A.37 Perturbation Theory

Most of the time in quantum mechanics, exact solution of the Hamiltonian eigenvalue problem of interest is not possible. To deal with that, approxi­mations are made.

Perturb­ation theory can be used when the Hamiltonian $H$ consists of two parts $H_0$ and $H_1$, where the problem for $H_0$ can be solved and where $H_1$ is small. The idea is then to adjust the found solutions for the “unperturbed Hamiltonian” $H_0$ so that they become approximately correct for $H_0+H_1$.

This addendum explains how perturb­ation theory works. It also gives a few simple but important examples: the helium atom and the Zeeman and Stark effects. Addendum,{A.38} will use the approach to study relativistic effects on the hydrogen atom.

A.37.1 Basic perturbation theory

To use perturb­ation theory, the eigen­functions and eigen­values of the unperturbed Hamiltonian $H_0$ must be known. These eigen­functions will here be indicated as $\psi_{{\vec n},0}$ and the corre­sponding eigen­values by $E_{{\vec n},0}$. Note the use of the generic ${\vec n}$ to indicate the quantum numbers of the eigen­functions. If the basic system is an hydrogen atom, as is often the case in textbook examples, and spin is unimportant, ${\vec n}$ would likely stand for the set of quantum numbers $n$, $l$, and $m$. But for a three-di­mensional harmonic oscillator, ${\vec n}$ might stand for the quantum numbers $n_x$, $n_y$, and $n_z$. In a three-di­mensional problem with one spinless particle, it takes three quantum numbers to describe an energy eigen­function. However, which three depends on the problem and your approach to it. The additional subscript 0 in $\psi_{{\vec n},0}$ and $E_{{\vec n},0}$ indicates that they ignore the perturb­ation Hamiltonian $H_1$. They are called the unperturbed wave functions and energies.

The key to perturb­ation theory are the “Hamiltonian perturb­ation coefficients” defined as

$$\fbox{$\displaystyle H_{\underline{\vec n}{\vec n},1} \e... ...derline{\vec n},0} \vert H_1 \psi_{{\vec n},0}\rangle $} %$$ (A.239)

If you can evaluate these for every pair of energy eigen­functions, you should be OK. Note that evaluating inner products is just summation or integr­ation; it is generally a lot simpler than trying to solve the eigenvalue problem $\left(H_0+H_1\right)\psi$ $\vphantom0\raisebox{1.5pt}{$=$}$ $E\psi$.

In the appli­cation of perturb­ation theory, the idea is to pick one unperturbed eigen­function $\psi_{{\vec n},0}$ of $H_0$ of interest and then correct it to account for $H_1$, and especially correct its energy $E_{{\vec n},0}$. Caution! If the energy $E_{{\vec n},0}$ is degenerate, i.e. there is more than one unperturbed eigen­function $\psi_{{\vec n},0}$ of $H_0$ with that energy, you must use a “good” eigen­function to correct the energy. How to do that will be discussed in subsection A.37.3.

For now just assume that the energy is not degenerate or that you picked a good eigen­function $\psi_{{\vec n},0}$. Then a first correction to the energy $E_{{\vec n},0}$ to account for the perturb­ation $H_1$ is very simple, {D.80}; just add the corre­sponding Hamiltonian perturb­ation coefficient:

$$\fbox{$\displaystyle E_{\vec n}= E_{{\vec n},0} + H_{{\vec n}{\vec n},1} + \ldots $} %$$ (A.240)

This is a quite user-friendly result, because it only involves the selected energy eigen­function $\psi_{{\vec n},0}$. The other energy eigen­functions are not involved. In a numerical solution, you might only have computed one state, say the ground state of $H_0$. Then you can use this result to correct the ground state energy for a perturb­ation even if you do not have data about any other energy states of $H_0$.

