# Creation and annihilation operators in the harmonic oscillator

References: Tom Lancaster and Stephen J. Blundell, Quantum Field Theory for the Gifted Amateur, (Oxford University Press, 2014) – Problems 2.1,2.2.

We’ve already seen a hint of quantum field theory when we studied the harmonic oscillator in non-relativistic quantum theory. We used raising and lowering operators to move between energy levels in an oscillator. We’ll quote the earlier results, but change the notation so that it matches that used in Lancaster & Blundell, where the raising operator is ${\hat{a}^{\dagger}}$ and the lowering operator is ${\hat{a}}$.

 $\displaystyle \hat{a}^{\dagger}$ $\displaystyle =$ $\displaystyle \frac{1}{\sqrt{2\hbar m\omega}}\left[-i\hat{p}+m\omega\hat{x}\right]\ \ \ \ \ (1)$ $\displaystyle \hat{a}$ $\displaystyle =$ $\displaystyle \frac{1}{\sqrt{2\hbar m\omega}}\left[i\hat{p}+m\omega\hat{x}\right] \ \ \ \ \ (2)$

Solving for ${\hat{x}}$ and ${\hat{p}}$ we get

 $\displaystyle \hat{x}$ $\displaystyle =$ $\displaystyle \sqrt{\frac{\hbar}{2m\omega}}\left(\hat{a}^{\dagger}+\hat{a}\right)\ \ \ \ \ (3)$ $\displaystyle \hat{p}$ $\displaystyle =$ $\displaystyle i\sqrt{\frac{\hbar m\omega}{2}}\left(\hat{a}^{\dagger}-\hat{a}\right) \ \ \ \ \ (4)$

We also showed that the oscillator hamiltonian can be written in terms of these operators as

$\displaystyle \hat{H}=\hbar\omega\left(\hat{a}^{\dagger}\hat{a}+\frac{1}{2}\right) \ \ \ \ \ (5)$

The operators have the commutator

$\displaystyle \left[\hat{a},\hat{a}^{\dagger}\right]=1 \ \ \ \ \ (6)$

Both operators also commute with themselves, since all operators commute with themselves.

We also showed that normalization of these operators gives the formulas:

 $\displaystyle \hat{a}^{\dagger}\left|n\right\rangle$ $\displaystyle =$ $\displaystyle \sqrt{n+1}\left|n+1\right\rangle \ \ \ \ \ (7)$ $\displaystyle \hat{a}\left|n\right\rangle$ $\displaystyle =$ $\displaystyle \sqrt{n}\left|n-1\right\rangle \ \ \ \ \ (8)$

When viewed from quantum field theory, the raising operator is seen as an operator that creates a particle of energy ${\hbar\omega}$ and is therefore called a creation operator. Similarly, the lowering operator annihilates a particle of energy ${\hbar\omega}$ and is called an annihilation operator. That is, the quantum of energy equal to the difference in adjacent levels in a harmonic oscillator is viewed as a particle.

We can use creation and annihilation operators together with perturbation theory to calculate energy levels of perturbed oscillators. For example, suppose we have a hamiltonian:

$\displaystyle \hat{H}=\frac{\hat{p}^{2}}{2m}+\frac{1}{2}m\omega^{2}\hat{x}^{2}+\lambda\hat{x}^{4} \ \ \ \ \ (9)$

where ${\lambda}$ is small. The first two terms are the unperturbed harmonic oscillator, so we can take the perturbation to be

$\displaystyle H'=\lambda\hat{x}^{4} \ \ \ \ \ (10)$

Then the perturbation in energy in the state${\left|n\right\rangle }$ of the original unperturbed hamiltonian is given by the matrix element of the perturbation:

$\displaystyle E'_{n}=\left\langle n\left|H'\right|n\right\rangle \ \ \ \ \ (11)$

Expressing ${H'}$ in terms of ${\hat{a}^{\dagger}}$ and ${\hat{a}}$ via 3 we get

$\displaystyle H'=\lambda\left(\frac{\hbar}{2m\omega}\right)^{2}\left(\hat{a}^{\dagger}+\hat{a}\right)^{4} \ \ \ \ \ (12)$

Because we’re interested only in terms in this operator that leave the number of particles unchanged (since we’re calculating ${\left\langle n\left|H'\right|n\right\rangle }$ and the states of the original hamiltonian are orthogonal) we can keep only those terms in the expansion of ${\left(\hat{a}^{\dagger}+\hat{a}\right)^{4}}$ that contain two of each operator. That is, we have

$\displaystyle \left\langle n\left|H'\right|n\right\rangle =\left\langle n\left|\hat{a}^{2}\left(\hat{a}^{\dagger}\right)^{2}+\hat{a}\left(\hat{a}^{\dagger}\right)^{2}\hat{a}+\left(\hat{a}^{\dagger}\right)^{2}\hat{a}^{2}+\hat{a}^{\dagger}\hat{a}^{2}\hat{a}^{\dagger}+\hat{a}\hat{a}^{\dagger}\hat{a}\hat{a}^{\dagger}+\hat{a}^{\dagger}\hat{a}\hat{a}^{\dagger}\hat{a}\right|n\right\rangle \ \ \ \ \ (13)$

Applying the equations 7 and 8 we get

 $\displaystyle \left\langle n\left|H'\right|n\right\rangle$ $\displaystyle =$ $\displaystyle \left(n+1\right)\left(n+2\right)+n\left(n+1\right)+n\left(n-1\right)+\left(n+1\right)n+\left(n+1\right)^{2}+n^{2}\ \ \ \ \ (14)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 6n^{2}+6n+3 \ \ \ \ \ (15)$

so the energy levels, correct to first order in ${\lambda}$, are

 $\displaystyle E_{n}$ $\displaystyle =$ $\displaystyle \hbar\omega\left(n+\frac{1}{2}\right)+\lambda\left(\frac{\hbar}{2m\omega}\right)^{2}\left(6n^{2}+6n+3\right)\ \ \ \ \ (16)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \hbar\omega\left(n+\frac{1}{2}\right)+\frac{3\lambda}{4}\left(\frac{\hbar}{m\omega}\right)^{2}\left(2n^{2}+2n+1\right) \ \ \ \ \ (17)$

# Perturbation theory and the variational principle

References: Griffiths, David J. (2005), Introduction to Quantum Mechanics, 2nd Edition; Pearson Education – Chapter 7, Post 5.

A simple application of the variational
principle

is that the first order correction to the energy given by perturbation theory provides an upper bound on the effect of the perturbation hamiltonian ${H'}$. The first order perturbation is given by

$\displaystyle E_{01}=\left\langle 00\left|H'\right|00\right\rangle \ \ \ \ \ (1)$

where ${\left|00\right\rangle }$ is the unperturbed ground state wave function. According to the variational principle, ${E_{01}\ge E_{0}^{\prime}}$ where ${E_{0}^{\prime}}$ is the ground state of ${H^{\prime}}$. Therefore ${E_{01}}$ can never underestimate the correction due to ${H^{\prime}}$ in the ground state.

Second-order perturbation theory says that the second-order correction to the ground state ${E_{02}}$ is

$\displaystyle E_{02}=\sum_{j\ne0}\frac{\left|\left\langle j0\right|H^{\prime}\left|00\right\rangle \right|^{2}}{E_{00}-E_{j0}} \ \ \ \ \ (2)$

Since the first order correction always overestimates the energy (unless it hits it bang on, in which case the second order correction will be zero), we would expect the second order correction to be negative in order to peg back the overestimate from the first order result. From the formula, this is always true, since ${E_{00} for all ${j\ne0}$ (assuming non-degenerate states).

# Perturbing the wave function (Stark effect and proton electric dipole moment)

References: Griffiths, David J. (2005), Introduction to Quantum Mechanics, 2nd Edition; Pearson Education – Problem 6.40.

