XI Perturbation theory

We want to find approximations for the energies $E_n$ and eigenstates $|{\psi }_n⟩$ of a Hamiltonian $\widehat{H}$,

$$E_n|{\psi }_n⟩=\widehat{H}|{\psi }_n⟩,$$

(139)

assuming that the Hamiltonian is of the form $\widehat{H}={\widehat{H}}^{(0)}+\widehat{w}$, where ${\widehat{H}}^{(0)}$ represents a simplified system with energies $E_n^{(0)}$ and eigenstates $|{\psi }_n^{(0)}⟩$ solving the Schrödinger equation

$$E_n^{(0)}|{\psi }_n^{(0)}⟩={\widehat{H}}^{(0)}|{\psi }_n^{(0)}⟩.$$

(140)

The difference $\widehat{w}=\widehat{H}-{\widehat{H}}^{(0)}$ between the real and the simplified system is called the perturbation, and the approximation scheme is called perturbation theory.

The idea of perturbation theory is to assume that the perturbation $\widehat{w}=\lambda \widehat{W}$ is the product of an operator $\widehat{W}$ and a small number $\lambda$ (determining the strength of the perturbation).

Since the Hamiltonian

$$\widehat{H}={\widehat{H}}^{(0)}+\lambda \widehat{W}$$

(141)

now depends on the parameter $\lambda$, the energies $E_n$ and eigenstates $|{\psi }_n⟩$ also depend on this parameter.

Consequently, we can expand the energies and eigenstates into a series:

	$E_n(\lambda )$	$=$	$E_n^{(0)}+\lambda E_n^{(1)}+{\lambda }^2{E}_n^{(2)}+\mathrm{\cdots }$		(142)
		$=$	${\displaystyle \sum _{l=0}^{\mathrm{\infty }}}{\lambda }^l{E}_n^{(l)},$
	$|{\psi }_n(\lambda )⟩$	$=$	$|{\psi }_n^{(0)}⟩+\lambda |{\psi }_n^{(1)}⟩+{\lambda }^2|{\psi }_n^{(2)}⟩+\mathrm{\cdots }$		(143)
		$=$	${\displaystyle \sum _{l=0}^{\mathrm{\infty }}}{\lambda }^l|{\psi }_n^{(l)}⟩.$

Perturbation theory provides a systematic scheme to express the quantities $E_n^{(l)}$ and $|{\psi }_n^{(l)}⟩$ in terms of the unperturbed energies $E_n^{(0)}$, the unperturbed eigenstates $|{\psi }_n^{(0)}⟩$, and the matrix elements of the perturbation

$$W_{mn}=⟨{\psi }_m^{(0)}|\widehat{W}|{\psi }_n^{(0)}⟩.$$

The scheme commences by introducing the series expansions (142) and (143) into the Schrödinger equation (139) and sorting the expressions order by order in $\lambda$:

	$E_n^{(0)}|{\psi }_n^{(0)}⟩+$
	$\lambda (E_n^{(0)}|{\psi }_n^{(1)}⟩+E_n^{(1)}|{\psi }_n^{(0)}⟩)+$
	${\lambda }^2(E_n^{(0)}|{\psi }_n^{(2)}⟩+E_n^{(1)}|{\psi }_n^{(1)}⟩+E_n^{(2)}|{\psi }_n^{(0)}⟩)+$
	$\mathrm{\cdots }$
	$=$
	${\widehat{H}}^{(0)}|{\psi }_n^{(0)}⟩+$
	$\lambda ({\widehat{H}}^{(0)}|{\psi }_n^{(1)}⟩+\widehat{W}|{\psi }_n^{(0)}⟩)+$
	${\lambda }^2({\widehat{H}}^{(0)}|{\psi }_n^{(2)}⟩+\widehat{W}|{\psi }_n^{(1)}⟩)+$
	$\mathrm{\cdots }$		(144)

In order to fulfill this equation, the expressions in front of the powers ${\lambda }^l$ have to be identical on both sides of the equation.

Since we assume that the perturbation $\lambda$ is small, we concentrate on the first few orders.

