there is no analogue of a bound particle in the unperturbed model and the energy of a soliton typically goes as the inverse of the expansion parameter. However, if we "integrate" over the solitonic phenomena, the nonperturbative corrections in this case will be tiny; of the order of exp(−1/g) or exp(−1/g²) in the perturbation parameter g. Perturbation theory can only detect solutions "close" to the unperturbed solution, even if there are other solutions for which the perturbative expansion is not valid.

The problem of non-perturbative systems has been somewhat alleviated by the advent of modern computers. It has become practical to obtain numerical non-perturbative solutions for certain problems, using methods such as density functional theory. These advances have been of particular benefit to the field of quantum chemistry.[3] Computers have also been used to carry out perturbation theory calculations to extraordinarily high levels of precision, which has proven important in particle physics for generating theoretical results that can be compared with experiment.

The process begins with an unperturbed Hamiltonian H₀, which is assumed to have no time dependence.[7] It has known energy levels and eigenstates, arising from the time-independent Schrödinger equation:

$${\displaystyle H_{0}\left|n^{(0)}\right\rangle =E_{n}^{(0)}\left|n^{(0)}\right\rangle ,\qquad n=1,2,3,\cdots }$$

For simplicity, it is assumed that the energies are discrete. The (0) superscripts denote that these quantities are associated with the unperturbed system. Note the use of bra–ket notation.

A perturbation is then introduced to the Hamiltonian. Let V be a Hamiltonian representing a weak physical disturbance, such as a potential energy produced by an external field. Thus, V is formally a Hermitian operator. Let λ be a dimensionless parameter that can take on values ranging continuously from 0 (no perturbation) to 1 (the full perturbation). The perturbed Hamiltonian is:

$${\displaystyle H=H_{0}+\lambda V}$$

The energy levels and eigenstates of the perturbed Hamiltonian are again given by the time-independent Schrödinger equation,

$${\displaystyle \left(H_{0}+\lambda V\right)|n\rangle =E_{n}|n\rangle .}$$

The objective is to express E_n and ${\displaystyle |n\rangle }$ in terms of the energy levels and eigenstates of the old Hamiltonian. If the perturbation is sufficiently weak, they can be written as a (Maclaurin) power series in λ,

$${\displaystyle {\begin{aligned}E_{n}&=E_{n}^{(0)}+\lambda E_{n}^{(1)}+\lambda ^{2}E_{n}^{(2)}+\cdots \\[1ex]|n\rangle &=\left|n^{(0)}\right\rangle +\lambda \left|n^{(1)}\right\rangle +\lambda ^{2}\left|n^{(2)}\right\rangle +\cdots \end{aligned}}}$$

where

$${\displaystyle {\begin{aligned}E_{n}^{(k)}&={\frac {1}{k!}}{\frac {d^{k}E_{n}}{d\lambda ^{k}}}{\bigg |}_{\lambda =0}\\[1ex]\left|n^{(k)}\right\rangle &=\left.{\frac {1}{k!}}{\frac {d^{k}|n\rangle }{d\lambda ^{k}}}\right|_{\lambda =0.}\end{aligned}}}$$

When k = 0, these reduce to the unperturbed values, which are the first term in each series. Since the perturbation is weak, the energy levels and eigenstates should not deviate too much from their unperturbed values, and the terms should rapidly become smaller as the order is increased.

Substituting the power series expansion into the Schrödinger equation produces:

Expanding this equation and comparing coefficients of each power of λ results in an infinite series of simultaneous equations. The zeroth-order equation is simply the Schrödinger equation for the unperturbed system,

$${\displaystyle H_{0}\left|n^{(0)}\right\rangle =E_{n}^{(0)}\left|n^{(0)}\right\rangle .}$$

The first-order equation is

$${\displaystyle H_{0}\left|n^{(1)}\right\rangle +V\left|n^{(0)}\right\rangle =E_{n}^{(0)}\left|n^{(1)}\right\rangle +E_{n}^{(1)}\left|n^{(0)}\right\rangle .}$$

Operating through by ${\displaystyle \langle n^{(0)}|}$, the first term on the left-hand side cancels the first term on the right-hand side. (Recall, the unperturbed Hamiltonian is Hermitian). This leads to the first-order energy shift,

$${\displaystyle E_{n}^{(1)}=\left\langle n^{(0)}\right|V\left|n^{(0)}\right\rangle .}$$

This is simply the expectation value of the perturbation Hamiltonian while the system is in the unperturbed eigenstate.