When we derived the equations for first order perturbation theory, we got an expression for the correction ${\psi_{n1}}$ to the wave function:

$\displaystyle H_{0}\psi_{n1}+H^{\prime}\psi_{n0}=E_{n0}\psi_{n1}+E_{n1}\psi_{n0} \ \ \ \ \ (1)$

In the earlier derivation, we expanded ${\psi_{n1}}$ as a series in terms of the unperturbed wave functions. However, in some cases it’s possible to solve this equation directly. Consider again the Stark effect in hydrogen. We know that the first order correction is ${E_{1,1}=0}$, and (expressing things in terms of the Bohr radius ${a=4\pi\epsilon_{0}\hbar^{2}/me^{2}}$):

 $\displaystyle H_{0}$ $\displaystyle =$ $\displaystyle -\frac{\hbar^{2}}{2m}\nabla^{2}-\frac{e^{2}}{4\pi\epsilon_{0}r}=-\frac{\hbar^{2}}{2m}\nabla^{2}-\frac{\hbar^{2}}{mar}\ \ \ \ \ (2)$ $\displaystyle H^{\prime}$ $\displaystyle =$ $\displaystyle eE_{ext}r\cos\theta\ \ \ \ \ (3)$ $\displaystyle E_{1,0}$ $\displaystyle =$ $\displaystyle -\frac{\hbar^{2}}{2ma^{2}}\ \ \ \ \ (4)$ $\displaystyle \psi_{1,0}$ $\displaystyle =$ $\displaystyle \frac{1}{\sqrt{\pi}a^{3/2}}e^{-r/a} \ \ \ \ \ (5)$

The equation to solve is then

$\displaystyle -\frac{\hbar^{2}}{2m}\nabla^{2}\psi_{1,1}-\frac{\hbar^{2}}{mar}\psi_{1,1}+\frac{\hbar^{2}}{2ma^{2}}\psi_{1,1}=-eE_{ext}r\cos\theta\frac{1}{\sqrt{\pi}a^{3/2}}e^{-r/a} \ \ \ \ \ (6)$

We try a solution of form

$\displaystyle \psi_{1,1}=\left(A+Br+Cr^{2}\right)e^{-r/a}\cos\theta \ \ \ \ \ (7)$

After calculating ${\nabla^{2}\psi_{1,1}}$ in spherical coordinates and cancelling off a common factor of ${e^{-r/a}\cos\theta}$ from both sides and multiplying through by ${r^{2}}$, we get

$\displaystyle aA+r^{2}B+\left(2r^{3}-2ar^{2}\right)C=-\frac{eE_{ext}m}{\sqrt{\pi a}\hbar^{2}}r^{3} \ \ \ \ \ (8)$

Equating powers of ${r}$ on each side, we get

 $\displaystyle A$ $\displaystyle =$ $\displaystyle 0\ \ \ \ \ (9)$ $\displaystyle B$ $\displaystyle =$ $\displaystyle 2aC\ \ \ \ \ (10)$ $\displaystyle C$ $\displaystyle =$ $\displaystyle -\frac{eE_{ext}m}{2\sqrt{\pi a}\hbar^{2}}\ \ \ \ \ (11)$ $\displaystyle B$ $\displaystyle =$ $\displaystyle -\frac{eE_{ext}m\sqrt{a}}{\sqrt{\pi}\hbar^{2}} \ \ \ \ \ (12)$

so the correction to the wave function is

$\displaystyle \psi_{1,1}=-\frac{eE_{ext}m}{2\sqrt{\pi a}\hbar^{2}}\left(2ar+r^{2}\right)e^{-r/a}\cos\theta \ \ \ \ \ (13)$

The second order correction to the energy can be found from this equation that we got in the process of deriving second order perturbations:

$\displaystyle \left\langle \psi_{1,0}\right|H^{\prime}\left|\psi_{1,1}\right\rangle =E_{1,1}\left\langle \psi_{1,0}\right.\left|\psi_{1,1}\right\rangle +E_{1,2} \ \ \ \ \ (14)$

Since ${E_{1,1}=0}$ we have

$\displaystyle E_{1,2}=\left\langle \psi_{1,0}\right|H^{\prime}\left|\psi_{1,1}\right\rangle \ \ \ \ \ (15)$

Plugging in the functions, we get

$\displaystyle E_{1,2}=-\frac{e^{2}E_{ext}^{2}m}{2\sqrt{\pi a}\hbar^{2}}\frac{1}{\sqrt{\pi}a^{3/2}}\int_{0}^{\infty}\int_{0}^{\pi}\int_{0}^{2\pi}\left(2ar+r^{2}\right)e^{-2r/a}\cos^{2}\theta r^{3}\sin\theta d\phi d\theta dr \ \ \ \ \ (16)$

This integral can be looked up or done with software, and the answer is

$\displaystyle E_{1,2}=-\frac{9a^{4}e^{2}E_{ext}^{2}m}{4\hbar^{2}} \ \ \ \ \ (17)$

As another example of a case where the equation can be solved exactly, suppose the proton had an electric dipole moment ${\mathbf{p}}$. The potential due to a dipole is

$\displaystyle V(\mathbf{r})=\frac{1}{4\pi\epsilon_{0}r^{2}}\mathbf{p}\cdot\hat{\mathbf{r}} \ \ \ \ \ (18)$

If ${\mathbf{p}}$ points in the ${z}$ direction, then

$\displaystyle V(\mathbf{r})=\frac{p\cos\theta}{4\pi\epsilon_{0}r^{2}} \ \ \ \ \ (19)$

and the perturbation on the electron’s hamiltonian is

$\displaystyle H^{\prime}=-eV=-\frac{ep\cos\theta}{4\pi\epsilon_{0}r^{2}} \ \ \ \ \ (20)$

The equation to solve has the same LHS as 6, but with a RHS using the new ${H^{\prime}}$:

$\displaystyle -\frac{\hbar^{2}}{2m}\nabla^{2}\psi_{1,1}-\frac{\hbar^{2}}{mar}\psi_{1,1}+\frac{\hbar^{2}}{2ma^{2}}\psi_{1,1}=-\frac{ep\cos\theta}{4\pi\epsilon_{0}r^{2}}\frac{1}{\sqrt{\pi}a^{3/2}}e^{-r/a} \ \ \ \ \ (21)$

Trying the same form for a solution as in 7 we get by the same calculation as above:

$\displaystyle aA+r^{2}B+\left(2r^{3}-2ar^{2}\right)C=\frac{mep}{4\pi\epsilon_{0}\hbar^{2}\sqrt{\pi a}} \ \ \ \ \ (22)$

This time, the RHS is a constant, so

 $\displaystyle A$ $\displaystyle =$ $\displaystyle \frac{mep}{4\pi\epsilon_{0}\hbar^{2}\sqrt{\pi a^{3}}}\ \ \ \ \ (23)$ $\displaystyle B$ $\displaystyle =$ $\displaystyle C=0 \ \ \ \ \ (24)$

and the correction to the wave function is

$\displaystyle \psi_{1,1}=\frac{mep}{4\pi\epsilon_{0}\hbar^{2}\sqrt{\pi a^{3}}}e^{-r/a}\cos\theta \ \ \ \ \ (25)$

The average dipole moment of the hydrogen atom is (treating the proton and electron as point charges):

 $\displaystyle \mathbf{p}_{H}$ $\displaystyle =$ $\displaystyle -e\mathbf{r}\ \ \ \ \ (26)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -er\left(\sin\theta\cos\phi\hat{\mathbf{x}}+\sin\theta\sin\phi\hat{\mathbf{y}}+\cos\theta\hat{\mathbf{z}}\right) \ \ \ \ \ (27)$

The average ${\left\langle \mathbf{p}_{H}\right\rangle }$ can be calculated using the unperturbed wave function plus the first order correction:

$\displaystyle \left\langle \mathbf{p}_{H}\right\rangle =\left\langle \psi_{1,0}+\psi_{1,1}\right|\mathbf{p}_{H}\left|\psi_{1,0}+\psi_{1,1}\right\rangle \ \ \ \ \ (28)$

The ${x}$ and ${y}$ components average out to zero due to the integrals over ${\phi}$, so we’re left with (using Maple for the integrals):

 $\displaystyle \left\langle \mathbf{p}_{H}\right\rangle$ $\displaystyle =$ $\displaystyle -e\hat{\mathbf{z}}\left\langle \psi_{1,0}+\psi_{1,1}\right|r\cos\theta\left|\psi_{1,0}+\psi_{1,1}\right\rangle \ \ \ \ \ (29)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\frac{me^{2}ap}{4\pi\epsilon_{0}\hbar^{2}}\hat{\mathbf{z}}\ \ \ \ \ (30)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\mathbf{p} \ \ \ \ \ (31)$

where we’ve used ${a=4\pi\epsilon_{0}\hbar^{2}/me^{2}}$ to get the last line. Thus the total electric dipole moment of the atom is zero.

The first-order correction to the energy is ${\left\langle \psi_{1,0}\right|H^{\prime}\left|\psi_{1,0}\right\rangle =0}$ (due to the integral over ${\theta}$ being zero), so as above, we can use 15 to calculate the second-order energy correction as before:

 $\displaystyle E_{1,2}$ $\displaystyle =$ $\displaystyle \left\langle \psi_{1,0}\right|H^{\prime}\left|\psi_{1,1}\right\rangle \ \ \ \ \ (32)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\frac{me^{2}p^{2}}{24\left(\pi\epsilon_{0}\hbar a\right)^{2}}\ \ \ \ \ (33)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\frac{2p^{2}\hbar^{2}}{3me^{2}a^{4}}\ \ \ \ \ (34)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{4}{3}\left(\frac{p}{ea}\right)^{2}E_{1,0} \ \ \ \ \ (35)$

# Van der Waals interaction

References: Griffiths, David J. (2005), Introduction to Quantum Mechanics, 2nd Edition; Pearson Education – Problem 6.31.

