Perturbation Theory - Quantum Mechanics for Chemistry

What you will learn

The big idea: Perturbation theory attacks analytically intractable problems by splitting the Hamiltonian into an exactly solvable part $\hat{H}^0$ and a small perturbation $\hat{H}^1$ . It is the quantum analog of a Taylor expansion: the key is finding the small parameter to expand in.
How it works: Expand the energies and eigenfunctions as power series in a bookkeeping parameter $\lambda$ , then collect terms order by order to get corrections of increasing order. In practice the first- and second-order corrections are usually enough for quantitatively accurate results.
The master formula: Perturbation theory lets us write all corrections entirely in terms of the eigenfunctions and eigenvalues of the exactly solved problem, through matrix elements $H_{nk} = \langle n^0|\hat{H}^1|k^0\rangle$ .
Where it shows up: Hydrogen in a magnetic field, spin-orbit coupling, a shifted particle in a box, the anharmonic oscillator, and a perturbative picture of chemical bonding.

The idea behind perturbation theory¶

Perturbation theory attempts to solve analytically intractable problems by identifying an exactly solvable part and a small perturbation to it.
The method is similar in spirit to the Taylor expansion of continuous functions familiar from calculus. Just as in a Taylor expansion, the key is identifying the relatively small parameter in the problem to expand in.
Application of perturbation theory proceeds in two steps. Step one: identify the solvable part and the perturbation. Step two: expand the energy and eigenfunctions as a series of corrections of increasing order. In practice the first- and second-order corrections to the energy are sufficient to get quantitatively accurate results.
Perturbation theory allows us to write down expressions entirely in terms of the eigenfunctions and eigenvalues of the exactly solved problem.

perturbation of energy levels — Fig.1 Perturbation theory quantifies how much the energy levels shift when a small deviation is added to an exactly solvable Hamiltonian. For many problems that are impossible to solve exactly, one can still identify part of the Hamiltonian as exactly solvable, $\hat{H}^0$ , with the rest treated as a perturbation.

Time-independent perturbations¶

We start with a Hamiltonian $\hat{H}^0$ for some exactly solvable problem, such as a particle in a box, a harmonic oscillator, and so on:

\hat{H}^0 \mid n^0\rangle=E^0_n \mid n^0\rangle

(1)

The 0 superscript indicates the exactly solvable Hamiltonian, its eigenfunctions, and its eigenvalues. The state $\mid n^0\rangle$ is the eigenfunction corresponding to the $n$ -th eigenvalue $E^0_n$ .
Consider a problem whose Hamiltonian is similar to an exactly solvable one, differing only by a small perturbation $\hat{H}^1$ . “Small” means the eigenvalue shifts of the two systems are small relative to the level spacing.
The parameter $\lambda$ turns the perturbation on ( $\lambda=1$ ) and off ( $\lambda=0$ ):

\hat{H}=\hat{H}^0+\lambda {\hat{H}^1}

(2)

The objective of perturbation theory is to solve the new problem, expressing everything in terms of the eigenvalues and eigenfunctions of the exactly solvable problem:

\hat{H}\mid n\rangle =E_n \mid n\rangle

(3)

It is just like a Taylor expansion¶

We assume that the eigenvalues and eigenfunctions can be expanded in a power series in the parameter $\lambda$ , which is set to 1 in the end:

E_n ={\color{green}E^0_n}+{\color{red}\lambda E^1_n}+{\color{blue}\lambda^2 E^2_n}+...

(4)

\mid n\rangle = {\color{green}\mid n^0\rangle}+{\color{red} \lambda\mid n^1\rangle}+{\color{blue} \lambda^2\mid n^2\rangle} ...

(5)

Plugging the expansions into $\hat{H}\mid n\rangle =E_n \mid n\rangle$ gives an expression with various powers of $\lambda$ . The next step is to expand the brackets and group terms according to $\lambda^0$ , $\lambda^1$ , and $\lambda^2$ :

\Big({\color{green}\hat{H}^0}+{\color{red}\lambda \hat{H}^1} \Big)\Big({\color{green}\mid n^0\rangle}+{\color{red}\lambda\mid n^1\rangle} +{\color{blue}\lambda^2\mid n^2\rangle}\Big) = \Big({\color{green}E^0_n}+{\color{red}\lambda E^1_n}+{\color{blue}\lambda^2 E^2_n}\Big) \Big({\color{green}\mid n^0\rangle}+{\color{red}\lambda\mid n^1\rangle}+{\color{blue}\lambda^2\mid n^2\rangle}\Big)

(6)

Perturbation equations of order 0, 1, and 2¶

Opening the brackets and collecting different orders of $\lambda$ , we get the zeroth-, first-, and second-order perturbation equations:

\color{green}{\hat{H}^0\mid n^0\rangle = E^0_n\mid n^0 \rangle}

(7)

\color{red}{\hat{H}^0\mid n^1\rangle +\hat{H}^1\mid n^0\rangle = E^0_n\mid n^1 \rangle+E^1_n\mid n^0 \rangle}

(8)

\color{blue}{\hat{H}^0\mid n^2\rangle+\hat{H}^1\mid n^1\rangle = E^0_n\mid n^2 \rangle + E^1_n\mid n^1 \rangle+E^2_n\mid n^0 \rangle}

(9)

Note how the sum of the superscript indices determines the order of the perturbation expansion.
The zeroth order is just the exact solution.
The Hamiltonian only has a first-order term, while the eigenfunctions and eigenvalues are expanded to infinitely many terms. Usually going to second order is enough for most problems.

