Time-independent perturbation theory

The basic setup is simple. We consider Hamiltonians of the form

where $\eps \ll 1$. Here we are assuming that the matrix elements of $H_1$ are of the same order or smaller than those of $H_0$. We wish to solve the time-independent Schroedinger equation $H \ket{\psi} = E \ket{\psi}$ when the solutions for $H_0$ are known. We will also assume that the spectra of $H_0$ are discrete. States in the regime with continuous spectra require a different treatment via scattering theory.

Since the eigenvalues and eigenvectors $\ket{n}$, $E_n$ should be those of $H_0$ as $\eps \to 0$, we make the following assumptions about the true eigenvalues and eigenvectors.

where $H_0 \ket{n^{(0)}} = E^{(0)}_n$, and the eigenvectors are taken to be orthonormal. We then write

We then demand that this equation be solved order by order in $\eps$. The solution at order $\eps^n$ is known as $n$th order perturbation theory.

First order perturbation theory (non-degenerate)¶

The $\eps^0$ terms cancel by supposition. At $\cO(\eps)$, the terms are:

We first take the inner product of this equation with \bra{n^{(0)}, and find

or

Next, we take the inner product with $\bra{m^{(0)}}$ for $m \neq n$. Then we have

or

We can see that this fails if $E^0_n$ is a degenerate eigenvalue, as there will be terms in the above expression with vanishing denominator. We will turn to this case later, and instead assume that there is no degeneracy. The techniques we explore here are called non-degenerate perturbation theory.

Now $\ket{m^{(0)}}$ is a complete basis in the Hilbert space, as $H_0$ is a Hermitian operator. To find $\ket{n^{(1)}}$, we will need to know $\brket{n^{(0)}}{n^{(1)}}$. I will demand that $\brket{n^{(0)}}{n} = 1$, which is not yet the standard normalization. We can rescale $\ket{n}$ after our calculation, but computig its norm and then dividing $\ket{n}$ by it. The upshot is that since $\brket{n^{(0)}}{n^{(0)}} = 1$, and $\ket{n} = \ket{n^{(0)}} + \ldots$, all corrections to $\ket{n^{(0)}}$ will be orthogonal to $\ket{n^{(0)}}$.

Given this, and that $\sum_n \ket{n^{(0)}}\bra{n^{(0)}} = 1$, we have

Note that at this order the norm of the state only differs from 1 by a factor of $\cO(\eps^2)$; we will need to go to second order in corrections to $\ket{k}$ to compute the norm, and to see what is needed to properly normalize the state.

Example: anharmonic oscillator¶

The classic example is the anharmonic quantum oscillator:

We will do perturbation series with the parameter $\lambda$. You should be objecting that this isn’t quite right. $\lambda$ has dimensions of $(energy)/(length)^4$. As a dimensionful paraneter one cannot technically call it “small”. But we can combine this with the parameters of the unperturbed Hamiltonian to make a small parameter. For example, we know that we have a length scale

and an energy scale ${\cal E} = \hbar\omega$. So by dimensional analysis, the combination

is a likely perturbation parameter. We will see that this does in fact control the correction terms in perturbation theory.

The correction to the energy $E^{(0)}_k = \hbar\omega(k + \half)$ in this case is taken from (6):

Now for $k \sim {\cal O}(1)$, the ratio of this to the energy gap $\hbar \omega$ is $\sim k^2 \eps$, confirming our point above. furthermore, note that once $k^2 \sim 1/\eps$, this correction is no longer small. This is a general feature of perturbation theory; it only works for low energies.

Similarly the first order correction to the state (9) gives:

We can see again how the dimensionless parameter $\eps$ appears. The numerator is straightforward but tedious to evaluate. $(a + a^{\dagger})^4$ clearly connects the level $k$ to the levels in the range $\ell \in \{k-4, k-2,k+2,k+4\}$ (you should think about why $k - 1, k-3$ do not appear). So the ratio will scale with $k^2$, and again if $\eps k^2 \sim \cO(1)$, perturbation theory will start to break down.