A simple one-dimensional model of the interaction between two electrically neutral hydrogen atoms can be constructed using the harmonic oscillator. Within each atom, we model the interaction between the proton and electron (in 1-d remember) as a spring with spring constant ${k}$:

$\displaystyle H_{i}=\frac{p_{i}^{2}}{2m}+\frac{1}{2}kx_{i}^{2}\mbox{ ; }i=1,2 \ \ \ \ \ (1)$

where ${x_{i}}$ is the separation between the proton and electron in atom ${i}$.

The unperturbed energy (ignoring the interaction between the two atoms) of the system is then

 $\displaystyle H_{0}$ $\displaystyle =$ $\displaystyle H_{1}+H_{2}\ \ \ \ \ (2)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{p_{1}^{2}}{2m}+\frac{1}{2}kx_{1}^{2}+\frac{p_{2}^{2}}{2m}+\frac{1}{2}kx_{2}^{2} \ \ \ \ \ (3)$

Now if we add in the Coulomb interaction between the four particles (taking ${R}$ to be the distance between the two protons, with atom 1 to the left of atom 2, and the proton to the left of the electron in each atom) we get the perturbation:

 $\displaystyle H^{\prime}$ $\displaystyle =$ $\displaystyle \frac{e^{2}}{4\pi\epsilon_{0}}\left[\frac{1}{R}-\frac{1}{R-x_{1}}-\frac{1}{R+x_{2}}+\frac{1}{R-x_{1}+x_{2}}\right]\ \ \ \ \ (4)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{e^{2}}{4\pi\epsilon_{0}R}\left[1-\frac{1}{1-x_{1}/R}-\frac{1}{1+x_{2}/R}-\frac{1}{1-\left(x_{1}-x_{2}\right)/R}\right] \ \ \ \ \ (5)$

where the first term is the interaction between the protons, the second is between proton 2 and electron 1, the third is between proton 1 and electron 2 and the last is between the two electrons. Note that the signs on the ${x}$s are opposite to that in Griffiths’s equation 6.97 which is needed to match his figure 6.14; however, this doesn’t affect the rest of the problem since only products of the ${x}$s occur.

If we assume that the spacing ${R}$ between the atoms is much greater than the proton-electron spacing within each atom, we can expand ${H^{\prime}}$ in a Taylor series up to second order (the first order terms cancel) and get (e.g. using Maple’s ‘mtaylor’ command, or by hand using ${1/\left(1-x\right)=1+x+x^{2}+\ldots}$) for the leading term:

$\displaystyle H^{\prime}=-\frac{e^{2}x_{1}x_{2}}{2\pi\epsilon_{0}R^{3}}+\frac{1}{R}\mathcal{O}\left(\frac{x_{i}^{3}}{R^{3}}\right) \ \ \ \ \ (6)$

Thus perturbed hamiltonian is approximately

$\displaystyle H_{1}=\frac{p_{1}^{2}}{2m}+\frac{1}{2}kx_{1}^{2}+\frac{p_{2}^{2}}{2m}+\frac{1}{2}kx_{2}^{2}-\frac{e^{2}x_{1}x_{2}}{2\pi\epsilon_{0}R^{3}} \ \ \ \ \ (7)$

We now make the change of variables:

$\displaystyle x_{\pm}=\frac{1}{\sqrt{2}}\left(x_{1}\pm x_{2}\right) \ \ \ \ \ (8)$

This has the inverse relation:

 $\displaystyle x_{1}$ $\displaystyle =$ $\displaystyle \frac{1}{\sqrt{2}}\left(x_{+}+x_{-}\right)\ \ \ \ \ (9)$ $\displaystyle x_{2}$ $\displaystyle =$ $\displaystyle \frac{1}{\sqrt{2}}\left(x_{+}-x_{-}\right) \ \ \ \ \ (10)$

This results in a change in the momentum variables:

 $\displaystyle p_{\pm}$ $\displaystyle =$ $\displaystyle -i\hbar\frac{d}{dx_{\pm}}\ \ \ \ \ (11)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -i\hbar\left(\frac{\partial}{\partial x_{1}}\frac{\partial x_{1}}{\partial x_{\pm}}+\frac{\partial}{\partial x_{2}}\frac{\partial x_{2}}{\partial x_{\pm}}\right)\ \ \ \ \ (12)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\frac{i\hbar}{\sqrt{2}}\left(\frac{\partial}{\partial x_{1}}\pm\frac{\partial}{\partial x_{2}}\right)\ \ \ \ \ (13)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{\sqrt{2}}\left(p_{1}\pm p_{2}\right) \ \ \ \ \ (14)$

From here, it is straightforward (although tedious) algebra to substitute for ${x_{1,2}}$ and ${p_{1,2}}$ (e.g. using Maple’s ‘subs’ command, followed by ‘simplify’ and ‘expand’) to get

$\displaystyle H_{1}=\frac{p_{+}^{2}}{2m}+\frac{1}{2}\left(k-\frac{e^{2}}{2\pi\epsilon_{0}R^{3}}\right)x_{+}^{2}+\frac{p_{-}^{2}}{2m}+\frac{1}{2}\left(k+\frac{e^{2}}{2\pi\epsilon_{0}R^{3}}\right)x_{-}^{2} \ \ \ \ \ (15)$

The perturbed hamiltonian is thus the sum of two harmonic oscillators with spring constants

$\displaystyle k_{\pm}=k\mp\frac{e^{2}}{2\pi\epsilon_{0}R^{3}} \ \ \ \ \ (16)$

Note that Griffiths’s equation 6.99 is wrong since the denominator in the second term should be ${2\pi\epsilon_{0}R^{3}}$ as shown here, not ${4\pi\epsilon_{0}R^{3}}$. This affects the rest of the question as well. Since the hamiltonian decouples into two independent oscillators, the ground state energy is

 $\displaystyle E$ $\displaystyle =$ $\displaystyle \frac{\hbar}{2}\left(\omega_{+}+\omega_{-}\right)\ \ \ \ \ (17)$ $\displaystyle \omega_{\pm}$ $\displaystyle =$ $\displaystyle \sqrt{\frac{k}{m}\mp\frac{e^{2}}{2\pi\epsilon_{0}mR^{3}}} \ \ \ \ \ (18)$

We define the parameter

$\displaystyle b\equiv\frac{e^{2}}{2\pi\epsilon_{0}R^{3}} \ \ \ \ \ (19)$

Then the energy is

 $\displaystyle E$ $\displaystyle =$ $\displaystyle \frac{\hbar}{2\sqrt{m}}\left(\sqrt{k-b}+\sqrt{k+b}\right)\ \ \ \ \ (20)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\hbar}{2}\sqrt{\frac{k}{m}}\left(\sqrt{1-b/k}+\sqrt{1+b/k}\right) \ \ \ \ \ (21)$

Expanding this in a Taylor series in ${b/k}$ we find that the first order term is zero so

 $\displaystyle E$ $\displaystyle =$ $\displaystyle \hbar\sqrt{\frac{k}{m}}\left(1-\frac{1}{8}\frac{b^{2}}{k^{2}}+\ldots\right)\ \ \ \ \ (22)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \hbar\omega_{0}-\frac{\hbar}{8m^{2}\omega_{0}^{3}}\left(\frac{e^{2}}{2\pi\epsilon_{0}R^{3}}\right)^{2}+\ldots \ \ \ \ \ (23)$

where ${\omega_{0}=\sqrt{k/m}}$ is the unperturbed frequency of oscillation. This correction to the energy is responsible for the Van der Waals interaction between two neutral atoms.

We can do the same calculation using first and second order perturbation theory. The first order perturbation is

$\displaystyle E_{1}=\left\langle 00\right|H^{\prime}\left|00\right\rangle \ \ \ \ \ (24)$

where the state ${\left|00\right\rangle }$ is the unperturbed ground state for both oscillators. To work this out, we use the result for mean values of position that we derived earlier.

