Chapter 11: Time-independent perturbation theory

Note

A farmer has a problem with foxes eating his hens. So he asks his physicist friend to help find a solution. The physicist spends a day thinking, then replies “Well, I’ve found a solution, but it will only work for spherical chickens in a vacuum”

Summary

Attention

So far, we have focused on solving quantum mechanical problems exactly. We realized that doing so is not an easy task, even for the simplest problems and we had to rely on special mathematical functions more often than not. Yet, most problems in physics are even more complicated than the select few we looked at so far. When we say more complicated, what we really mean is that the inclusion of realistic terms in the Hamiltonian usually leads to equations that cannot be solved analytically. However, often the problems can be approximated by a simpler Hamiltonian for which a solution is known. The idea here is to treat the part of the Hamiltonian that makes it unsolvable as a (time-independent) perturbation to the known problem. This is what this chapter is about (time-independent perturbation is treated in a separate chapter). Formally, this can be summarized as follows:

We start by writing the actual Hamiltonian as a sum of Hamiltonian corresponding to a known problem ( $\hat{H}_0$ ) and a perturbative Hamiltonian ( $\hat{H}_1$ ). The assumption is that this perturbation is small compared to the energy scale corresponding to the unperturbed system (this translates into the fact that the matrix elements of the perturbation are small compared to those of $\hat{H}_0$ ):

$\hat{H}=\hat{H}_{0}+\hat{H}_{1}$

Likewise, we express the solution of the perturbed Hamiltonian as a series in power of $\lambda$ :

$\left|\psi_{n}\right\rangle=\left|\varphi_{n}^{(0)}\right\rangle+\lambda\left|\varphi_{n}^{(1)}\right\rangle+\lambda^{2}\left|\varphi_{n}^{(2)}\right\rangle+\cdots$

We then put everything together so that the full eigenvalue problem becomes:

$\left(\hat{H}_{0}+\lambda \hat{H}_{1}\right)\left(\left|\varphi_{n}^{(0)}\right\rangle+\lambda\left|\varphi_{n}^{(1)}\right\rangle+\lambda^{2}\left|\varphi_{n}^{(2)}\right\rangle+\cdots\right)=\left(E_{n}^{(0)}+\lambda E_{n}^{(1)}+\lambda^{2} E_{n}^{(2)}+\cdots\right)\left(\left|\varphi_{n}^{(0)}\right\rangle+\lambda\left|\varphi_{n}^{(1)}\right\rangle+\lambda^{2}\left|\varphi_{n}^{(2)}\right\rangle+\cdots\right)$

The energy of each state (indexed by integer n) is:

$E_{n}=E_{n}^{(0)}+\lambda E_{n}^{(1)}+\lambda^{2} E_{n}^{(2)}+\cdots$

We spend quite a bit of time working out the different orders of the solution and came up with solutions at various orders, as expressed in the Key Learning Points box below.

In this chapter, equipped with all the equations necessary for the application of perturbation theory, we first focused on the Stark effect (the effect of an electric field added as a perturbation to a known problem such as the hydrogen atom). This is a very interesting example because it allows to study the effect of the perturbation of a non-degenerate state (ground state) and a degenerate state (first excited state). This is explained in the blue box below.

The second half of the Chapter is devoted to relativistic corrections to the spectrum of the hydrogen atoms. We looked at three different corrections:

Correction due to the fact the electron can take on velocities that are close to the speed of light, thus leading to the relativistic expression for the kinetic energy.

Correction due to the interaction of the magnetic moment (due to the presence of the intrinsic spin) with the magnetic field induced by the rotation (relative o the electron) of a charged particle (i.e., the proton) in the reference frame of the electron. This is called the spin-orbit coupling.

The Darwin term: this term is needed to correct the spin-orbit coupling correction as it is expected that there is no correction for the $l=0$ state.

In all those corrections (which describe the fine structure o the hydrogen atom), we are reminded that a proper description of those effects should be performed using Dirac equation (which is the relativistic version of Schrodinger equation). We also mention that an additional correction is due to the quantization to the fields themselves (this is called the Lamb shift), in addition to the hyperfine correction, due to the interaction between the spin of the proton and that of the electron (we already studied it in Chapter 5: Combining two spin-1/2 particles).

Note

Electric field effect on hydrogen atom: Stark Effect

One application of the theory of time-independent perturbation theory is the effect of a static electric field on the states of the hydrogen atom. This is a good example of a problem for which we know exactly the solution of the unperturbed Hamiltonian (i.e., in the absence of the elective field). The effect of the electric is usually calculated on the ground state (non-degenerate) and on the first excited state (fourfold degenerate).

For the ground state, one can directly apply the formulae of the non-degenerate perturbation theory. We see that because the ground state is spherically symmetric (i.e., it is an s state), it does not have a dipole moment (the way to see this is that there is no preferred direction in the problem). As a result, the first order correction is zero. In contrast, the second order correction is non-zero as, in that case, one considers the effect of the perturbation on the state at first order first and the effect is to polarize the charge distribution.