Second order perturbation theory (non-degenerate)¶

It is worth going out one more order so that you can see how things develop and how complicated they get. Also, it is often the case that corrections to the energy vanishes at first order. For example, let us say that we perturbed the harmonic oscillator by $-\alpha x$. (In fact this is exactly solvable; but we can still do perturbatiob theory and I’m trying to make a point. By the way, what might you worry about if the perturbation was by $g x^3$?). The first order energy shift \bra{k^{(0)} x \ket{k^{(0)} = 0 because the states have definite parity but $x$ has odd parity. Thus $x\ket{k}$, ket{k have opposite parity, and so zero overlap.

The $\cO(\eps^2)$ terms in (3) are:

Taking the inner product with $\bra{n^{(0)}}$, the first terms on each side cancel off of each other. The second term on the RHS canishes by our normalization convention. Thus, we find

We can similarly work out the second order correction to the state, again assuming that it is orthoginal to the zeroth order correction:

Now let us work out the normalization of this state. We define

Note that the overlap of the zeroth and second order correction does not appear, by our normalization assumption. Note that at this order, we can write

so that the following state has unit norm

where both terms are understood to go to $\cO(\eps^2)$.

There is a nice expression for $Z$ in this case. Note that to $\cO(\eps^2)$,

In fact, the formula

holds to all orders in $\eps$.

Validity of perturbation theory¶

Perturbation theory can break down in two ways. First, the corrections can fail to be smaller because the ratio of matrix elements (times the appropriate power of $\eps$) to energy differences in Equations (9), (16), (17) fail to be small. For example, we may have a group of states which are degenerate or nearly degenerate, so that the denominators are small compared to the matrix elements. We can often deal with this by using (almost) degenerate perturbation theory.

At sufficiently high energies, however, perturbation theory can fail for a different reason. Consider the anharmonic oscillator. One has ratios like

where $k \sim 1-4$. But the matrix element in the numerator can be shown to scale as $\sim n^2$. The point is that the unperturbed eigenstates at higher order are more spread out -- you should convince yourself that $\bra{n} x^2 \ket{n} \sim n L^2= \frac{n \hbar}{m\omega}$. At some $n$ they will extend well into the region where $\lambda x^4/4! \gg \half m \omega^2$, and the perturbation is no longer small.

In fact, even for low-energy eigenstates this ends up being a problem. As we can start to see in (16), (17), as we go to higher and higher orders, we start connecting states at higher and higher energies with respect to the ground state. One can show (see for example Bender & Wu, 1971) that if $E_0 = \sum_n A_n \lambda^n$, then for large $n$, $A_n \sim n!$. Since $A_n/A_{n-1} \sim n$, for $n\lambda \sim 1$ the higher orders in perturbation theory will cease to be smaller: the series is an asymptotic series. In fact low orders in perturbation theory ofen give you very accurate answers! But they can only be so accurate.

An essential problem is that there are general terms in the exact answer which scale as $e^{-1/\hbar\lambda}$. This is not an analytic function near $\lambda = 0$, even though for positive real $\lambda$, it is very small. So the radius of converrgence of pertrubation theory is zero! A sign of this is that if we make $\lambda$ negative, even small and negative, the theory becomes unstable and ill-defined (it is not just that the potential is unbounded below, but that it falls off more rapidly than $-x^2$). The radius of convergence of a power series in $\lambda$ can be understood by studyying the behavior of the actual funcion in the complex plane. Singular behavior along the negative axis, up to the origin, means that a Taylor series in small $\lambda$ will not converge.

That being said, we will continue to study perturbation theory because while it may be accurate only up to a point, that point is still pretty accurate! Quantum Electrodynamics can be treated in perturbation theory with respect to the fine structure constant $\alpha = \frac{e^2}{\hbar c} \sim \frac{1}{137}$. This series is asymptotic, but nonetheless perturbation theory has given answers exquisitely confirmed by experiment.