Unfortunately, it does happen quite a lot that the above correction $H_{{\vec n}{\vec n},1}$ is zero because of some symmetry or the other. Or it may simply not be accurate enough. In that case, to find the energy change you have to use what is called “second order perturb­ation theory:”

$$\fbox{$\displaystyle E_{\vec n}= E_{{\vec n},0} + H_{{\v... ...}{E_{\underline{\vec n},0}-E_{{\vec n},0}} + \ldots $} %$$ (A.241)

Now all eigen­functions of $H_0$ will be needed, which makes second order theory a lot nastier. Then again, even if the “first order” correction $H_{{\vec n}{\vec n},1}$ to the energy is non­zero, the second order formula will likely give a much more accurate result.

Sometimes you may also be inter­ested in what happens to the energy eigen­functions, not just the energy eigen­values. The corre­sponding formula is

$$\fbox{$\displaystyle \psi_{\vec n}= \psi_{{\vec n},0} ... ...rline{\vec n}} \psi_{\underline{\vec n},0} + \ldots $} %$$ (A.242)

That is the first order result. The second sum is zero if the problem is not degenerate. Otherwise its coefficients $c_{\underline{\vec n}}$ are determined by consid­erations found in derivation {D.80}.

In some cases, instead of using second order theory as above, it may be simpler to compute the first order wave function perturb­ation and the second order energy change from

$$\fbox{$\displaystyle (H_0-E_{{\vec n},0})\psi_{{\vec n},... ...},0}\vert(H_1-E_{{\vec n},1})\psi_{{\vec n},1}\rangle $} %$$ (A.243)

Eigen­function $\psi_{{\vec n},0}$ must be good. The good news is that this does not require all the unperturbed eigen­functions. The bad news is that it requires solution of a non­trivial equation involving the unperturbed Hamiltonian instead of just integr­ation. It may be the best way to proceed for a perturb­ation of a numerical solution.

One appli­cation of perturb­ation theory is the “Hellmann-Feynman theorem.” Here the perturb­ation Hamiltonian is an infini­tesimal change $\partial{H}$ in the unperturbed Hamiltonian caused by an infini­tesimal change in some parameter that it depends on. If the parameter is called $\lambda$, perturb­ation theory says that the first order energy change is

$$\fbox{$\displaystyle \frac{\partial E_{\vec n}}{\partial... ...al H}{\partial\lambda} \psi_{{\vec n},0} \Bigg\rangle $} %$$ (A.244)

when divided by the change in parameter $\partial\lambda$. If you can figure out the inner product, you can figure out the change in energy. But more important is the reverse: if you can find the derivative of the energy with respect to the parameter, you have the inner product. For example, the Hellmann-Feynman theorem is helpful for finding the expec­tation value of 1$\raisebox{.5pt}{$/$}$$r^2$ for the hydrogen atom, a nasty problem, {D.84}. Of course, always make sure the eigen­function $\psi_{{\vec n},0}$ is a good one for the derivative of the Hamiltonian.

A.37.2 Ionization energy of helium

One prominent deficiency in the approximate analysis of the heavier atoms in chapter 5.9 was the poor ionization energy that it gave for helium. The purpose of this example is to derive a much more reasonable value using perturb­ation theory.

Exactly speaking, the ionization energy is the difference between the energy of the helium atom with both its electrons in the ground state and the helium ion with its second electron removed. Now the energy of the helium ion with electron 2 removed is easy; the Hamiltonian for the remaining electron 1 is

$$H_{\rm He\ ion} = \mbox{} - \frac{\hbar^2}{2m_{\rm e}} \nabla_1^2 - 2 \frac{e^2}{4\pi\epsilon_0} \frac{1}{r_1}$$

where the first term represents the kinetic energy of the electron and the second its attraction to the two-proton nucleus. The helium nucleus normally also contains two neutrons, but they do not attract the electron.

