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In perturbation theory for quantum mechanics, using Schrodinger equation and the expansions $$H=H_0+\varepsilon H_1+\varepsilon^2 H_2+\cdots$$ and $$E_n=E_n^{(0)}+\varepsilon E_n^{(1)}+\varepsilon^2 E_n^{(2)}+\cdots$$ one can get the $1$st, $2$nd and even $3$rd, $4$th, $5$th order corrections of energy-level $E_n$ and the corresponding wave function corrections.

(Non-degenerate, time-independent.)

But does there exist systematic(?) series (not Dyson series) to attain higher order (6th, 7th and more) corrections of them? The method is clear, but the results are too complicated, are there any simplified results?

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2 Answers 2

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Dyson series are specifically geared for many body problems. However, if you have been that far, you are probably also familiar with the Brueckner-Goldstone expansion for the ground state energy that in principle is applicable for any type of system.

The perturbation theory that you cite is known as Rayleigh-Schrödinger perturbation theory - it is used for eigenvalue problems, where we try to determine the eigenstates and eigenenergies. There is its close relative Brilluoin-Wigner perturbation theory, which is used for dealing with scattering problems, and where the energy is usually given, which greatly simplifies obtaining general formulas. Dyson expansion is essentially a more sophisticated form of this perturbation theory - even when applied in condensed matter. One then uses some additional relations to determine the ground state energy, thermodynamic potentials, etc. - these can be again teased out from the many-body texts.

Caveat
Nice closed form expressions for infinite series do not necessarily make life easier. E.g., using Goldstone theorem to rederive ground state of an exactly solvable system can prove to be quite a pain.

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  • $\begingroup$ ... with the curse that you still need to find the energy from the zeroes of a polynomial. $\endgroup$ Jul 6, 2022 at 14:13
  • $\begingroup$ Dec. 27th 2023: 404 Link not found. $\endgroup$
    – Qmechanic
    Dec 28, 2023 at 11:41
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  1. Let us start with time-independent Brillouin-Wigner (BW) perturbation theory for a non-degenerate isolated eigenstate, i.e. it is assumed to not be part of a continuous spectrum. Here we will consider a (not necessarily convergent) formal power series in a coupling constant $\lambda$.

  2. Using standard notation we have $$\begin{array}{lrcll}\text{Hamiltonian}& \hat{H}&=&\hat{H}_0+\lambda\hat{V}&\cr \text{Unperturbed eigenstate}& \hat{H}_0|\psi_0\rangle&=& E_0|\psi_0\rangle,& \langle\psi_0|\psi_0\rangle~=~1,\cr \text{Perturbed eigenstate}& \hat{H}|\psi\rangle&=& E|\psi\rangle,&\cr \text{Energy shift}&\Delta&:=&E-E_0&\cr & &=&\sum_{n\in\mathbb{N}} E_n\lambda^n&~=~{\cal O}(\lambda),\cr &E&=&\sum_{n\in\mathbb{N}_0} E_n\lambda^n,& \cr \text{Projection operator}& \hat{P}&:=&|\psi_0\rangle\langle\psi_0|,& [\hat{P},\hat{H}_0]~=~0,\cr \text{Projection operator}& \hat{Q}&:=&\hat{\bf 1}-\hat{P}~=~&(E-\hat{H}_0)\hat{R},\cr \text{Resolvent}& \hat{R}&:=&\frac{1}{E-\hat{H}_0}\hat{Q}~=~&\frac{1}{E_0-\hat{H}_0+\Delta}\hat{Q}\cr & &=&\frac{1}{1+\hat{R}_0\Delta}\hat{R}_0 ~=~&\sum_{m\in\mathbb{N}_0}(-\Delta)^m\hat{R}_0^{m+1},\cr \text{Resolvent}& \hat{R}_0&:=&\frac{1}{E_0-\hat{H}_0}\hat{Q}. \end{array}\tag{1}$$

  3. We now assume that the perturbed state $|\psi\rangle$ has a non-zero overlap $$ \langle\psi_0|\psi\rangle~\neq~0 \tag{2}$$ with the unperturbed state $|\psi_0\rangle$. We may then rescale $|\psi\rangle$ such that $$ \hat{P}|\psi\rangle~=~|\psi_0\rangle \quad\Rightarrow\quad \langle\psi_0|\psi\rangle~=~1 \quad\wedge\quad \langle\psi|\psi\rangle~>~1. \tag{3}$$