In order to use this property, take the scalar product of an unperturbed eigenstate $|{\psi }_m^{(0)}⟩$ with both sides of Eq. (146):

$$E_n^{(0)}⟨{\psi }_m^{(0)}|{\psi }_n^{(1)}⟩+E_n^{(1)}{\delta }_{nm}=E_m^{(0)}⟨{\psi }_m^{(0)}|{\psi }_n^{(1)}⟩+W_{mn}.$$

(147)

Here we used $⟨{\psi }_m^{(0)}|{\widehat{H}}^{(0)}{\psi }_n^{(1)}⟩=⟨{\widehat{H}}^{(0)}{\psi }_m^{(0)}|{\psi }_n^{(1)}⟩=E_m^{(0)}⟨{\psi }_m^{(0)}|{\psi }_n^{(1)}⟩$ and the definition of the matrix element $W_{mn}$.

For $n=m$, Eq. (147) reduces to

$$E_n^{(1)}=W_{nn},$$

(148)

while for $n\ne m$ we find

$$⟨{\psi }_m^{(0)}|{\psi }_n^{(1)}⟩=\frac{W_{mn}}{E_n^{(0)}-E_m^{(0)}}.$$

(149)

Since the unperturbed eigenstates form a complete basis this gives

$$|{\psi }_n^{(1)}⟩=\sum _{m\ne n}\frac{W_{mn}}{E_n^{(0)}-E_m^{(0)}}|{\psi }_m^{(0)}⟩,$$

(150)

where the sum is over all indices $m$, with the exception of $m=n$.

In this course, we will not need $|{\psi }_n^{(2)}⟩$, $E_n^{(3)}$, or higher terms of the perturbation series.

• •

  Perturbed energy in first order of $\lambda$:

  $$E_n\approx E_n^{(0)}+\lambda W_{nn}.$$

  (152)

  Hence, the energy shift $E_n-E_n^{(0)}=\lambda ⟨{\psi }_n^{(0)}|\widehat{W}|{\psi }_n^{(0)}⟩=⟨\widehat{w}⟩$ is given by the expectation value of the perturbation $\widehat{w}$.

• •

  Perturbed eigenstate in first order of $\lambda$:

  $$|{\psi }_n⟩\approx |{\psi }_n^{(0)}⟩+\lambda \sum _{m\ne n}\frac{W_{mn}}{E_n^{(0)}-E_m^{(0)}}|{\psi }_m^{(0)}⟩.$$

  (153)

• •

  Perturbed energy in second order of $\lambda$:

  $$E_n\approx E_n^{(0)}+\lambda W_{nn}+{\lambda }^2\sum _{m\ne n}\frac{{|W_{mn}|}^2}{E_n^{(0)}-E_m^{(0)}}.$$

  (154)

In the following we only need these results, not any details of their derivation.

We now discuss various simple examples of perturbed systems, including cases where we can compare to exact solutions. The first example is a worn-out oscillator, described by a reduced restoring force $F=-m{\omega }_0^{2}x+\lambda x$. This corresponds to a potential $V=\frac{1}{2}m{\omega }_0^2{x}^2-\frac{1}{2}\lambda x^2$, which is still parabolic, and hence still describes a harmonic oscillator.

The Hamiltonian

$$\widehat{H}=\frac{1}{2m}{\widehat{p}}^2+\frac{1}{2}m{\omega }_0^2{x}^2-\frac{1}{2}\lambda x^2$$

(159)

differs from the unperturbed harmonic oscillator by the perturbation $\widehat{w}=-\frac{1}{2}\lambda x^2$. According to first-order perturbation theory, the energy shift of the states is given by the expectation value of this perturbation, calculated with the unperturbed states. Hence, the ground-state energy shift is

$$E_0-E_0^{(0)}\approx ⟨{\psi }_0^{(0)}|\widehat{w}|{\psi }_0^{(0)}⟩=-\frac{1}{2}\lambda ⟨{\psi }_0^{(0)}|{\widehat{x}}^2|{\psi }_0^{(0)}⟩,$$

(160)

where ${\psi }_0^{(0)}(x)={\left(\frac{m{\omega }_0}{\pi \mathrm{\hslash }}\right)}^{1/4}\exp \left(-x^2\frac{m{\omega }_0}{2\mathrm{\hslash }}\right)$ is the ground state wave function of the unperturbed harmonic oscillator (see section VI). This state has the form of a minimal-uncertainty wave packet (see section VII.5), for which we already obtained $⟨{\psi }_0^{(0)}|{\widehat{x}}^2|{\psi }_0^{(0)}⟩=\frac{\mathrm{\hslash }}{2m{\omega }_0}$. Consequently, first-order perturbation theory predicts

$$E_0-E_0^{(0)}\approx -\frac{1}{2}\lambda \frac{\mathrm{\hslash }}{2m{\omega }_0},$$

(161)

which indeed agrees with Eq. (158).