This Hamiltonian is exactly the same as the one for the hydrogen atom in chapter 4.3, except that it has $2e^2$ where the hydrogen one, with just one proton in its nucleus, has $e^2$. So the solution for the helium ion is simple: just take the hydrogen solution, and everywhere where there is an $e^2$ in that solution, replace it by $2e^2$. In particular, the Bohr radius $a$ for the helium ion is half the Bohr radius $a_0$ for hydrogen,

$$a=\frac{4\pi\epsilon_0\hbar^2}{m_{\rm e}2e^2} = {\textstyle\frac{1}{2}} a_0$$

and so its energy and wave function become

$$E_{\rm gs,ion} = - \frac{\hbar^2}{2 m_{\rm e}a^2} = 4 E_1 ... ...m gs,ion}({\skew0\vec r}) = \frac{1}{\sqrt{\pi a^3}} e^{-r/a}$$

where $E_1$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\vphantom0\raisebox{1.5pt}{$-$}$13.6 eV is the energy of the hydrogen atom.

It is inter­esting to see that the helium ion has four times the energy of the hydrogen atom. The reasons for this much higher energy are both that the nucleus is twice as strong, and that the electron is twice as close to it: the Bohr radius is half the size. More generally, in heavy atoms the electrons that are poorly shielded from the nucleus, which means the inner electrons, have energies that scale with the square of the nuclear strength. For such electrons, relativistic effects are much more important than they are for the electron in a hydrogen atom.

The neutral helium atom is not by far as easy to analyze as the ion. Its Hamiltonian is, from (5.34):

$$H_{\rm He} = \mbox{} - \frac{\hbar^2}{2m_{\rm e}} \nab... ...silon_0} \frac1{\vert{\skew0\vec r}_2 -{\skew0\vec r}_1\vert}$$

The first two terms are the kinetic energy and nuclear attraction of electron 1, and the next two the same for electron 2. The final term is the electron to electron repulsion, the curse of quantum mechanics. This final term is the reason that the ground state of helium cannot be found analyti­cally.

Note however that the repulsion term is quali­tatively similar to the nuclear attraction terms, except that there are four of these nuclear attraction terms versus a single repulsion term. So maybe then, it may work to treat the repulsion term as a small perturb­ation, call it $H_1$, to the Hamiltonian $H_0$ given by the first four terms? Of course, if you ask mathema­ticians whether 25% is a small amount, they are going to vehemently deny it; but then, so they would for any amount if there is no limit process involved, so just don’t ask them, OK?

The solution of the eigenvalue problem $H_0\psi$ $\vphantom0\raisebox{1.5pt}{$=$}$ $E\psi$ is simple: since the electrons do not interact with this Hamiltonian, the ground state wave function is the product of the ground state wave functions for the individual electrons, and the energy is the sum of their energies. And the wave functions and energies for the separate electrons are given by the solution for the ion above, so

$$\psi_{\rm gs,0} = \frac{1}{\pi a^3} e^{-(r_1+r_2)/a} \qquad E_{\rm gs,0} = 8 E_1$$

According to this result, the energy of the atom is $8E_1$ while the ion had $4E_1$, so the ionization energy would be $4\vert E_1\vert$, or 54.4 eV. Since the experi­mental value is 24.6 eV, this is no better than the 13.6 eV chapter 5.9 came up with.

To get a better ionization energy, try perturb­ation theory. According to first order perturb­ation theory, a better value for the energy of the hydrogen atom should be

$$E_{\rm gs} = E_{\rm gs,0} + \langle\psi_{\rm gs,0}\vert H_1\psi_{\rm gs,0}\rangle$$

or substituting in from above,

$$E_{\rm gs} = 8 E_1 + \frac{e^2}{4\pi\epsilon_0} \bigg\la... ...\vec r}_1\vert}\frac{1}{\pi a^3} e^{-(r_1+r_2)/a}\bigg\rangle$$