This result can be interpreted in the following way: supposing that the perturbation is applied, but the system is kept in the quantum state ${\displaystyle |n^{(0)}\rangle }$, which is a valid quantum state though no longer an energy eigenstate. The perturbation causes the average energy of this state to increase by ${\displaystyle \langle n^{(0)}|V|n^{(0)}\rangle }$. However, the true energy shift is slightly different, because the perturbed eigenstate is not exactly the same as ${\displaystyle |n^{(0)}\rangle }$. These further shifts are given by the second and higher order corrections to the energy.

Before corrections to the energy eigenstate are computed, the issue of normalization must be addressed. Supposing that

$${\displaystyle \left\langle n^{(0)}\right|\left.n^{(0)}\right\rangle =1,}$$

but perturbation theory also assumes that ${\displaystyle \langle n|n\rangle =1}$.

Then at first order in λ, the following must be true:

$${\displaystyle \left(\left\langle n^{(0)}\right|+\lambda \left\langle n^{(1)}\right|\right)\left(\left|n^{(0)}\right\rangle +\lambda \left|n^{(1)}\right\rangle \right)=1}$$

$${\displaystyle \left\langle n^{(0)}\right|\left.n^{(0)}\right\rangle +\lambda \left\langle n^{(0)}\right|\left.n^{(1)}\right\rangle +\lambda \left\langle n^{(1)}\right|\left.n^{(0)}\right\rangle +{\cancel {\lambda ^{2}\left\langle n^{(1)}\right|\left.n^{(1)}\right\rangle }}=1}$$

$${\displaystyle \left\langle n^{(0)}\right|\left.n^{(1)}\right\rangle +\left\langle n^{(1)}\right|\left.n^{(0)}\right\rangle =0.}$$

Since the overall phase is not determined in quantum mechanics, without loss of generality, in time-independent theory it can be assumed that ${\displaystyle \langle n^{(0)}|n^{(1)}\rangle }$ is purely real. Therefore,

$${\displaystyle \left\langle n^{(0)}\right|\left.n^{(1)}\right\rangle =\left\langle n^{(1)}\right|\left.n^{(0)}\right\rangle =-\left\langle n^{(1)}\right|\left.n^{(0)}\right\rangle ,}$$

leading to

$${\displaystyle \left\langle n^{(0)}\right|\left.n^{(1)}\right\rangle =0.}$$

To obtain the first-order correction to the energy eigenstate, the expression for the first-order energy correction is inserted back into the result shown above, equating the first-order coefficients of λ. Then by using the resolution of the identity:

$${\displaystyle {\begin{aligned}V\left|n^{(0)}\right\rangle &=\left(\sum _{k\neq n}\left|k^{(0)}\right\rangle \left\langle k^{(0)}\right|\right)V\left|n^{(0)}\right\rangle +\left(\left|n^{(0)}\right\rangle \left\langle n^{(0)}\right|\right)V\left|n^{(0)}\right\rangle \\&=\sum _{k\neq n}\left|k^{(0)}\right\rangle \left\langle k^{(0)}\right|V\left|n^{(0)}\right\rangle +E_{n}^{(1)}\left|n^{(0)}\right\rangle ,\end{aligned}}}$$

where the ${\displaystyle |k^{(0)}\rangle }$ are in the orthogonal complement of ${\displaystyle |n^{(0)}\rangle }$, i.e., the other eigenvectors.

The first-order equation may thus be expressed as

$${\displaystyle \left(E_{n}^{(0)}-H_{0}\right)\left|n^{(1)}\right\rangle =\sum _{k\neq n}\left|k^{(0)}\right\rangle \left\langle k^{(0)}\right|V\left|n^{(0)}\right\rangle .}$$

Supposing that the zeroth-order energy level is not degenerate, i.e. that there is no eigenstate of H₀ in the orthogonal complement of ${\displaystyle |n^{(0)}\rangle }$ with the energy ${\displaystyle E_{n}^{(0)}}$. After renaming the summation dummy index above as ${\displaystyle k'}$, any ${\displaystyle k\neq n}$ can be chosen and multiplying the first-order equation through by ${\displaystyle \langle k^{(0)}|}$ gives

$${\displaystyle \left(E_{n}^{(0)}-E_{k}^{(0)}\right)\left\langle k^{(0)}\right.\left|n^{(1)}\right\rangle =\left\langle k^{(0)}\right|V\left|n^{(0)}\right\rangle .}$$

The above ${\displaystyle \langle k^{(0)}|n^{(1)}\rangle }$ also gives us the component of the first-order correction along ${\displaystyle |k^{(0)}\rangle }$.