$\displaystyle \langle n|x|n'\rangle=\sqrt{\frac{\hbar}{2m\omega}}\left(\sqrt{n'+1}\delta_{n,n'+1}+\sqrt{n'}\delta_{n,n'-1}\right) \ \ \ \ \ (25)$

We get

 $\displaystyle \left\langle 00\right|H^{\prime}\left|00\right\rangle$ $\displaystyle =$ $\displaystyle -\left\langle 00\right|\frac{e^{2}x_{1}x_{2}}{2\pi\epsilon_{0}R^{3}}\left|00\right\rangle \ \ \ \ \ (26)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\frac{e^{2}}{2\pi\epsilon_{0}R^{3}}\left\langle 00\right|x_{1}x_{2}\left|00\right\rangle \ \ \ \ \ (27)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\frac{e^{2}}{2\pi\epsilon_{0}R^{3}}\left\langle 0\right|x_{1}\left|0\right\rangle \left\langle 0\right|x_{2}\left|0\right\rangle \ \ \ \ \ (28)$

We can split ${\left\langle 00\right|x_{1}x_{2}\left|00\right\rangle }$ into ${\left\langle 0\right|x_{1}\left|0\right\rangle \left\langle 0\right|x_{2}\left|0\right\rangle }$ since ${x_{1}}$ and ${x_{2}}$ are independent coordinates so the average of one of them doesn’t affect the other. From 25, both these averages are zero, so

$\displaystyle E_{1}=\left\langle 00\right|H^{\prime}\left|00\right\rangle =0 \ \ \ \ \ (29)$

and there is no first order correction to the energy, in agreement with the solution above.

For the second order correction we need ${\left\langle n_{1}n_{2}\right|x_{1}x_{2}\left|00\right\rangle =\left\langle n_{1}\right|x_{1}\left|0\right\rangle \left\langle n_{2}\right|x_{2}\left|0\right\rangle }$. Again, from 25, the only non-zero such matrix element is ${\left\langle 1\right|x\left|0\right\rangle =\sqrt{\frac{\hbar}{2m\omega_{0}}}}$, so we get

 $\displaystyle E_{2}$ $\displaystyle =$ $\displaystyle \left(-\frac{e^{2}}{2\pi\epsilon_{0}R^{3}}\right)^{2}\sum_{n_{1},n_{2}\ne0}\frac{\left|\left\langle n_{1}\right|x_{1}\left|0\right\rangle \left\langle n_{2}\right|x_{2}\left|0\right\rangle \right|^{2}}{\hbar\omega_{0}-E_{n_{1}n_{2},0}}\ \ \ \ \ (30)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left(\frac{e^{2}}{2\pi\epsilon_{0}R^{3}}\right)^{2}\left(\frac{\hbar}{2m\omega_{0}}\right)^{2}\frac{1}{\hbar\omega_{0}-E_{11,0}} \ \ \ \ \ (31)$

The unperturbed energy ${E_{11,0}}$ is

$\displaystyle E_{11,0}=\frac{3}{2}\hbar\omega_{0}+\frac{3}{2}\hbar\omega_{0}=3\hbar\omega_{0} \ \ \ \ \ (32)$

so the second order perturbation is

 $\displaystyle E_{2}$ $\displaystyle =$ $\displaystyle \left(\frac{e^{2}}{2\pi\epsilon_{0}R^{3}}\right)^{2}\left(\frac{\hbar}{2m\omega_{0}}\right)^{2}\left(-\frac{1}{2\hbar\omega_{0}}\right)\ \ \ \ \ (33)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\frac{\hbar}{8m^{2}\omega_{0}^{3}}\left(\frac{e^{2}}{2\pi\epsilon_{0}R^{3}}\right)^{2} \ \ \ \ \ (34)$

which agrees with the earlier result.

# Perturbing the 3-d harmonic oscillator

References: Griffiths, David J. (2005), Introduction to Quantum Mechanics, 2nd Edition; Pearson Education – Problem 6.30.

Here’s an example of perturbation theory applied to the 3-d harmonic oscillator. Using separation of variables, the stationary states in 3-d are just products of the 1-d states:

$\displaystyle \psi_{n}(x)=\left(\frac{m\omega}{\pi\hbar}\right)^{1/4}\frac{1}{\sqrt{2^{n}n!}}H_{n}\left(\sqrt{\frac{m\omega}{\hbar}}x\right)e^{-m\omega x^{2}/2\hbar} \ \ \ \ \ (1)$

where ${H_{n}}$ is the ${n}$th Hermite polynomial. The first two Hermite polynomials are

 $\displaystyle H_{0}\left(x\right)$ $\displaystyle =$ $\displaystyle 1\ \ \ \ \ (2)$ $\displaystyle H_{1}\left(x\right)$ $\displaystyle =$ $\displaystyle 2x \ \ \ \ \ (3)$

Thus the ground state in 3-d is (where the notation is ${\psi_{n_{x}n_{y}n_{z}}):}$

$\displaystyle \psi_{000}=\left(\frac{m\omega}{\pi\hbar}\right)^{3/4}e^{-m\omega x^{2}/2\hbar}e^{-m\omega y^{2}/2\hbar}e^{-m\omega z^{2}/2\hbar} \ \ \ \ \ (4)$

The first excited state is triply degenerate, since the excitation can be in any of the three coordinates. We get

 $\displaystyle \psi_{100}$ $\displaystyle =$ $\displaystyle \frac{2}{\pi^{3/4}}\left(\frac{m\omega}{\hbar}\right)^{5/4}xe^{-m\omega x^{2}/2\hbar}e^{-m\omega y^{2}/2\hbar}e^{-m\omega z^{2}/2\hbar}\ \ \ \ \ (5)$ $\displaystyle \psi_{010}$ $\displaystyle =$ $\displaystyle \frac{2}{\pi^{3/4}}\left(\frac{m\omega}{\hbar}\right)^{5/4}ye^{-m\omega x^{2}/2\hbar}e^{-m\omega y^{2}/2\hbar}e^{-m\omega z^{2}/2\hbar}\ \ \ \ \ (6)$ $\displaystyle \psi_{001}$ $\displaystyle =$ $\displaystyle \frac{2}{\pi^{3/4}}\left(\frac{m\omega}{\hbar}\right)^{5/4}ze^{-m\omega x^{2}/2\hbar}e^{-m\omega y^{2}/2\hbar}e^{-m\omega z^{2}/2\hbar} \ \ \ \ \ (7)$

Now suppose we introduce a perturbation given by

$\displaystyle H^{\prime}=\lambda x^{2}yz \ \ \ \ \ (8)$

for some constant ${\lambda}$. We can use non-degenerate perturbation theory to calculate the correction to the energy in the ground state. This gives

 $\displaystyle E_{000,1}$ $\displaystyle =$ $\displaystyle \left\langle \psi_{000}\right|H^{\prime}\left|\psi_{000}\right\rangle \ \ \ \ \ (9)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \lambda\int\int\int\left|\psi_{000}\left(x,y,z\right)\right|^{2}x^{2}yzdxdydz \ \ \ \ \ (10)$

The integrals over ${y}$ and ${z}$ are zero, since ${\psi_{000}}$ is an even function over all three coordinates and ${H^{\prime}}$ is odd in ${y}$ and ${z}$. Thus

$\displaystyle E_{000,1}=0 \ \ \ \ \ (11)$

and there is no correction to the ground state energy to first order.

For the first excited state, we need to use the degenerate theory which means we need to calculate the matrix elements ${W_{ab}=\left\langle n_{x}n_{y}n_{z}\right|H^{\prime}\left|m_{x}m_{y}m_{z}\right\rangle }$ for the 3 degenerate first excited states. These matrix elements are just the means of the coordinates or their squares, that is

$\displaystyle W_{ab}=\lambda\left\langle n_{x}\right|x^{2}\left|m_{x}\right\rangle \left\langle n_{y}\right|y\left|m_{y}\right\rangle \left\langle n_{z}\right|z\left|m_{z}\right\rangle \ \ \ \ \ (12)$

where the ${n}$s and ${m}$s must be chosen so that they make up one of the three degenerate states above. (That is, one each of the ${n}$s and ${m}$s must be 1 and the other two are 0.)