  4. Then $$|\psi\rangle-|\psi_0\rangle~\stackrel{(3)}{=}~\hat{Q}|\psi\rangle ~\stackrel{(1)}{=}~\hat{R}(E-\hat{H}_0)|\psi\rangle ~\stackrel{(1)}{=}~\hat{R}\lambda\hat{V}|\psi\rangle,\tag{4}$$ which leads to a fixed-point equation for the perturbed state $$ |\psi\rangle~\stackrel{(4)}{=}~|\psi_0\rangle+\hat{R}\lambda\hat{V}|\psi\rangle~\stackrel{(4)}{=}~\sum_{n\in\mathbb{N}_0}(\hat{R}\lambda\hat{V})^n|\psi_0\rangle~=~\frac{1}{1-\hat{R}\lambda\hat{V}}|\psi_0\rangle.\tag{5}$$

  5. This leads to the BW fixed-point equation for the shifted energy $$\begin{align}\Delta~\stackrel{(1)}{:=}~&E-E_0\cr ~\stackrel{(3)}{=}~&\langle\psi_0|(E-E_0)|\psi\rangle\cr ~\stackrel{(1)}{=}~&\langle\psi_0|(E-\hat{H}_0)|\psi\rangle\cr ~\stackrel{(1)}{=}~&\langle\psi_0|\lambda\hat{V}|\psi\rangle\cr ~\stackrel{(5)}{=}~&\langle\psi_0|\lambda\hat{V}\frac{1}{1-\hat{R}\lambda\hat{V}}|\psi_0\rangle\cr ~\stackrel{(5)}{=}~&\sum_{n\in\mathbb{N}_0}\langle\psi_0|\lambda\hat{V}(\hat{R}\lambda\hat{V})^n|\psi_0\rangle\cr ~\stackrel{(1)}{=}~&\langle\psi_0|\lambda\hat{V}\frac{1}{1-\frac{1}{1+\hat{R}_0\Delta}\hat{R}_0\lambda\hat{V}}|\psi_0\rangle\cr ~=~&\sum_{n\in\mathbb{N}_0}\lambda^{n+1}\sum_{m_1,\ldots,m_n\in\mathbb{N}_0}(-\Delta)^{\sum_{i=1}^nm_i}\cr &\underbrace{\langle\psi_0|\hat{V}\hat{R}_0^{m_1+1}\hat{V}\hat{R}_0^{m_2+1}\hat{V}\ldots\hat{V}\hat{R}_0^{m_n+1}\hat{V}|\psi_0\rangle}_{~=:~ \langle m_1,m_2,\ldots,m_n\rangle}. \end{align}\tag{6}$$

  6. We may then systematically obtain a formula for the $N$th-order energy correction $$\begin{align}E_N~\stackrel{(6)+(8)}{=}~& \sum_{n=0}^{N-1}\sum_{M=0}^{N-(n+1)}(-1)^MP_{N-(n+1),M}\cr &\sum_{m_1,\ldots,m_n\in\mathbb{N}_0}^{\sum_{i=1}^nm_i=M} \underbrace{\langle\psi_0|\hat{V}\hat{R}_0^{m_1+1}\hat{V}\hat{R}_0^{m_2+1}\hat{V}\ldots\hat{V}\hat{R}_0^{m_n+1}\hat{V}|\psi_0\rangle}_{~=:~ \langle m_1,m_2,\ldots,m_n\rangle}.\end{align} \tag{7}$$ in Rayleigh-Schrödinger (RS) perturbation theory.

  7. Here the polynomials $$P_{N,M}(E_1,\ldots,E_{N-M+1})~:=~\sum_{n_1,\ldots,n_M\in\mathbb{N}}^{\sum_{i=1}^M n_i=N}\prod_{j=1}^ME_{n_j}\tag{8}$$ have a generating function $$ P_{N,M}\lambda^Nq^M~=~\sum_{M\in\mathbb{N}_0}(q\Delta)^M~=~\frac{1}{1-q\Delta}.\tag{9}$$ The first few polynomials read $$\begin{align} P_{N,0}~=~&\delta_N^0,\cr P_{N,1}~=~&E_N,\cr P_{N,2}~=~&\sum_{n-1}^{N-1}E_nE_{N-n},\cr ~\vdots&\cr P_{N,N}~=~&E_1^N. \end{align}\tag{10}$$ One can check that the polynomial in eq. (7) only depends on the previous energy corrections $E_1,\ldots, E_{N-2}$, so that eq. (7) constitutes a recurrence relation.