Thus, in total, the result is,

$${\displaystyle \left|n^{(1)}\right\rangle =\sum _{k\neq n}{\frac {\left\langle k^{(0)}\right|V\left|n^{(0)}\right\rangle }{E_{n}^{(0)}-E_{k}^{(0)}}}\left|k^{(0)}\right\rangle .}$$

The first-order change in the n-th energy eigenket has a contribution from each of the energy eigenstates k ≠ n. Each term is proportional to the matrix element ${\displaystyle \langle k^{(0)}|V|n^{(0)}\rangle }$, which is a measure of how much the perturbation mixes eigenstate n with eigenstate k; it is also inversely proportional to the energy difference between eigenstates k and n, which means that the perturbation deforms the eigenstate to a greater extent if there are more eigenstates at nearby energies. The expression is singular if any of these states have the same energy as state n, which is why it was assumed that there is no degeneracy. The above formula for the perturbed eigenstates also implies that the perturbation theory can be legitimately used only when the absolute magnitude of the matrix elements of the perturbation is small compared with the corresponding differences in the unperturbed energy levels, i.e., ${\displaystyle |\langle k^{(0)}|\lambda V|n^{(0)}\rangle |\ll |E_{n}^{(0)}-E_{k}^{(0)}|.}$

We can find the higher-order deviations by a similar procedure, though the calculations become quite tedious with our current formulation. Our normalization prescription gives that

$${\displaystyle 2\left\langle n^{(0)}\right|\left.n^{(2)}\right\rangle +\left\langle n^{(1)}\right|\left.n^{(1)}\right\rangle =0.}$$

Up to second order, the expressions for the energies and (normalized) eigenstates are:

$${\displaystyle E_{n}(\lambda )=E_{n}^{(0)}+\lambda \left\langle n^{(0)}\right|V\left|n^{(0)}\right\rangle +\lambda ^{2}\sum _{k\neq n}{\frac {\left|\left\langle k^{(0)}\right|V\left|n^{(0)}\right\rangle \right|^{2}}{E_{n}^{(0)}-E_{k}^{(0)}}}+O(\lambda ^{3})}$$

$${\displaystyle {\begin{aligned}|n(\lambda )\rangle =\left|n^{(0)}\right\rangle &+\lambda \sum _{k\neq n}\left|k^{(0)}\right\rangle {\frac {\left\langle k^{(0)}\right|V\left|n^{(0)}\right\rangle }{E_{n}^{(0)}-E_{k}^{(0)}}}+\lambda ^{2}\sum _{k\neq n}\sum _{\ell \neq n}\left|k^{(0)}\right\rangle {\frac {\left\langle k^{(0)}\right|V\left|\ell ^{(0)}\right\rangle \left\langle \ell ^{(0)}\right|V\left|n^{(0)}\right\rangle }{\left(E_{n}^{(0)}-E_{k}^{(0)}\right)\left(E_{n}^{(0)}-E_{\ell }^{(0)}\right)}}\\[1ex]&-\lambda ^{2}\sum _{k\neq n}\left|k^{(0)}\right\rangle {\frac {\left\langle k^{(0)}\right|V\left|n^{(0)}\right\rangle \left\langle n^{(0)}\right|V\left|n^{(0)}\right\rangle }{\left(E_{n}^{(0)}-E_{k}^{(0)}\right)^{2}}}-{\frac {1}{2}}\lambda ^{2}\left|n^{(0)}\right\rangle \sum _{k\neq n}{\frac {|\left\langle k^{(0)}\right|V\left|n^{(0)}\right\rangle |^{2}}{\left(E_{n}^{(0)}-E_{k}^{(0)}\right)^{2}}}+O(\lambda ^{3}).\end{aligned}}}$$

If an intermediate normalization is taken (it means, if we require that ${\displaystyle \langle n^{(0)}|n(\lambda )\rangle =1}$), we obtain the same expression for the second-order correction to the wave function, except for the last term.

Extending the process further, the third-order energy correction can be shown to be [8]

$${\displaystyle E_{n}^{(3)}=\sum _{k\neq n}\sum _{m\neq n}{\frac {\langle n^{(0)}|V|m^{(0)}\rangle \langle m^{(0)}|V|k^{(0)}\rangle \langle k^{(0)}|V|n^{(0)}\rangle }{\left(E_{n}^{(0)}-E_{m}^{(0)}\right)\left(E_{n}^{(0)}-E_{k}^{(0)}\right)}}-\langle n^{(0)}|V|n^{(0)}\rangle \sum _{m\neq n}{\frac {|\langle n^{(0)}|V|m^{(0)}\rangle |^{2}}{\left(E_{n}^{(0)}-E_{m}^{(0)}\right)^{2}}}.}$$