We’ve worked out some of these elements before, in particular in the 1-d case:

$\displaystyle \langle n|x|n'\rangle=\sqrt{\frac{\hbar}{2m\omega}}\left(\sqrt{n'+1}\delta_{n,n'+1}+\sqrt{n'}\delta_{n,n'-1}\right) \ \ \ \ \ (13)$

This shows that all the diagonal elements of ${W}$ are zero, so we need look only at the off-diagonal elements. The quantity ${\left\langle n_{x}\right|x^{2}\left|m_{x}\right\rangle =0}$ if ${n_{x}\ne m_{x}}$ since the integrand contains a factor of ${x^{3}}$ which is an odd function. Thus the only non-zero elements must have ${n_{x}=m_{x}}$, ${n_{y}\ne m_{y}}$ and ${n_{z}\ne m_{z}}$. The only elements that satisfy these conditions are ${W_{010,001}=\lambda\left\langle 0\right|x^{2}\left|0\right\rangle \left\langle 1\right|y\left|0\right\rangle \left\langle 0\right|z\left|1\right\rangle }$ and ${W_{001,010}=\lambda\left\langle 0\right|x^{2}\left|0\right\rangle \left\langle 0\right|y\left|1\right\rangle \left\langle 1\right|z\left|0\right\rangle }$. For the last two factors we can use the formula above, while for ${\left\langle 0\right|x^{2}\left|0\right\rangle }$ we can use an earlier result:

 $\displaystyle \left\langle 0\right|x^{2}\left|0\right\rangle$ $\displaystyle =$ $\displaystyle \frac{\hbar}{2m\omega}\ \ \ \ \ (14)$ $\displaystyle \left\langle 0\right|y\left|1\right\rangle$ $\displaystyle =$ $\displaystyle \sqrt{\frac{\hbar}{2m\omega}}\ \ \ \ \ (15)$ $\displaystyle \left\langle 1\right|z\left|0\right\rangle$ $\displaystyle =$ $\displaystyle \sqrt{\frac{\hbar}{2m\omega}} \ \ \ \ \ (16)$

Thus

$\displaystyle W=\lambda\left[\begin{array}{ccc} 0 & 0 & 0\\ 0 & 0 & \left(\frac{\hbar}{2m\omega}\right)^{2}\\ 0 & \left(\frac{\hbar}{2m\omega}\right)^{2} & 0 \end{array}\right] \ \ \ \ \ (17)$

The energy corrections are the eigenvalues of ${W}$, which are

$\displaystyle E_{1}=0,\pm\lambda\left(\frac{\hbar}{2m\omega}\right)^{2} \ \ \ \ \ (18)$

Thus the perturbation splits the degeneracy, leaving one energy the same and moving the other two on either side of the original energy.

# Harmonic oscillator: relativistic correction

References: Griffiths, David J. (2005), Introduction to Quantum Mechanics, 2nd Edition; Pearson Education – Problem 6.14.

Shankar, R. (1994), Principles of Quantum Mechanics, Plenum Press. Section 7.4, Exercise 7.4.4.

We can apply the relativistic correction to the one-dimensional harmonic oscillator as another example. When analyzing the hydrogen atom, we arrived at this formula for the first order correction to the energy:

$\displaystyle E_{n1}=-\frac{1}{2mc^{2}}\left\langle n0\right|\left(E_{n0}-V\right)^{2}\left|n0\right\rangle \ \ \ \ \ (1)$

where we’ve adjusted the wave functions so they apply to the harmonic oscillator.

Before applying this formula, we should check a couple of things. First, this formula was derived using non-degenerate perturbation theory. In the one-dimensional oscillator this is fine, since there are no degenerate states.

Second, we assumed that the operator ${p^{4}}$ was hermitian, and to check this it is easiest to use the raising and lowering operators. We have

$\displaystyle p=i\sqrt{\frac{\hbar m\omega}{2}}(a_{+}-a_{-}) \ \ \ \ \ (2)$

The raising and lowering operators transform one wave function into another:

 $\displaystyle a_{+}\left|n0\right\rangle$ $\displaystyle =$ $\displaystyle \sqrt{n+1}\left|n+1,0\right\rangle \ \ \ \ \ (3)$ $\displaystyle a_{-}\left|n0\right\rangle$ $\displaystyle =$ $\displaystyle \sqrt{n}\left|n-1,0\right\rangle \ \ \ \ \ (4)$

Therefore, each application of ${p}$ transforms the original wave function into a linear combination of other wave functions and since ${p}$ itself must be hermitian (it represents an observable: the momentum) when applied to any oscillator wave function, any power of ${p}$ is also hermitian in the same situation.

Having verified that the first order energy correction may be applied to the harmonic oscillator, we can now plug in the values. The unperturbed energies are

$\displaystyle E_{n0}=\left(n+\frac{1}{2}\right)\hbar\omega \ \ \ \ \ (5)$

From the virial theorem we know that ${\left\langle T\right\rangle =\left\langle V\right\rangle =\frac{1}{2}E_{n0}}$ so

 $\displaystyle E_{n1}$ $\displaystyle =$ $\displaystyle -\frac{1}{2mc^{2}}\left(E_{n0}^{2}-2E_{n0}\left\langle V\right\rangle +\left\langle V^{2}\right\rangle \right)\ \ \ \ \ (6)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\frac{1}{2mc^{2}}\left[\left(\left(n+\frac{1}{2}\right)\hbar\omega\right)^{2}-\left(\left(n+\frac{1}{2}\right)\hbar\omega\right)^{2}+\frac{1}{4}m^{2}\omega^{4}\left\langle x^{4}\right\rangle \right]\ \ \ \ \ (7)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\frac{m\omega^{4}}{8c^{2}}\left\langle x^{4}\right\rangle \ \ \ \ \ (8)$

To calculate ${\left\langle x^{4}\right\rangle }$, we can use the raising and lowering operators again. We have

 $\displaystyle x$ $\displaystyle =$ $\displaystyle \sqrt{\frac{\hbar}{2m\omega}}(a_{+}+a_{-})\ \ \ \ \ (9)$ $\displaystyle x^{2}$ $\displaystyle =$ $\displaystyle \frac{\hbar}{2m\omega}\left(a_{+}^{2}+a_{+}a_{-}+a_{-}a_{+}+a_{-}^{2}\right) \ \ \ \ \ (10)$

Since ${\left\langle x^{4}\right\rangle =\left\langle n0\right|x^{4}\left|n0\right\rangle }$, the two wave functions involved in calculating the mean value are the same (both ${\left|n0\right\rangle }$) and ${\left\langle n0\right|\left.m0\right\rangle =\delta_{mn}}$, any combination of ${a_{+}}$ and ${a_{-}}$ that converts ${\left|n0\right\rangle }$ into a different wave function will not contribute to the overall integral, so we need consider only those terms in the operator ${x^{4}}$ with equal numbers of ${a_{+}}$ and ${a_{-}}$. Retaining only these terms, we get

$\displaystyle x^{4}=\left(\frac{\hbar}{2m\omega}\right)^{2}\left(a_{+}^{2}a_{-}^{2}+a_{+}a_{-}a_{+}a_{-}+a_{+}a_{-}^{2}a_{+}+a_{-}a_{+}a_{-}a_{+}+a_{-}a_{+}^{2}a_{-}+a_{-}^{2}a_{+}^{2}\right) \ \ \ \ \ (11)$

Applying the operators according to the formulas above, we get

 $\displaystyle \left\langle x^{4}\right\rangle$ $\displaystyle =$ $\displaystyle \left(\frac{\hbar}{2m\omega}\right)^{2}\left[n\left(n-1\right)+n^{2}+n\left(n+1\right)+\left(n+1\right)^{2}+n\left(n+1\right)+\left(n+1\right)\left(n+2\right)\right]\ \ \ \ \ (12)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left(\frac{\hbar}{2m\omega}\right)^{2}\left(6n^{2}+6n+3\right) \ \ \ \ \ (13)$

The energy correction is then

$\displaystyle E_{n1}=-\frac{3\hbar^{2}\omega^{2}}{32mc^{2}}\left(2n^{2}+2n+1\right) \ \ \ \ \ (14)$

# Degenerate perturbation in 3 state system

References: Griffiths, David J. (2005), Introduction to Quantum Mechanics, 2nd Edition; Pearson Education – Problem 6.9.

Here’s another example of multiple degenerate perturbation theory. This time, the system has only 3 linearly independent states, with the hamiltonian given by

$\displaystyle H=V_{0}\left[\begin{array}{ccc} 1-\epsilon & 0 & 0\\ 0 & 1 & \epsilon\\ 0 & \epsilon & 2 \end{array}\right] \ \ \ \ \ (1)$

where ${\epsilon\ll1}$ and can be regarded as a perturbation. The unperturbed hamiltonian is then

$\displaystyle H=V_{0}\left[\begin{array}{ccc} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 2 \end{array}\right] \ \ \ \ \ (2)$

and has one state with energy ${2V_{0}}$ and a two-fold degenerate state with energy ${V_{0}}$. The normalized eigenvectors are

$\displaystyle \left[\begin{array}{c} 1\\ 0\\ 0 \end{array}\right],\left[\begin{array}{c} 0\\ 1\\ 0 \end{array}\right],\left[\begin{array}{c} 0\\ 0\\ 1 \end{array}\right] \ \ \ \ \ (3)$

In this case, we can solve the perturbed system exactly by finding the eigenvalues of the full matrix from the equation

$\displaystyle \left(1-\epsilon-\lambda\right)\left[\left(1-\lambda\right)\left(2-\lambda\right)-\epsilon^{2}\right]=0 \ \ \ \ \ (4)$

The solutions of this equation are

$\displaystyle \lambda=\begin{cases} 1-\epsilon\\ \frac{3}{2}+\frac{1}{2}\sqrt{1+4\epsilon^{2}}\\ \frac{3}{2}-\frac{1}{2}\sqrt{1+4\epsilon^{2}} \end{cases} \ \ \ \ \ (5)$

so the three energies are ${V_{0}\lambda}$. We can expand the square root in the last two energies in a Taylor series in ${\epsilon^{2}}$ using ${\sqrt{1+4\epsilon^{2}}=1+2\epsilon^{2}+\mathcal{O}\left(\epsilon^{4}\right)}$ so we get

$\displaystyle E_{\epsilon}=\begin{cases} V_{0}\left(1-\epsilon\right)\\ V_{0}\left(2+\epsilon^{2}+\mathcal{O}\left(\epsilon^{4}\right)\right)\\ V_{0}\left(1-\epsilon^{2}+\mathcal{O}\left(\epsilon^{4}\right)\right) \end{cases} \ \ \ \ \ (6)$

We can see that two of these energies are perturbations on the original degenerate state with ${E_{0}=V_{0}}$ and the third is a perturbation on ${E_{0}=2V_{0}}$.