The inner product of the final term can be written out as

$$\frac{e^2}{4\pi\epsilon_0} \frac{1}{\pi^2a^6} \int_{{\rm... ...\vert} {\,\rm d}^3{\skew0\vec r}_1{\rm d}^3{\skew0\vec r}_2$$

This integral can be done analyti­cally. Try it, if you are so inclined; integrate ${\rm d}^3{\skew0\vec r}_1$ first, using spherical coordinates with ${\skew0\vec r}_2$ as their axis and doing the azimuthal and polar angles first. Be careful, $\sqrt{(r_1-r_2)^2}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\vert r_1-r_2\vert$, not $r_1-r_2$, so you will have to integrate $r_1$ $\raisebox{.3pt}{$<$}$ $r_2$ and $r_1$ $\raisebox{.3pt}{$>$}$ $r_2$ separately in the final integr­ation over ${\rm d}{r}_1$. Then integrate ${\rm d}^3{\skew0\vec r}_2$.

The result of the integr­ation is

$$\frac{e^2}{4\pi\epsilon_0} \bigg\langle\frac{1}{\pi a^3}... ...}{4\pi\epsilon_0} \frac{5}{8a} = \frac{5}{2} \vert E_1\vert$$

Therefore, the helium atom energy increases by $2.5\vert E_1\vert$ due to the electron repulsion, and with it, the ionization energy decreases to $1.5\vert E_1\vert$, or 20.4 eV. It is not 24.6 eV, but it is clearly much more reasonable than 54 or 13.6 eV were.

The second order perturb­ation result should give a much more accurate result still. However, if you did the integral above, you may feel little inclin­ation to try the ones involving all possible products of hydrogen energy eigen­functions.

Instead, the result can be improved using a varia­tional approach, like the ones that were used earlier for the hydrogen molecule and molecular ion, and this requires almost no additional work. The idea is to accept the hint from perturb­ation theory that the wave function of helium can be approximated as $\psi_a({\skew0\vec r}_1)\psi_a({\skew0\vec r}_2)$ where $\psi_a$ is the hydrogen ground state wave function using a modified Bohr radius $a$ instead of $a_0$:

$$\psi_{\rm gs} = \psi_a({\skew0\vec r}_1)\psi_a({\skew0\vec... ...si_a({\skew0\vec r}) \equiv \frac{1}{\sqrt{\pi a^3}} e^{-r/a}$$

However, instead of accepting the perturb­ation theory result that $a$ should be half the normal Bohr radius $a_0$, let $a$ be optimized to make the expec­tation energy for the ground state

$$E_{\rm gs} = \langle\psi_{\rm gs}\vert H_{\rm He}\psi_{\rm gs}\rangle$$

as small as possible. This will produce the most accurate ground state energy possible for a ground state wave function of this form, guaranteed no worse than assuming that $a$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\frac12a_0$, and probably better.

No new integrals need to be done to evaluate the inner product above. Instead, noting that for the hydrogen atom according to the virial theorem of chapter 7.2 the expec­tation kinetic energy equals $-E_1$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\hbar^2$$\raisebox{.5pt}{$/$}$$2{m_{\rm e}}a_0^2$ and the potential energy equals $2E_1$, two of the needed integrals can be inferred from the hydrogen solution: chapter 4.3,

$$\bigg\langle\psi_a\bigg\vert{-}\frac{\hbar^2}{2m_{\rm e}}\nabla^2\psi_a\bigg\rangle = \frac{\hbar^2}{2m_{\rm e}a^2}$$

$$- \frac{e^2}{4\pi\epsilon_0} \bigg\langle\psi_a\bigg\ver... ..._a\bigg\rangle = - \frac{\hbar^2}{m_{\rm e}a_0} \frac{1}{a}$$

and this subsection added

$$\bigg\langle\psi_a\psi_a\bigg\vert\frac{1}{\vert{\skew0\ve... ...skew0\vec r}_1\vert}\psi_a\psi_a\bigg\rangle = \frac{5}{8a}$$