Corrections to fifth order (energies) and fourth order (states) in compact notation

If we introduce the notation,

$${\displaystyle V_{nm}\equiv \langle n^{(0)}|V|m^{(0)}\rangle ,}$$

$${\displaystyle E_{nm}\equiv E_{n}^{(0)}-E_{m}^{(0)},}$$

then the energy corrections to fifth order can be written

$${\displaystyle {\begin{aligned}E_{n}^{(1)}&=V_{nn}\\E_{n}^{(2)}&={\frac {|V_{nk_{2}}|^{2}}{E_{nk_{2}}}}\\E_{n}^{(3)}&={\frac {V_{nk_{3}}V_{k_{3}k_{2}}V_{k_{2}n}}{E_{nk_{2}}E_{nk_{3}}}}-V_{nn}{\frac {|V_{nk_{3}}|^{2}}{E_{nk_{3}}^{2}}}\\E_{n}^{(4)}&={\frac {V_{nk_{4}}V_{k_{4}k_{3}}V_{k_{3}k_{2}}V_{k_{2}n}}{E_{nk_{2}}E_{nk_{3}}E_{nk_{4}}}}-{\frac {|V_{nk_{4}}|^{2}}{E_{nk_{4}}^{2}}}{\frac {|V_{nk_{2}}|^{2}}{E_{nk_{2}}}}-V_{nn}{\frac {V_{nk_{4}}V_{k_{4}k_{3}}V_{k_{3}n}}{E_{nk_{3}}^{2}E_{nk_{4}}}}-V_{nn}{\frac {V_{nk_{4}}V_{k_{4}k_{2}}V_{k_{2}n}}{E_{nk_{2}}E_{nk_{4}}^{2}}}+V_{nn}^{2}{\frac {|V_{nk_{4}}|^{2}}{E_{nk_{4}}^{3}}}\\&={\frac {V_{nk_{4}}V_{k_{4}k_{3}}V_{k_{3}k_{2}}V_{k_{2}n}}{E_{nk_{2}}E_{nk_{3}}E_{nk_{4}}}}-E_{n}^{(2)}{\frac {|V_{nk_{4}}|^{2}}{E_{nk_{4}}^{2}}}-2V_{nn}{\frac {V_{nk_{4}}V_{k_{4}k_{3}}V_{k_{3}n}}{E_{nk_{3}}^{2}E_{nk_{4}}}}+V_{nn}^{2}{\frac {|V_{nk_{4}}|^{2}}{E_{nk_{4}}^{3}}}\\E_{n}^{(5)}&={\frac {V_{nk_{5}}V_{k_{5}k_{4}}V_{k_{4}k_{3}}V_{k_{3}k_{2}}V_{k_{2}n}}{E_{nk_{2}}E_{nk_{3}}E_{nk_{4}}E_{nk_{5}}}}-{\frac {V_{nk_{5}}V_{k_{5}k_{4}}V_{k_{4}n}}{E_{nk_{4}}^{2}E_{nk_{5}}}}{\frac {|V_{nk_{2}}|^{2}}{E_{nk_{2}}}}-{\frac {V_{nk_{5}}V_{k_{5}k_{2}}V_{k_{2}n}}{E_{nk_{2}}E_{nk_{5}}^{2}}}{\frac {|V_{nk_{2}}|^{2}}{E_{nk_{2}}}}-{\frac {|V_{nk_{5}}|^{2}}{E_{nk_{5}}^{2}}}{\frac {V_{nk_{3}}V_{k_{3}k_{2}}V_{k_{2}n}}{E_{nk_{2}}E_{nk_{3}}}}\\&\quad -V_{nn}{\frac {V_{nk_{5}}V_{k_{5}k_{4}}V_{k_{4}k_{3}}V_{k_{3}n}}{E_{nk_{3}}^{2}E_{nk_{4}}E_{nk_{5}}}}-V_{nn}{\frac {V_{nk_{5}}V_{k_{5}k_{4}}V_{k_{4}k_{2}}V_{k_{2}n}}{E_{nk_{2}}E_{nk_{4}}^{2}E_{nk_{5}}}}-V_{nn}{\frac {V_{nk_{5}}V_{k_{5}k_{3}}V_{k_{3}k_{2}}V_{k_{2}n}}{E_{nk_{2}}E_{nk_{3}}E_{nk_{5}}^{2}}}+V_{nn}{\frac {|V_{nk_{5}}|^{2}}{E_{nk_{5}}^{2}}}{\frac {|V_{nk_{3}}|^{2}}{E_{nk_{3}}^{2}}}+2V_{nn}{\frac {|V_{nk_{5}}|^{2}}{E_{nk_{5}}^{3}}}{\frac {|V_{nk_{2}}|^{2}}{E_{nk_{2}}}}\\&\quad +V_{nn}^{2}{\frac {V_{nk_{5}}V_{k_{5}k_{4}}V_{k_{4}n}}{E_{nk_{4}}^{3}E_{nk_{5}}}}+V_{nn}^{2}{\frac {V_{nk_{5}}V_{k_{5}k_{3}}V_{k_{3}n}}{E_{nk_{3}}^{2}E_{nk_{5}}^{2}}}+V_{nn}^{2}{\frac {V_{nk_{5}}V_{k_{5}k_{2}}V_{k_{2}n}}{E_{nk_{2}}E_{nk_{5}}^{3}}}-V_{nn}^{3}{\frac {|V_{nk_{5}}|^{2}}{E_{nk_{5}}^{4}}}\\&={\frac {V_{nk_{5}}V_{k_{5}k_{4}}V_{k_{4}k_{3}}V_{k_{3}k_{2}}V_{k_{2}n}}{E_{nk_{2}}E_{nk_{3}}E_{nk_{4}}E_{nk_{5}}}}-2E_{n}^{(2)}{\frac {V_{nk_{5}}V_{k_{5}k_{4}}V_{k_{4}n}}{E_{nk_{4}}^{2}E_{nk_{5}}}}-{\frac {|V_{nk_{5}}|^{2}}{E_{nk_{5}}^{2}}}{\frac {V_{nk_{3}}V_{k_{3}k_{2}}V_{k_{2}n}}{E_{nk_{2}}E_{nk_{3}}}}\\&\quad +V_{nn}\left(-2{\frac {V_{nk_{5}}V_{k_{5}k_{4}}V_{k_{4}k_{3}}V_{k_{3}n}}{E_{nk_{3}}^{2}E_{nk_{4}}E_{nk_{5}}}}-{\frac {V_{nk_{5}}V_{k_{5}k_{4}}V_{k_{4}k_{2}}V_{k_{2}n}}{E_{nk_{2}}E_{nk_{4}}^{2}E_{nk_{5}}}}+{\frac {|V_{nk_{5}}|^{2}}{E_{nk_{5}}^{2}}}{\frac {|V_{nk_{3}}|^{2}}{E_{nk_{3}}^{2}}}+2E_{n}^{(2)}{\frac {|V_{nk_{5}}|^{2}}{E_{nk_{5}}^{3}}}\right)\\&\quad +V_{nn}^{2}\left(2{\frac {V_{nk_{5}}V_{k_{5}k_{4}}V_{k_{4}n}}{E_{nk_{4}}^{3}E_{nk_{5}}}}+{\frac {V_{nk_{5}}V_{k_{5}k_{3}}V_{k_{3}n}}{E_{nk_{3}}^{2}E_{nk_{5}}^{2}}}\right)-V_{nn}^{3}{\frac {|V_{nk_{5}}|^{2}}{E_{nk_{5}}^{4}}}\end{aligned}}}$$