We can now analyze the system using perturbation theory and compare the results with the exact solutions above. The perturbation is

$\displaystyle V=V_{0}\left[\begin{array}{ccc} -\epsilon & 0 & 0\\ 0 & 0 & \epsilon\\ 0 & \epsilon & 0 \end{array}\right] \ \ \ \ \ (7)$

Since the unperturbed state with ${E_{0}=2V_{0}}$ is non-degenerate, we can use non-degenerate perturbation theory to find the change in energy. The state corresponding to this energy is given by the eigenvector ${\left[\begin{array}{c} 0\\ 0\\ 1 \end{array}\right]}$, so the energy perturbation is given by

$\displaystyle E_{1}=\left[\begin{array}{ccc} 0 & 0 & 1\end{array}\right]V_{0}\left[\begin{array}{ccc} -\epsilon & 0 & 0\\ 0 & 0 & \epsilon\\ 0 & \epsilon & 0 \end{array}\right]\left[\begin{array}{c} 0\\ 0\\ 1 \end{array}\right]=0 \ \ \ \ \ (8)$

This is consistent with the result above, since there is no first order term in ${V_{0}\left(2+\epsilon^{2}+\mathcal{O}\left(\epsilon^{4}\right)\right)}$. We can work out the second order correction using our earlier formula:

$\displaystyle E_{n2}=\sum_{j\ne n}\frac{\left|\left\langle j0\right|V\left|n0\right\rangle \right|^{2}}{E_{n0}-E_{j0}} \ \ \ \ \ (9)$

For this, we need the off-diagonal matrix elements. If we number the eigenvectors of the unperturbed system above in order, then the ${E_{0}=2}$ state has ${n=3}$, so we need

$\displaystyle \left\langle 20\right|V\left|30\right\rangle =\left[\begin{array}{ccc} 0 & 1 & 0\end{array}\right]V_{0}\left[\begin{array}{ccc} -\epsilon & 0 & 0\\ 0 & 0 & \epsilon\\ 0 & \epsilon & 0 \end{array}\right]\left[\begin{array}{c} 0\\ 0\\ 1 \end{array}\right]=\epsilon V_{0} \ \ \ \ \ (10)$

$\displaystyle \left\langle 10\right|V\left|30\right\rangle =\left[\begin{array}{ccc} 1 & 0 & 0\end{array}\right]V_{0}\left[\begin{array}{ccc} -\epsilon & 0 & 0\\ 0 & 0 & \epsilon\\ 0 & \epsilon & 0 \end{array}\right]\left[\begin{array}{c} 0\\ 0\\ 1 \end{array}\right]=0 \ \ \ \ \ (11)$

Since ${E_{3,0}-E_{j,0}=V_{0}}$ for ${j=1,2}$ we get

$\displaystyle E_{3,2}=\frac{\left|\left\langle 20\right|V\left|30\right\rangle \right|^{2}}{V_{0}}=V_{0}\epsilon^{2} \ \ \ \ \ (12)$

This correction term matches the ${\epsilon^{2}}$ term in the expansion above: ${V_{0}\left(2+\epsilon^{2}+\mathcal{O}\left(\epsilon^{4}\right)\right)}$.

For the degenerate energy ${E_{0}=V_{0}}$, we can use degenerate perturbation theory, but only up to first order since we haven’t worked out the higher order cases. In this case, the matrix ${W}$ is a ${2\times2}$ matrix, since the unperturbed state is only doubly degenerate. We get

$\displaystyle W_{11}=\left\langle 10\right|V\left|10\right\rangle =\left[\begin{array}{ccc} 1 & 0 & 0\end{array}\right]V_{0}\left[\begin{array}{ccc} -\epsilon & 0 & 0\\ 0 & 0 & \epsilon\\ 0 & \epsilon & 0 \end{array}\right]\left[\begin{array}{c} 1\\ 0\\ 0 \end{array}\right]=-\epsilon V_{0} \ \ \ \ \ (13)$

$\displaystyle W_{21}=W_{12}=\left\langle 10\right|V\left|20\right\rangle =\left[\begin{array}{ccc} 1 & 0 & 0\end{array}\right]V_{0}\left[\begin{array}{ccc} -\epsilon & 0 & 0\\ 0 & 0 & \epsilon\\ 0 & \epsilon & 0 \end{array}\right]\left[\begin{array}{c} 0\\ 1\\ 0 \end{array}\right]=0 \ \ \ \ \ (14)$

$\displaystyle W_{22}=\left\langle 20\right|V\left|20\right\rangle =\left[\begin{array}{ccc} 0 & 1 & 0\end{array}\right]V_{0}\left[\begin{array}{ccc} -\epsilon & 0 & 0\\ 0 & 0 & \epsilon\\ 0 & \epsilon & 0 \end{array}\right]\left[\begin{array}{c} 0\\ 1\\ 0 \end{array}\right]=0 \ \ \ \ \ (15)$

$\displaystyle W=\left[\begin{array}{cc} -\epsilon V_{0} & 0\\ 0 & 0 \end{array}\right] \ \ \ \ \ (16)$

Thus, to first order, one state gets an adjustment to the energy of ${-\epsilon V_{0}}$ and the other gets zero, which is consistent with the exact results above, since there is no first order term in ${V_{0}\left(1-\epsilon^{2}+\mathcal{O}\left(\epsilon^{4}\right)\right)}$. Since ${W}$ is diagonal, the special states are just the original states.

# Perturbation of 3-d square well

References: Griffiths, David J. (2005), Introduction to Quantum Mechanics, 2nd Edition; Pearson Education – Problem 6.8.

As a simple example of multiple degenerate perturbation theory, we start with the 3-d infinite square well and add a perturbation:

$\displaystyle V=a^{3}V_{0}\delta\left(x-\frac{a}{4}\right)\delta\left(y-\frac{a}{2}\right)\delta\left(z-\frac{3a}{4}\right) \ \ \ \ \ (1)$

The wave functions of the unperturbed hamiltonian are

$\displaystyle \left|0,n_{x}n_{y}n_{z}\right\rangle =\left(\frac{2}{a}\right)^{3/2}\sin\frac{n_{x}\pi x}{a}\sin\frac{n_{y}\pi y}{a}\sin\frac{n_{z}\pi z}{a} \ \ \ \ \ (2)$

The ground state ${n_{x}=n_{y}=n_{z}=1}$ is non-degenerate, so we can use non-degenerate perturbation theory to find the change in energy. We get

 $\displaystyle E_{1,111}$ $\displaystyle =$ $\displaystyle \left\langle 0,111\right|V\left|0,111\right\rangle \ \ \ \ \ (3)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left(\frac{2}{a}\right)^{3}V_{0}a^{3}\left[\sin\frac{\pi}{4}\sin\frac{\pi}{2}\sin\frac{3\pi}{4}\right]^{2}\ \ \ \ \ (4)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 2V_{0} \ \ \ \ \ (5)$

The first excited state is triply degenerate, since two of the ${n}$s are 1 and the other is 2. We need to calculate the matrix ${W_{ij}=\left\langle 0,i_{x}i_{y}i_{z}\right|V\left|0,j_{x}j_{y}j_{z}\right\rangle }$. We take the elements in the order ${\left|0,211\right\rangle ,}$ ${\left|0,121\right\rangle }$ and ${\left|0,112\right\rangle }$. We then get