Using these results with the helium Hamiltonian, the expec­tation energy of the helium atom can be written out to be

$$\langle\psi_a\psi_a\vert H_{\rm He}\psi_a\psi_a\rangle =... ...2}{m_{\rm e}a^2} - \frac{27}{8} \frac{\hbar^2}{m_{\rm e}a_0a}$$

Setting the derivative with respect to $a$ to zero locates the minimum at $a$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\frac{16}{27}a_0$, rather than $\frac12a_0$. Then the corre­sponding expec­tation energy is $\vphantom0\raisebox{1.5pt}{$-$}$$3^6\hbar^2$$\raisebox{.5pt}{$/$}$$2^8{m_{\rm e}}a_0^2$, or $3^6E_1$$\raisebox{.5pt}{$/$}$$2^7$. Putting in the numbers, the ionization energy is now found as 23.1 eV, in quite good agreement with the experi­mental 24.6 eV.

A.37.3 Degenerate perturbation theory

Energy eigen­values are degenerate if there is more than one independent eigen­function with that energy. Now, if you try to use perturb­ation theory to correct a degenerate eigenvalue of a Hamiltonian $H_0$ for a perturb­ation $H_1$, there may be a problem. Assume that there are $d$ $\raisebox{.3pt}{$>$}$ 1 independent eigen­functions with energy $E_{{\vec n},0}$ and that they are numbered as

$$\psi_{{\vec n}_1,0},\psi_{{\vec n}_2,0},\ldots,\psi_{{\vec n}_d,0}$$

Then as far as $H_0$ is concerned, any combin­ation

$$\psi_{{\vec n},0} = c_1 \psi_{{\vec n}_1,0} + c_2\psi_{{\vec n}_2,0} + \ldots + c_d\psi_{{\vec n}_d,0}$$

with arbitrary coefficients $c_1,c_2,\ldots,c_d$, (not all zero, of course), is just as good an eigen­function with energy $E_{{\vec n},0}$ as any other.

Unfortunately, the full Hamiltonian $H_0+H_1$ is not likely to agree with $H_0$ about that. As far as the full Hamiltonian is concerned, normally only very specific combin­ations are acceptable, the “good” eigen­functions. It is said that the perturb­ation $H_1$ “breaks the degeneracy” of the energy eigenvalue. The single energy eigenvalue splits into several eigen­values of different energy. Only good combin­ations will show up these changed energies; the bad ones will pick up uncertainty in energy that hides the effect of the perturb­ation.

The various ways of ensuring good eigen­functions are illustrated in the following subsections for example perturb­ations of the energy levels of the hydrogen atom. Recall that the unperturbed energy eigen­functions of the hydrogen atom electron, as derived in chapter 4.3, and also including spin, are given as $\psi_{nlm}{\uparrow}$ and $\psi_{nlm}{\downarrow}$. They are highly degenerate: all the eigen­functions with the same value of $n$ have the same energy $E_n$, regardless of what is the value of the azimuthal quantum number 0 $\raisebox{-.3pt}{$\leqslant$}$ $l$ $\raisebox{-.3pt}{$\leqslant$}$ $n-1$ corre­sponding to the square orbital angular momentum $L^2$ $\vphantom0\raisebox{1.5pt}{$=$}$ $l(l+1)\hbar^2$; regardless of what is the magnetic quantum number $\vert m\vert$ $\raisebox{-.3pt}{$\leqslant$}$ $l$ corre­sponding to the orbital angular momentum $L_z$ $\vphantom0\raisebox{1.5pt}{$=$}$ $m\hbar$ in the $z$-direction; and regardless of what is the spin quantum number $m_s$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\pm\frac12$ corre­sponding to the spin angular momentum $m_s\hbar$ in the $z$-direction. In particular, the ground state energy level $E_1$ is two-fold degenerate, it is the same for both $\psi_{100}{\uparrow}$, i.e. $m_s$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\frac12$ and $\psi_{100}{\downarrow}$, $m_s$ $\vphantom0\raisebox{1.5pt}{$=$}$ $-\frac12$. The next energy level $E_2$ is eight-fold degenerate, it is the same for $\psi_{200}{\updownarrow}$, $\psi_{211}{\updownarrow}$, $\psi_{210}{\updownarrow}$, and $\psi_{21{-1}}{\updownarrow}$, and so on for higher values of $n$.