and the states to fourth order can be written

$${\displaystyle {\begin{aligned}|n^{(1)}\rangle &={\frac {V_{k_{1}n}}{E_{nk_{1}}}}|k_{1}^{(0)}\rangle \\|n^{(2)}\rangle &=\left({\frac {V_{k_{1}k_{2}}V_{k_{2}n}}{E_{nk_{1}}E_{nk_{2}}}}-{\frac {V_{nn}V_{k_{1}n}}{E_{nk_{1}}^{2}}}\right)|k_{1}^{(0)}\rangle -{\frac {1}{2}}{\frac {V_{nk_{1}}V_{k_{1}n}}{E_{k_{1}n}^{2}}}|n^{(0)}\rangle \\|n^{(3)}\rangle &={\Bigg [}-{\frac {V_{k_{1}k_{2}}V_{k_{2}k_{3}}V_{k_{3}n}}{E_{k_{1}n}E_{nk_{2}}E_{nk_{3}}}}+{\frac {V_{nn}V_{k_{1}k_{2}}V_{k_{2}n}}{E_{k_{1}n}E_{nk_{2}}}}\left({\frac {1}{E_{nk_{1}}}}+{\frac {1}{E_{nk_{2}}}}\right)-{\frac {|V_{nn}|^{2}V_{k_{1}n}}{E_{k_{1}n}^{3}}}+{\frac {|V_{nk_{2}}|^{2}V_{k_{1}n}}{E_{k_{1}n}E_{nk_{2}}}}\left({\frac {1}{E_{nk_{1}}}}+{\frac {1}{2E_{nk_{2}}}}\right){\Bigg ]}|k_{1}^{(0)}\rangle \\&\quad +{\Bigg [}-{\frac {V_{nk_{2}}V_{k_{2}k_{1}}V_{k_{1}n}+V_{k_{2}n}V_{k_{1}k_{2}}V_{nk_{1}}}{2E_{nk_{2}}^{2}E_{nk_{1}}}}+{\frac {|V_{nk_{1}}|^{2}V_{nn}}{E_{nk_{1}}^{3}}}{\Bigg ]}|n^{(0)}\rangle \\|n^{(4)}\rangle &={\Bigg [}{\frac {V_{k_{1}k_{2}}V_{k_{2}k_{3}}V_{k_{3}k_{4}}V_{k_{4}k_{2}}+V_{k_{3}k_{2}}V_{k_{1}k_{2}}V_{k_{4}k_{3}}V_{k_{2}k_{4}}}{2E_{k_{1}n}E_{k_{2}k_{3}}^{2}E_{k_{2}k_{4}}}}-{\frac {V_{k_{2}k_{3}}V_{k_{3}k_{4}}V_{k_{4}n}V_{k_{1}k_{2}}}{E_{k_{1}n}E_{k_{2}n}E_{nk_{3}}E_{nk_{4}}}}+{\frac {V_{k_{1}k_{2}}}{E_{k_{1}n}}}\left({\frac {|V_{k_{2}k_{3}}|^{2}V_{k_{2}k_{2}}}{E_{k_{2}k_{3}}^{3}}}-{\frac {|V_{nk_{3}}|^{2}V_{k_{2}n}}{E_{k_{3}n}^{2}E_{k_{2}n}}}\right)\\&\quad +{\frac {V_{nn}V_{k_{1}k_{2}}V_{k_{3}n}V_{k_{2}k_{3}}}{E_{k_{1}n}E_{nk_{3}}E_{k_{2}n}}}\left({\frac {1}{E_{nk_{3}}}}+{\frac {1}{E_{k_{2}n}}}+{\frac {1}{E_{k_{1}n}}}\right)+{\frac {|V_{k_{2}n}|^{2}V_{k_{1}k_{3}}}{E_{nk_{2}}E_{k_{1}n}}}\left({\frac {V_{k_{3}n}}{E_{nk_{1}}E_{nk_{3}}}}-{\frac {V_{k_{3}k_{1}}}{E_{k_{3}k_{1}}^{2}}}\right)-{\frac {V_{nn}\left(V_{k_{3}k_{2}}V_{k_{1}k_{3}}V_{k_{2}k_{1}}+V_{k_{3}k_{1}}V_{k_{2}k_{3}}V_{k_{1}k_{2}}\right)}{2E_{k_{1}n}E_{k_{1}k_{3}}^{2}E_{k_{1}k_{2}}}}\\&\quad +{\frac {|V_{nn}|^{2}}{E_{k_{1}n}}}\left({\frac {V_{k_{1}n}V_{nn}}{E_{k_{1}n}^{3}}}+{\frac {V_{k_{1}k_{2}}V_{k_{2}n}}{E_{k_{2}n}^{3}}}\right)-{\frac {|V_{k_{1}k_{2}}|^{2}V_{nn}V_{k_{1}n}}{E_{k_{1}n}E_{k_{1}k_{2}}^{3}}}{\Bigg ]}|k_{1}^{(0)}\rangle +{\frac {1}{2}}\left[{\frac {V_{nk_{1}}V_{k_{1}k_{2}}}{E_{nk_{1}}E_{k_{2}n}^{2}}}\left({\frac {V_{k_{2}n}V_{nn}}{E_{k_{2}n}}}-{\frac {V_{k_{2}k_{3}}V_{k_{3}n}}{E_{nk_{3}}}}\right)\right.\\&\quad \left.-{\frac {V_{k_{1}n}V_{k_{2}k_{1}}}{E_{k_{1}n}^{2}E_{nk_{2}}}}\left({\frac {V_{k_{3}k_{2}}V_{nk_{3}}}{E_{nk_{3}}}}+{\frac {V_{nn}V_{nk_{2}}}{E_{nk_{2}}}}\right)+{\frac {|V_{nk_{1}}|^{2}}{E_{k_{1}n}^{2}}}\left({\frac {3|V_{nk_{2}}|^{2}}{4E_{k_{2}n}^{2}}}-{\frac {2|V_{nn}|^{2}}{E_{k_{1}n}^{2}}}\right)-{\frac {V_{k_{2}k_{3}}V_{k_{3}k_{1}}|V_{nk_{1}}|^{2}}{E_{nk_{3}}^{2}E_{nk_{1}}E_{nk_{2}}}}\right]|n^{(0)}\rangle \end{aligned}}}$$

All terms involved k_j should be summed over k_j such that the denominator does not vanish.