Another exactly solvable perturbed problem describes a harmonic oscillator subjected to an additional constant force of strength $\lambda$, $F=-m{\omega }^{2}x+\lambda$. A constant force simply shifts the equilibrium position (the position where the force vanishes) to

$$x_0=\lambda /(m{\omega }^2).$$

(162)

The potential energy can be written as

$$V(x)=\frac{1}{2}m{\omega }^2{x}^2-\lambda x,$$

(163)

hence, the perturbation is $\widehat{w}=-\lambda x$.

We again concentrate on the ground-state energy $E_0$. The first-order energy shift $E_0-E_0^{(0)}\approx \lambda W_{00}$ is given by the expectation value of the perturbation $\widehat{W}=-\widehat{x}$,

$$W_{00}=-⟨{\psi }_0^{(0)}|\widehat{x}|{\psi }_0^{(0)}⟩.$$

(167)

Now notice that $x{\psi }_0^{(0)}(x)=\sqrt{\mathrm{\hslash }/(2m\omega )}{\psi }_1^{(0)}(x)$ is proportional to the wave function

$${\psi }_1^{(0)}(x)=c_{1}x\exp \left(-x^2\frac{m\omega }{2\mathrm{\hslash }}\right),$$

of the first excited state of the unperturbed system, calculated in section VI.

Hence

$$W_{00}=-⟨{\psi }_0^{(0)}|x|{\psi }_0^{(0)}⟩=-\sqrt{\mathrm{\hslash }/(2m\omega )}⟨{\psi }_0^{(0)}|{\psi }_1^{(0)}⟩=0$$

(168)

due to the orthogonality of the eigenstates $|{\psi }_0^{(0)}⟩$ and $|{\psi }_1^{(0)}⟩$. This agrees with the absence of a term linear in $\lambda$ in the exact expression (166) of the energy shift.

In second-order perturbation theory,

$$E_0\approx E_0^{(0)}+\lambda W_{00}+{\lambda }^2\sum _{m\ne n}\frac{{|W_{m0}|}^2}{E_0^{(0)}-E_m^{(0)}},$$

(169)

we have to calculate the matrix elements

	$W_{m0}$	$=$	$⟨{\psi }_m^{(0)}|\widehat{W}|{\psi }_0^{(0)}⟩$		(170)
		$=$	$-⟨{\psi }_m^{(0)}|\widehat{x}|{\psi }_0^{(0)}⟩$		(171)
		$=$	$-\sqrt{\mathrm{\hslash }/(2m\omega )}⟨{\psi }_m^{(0)}|{\psi }_1^{(0)}⟩$		(172)

Again because of the orthogonality of the eigenstates $|{\psi }_n^{(0)}⟩$, all of these matrix elements vanish, with the exception of the element

$W_{10}=-\sqrt{\mathrm{\hslash }/(2m\omega )}.$

(173)

Hence the energy shift reduces to

$$E_0-E_0^{(0)}\approx \frac{{|W_{10}|}^2}{E_0^{(0)}-E_1^{(0)}}.$$

(174)

Moreover, since $E_n^{(0)}=\mathrm{\hslash }\omega (n+1/2)$, we have $E_0^{(0)}-E_1^{(0)}=-\mathrm{\hslash }\omega$. Collecting all results we obtain

$$E_0-E_0^{(0)}\approx {\lambda }^2\frac{\mathrm{\hslash }/(2m\omega )}{-\mathrm{\hslash }\omega }=-\frac{{\lambda }^2}{2m{\omega }^2},$$

(175)

which agrees with the exact result (166).