For the first excited state, one examines if the electric field can lift the degeneracy. For this, we build a representation of the perturbation in the sub-space of degenerate states. We find that the 4x4 matrix has four eigenvalues: one positive, one negative, and two zeros. This shows that the electric field partially lifts the degeneracy of the first excited state.

Note that the same approach can be used to study the effect of a static magnetic field: this is called the Zeeman effect.

Learning Material

Copy of Slides

The slides for Chapter 11 are available in pdf format here: 📂.

Screencast

Key Learning Points

Non-degenerate perturbation theory:

Zeroth-order: $\hat{H}_{0}\left|\varphi_{n}^{(0)}\right\rangle=E_{n}^{(0)}\left|\varphi_{n}^{(0)}\right\rangle$
First-order:

Energy: $E_{n}^{(1)}=\left\langle\varphi_{n}^{(0)}\left|\hat{H}_{1}\right| \varphi_{n}^{(0)}\right\rangle$ (expectation value of the perturbation for the unperturbed state)

State: $\left|\psi_{n}\right\rangle=\left|\varphi_{n}^{(0)}\right\rangle+\lambda \sum_{k \neq n}\left|\varphi_{k}^{(0)}\right\rangle \frac{\left\langle\varphi_{k}^{(0)}\left|\hat{H}_{1}\right| \varphi_{n}^{(0)}\right\rangle}{E_{n}^{(0)}-E_{k}^{(0)}}+O\left(\lambda^{2}\right)$
Second-order:

Energy: $E_{n}^{(2)}=\sum_{k \neq n} \frac{\left|\left\langle\varphi_{k}^{(0)}\left|\hat{H}_{1}\right| \varphi_{n}^{(0)}\right\rangle\right|^{2}}{E_{n}^{(0)}-E_{k}^{(0)}}$

Degenerate perturbation theory:

We create first-order states as linear combinations of states that were degenerate before the perturbation. The coefficients of the linear combinations are obtained with:

$\left(\begin{array}{ll} \left(H_{1}\right)_{11} & \left(H_{1}\right)_{12} \\ \left(H_{1}\right)_{21} & \left(H_{1}\right)_{22} \end{array}\right)\left(\begin{array}{l} c_{1} \\ c_{2} \end{array}\right)=E_{n}^{(1)}\left(\begin{array}{l} c_{1} \\ c_{2} \end{array}\right)$

The correction to the energy at first order is the eigenvalue of that equation.

Test your knowledge

In the Stark effect, we apply a constant electric field (of low intensity) to a hydrogen atom. Consider the first-order correction to the ground state of the hydrogen atom. Here we focus on the energy.
1. The correction to the energy at first order is proportional to the intensity of the magnitude of the electric field.
2. The correction to the energy at first order is proportional to the square of the magnitude of the electric field.
3. The correction to the energy at first order is zero.
In the Stark effect, we apply a constant electric field (of low intensity) to a hydrogen atom. Consider the first-order correction to the ground state of the hydrogen atom. Here we focus on the eigenstate.
1. The correction to the 1s state of hydrogen is zero.
2. The correction to the 1s state of hydrogen consists in creating a dipole oriented along the electric field.
3. The correction to the 1s state of hydrogen consists in promoting the ground state into the 2s state of the hydrogen atom.
In the Stark effect, we apply a constant electric field (of low intensity) to a hydrogen atom. Consider the second-order correction to the ground state of the hydrogen atom.
1. The correction to the energy is proportional to the intensity of the electric field.
2. The correction to the energy is proportional to the square of the magnitude of the electric field.
3. The correction to the energy at first order is zero.
We have a Hamiltonian that has a doubly-degenerate state. Now, we turn on a small perturbation, whose representation in the subspace of the degenerate states is given by

$\left(\begin{array}{ll} a & 0 \\ 0 & 0 \end{array}\right)$

where $a$ is a real number.

I know that the perturbation lifts the degeneracy of the state, so long as $a\ne0$ .

I do not have enough information to tell if this perturbation can lift the degeneracy.

Because the perturbation is diagonal in the subspace of the degenerate state, it cannot lift the degeneracy.

We have a Hamiltonian that has a triply-degenerate state. Now, we turn on a small perturbation, whose representation in the subspace of the degenerate states is given by

$\left(\begin{array}{lll} 1 & 0 &0\\ 0 & 0 &0\\ 0 & 0 &2\\ \end{array}\right).$

What can I say about the correction to the states at lowest (non-zero) order?

The states are still degenerate.

The states are no longer degenerate and the corresponding eigenvectors have also been changed (linear combinations of the unperturbed states)

The states are no longer degenerate but the eigen-vectors have not changed

Learning online has been a dream come true to me.
1. Can you repeat the question?
2. It is actually better than my wildest dream.
3. With dreams like these, I do not need nightmares.

Hint

Find the answer keys on this page: Answers to selected test your knowledge questions. Don’t cheat! Try solving the problems on your own first!

Homework Assignment

Solve the following problem in the pdf: pdf

Recitation Assignment

Solve the following problems from the textbook: 11.3; 11.6