It is possible to relate the k-th order correction to the energy E_n to the k-point connected correlation function of the perturbation V in the state ${\displaystyle |n^{(0)}\rangle }$. For ${\displaystyle k=2}$, one has to consider the inverse Laplace transform ${\displaystyle \rho _{n,2}(s)}$ of the two-point correlator:

$${\displaystyle \langle n^{(0)}|V(\tau )V(0)|n^{(0)}\rangle -\langle n^{(0)}|V|n^{(0)}\rangle ^{2}=\mathrel {\mathop {:} } \int _{\mathbb {R} }\!ds\;\rho _{n,2}(s)\,e^{-(s-E_{n}^{(0)})\tau }}$$

where ${\displaystyle V(\tau )=e^{H_{0}\tau }Ve^{-H_{0}\tau }}$ is the perturbing operator V in the interaction picture, evolving in Euclidean time. Then

$${\displaystyle E_{n}^{(2)}=-\int _{\mathbb {R} }\!{\frac {ds}{s-E_{n}^{(0)}}}\,\rho _{n,2}(s).}$$

Similar formulas exist to all orders in perturbation theory, allowing one to express ${\displaystyle E_{n}^{(k)}}$ in terms of the inverse Laplace transform ${\displaystyle \rho _{n,k}}$ of the connected correlation function

$${\displaystyle \langle n^{(0)}|V(\tau _{1}+\ldots +\tau _{k-1})\dotsm V(\tau _{1}+\tau _{2})V(\tau _{1})V(0)|n^{(0)}\rangle _{\text{conn}}=\langle n^{(0)}|V(\tau _{1}+\ldots +\tau _{k-1})\dotsm V(\tau _{1}+\tau _{2})V(\tau _{1})V(0)|n^{(0)}\rangle -{\text{subtractions}}.}$$

To be precise, if we write

$${\displaystyle \langle n^{(0)}|V(\tau _{1}+\ldots +\tau _{k-1})\dotsm V(\tau _{1}+\tau _{2})V(\tau _{1})V(0)|n^{(0)}\rangle _{\text{conn}}=\int _{\mathbb {R} }\,\prod _{i=1}^{k-1}ds_{i}\,e^{-(s_{i}-E_{n}^{(0)})\tau _{i}}\,\rho _{n,k}(s_{1},\ldots ,s_{k-1})\,}$$

then the k-th order energy shift is given by [9]

$${\displaystyle E_{n}^{(k)}=(-1)^{k-1}\int _{\mathbb {R} }\,\prod _{i=1}^{k-1}{\frac {ds_{i}}{s_{i}-E_{n}^{(0)}}}\,\rho _{n,k}(s_{1},\ldots ,s_{k-1}).}$$

Suppose that two or more energy eigenstates of the unperturbed Hamiltonian are degenerate. The first-order energy shift is not well defined, since there is no unique way to choose a basis of eigenstates for the unperturbed system. The various eigenstates for a given energy will perturb with different energies, or may well possess no continuous family of perturbations at all.

This is manifested in the calculation of the perturbed eigenstate via the fact that the operator

$${\displaystyle E_{n}^{(0)}-H_{0}}$$

does not have a well-defined inverse.

Let D denote the subspace spanned by these degenerate eigenstates. No matter how small the perturbation is, in the degenerate subspace D the energy differences between the eigenstates of H are non-zero, so complete mixing of at least some of these states is assured. Typically, the eigenvalues will split, and the eigenspaces will become simple (one-dimensional), or at least of smaller dimension than D.

The successful perturbations will not be "small" relative to a poorly chosen basis of D. Instead, we consider the perturbation "small" if the new eigenstate is close to the subspace D. The new Hamiltonian must be diagonalized in D, or a slight variation of D, so to speak. These perturbed eigenstates in D are now the basis for the perturbation expansion,

$${\displaystyle |n\rangle =\sum _{k\in D}\alpha _{nk}|k^{(0)}\rangle +\lambda |n^{(1)}\rangle .}$$