 $\displaystyle W_{11}$ $\displaystyle =$ $\displaystyle \left\langle 0,211\right|V\left|0,211\right\rangle \ \ \ \ \ (6)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left(\frac{2}{a}\right)^{3}V_{0}a^{3}\left[\sin\frac{2\pi}{4}\sin\frac{\pi}{2}\sin\frac{3\pi}{4}\right]^{2}\ \ \ \ \ (7)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 4V_{0}\ \ \ \ \ (8)$ $\displaystyle W_{12}$ $\displaystyle =$ $\displaystyle \left\langle 0,211\right|V\left|0,121\right\rangle \ \ \ \ \ (9)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left(\frac{2}{a}\right)^{3}V_{0}a^{3}\left[\sin\frac{2\pi}{4}\sin\frac{\pi}{2}\sin\frac{3\pi}{4}\right]\left[\sin\frac{\pi}{4}\sin\frac{2\pi}{2}\sin\frac{3\pi}{4}\right]\ \ \ \ \ (10)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 0 \ \ \ \ \ (11)$

We can work out the other elements the same way and get

$\displaystyle W=4V_{0}\left[\begin{array}{ccc} 1 & 0 & -1\\ 0 & 0 & 0\\ -1 & 0 & 1 \end{array}\right] \ \ \ \ \ (12)$

The eigenvalues of the matrix ${W/4V_{0}}$ are obtained from

 $\displaystyle \left(1-\lambda\right)\left(-\lambda\right)\left(1-\lambda\right)+\lambda$ $\displaystyle =$ $\displaystyle 0\ \ \ \ \ (13)$ $\displaystyle \lambda$ $\displaystyle =$ $\displaystyle 0,0,2 \ \ \ \ \ (14)$

Thus the triply degenerate unperturbed state splits into a non-degenerate state with first order energy correction of ${8V_{0}}$ and a doubly degenerate state with the original energy.

The special states are found from the normalized eigenvectors of ${W}$ which are

$\displaystyle \frac{1}{\sqrt{2}}\left[\begin{array}{c} -1\\ 0\\ 1 \end{array}\right],\frac{1}{\sqrt{2}}\left[\begin{array}{c} 1\\ 0\\ 1 \end{array}\right],\left[\begin{array}{c} 0\\ 1\\ 0 \end{array}\right] \ \ \ \ \ (15)$

So the states are

 $\displaystyle \frac{1}{\sqrt{2}}\left(-\left|0,211\right\rangle +\left|0,112\right\rangle \right)$ $\displaystyle$ $\displaystyle \left(E_{1}=8V_{0}\right)\ \ \ \ \ (16)$ $\displaystyle \frac{1}{\sqrt{2}}\left(\left|0,211\right\rangle +\left|0,112\right\rangle \right)$ $\displaystyle$ $\displaystyle \left(E_{1}=0\right)\ \ \ \ \ (17)$ $\displaystyle \left|0,121\right\rangle$ $\displaystyle$ $\displaystyle \left(E_{1}=0\right) \ \ \ \ \ (18)$

# Perturbation theory for higher-level degenerate systems

References: Griffiths, David J. (2005), Introduction to Quantum Mechanics, 2nd Edition; Pearson Education – Problem 6.10.

The analysis of degenerate pertrubation theory for a doubly degenerate state generalizes fairly easily to higher levels of degeneracy. The equations we got for the doubly degenerate case are

 $\displaystyle \alpha W_{aa}+\beta W_{ab}$ $\displaystyle =$ $\displaystyle \alpha E_{n1}\ \ \ \ \ (1)$ $\displaystyle \alpha W_{ba}+\beta W_{bb}$ $\displaystyle =$ $\displaystyle \beta E_{n1} \ \ \ \ \ (2)$

where

$\displaystyle W_{ij}\equiv\left\langle i0\right|V\left|j0\right\rangle \ \ \ \ \ (3)$

and ${V}$ is the perturbation in the potential and ${E_{n1}}$ is the first order perturbation in the energy for state ${n}$. The coefficients ${\alpha}$ and ${\beta}$ determine the linear combination of the unperturbed wave functions according to

$\displaystyle \left|n0\right\rangle =\alpha\left|a0\right\rangle +\beta\left|b0\right\rangle \ \ \ \ \ (4)$

The generalization is more easily seen if we rewrite the first pair of equations in matrix form

$\displaystyle \left[\begin{array}{cc} W_{aa} & W_{ab}\\ W_{ba} & W_{bb} \end{array}\right]\left[\begin{array}{c} \alpha\\ \beta \end{array}\right]=E_{n1}\left[\begin{array}{c} \alpha\\ \beta \end{array}\right] \ \ \ \ \ (5)$

That is, the energy perturbations are the eigenvalues of the matrix ${W}$ and the vectors ${\left[\begin{array}{c} \alpha\\ \beta \end{array}\right]}$ are the eigenvectors.

In the general case where we have a ${d}$-fold degeneracy with unperturbed energy ${E_{n0}}$, the most general unperturbed wave function is

$\displaystyle \left|n0\right\rangle =\sum_{j=1}^{d}\alpha_{j}\left|j0\right\rangle \ \ \ \ \ (6)$

The derivation in the ${d}$-fold case is exactly the same as in the 2-fold case up until we get the equation

$\displaystyle \left\langle a0\right|V\left|n0\right\rangle =E_{n1}\left\langle a0\right.\left|n0\right\rangle \ \ \ \ \ (7)$

where ${\left|a0\right\rangle }$ is one of the degenerate unperturbed states. We now substitute for ${\left|n0\right\rangle }$ and using the orthogonality of the unperturbed states we get

 $\displaystyle \sum_{j=1}^{d}\alpha_{j}\left\langle a0\right|V\left|j0\right\rangle$ $\displaystyle =$ $\displaystyle \alpha_{a}E_{n1}\ \ \ \ \ (8)$ $\displaystyle \sum_{j=1}^{d}\alpha_{j}W_{aj}$ $\displaystyle =$ $\displaystyle \alpha_{a}E_{n1} \ \ \ \ \ (9)$

This is just the ${a}$th row of the eigenvalue equation for the ${d\times d}$ matrix ${W}$ so the eigenvalue equation applies to any level of degeneracy.

In systems that have some states with one level of degeneracy and other states with different levels of degeneracy, we would apply the appropriate eigenvalue equation to each set of degenerate states separately. For example, in the particle on a circular wire problem, all states except ${n=0}$ are doubly degenerate so we use the ${2\times2}$ matrix for each value of ${n>0}$ and the nondegenerate perturbation theory for ${n=0}$.

# Perturbing a particle on a circular wire

References: Griffiths, David J. (2005), Introduction to Quantum Mechanics, 2nd Edition; Pearson Education – Problem 6.7.

Here’s another example of applying degenerate pertrubation theory. Suppose we revisit the problem of a particle with a periodic wave function: the particle of mass ${m}$ that slides on a frictionless circular wire of circumference ${L}$. We saw that the stationary states are

$\displaystyle \left|n0\right\rangle =\frac{1}{\sqrt{L}}e^{2n\pi xi/L} \ \ \ \ \ (1)$

and the energies are

$\displaystyle E_{\pm n0}=\frac{2n^{2}\pi^{2}\hbar^{2}}{mL^{2}} \ \ \ \ \ (2)$

The system is doubly degenerate for all values of ${n}$ except 0, so the degenerate perturbation theory applies.

Now we introduce the perturbation

$\displaystyle V=-V_{0}e^{-x^{2}/a^{2}} \ \ \ \ \ (3)$

where ${V_{0}}$ is assumed to be small compared with the unperturbed energies, and ${a\ll L}$. Since the unperturbed hamiltonian is that of the free particle (with the periodic constraint), the perturbation is the only non-zero part of the potential.