There are two important rules to identify the good eigen­functions, {D.80}:

1. Look for good quantum numbers. The quantum numbers that make the energy eigen­functions of the unperturbed Hamiltonian $H_0$ unique corre­spond to the eigen­values of additional operators besides the Hamiltonian. If the perturb­ation Hamiltonian $H_1$ commutes with one of these additional operators, the corre­sponding quantum number is good. You do not need to combine eigen­functions with different values of that quantum number.

   In particular, if the perturb­ation Hamiltonian commutes with all additional operators that make the eigen­functions of $H_0$ unique, stop worrying: every eigen­function is good already.

   For example, for the usual hydrogen energy eigen­functions $\psi_{nlm}{\updownarrow}$, the quantum numbers $l$, $m$, and $m_s$ make the eigen­functions at a given unperturbed energy level $n$ unique. They corre­spond to the operators $\L ^2$, $\L _z$, and ${\widehat S}_z$. If the perturb­ation Hamiltonian $H_1$ commutes with any one of these operators, the corre­sponding quantum number is good. If the perturb­ation commutes with all three, all eigen­functions are good already.

2. Even if some quantum numbers are bad because the perturb­ation does not commute with that operator, eigen­functions are still good if there are no other eigen­functions with the same unperturbed energy and the same good quantum numbers.

   Otherwise linear algebra is required. For each set of energy eigen­functions
   

   $$\psi_{{\vec n}_1,0},\psi_{{\vec n}_2,0},\ldots$$
   
   
   with the same unperturbed energy and the same good quantum numbers, but different bad ones, form the matrix of Hamiltonian perturb­ation coefficients
   
   $$\left( \begin{array}{ccc} \langle\psi_{{\vec n}_1,0}\v... ...cdots \\ \vdots & \vdots & \ddots \end{array} \right)$$
   
   
   The eigen­values of this matrix are the first order energy corrections. Also, the coefficients $c_1,c_2,\ldots$ of each good eigen­function
   
   $$c_1 \psi_{{\vec n}_1,0} + c_2 \psi_{{\vec n}_2,0} + \ldots$$
   
   
   must be an eigen­vector of the matrix.

   Unfortunately, if the eigen­values of this matrix are not all different, the eigen­vectors are not unique, so you remain unsure about what are the good eigen­functions. In that case, if the second order energy corrections are needed, the detailed analysis of derivation {D.80} will need to be followed.

   If you are not familiar with linear algebra at all, in all cases mentioned here the matrices are just two by two, and you can find that solution spelled out in the notations under “eigen­vector.”

The following, related, practical observ­ation can also be made:

Hamiltonian perturb­ation coefficients can only be non­zero if all the good quantum numbers are the same.

A.37.4 The Zeeman effect

If you put an hydrogen atom in an external magnetic field $\skew2\vec{\cal B}_{\rm {ext}}$, the energy levels of the electron change. That is called the “Zeeman effect.”

If for simplicity a coordinate system is used with its $z$-axis aligned with the magnetic field, then according to chapter 13.4, the Hamiltonian of the hydrogen atom acquires an additional term

$$H_1 = \frac{e}{2m_{\rm e}}{\cal B}_{\rm ext}\left(\L _z + 2 {\widehat S}_z\right)$$ (A.245)

beyond the basic hydrogen atom Hamiltonian $H_0$ of chapter 4.3.1. Quali­tatively, it expresses that a spinning charged particle is equivalent to a tiny electro­magnet, and a magnet wants to align itself with a magnetic field, just like a compass needle aligns itself with the magnetic field of earth.