For the first-order perturbation, we need solve the perturbed Hamiltonian restricted to the degenerate subspace D,

$${\displaystyle V|k^{(0)}\rangle =\epsilon _{k}|k^{(0)}\rangle +{\text{small}}\qquad \forall |k^{(0)}\rangle \in D,}$$

simultaneously for all the degenerate eigenstates, where ${\displaystyle \epsilon _{k}}$ are first-order corrections to the degenerate energy levels, and "small" is a vector of ${\displaystyle O(\lambda )}$ orthogonal to D. This amounts to diagonalizing the matrix

$${\displaystyle \langle k^{(0)}|V|l^{(0)}\rangle =V_{kl}\qquad \forall \;|k^{(0)}\rangle ,|l^{(0)}\rangle \in D.}$$

This procedure is approximate, since we neglected states outside the D subspace ("small"). The splitting of degenerate energies ${\displaystyle \epsilon _{k}}$ is generally observed. Although the splitting may be small, ${\displaystyle O(\lambda )}$, compared to the range of energies found in the system, it is crucial in understanding certain details, such as spectral lines in Electron Spin Resonance experiments.

Higher-order corrections due to other eigenstates outside D can be found in the same way as for the non-degenerate case,

$${\displaystyle \left(E_{n}^{(0)}-H_{0}\right)|n^{(1)}\rangle =\sum _{k\not \in D}\left(\langle k^{(0)}|V|n^{(0)}\rangle \right)|k^{(0)}\rangle .}$$

The operator on the left-hand side is not singular when applied to eigenstates outside D, so we can write

$${\displaystyle |n^{(1)}\rangle =\sum _{k\not \in D}{\frac {\langle k^{(0)}|V|n^{(0)}\rangle }{E_{n}^{(0)}-E_{k}^{(0)}}}|k^{(0)}\rangle ,}$$

but the effect on the degenerate states is of ${\displaystyle O(\lambda )}$.

Near-degenerate states should also be treated similarly, when the original Hamiltonian splits aren't larger than the perturbation in the near-degenerate subspace. An application is found in the nearly free electron model, where near-degeneracy, treated properly, gives rise to an energy gap even for small perturbations. Other eigenstates will only shift the absolute energy of all near-degenerate states simultaneously.

Let us consider degenerate energy eigenstates and a perturbation that completely lifts the degeneracy to first order of correction.

The perturbed Hamiltonian is denoted as

$${\displaystyle {\hat {H}}={\hat {H}}_{0}+\lambda {\hat {V}}\,,}$$

where ${\displaystyle {\hat {H}}_{0}}$ is the unperturbed Hamiltonian, ${\displaystyle {\hat {V}}}$ is the perturbation operator, and ${\displaystyle 0<\lambda <1}$ is the parameter of the perturbation.

Let us focus on the degeneracy of the ${\displaystyle n}$-th unperturbed energy ${\displaystyle E_{n}^{(0)}}$. We will denote the unperturbed states in this degenerate subspace as ${\displaystyle \left|\psi _{nk}^{(0)}\right\rangle }$ and the other unperturbed states as ${\displaystyle \left|\psi _{m}^{(0)}\right\rangle }$, where ${\displaystyle k}$ is the index of the unperturbed state in the degenerate subspace and ${\displaystyle m\neq n}$ represents all other energy eigenstates with energies different from ${\displaystyle E_{n}^{(0)}}$. The eventual degeneracy among the other states with ${\displaystyle \forall m\neq n}$ does not change our arguments. All states ${\displaystyle \left|\psi _{nk}^{(0)}\right\rangle }$ with various values of ${\displaystyle k}$ share the same energy ${\displaystyle E_{n}^{(0)}}$ when there is no perturbation, i.e., when ${\displaystyle \lambda =0}$. The energies ${\displaystyle E_{m}^{(0)}}$ of the other states ${\displaystyle \left|\psi _{m}^{(0)}\right\rangle }$ with ${\displaystyle m\neq n}$ are all different from ${\displaystyle E_{n}^{(0)}}$, but not necessarily unique, i.e. not necessarily always different among themselves.

By ${\displaystyle V_{nl,nk}}$ and ${\displaystyle V_{m,nk}}$, we denote the matrix elements of the perturbation operator ${\displaystyle {\hat {V}}}$ in the basis of the unperturbed eigenstates. We assume that the basis vectors ${\displaystyle \left|\psi _{nk}^{(0)}\right\rangle }$ in the degenerate subspace are chosen such that the matrix elements ${\displaystyle V_{nl,nk}\equiv \left\langle \psi _{nl}^{(0)}\right|{\hat {V}}\left|\psi _{nk}^{(0)}\right\rangle }$ are diagonal. Assuming also that the degeneracy is completely lifted to the first order, i.e. that ${\displaystyle E_{nl}^{(1)}\neq E_{nk}^{(1)}}$ if ${\displaystyle l\neq k}$, we have the following formulae for the energy correction to the second order in ${\displaystyle \lambda }$