We can find the first-order perturbations to the energy using the formula we derived in the last post, with the suffixes on the matrix elements customized to the two degenerate states for ${E_{\pm n0}}$:

$\displaystyle E_{n1}=\frac{1}{2}\left(W_{nn}+W_{-n,-n}\pm\sqrt{\left(W_{nn}-W_{-n,-n}\right)^{2}+4\left|W_{n,-n}\right|^{2}}\right) \ \ \ \ \ (4)$

The matrix elements can be calculated as:

 $\displaystyle W_{qr}$ $\displaystyle =$ $\displaystyle \left\langle q0\right|V\left|r0\right\rangle \ \ \ \ \ (5)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\frac{V_{0}}{L}\int_{-L/2}^{L/2}e^{2\left(r-q\right)\pi xi/L}e^{-x^{2}/a^{2}}dx \ \ \ \ \ (6)$

Because of the assumption ${a\ll L}$, the ${e^{-x^{2}/a^{2}}}$ term becomes very small for relatively small values of ${x}$, so we can approximate the integral by extending the limits to infinity in both directions. Using software, this integral comes out to

$\displaystyle W_{qr}=-\frac{\sqrt{\pi}aV_{0}}{L}e^{-a^{2}\pi^{2}\left(r-q\right)^{2}/L^{2}} \ \ \ \ \ (7)$

The required matrix elements are then

 $\displaystyle W_{nn}=W_{-n,-n}$ $\displaystyle =$ $\displaystyle -\frac{\sqrt{\pi}aV_{0}}{L}\ \ \ \ \ (8)$ $\displaystyle W_{n,-n}$ $\displaystyle =$ $\displaystyle -\frac{\sqrt{\pi}aV_{0}}{L}e^{-4a^{2}\pi^{2}n^{2}/L^{2}} \ \ \ \ \ (9)$

The two perturbations on the unperturbed energy ${E_{\pm n0}}$ are

$\displaystyle E_{n1}=-\frac{\sqrt{\pi}aV_{0}}{L}\left(1\pm e^{-4a^{2}\pi^{2}n^{2}/L^{2}}\right) \ \ \ \ \ (10)$

Since ${W_{n,-n}\ne0}$, the states ${\psi_{n0}}$ are not the ‘special’ states we mentioned in the last post. We can find the special states as follows. We know that these special states are linear combinations of ${\psi_{+n0}}$ and ${\psi_{-n0}}$, so let’s define the special states as

 $\displaystyle \left|a0\right\rangle$ $\displaystyle =$ $\displaystyle \alpha_{a}\left|+n0\right\rangle +\beta_{a}\left|-n0\right\rangle \ \ \ \ \ (11)$ $\displaystyle \left|b0\right\rangle$ $\displaystyle =$ $\displaystyle \alpha_{b}\left|+n0\right\rangle +\beta_{b}\left|-n0\right\rangle \ \ \ \ \ (12)$

Since ${W_{ab}=0}$ for these states, we have

 $\displaystyle W_{ab}$ $\displaystyle =$ $\displaystyle \left\langle a0\right|V\left|b0\right\rangle \ \ \ \ \ (13)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \alpha_{a}^*\alpha_{b}W_{nn}+\alpha_{a}^*\beta_{b}W_{n,-n}+\beta_{a}^*\alpha_{b}W_{-n,n}+\beta_{a}^*\beta_{b}W_{-n,-n}\ \ \ \ \ (14)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 0 \ \ \ \ \ (15)$

Substituting the values for the ${W}$s from above, we get

$\displaystyle \alpha_{a}^*\alpha_{b}+\beta_{a}^*\beta_{b}+e^{-4a^{2}\pi^{2}n^{2}/L^{2}}\left(\alpha_{a}^*\beta_{b}+\beta_{a}^*\alpha_{b}\right)=0 \ \ \ \ \ (16)$

We can satisfy this equation for all ${n}$ if we choose

 $\displaystyle \alpha_{a}$ $\displaystyle =$ $\displaystyle -\beta_{a}\ \ \ \ \ (17)$ $\displaystyle \alpha_{b}$ $\displaystyle =$ $\displaystyle \beta_{b} \ \ \ \ \ (18)$

With the normalization condition ${\left|\alpha_{i}\right|^{2}+\left|\beta_{i}\right|^{2}=1}$ we get the final form of the special states:

 $\displaystyle \left|a0\right\rangle$ $\displaystyle =$ $\displaystyle \frac{1}{\sqrt{2}}\left(\left|+n0\right\rangle -\left|-n0\right\rangle \right)\ \ \ \ \ (19)$ $\displaystyle \left|b0\right\rangle$ $\displaystyle =$ $\displaystyle \frac{1}{\sqrt{2}}\left(\left|+n0\right\rangle +\left|-n0\right\rangle \right) \ \ \ \ \ (20)$

For the special states, we can get the energy perturbations using the formula from nondegenerate theory

 $\displaystyle E_{n1,a}$ $\displaystyle =$ $\displaystyle W_{aa}\ \ \ \ \ (21)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{2}\left[W_{nn}+W_{-n,-n}-2W_{n,-n}\right]\ \ \ \ \ (22)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\frac{\sqrt{\pi}aV_{0}}{L}\left(1-e^{-4a^{2}\pi^{2}n^{2}/L^{2}}\right)\ \ \ \ \ (23)$ $\displaystyle E_{n1,b}$ $\displaystyle =$ $\displaystyle W_{bb}\ \ \ \ \ (24)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{2}\left[W_{nn}+W_{-n,-n}+2W_{n,-n}\right]\ \ \ \ \ (25)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\frac{\sqrt{\pi}aV_{0}}{L}\left(1+e^{-4a^{2}\pi^{2}n^{2}/L^{2}}\right) \ \ \ \ \ (26)$

These are the same energies as before.

Finally, it’s worth quoting a theorem that makes finding these special states a little easier in some cases. If you can find some hermitian operator ${A}$ that commutes with both the unperturbed hamiltonian ${H_{0}}$ and the perturbation ${V}$ and you can find two eigenfunctions that have the same eigenvalue under ${H_{0}}$ but different eigenvalues under ${A}$, then these eigenfunctions are the special states. That is, we look for functions ${\left|a0\right\rangle }$ and ${\left|b0\right\rangle }$ Such that

 $\displaystyle H_{0}\left|a0\right\rangle$ $\displaystyle =$ $\displaystyle E\left|a0\right\rangle \ \ \ \ \ (27)$ $\displaystyle H_{0}\left|b0\right\rangle$ $\displaystyle =$ $\displaystyle E\left|b0\right\rangle \ \ \ \ \ (28)$ $\displaystyle A\left|a0\right\rangle$ $\displaystyle =$ $\displaystyle A_{a}\left|a0\right\rangle \ \ \ \ \ (29)$ $\displaystyle A\left|b0\right\rangle$ $\displaystyle =$ $\displaystyle A_{b}\left|b0\right\rangle \ \ \ \ \ (30)$

where ${A_{a}\ne A_{b}}$. To see this, we use the assumption that ${\left[V,A\right]=0}$ so that

 $\displaystyle \left\langle a0\right|\left[V,A\right]\left|b0\right\rangle$ $\displaystyle =$ $\displaystyle \left\langle a0\right|VA\left|b0\right\rangle -\left\langle a0\right|AV\left|b0\right\rangle \ \ \ \ \ (31)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left(A_{b}-A_{a}\right)\left\langle a0\right|V\left|b0\right\rangle \ \ \ \ \ (32)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left(A_{b}-A_{a}\right)W_{ab}\ \ \ \ \ (33)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 0 \ \ \ \ \ (34)$

Since we’re assuming ${A_{a}\ne A_{b}}$, we must have ${W_{ab}=0}$.

Actually finding such an operator ${A}$ can be a bit tricky however. In the current example, we can’t use an operator that has a derivative with respect to ${x}$, since that wouldn’t commute with ${V}$. Likewise, we can’t use an operator that involves just multiplying by a function of ${x}$, since that wouldn’t commute with ${H_{0}}$ (which has a derivative in it). After some searching, it seems that we can use the parity operator which has the effect of reversing the sign of ${x}$: ${Af\left(x\right)=f\left(-x\right)}$. It’s not at all clear to me how you can arrive at this answer through any process of logical deduction, however.

To see that it works, we can look at the special functions we derived above:

 $\displaystyle \left|a0\right\rangle$ $\displaystyle =$ $\displaystyle \frac{1}{\sqrt{2}}\left(\left|+n0\right\rangle -\left|-n0\right\rangle \right)=\sqrt{\frac{2}{L}}i\sin\frac{2\pi nx}{L}\ \ \ \ \ (35)$ $\displaystyle \left|b0\right\rangle$ $\displaystyle =$ $\displaystyle \frac{1}{\sqrt{2}}\left(\left|+n0\right\rangle +\left|-n0\right\rangle \right)=\sqrt{\frac{2}{L}}\cos\frac{2\pi nx}{L} \ \ \ \ \ (36)$

Under the parity operator, we have

 $\displaystyle A\left|a0\right\rangle$ $\displaystyle =$ $\displaystyle -\left|a0\right\rangle \ \ \ \ \ (37)$ $\displaystyle A\left|b0\right\rangle$ $\displaystyle =$ $\displaystyle \left|b0\right\rangle \ \ \ \ \ (38)$

Thus the eigenvalues of ${A}$ are ${\pm1}$ and the conditions are satisfied. However, it’s not exactly clear that if we didn’t know ${\left|a0\right\rangle }$ and ${\left|b0\right\rangle }$ beforehand that we would hit on the idea of using the parity operator, and then manage to find its eigenfunctions. The straightforward route taken above (using the quadratic formula to find the energies) is at least always reliable, even if it might take a bit more calculation.