Computing perturbation corrections to energy levels¶

Example: first- and second-order corrections to the ground state

The first-order correction to the ground state requires computing the diagonal matrix element only:

E_0^{(1)} = \langle 0|\hat{H}^1|0\rangle

(13)

The second-order correction to the ground state requires the off-diagonal elements $H_{0k}$ , where $n=0$ and $k$ runs over all excited states:

E_0^{(2)} = {\sum_{k \neq 0} \frac{\mid H_{0k}\mid^2}{E^0_0-E^0_k}}

(14)

The denominator of the second-order term involves the difference between the energy of the given state $E_n$ and all other states $E_k$ (the summation index).
Key insight: if the matrix elements are of comparable magnitude, neighboring energy levels make the larger contributions to the perturbation expression, because their small energy denominators amplify the term.

Derivations of the first- and second-order corrections¶

Deriving the first-order energy correction $E^1_n$

Fixing the normalization. If the zeroth-order eigenfunctions are normalized, the unperturbed eigenfunction is orthogonal to all higher-order corrections:

\langle n^0 \mid n^0 \rangle=1

(15)

\langle n^0\mid n\rangle = \langle n^0\mid n^0\rangle + \lambda\langle n^0\mid n^1\rangle=1+0+...

(16)

\langle n^0 \mid n^{k} \rangle=0, \quad k=1,2,\dots

(17)

Using the orthogonality. Start from the first-order perturbation equation:

{\hat{H}^0\mid n^1\rangle +\hat{H}^1\mid n^0\rangle = E^0_n\mid n^1 \rangle+E^1_n\mid n^0 \rangle}

(18)

Multiply by $\langle n^0 \mid$ and use the Hermitian property of the Hamiltonian:

{{\langle n^0} \mid \hat{H}^0\mid n^1\rangle +{\langle n^0} \mid \hat{H}^1\mid n^0\rangle = E^0_n{\langle n^0} \mid n^1 \rangle+E^1_n {\langle n^0} \mid n^0 \rangle}

(19)

The first terms on each side cancel by orthogonality, since $\langle n^0 \mid \hat{H}^0\mid n^1\rangle = E^0_n \langle n^0 \mid n^1\rangle = 0$ .

First-order correction. We obtain the central result of perturbation theory, the first-order correction to the energy:

\color{red}{E_n^1 = \langle n^0 \mid \hat{H}^1\mid n^0 \rangle}

(20)

\boxed{E_n=\color{green}{E^0_n}+\color{red}{E^1_n}=\color{green}{E^0_n}+\color{red}{\langle n^0 \mid \hat{H}^1\mid n^0 \rangle}}

(21)

This expression looks like an expectation value but is different: the eigenfunctions of $\hat{H}^0$ sandwich the perturbation $\hat{H}^1$ . The two Hamiltonians in general do not share eigenfunctions.

Deriving the first-order correction to the eigenfunction $\mid n^1 \rangle$

We express the unknown first-order eigenfunction $\mid n^1 \rangle$ in terms of the known eigenfunctions $\mid k^0 \rangle$ , which form a complete basis set because $\hat{H}^0$ is Hermitian:

\mid n^1 \rangle = \sum_{k \neq n} c_{k} \mid k^0 \rangle

(22)

The coefficients are $c_k =\langle k^0 \mid n^1 \rangle$ . By orthogonality, the $k=n$ term drops out, since $c_n=\langle n^0 \mid n^1 \rangle=0$ ; hence the $k\neq n$ condition in the sum.

Insert the expansion into the first-order equation and take the dot product with $\langle k^0 \mid$ :

{\hat{H}^0\mid n^1\rangle +\hat{H}^1\mid n^0\rangle = E^0_n\mid n^1 \rangle+E^1_n\mid n^0 \rangle}

(23)

c_k E^0_k + \langle k^0 \mid \hat{H}^1 \mid n^0\rangle = c_k E^0_n

(24)

c_k = \frac{ \langle k^0 \mid \hat{H}^1 \mid n^0\rangle}{E^0_n-E^0_k}=\frac{H_{nk}}{E^0_n-E^0_k}

(25)

\boxed{\mid n^1 \rangle = \sum_{k \neq n} c_{k} \mid k^0 \rangle = \sum_{k \neq n} \frac{H_{nk}}{E^0_n-E^0_k} \mid k^0 \rangle}

(26)

Matrix element notation. We have introduced the convenient notation $H_{nk} = \langle k^0 \mid \hat{H}^1 \mid n^0\rangle$ . Note that the Hamiltonian inside the matrix element is always the perturbation part.