Example: the quadratic Stark effect¶

A more nontrivial example is the response of the hydrogen atom to an electric field, for which $H_0$ is the usual Coulomb Hamiltonian, and $H_1 = - e E z$, corresponding to a charged particle in a constant electric field ${\vec E} = E {\hat z}$.

Problem for the student: Can you give an estimate of the dimensionless parameter that controls the size of the corrections?

We will focus on corrections to the ground state energy -- the perturbing Hamiltonian doesn’t act on spin, so we can ignore that degree of freedom, and otherwise the ground state is nondegenerate.

The first thing to note is that $\bra{n = 1, \ell = 0, m = 0} H_1 \ket{100} = 0$. This is almost immediately apparent because $H_1$ is odd under $z \to - z$ whereas $\ket{100}$ is even. Thus $z\ket{100}$ most be odd; since parity is implemented by a Hermitian operator, with eigenvalues $\pm 1$ for even and odd states, the overlap of $z\ket{100}$ with $\bra{100}$ must vanish.

Another way to see this is to recall that for tensor operators, ${\cal O}_{\ell,m}\ket{n \ell' m'}$ will transform in a representation in $D_{\ell}\otimes D_{\ell'}$. Now $z$ is a tensor operator with $\ell = 1, m = 0$. This $z \ket{100}$ must transform as $\ell = 1$, and so be orthogonal to the $\ell = 0$ state $\ket{100}$.

The result is that the leading behavior of the energy for small $E$ is quadratic in $E$ (because we have to go to second order in perturbation theory). As it happens, for excited states, when we have to use degenerate perturbation theory, there can be linear shifts, an effect called the linear Stark effect which we will describe later.

Equation (16) gives

The last step is nontrivial, I am just asserting it. One simplification is that we know that only $m = 0, \ell = 1$ terms can enter in. Relatedly, this means for the corrections to the wavefunction (9), we get $\ell = 1,m=0$ component which means that the wavefunction elongates in the electric field, due to the response of the charges in the atom.

Note that the force of an electric field on an electric dipole is:

This can describe the present situation if we set

The electric field, by elongating the electron wavefunction, induces an electric dipole moment in the atom. We can define the polarizability as the coefficient $\alpha$ in

Here $\alpha = \frac{9}{2} a_0^3$.

(Almost) degenerate perturbation theory¶

We now want to turn to the case that there are groups of states for which the energy splitting is small compared to the matrix elements of $\eps H_1$. More precisely, consider the diagram of energy levels below with the condition

The unperturbed Hamiltonian will have the block diagonal structure

while the perturbed Hamiltonian will have the structure

Here the $n,m$ indices in $H_{k,mn}$ refer to the subspaces $\cH_m, \cH_m$ spanned by the nearly-degenerate unperturbed eigenstates as labelled in the figure. The problems arise when $\eps H_{1,nn}$ have nonvanishing matrix elements between the unperturbed eigenstates, which are larger than $\delta \eps$.

Degenerate perturbation theory¶

Let us first consider the case that $\delta \eps = 0$, so that $\cH_{0,nn} = {E_N^{(0)} \bf 1}$ in this subspace . Since $\cH_m$ is often finite dimensional and $H_{1,nn}$ comparatively simple, a solution is to first choose bases in $\cH_n$ that diagonalize $H_{1,nn}$. $H_0$ remains diagonal, since the identity remains the identity in any orthonormal basis. The eigenvalues of $\eps H_1$ then give the first-order corrections to the energy, as expressed in this basis. If this resolves the degeneracy, one can then compute further corrections by using standard perturbation theory: for this, only off-diagonal matrix elements appear, and $H_1$ will only have nonvanishing off-diagonal matrix elements between different subspaces $\cH_n$.