The parabolic potential $V_{\mathrm{osc}}(x)=\frac{1}{2}m{\omega }^2{x}^2$ is often used as an approximation of the motion in around an equilibrium position $x-x_0$, such as in the vibration of molecules or solids. The exact potential $V(x)$ can be expanded in a Taylor series around the equilibrium position,

	$V(x)$	$=$	$V_0+{\displaystyle \frac{1}{2}}m{\omega }^2{(x-x_0)}^2$
			$+\text{terms of order }{(x-x_0)}^3\text{ and higher.}$

In this expansion there are no terms linear in $(x-x_0)$ since the force $F=-V^{\prime }$ vanishes at the equilibrium position $x=x_0$.

As we have seen in example II, the constant $V_0$ simply shifts the energy, and the displacement by $x_0$ does not change the energies at all. In the following we set $x_0=0$, $V_0=0$.

The terms of order $x^3$ and higher are called the anharmonicity of the potential. Classically, they perturb the motion of the oscillator so that the oscillation period $T$ depends on the energy of the oscillator (recall the period $T$ of a harmonic oscillator is independent of the oscillation amplitude $\mathrm{\Delta }x$). Quantum mechanical, this results in a perturbation of the bound-state energies. For small energies, the classical oscillation amplitude $\mathrm{\Delta }x$ is small, and hence the effect of the anharmonic terms $x^n\propto \mathrm{\Delta }{x}^n$ should be small and rapidly decrease with $n$. We hence only account for the anharmonic term of lowest order in the Taylor series,

$$V(x)=\frac{1}{2}m{\omega }^2{x}^2+\lambda \frac{x^n}{n!}$$

(177)

where typically $n=3$, but sometimes also $n=4$ or larger due to symmetries of the problem.

Now we apply perturbation theory in order to estimate the resulting energy shift of the ground state.

In first-order perturbation theory, the energy shift is given by the expectation value of the perturbation $\lambda x^n/n!$,

$$E_0-E_0^{(0)}=\frac{\lambda }{n!}⟨{\psi }_0^{(0)}|x^n|{\psi }_0^{(0)}⟩.$$

(178)

We hence have to calculate the overlap integral

	$⟨{\psi }_0^{(0)}|x^n|{\psi }_0^{(0)}⟩$	$=$	${\displaystyle {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}}{x}^n{|{\psi }_0^{(0)}(x)|}^2𝑑x$
		$=$	${\displaystyle {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}}{x}^n{\left({\displaystyle \frac{m\omega }{\pi \mathrm{\hslash }}}\right)}^{1/2}\exp \left(-x^2{\displaystyle \frac{m\omega }{\mathrm{\hslash }}}\right)𝑑x.$

This integral vanishes if $n$ is an odd integer, since the integrand is antisymmetric (the integral from $-\mathrm{\infty }$ to 0 exactly cancels the integral from $0$ to $\mathrm{\infty }$). Hence, in first-order perturbation theory

$$E_0-E_0^{(0)}\approx 0\mathit{\hspace{1em}}(n\text{ odd}).$$

(181)

An energy shift is only found in second-order perturbation theory, which we however do not pursue for the present problem.

For $n$ and even integer we use the standard integral

$${\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{x}^n\exp (-a{x}^2)𝑑x=\sqrt{\pi }{a}^{-(n+1)/2}{2}^{-n}\frac{n!}{(n/2)!},$$

where we set $a=\frac{m\omega }{\mathrm{\hslash }}$. This gives the first order energy shift

$$E_0-E_0^{(0)}\approx \lambda {\left(\frac{\mathrm{\hslash }}{m\omega }\right)}^{n/2}\frac{2^{-n}}{(n/2)!}.$$

(182)

It is convenient to express this result in terms of the uncertainty $\mathrm{\Delta }x=\sqrt{\frac{\mathrm{\hslash }}{2m\omega }}$ of position in the ground state, which is the equivalent of the classical oscillation amplitude. One then obtains

$$E_0-E_0^{(0)}\approx \lambda {(\mathrm{\Delta }x)}^n\frac{2^{-n/2}}{(n/2)!},(n\text{ even}).$$

(183)

Note that the numerical coefficient rapidly falls of with increasing $n$.