Deriving the second-order correction $E^2_n$

{\hat{H}^0\mid n^2\rangle+\hat{H}^1\mid n^1\rangle = E_n^0\mid n^2\rangle+ E^1_n\mid n^1 \rangle+E^2_n\mid n^0 \rangle}

(27)

Take the dot product with $\langle n^0 \mid$ :

{{\langle n^0 \mid }\hat{H}^0 \mid n^2\rangle+{\langle n^0 \mid }\hat{H}^1\mid n^1\rangle = E^0{\langle n^0}\mid n^2\rangle+ E^1_n{\langle n^0 \mid } n^1 \rangle+E^2_n {\langle n^0}\mid n^0 \rangle}

(28)

The first term on the left vanishes (Hermitian plus orthogonality), and the first two terms on the right vanish by orthogonality, leaving:

\color{blue}{E^2_n = \langle n^0 \mid \hat{H}^1 \mid n^1 \rangle}

(29)

We are not done: the expression still contains $\mid n^1 \rangle$ , which we express in terms of the known solutions:

\color{black}{E^2_n = \langle n^0 \mid \hat{H}^1 \mid n^1 \rangle}= \sum_{k \neq n} c_k \langle n^0 \mid \hat{H}^1 \mid k^0 \rangle = \sum_{k \neq n} c_k H_{nk}

(30)

\color{blue}{E^2_n = \sum_{k \neq n} \frac{\mid H_{nk}\mid^2}{E^0_n-E^0_k}}

(31)

\boxed{E_n = \color{green}{E^0_n} + \color{red}{\langle n^0\mid \hat{H}^1\mid n^0\rangle} + \color{blue}{\sum_{k \neq n} \frac{\mid H_{nk}\mid^2}{E^0_n-E^0_k}}}

(32)

Applications¶

Example 1: Second-order correction to the ground state

Write the second-order correction explicitly for the ground state of some exactly solvable Hamiltonian $\hat{H}^0$ perturbed by $\hat{H}^1$ :

E_n = E^0_n+ H_{nn} + \sum_{k \neq n} \frac{\mid H_{nk}\mid^2}{E^0_n-E^0_k}

(33)

E_0 =E^0_0+ H_{00} + \frac{\mid H_{01}\mid^2}{E^0_0-E^0_1}+\frac{\mid H_{02}\mid^2}{E^0_0-E^0_2}+ \frac{\mid H_{03}\mid^2}{E^0_0-E^0_3}+ ...

(34)

Notice that for the ground state the second-order correction is always negative, because $\Delta E_{0k}=E_0-E_k<0$ for every excited state.

Example 2: Magnetic field and spin-orbit coupling

A hydrogen atom in a magnetic field can be viewed as the H-atom Hamiltonian plus a small perturbation from the interaction with the field:

\hat{H}=\hat{H}_0 + \frac{e}{2m_e} B \hat{L}_z = \hat{H}_0 + \hat{H}^1

(35)

Using the first-order expression, the ground-state energy shift is (with $R_H$ the Rydberg constant and $\beta_B$ the Bohr magneton):

E_0=E^0_0 + \langle 0\mid \hat{H}^1 \mid 0\rangle = -\frac{R_H}{n^2}+m_l \beta_B B

(36)

In a similar way, the effect of spin-orbit coupling ( $LS$ ) is:

\hat{H} = \hat{H}_0 + A_{SO}\hat{L}\hat{S}, \qquad E=E_0+ A_{SO} \langle 0 \mid \hat{L} \hat{S}\mid 0 \rangle

(37)

Example 4: Anharmonic oscillator

The anharmonic oscillator is a harmonic oscillator plus a cubic perturbation:

\hat{H} = \hat{K}+ \frac{kx^2}{2} +\gamma x^3 = \hat{H}_0+\gamma x^3

(41)

The first-order correction vanishes by symmetry, since $x^3$ is odd and $|\psi_n|^2$ is even:

E_n^1 = \langle n\mid \gamma x^3 \mid n\rangle = \langle \text{even/odd} \mid \text{odd} \mid \text{even/odd}\rangle = 0

(42)

The energy levels must still change, so we turn to second order. For the ground state:

E^2_0 = \sum_{k \neq 0} \frac{\mid \langle 0\mid \gamma x^3 \mid k\rangle \mid^2}{E^0_0-E^0_k} = \frac{\mid \langle 0\mid \gamma x^3 \mid 1\rangle \mid^2}{E^0_0-E^0_1}+\frac{\mid \langle 0\mid \gamma x^3 \mid 3\rangle \mid^2}{E^0_0-E^0_3}+ ...

(43)

E_0 = \frac{\hbar \omega}{2} + 0 + \frac{H^2_{01}}{\hbar \omega}+\frac{H^2_{03}}{2\hbar \omega}+ ...

(44)

Only odd terms in the sum contribute. The matrix elements must be evaluated explicitly using Hermite polynomials.

The idea behind perturbation theory¶

Time-independent perturbations¶

It is just like a Taylor expansion¶

Perturbation equations of order 0, 1, and 2¶

Computing perturbation corrections to energy levels¶

Derivations of the first- and second-order corrections¶

Applications¶

Problems¶

Problem 1¶

Problem 2¶

Problem 3¶

Problem 4¶

Problem 5¶