  8. The first few orders of energy corrections read $$\begin{align} E_1~\stackrel{(7)}{=}~&P_{0,0}\langle\rangle ~=~ \langle\psi_0|\hat{V}|\psi_0\rangle, \cr E_2~\stackrel{(7)}{=}~&P_{0,0}\langle 0\rangle ~=~\langle\psi_0|\hat{V}\hat{R}_0\hat{V}|\psi_0\rangle,\cr E_3~\stackrel{(7)}{=}~&P_{0,0}\langle 00\rangle-P_{1,1}\langle 0\rangle\cr ~=~&\langle\psi_0|\hat{V}\hat{R}_0\hat{V}\hat{R}_0\hat{V}|\psi_0\rangle-E_1\langle\psi_0|\hat{V}\hat{R}_0^2\hat{V}|\psi_0\rangle,\cr E_4~\stackrel{(7)}{=}~&P_{0,0}\langle 000\rangle-2P_{1,1}\langle 10\rangle -P_{2,1}\langle 1\rangle +P_{2,2}\langle 2\rangle\cr ~=~&\langle\psi_0|\hat{V}\hat{R}_0\hat{V}\hat{R}_0\hat{V}\hat{R}_0\hat{V}|\psi_0\rangle -2E_1\langle\psi_0|\hat{V}\hat{R}_0^2\hat{V}\hat{R}_0\hat{V}|\psi_0\rangle\cr &-E_2\langle\psi_0|\hat{V}\hat{R}_0^2\hat{V}|\psi_0\rangle +E_1^2\langle\psi_0|\hat{V}\hat{R}_0^3\hat{V}|\psi_0\rangle,\cr E_5~\stackrel{(7)}{=}~&P_{0,0}\langle 0000\rangle -P_{1,1}(2\langle 100\rangle+\langle 010\rangle) -2P_{2,1}\langle 10\rangle\cr &+P_{2,2}(2\langle 20\rangle+\langle 11\rangle) -P_{3,1}\langle 1\rangle +P_{3,2}\langle 2\rangle-P_{3,3}\langle 3\rangle,\cr E_6~\stackrel{(7)}{=}~&P_{0,0}\langle 00000\rangle -P_{1,1}(2\langle 1000\rangle+\langle 0100\rangle)\cr &-P_{2,1}(2\langle 100\rangle+\langle 010\rangle)\cr &+P_{2,2}(2\langle 200\rangle+\langle 020\rangle+2\langle 110\rangle+\langle 101\rangle)\cr &-2P_{3,1}\langle 10\rangle +P_{3,2}(2\langle 20\rangle+\langle 11\rangle) -2P_{3,3}(\langle 30\rangle+\langle 21\rangle)\cr &-P_{4,1}\langle 1\rangle +\underbrace{P_{4,2}}_{=2E_3E_1+E_2^2}\langle 2\rangle -\underbrace{P_{4,3}}_{=3E_2E_1^2}\langle 3\rangle +P_{4,4}\langle 4\rangle, \end{align}\tag{11}$$ and so forth.

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  • $\begingroup$ If $|\psi^{\prime}_0\rangle\neq|\psi_0\rangle$ is another non-degenerate isolated eigenstate with energy $E^{\prime}_0\neq E_0$, and knowing the perturbed energy $E$ (to some finite order), then we can calculate the overlap $\langle\psi^{\prime}_0|\psi\rangle$ $~=~\langle\psi^{\prime}_0|\hat{R}\lambda\hat{V}|\psi\rangle$ $~=~\langle\psi^{\prime}_0|\sum_{n\in\mathbb{N}}(\hat{R}\lambda\hat{V})^n|\psi_0\rangle$ to some finite order. $\endgroup$
    – Qmechanic
    Jan 16 at 11:41

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