For this perturb­ation, the $\psi_{nml}{\updownarrow}$ energy eigen­functions are already good ones, because $H_1$ commutes with all of $\L ^2$, $\L _z$ and ${\widehat S}_z$. So, according to perturb­ation theory, the energy eigen­values of an hydrogen atom in a magnetic field are approximately

$$E_n + \langle \psi_{nml}{\updownarrow}\vert H_1\vert\psi_{... ... E_n + \frac{e}{2m_{\rm e}}{\cal B}_{\rm ext}(m + 2 m_s)\hbar$$

Actually, this is not approximate at all; it is the exact eigenvalue of $H_0+H_1$ corre­sponding to the exact eigen­function $\psi_{nml}{\updownarrow}$.

The Zeeman effect can be seen in an experi­mental spectrum. Consider first the ground state. If there is no electro­magnetic field, the two ground states $\psi_{100}{\uparrow}$ and $\psi_{100}{\downarrow}$ would have exactly the same energy. Therefore, in an experi­mental spectrum, they would show up as a single line. But with the magnetic field, the two energy levels are different,

$$E_{100{\downarrow}} = E_1 - \frac{e\hbar}{2m_{\rm e}}{\cal... ...2m_{\rm e}}{\cal B}_{\rm ext} \qquad E_1 = - 13.6\;{\rm eV}$$

so the single line splits into two! Do note that the energy change due to even an extremely strong magnetic field of 100 Tesla is only 0.006 eV or so, chapter 13.4, so it is not like the spectrum would become unrecog­nizable. The single spectral line of the eight $\psi_{2lm}{\updownarrow}$ “L” shell states will similarly split in five closely spaced but separate lines, corre­sponding to the five possible values $\vphantom0\raisebox{1.5pt}{$-$}$2, $\vphantom0\raisebox{1.5pt}{$-$}$1, 0, 1 and 2 for the factor $m+2m_s$ above.

Some disclaimers should be given here. First of all, the 2 in $m+2m_s$ is only equal to 2 up to about 0.1% accuracy. More importantly, even in the absence of a magnetic field, the energy levels at a given value of $n$ do not really form a single line in the spectrum if you look closely enough. There are small errors in the solution of chapter 4.3 due to relativistic effects, and so the theoretical lines are already split. That is discussed in addendum {A.38}. The description given above is a good one for the “strong” Zeeman effect, in which the magnetic field is strong enough to swamp the relativistic errors.

A.37.5 The Stark effect

If an hydrogen atom is placed in an external electric field $\skew3\vec{\cal E}_{\rm {ext}}$ instead of the magnetic one of the previous subsection, its energy levels will change too. That is called the “Stark effect.” Of course a Zeeman, Dutch for sea-man, would be most inter­ested in magnetic fields. A Stark, maybe in a spark? (Apologies.)

If the $z$-axis is taken in the direction of the electric field, the contribution of the electric field to the Hamiltonian is given by:

$$H_1 = e{\cal E}_{\rm ext}z$$ (A.246)

It is much like the potential energy $mgh$ of gravity, with the electron charge $e$ taking the place of the mass $m$, ${\cal E}_{\rm {ext}}$ that of the gravity strength $g$, and $z$ that of the height $h$.

Since the typical magnitude of $z$ is of the order of a Bohr radius $a_0$, you would expect that the energy levels will change due to the electric field by an amount of rough size $e{\cal E}_{\rm {ext}}a_0$. A strong laboratory electric field might have $e{\cal E}_{\rm {ext}}a_0$ of the order of 0.000,5 eV, [24, p. 339]. That is really small compared to the typical electron energy levels.

And addi­tionally, it turns out that for many eigen­functions, including the ground state, the first order correction to the energy is zero. To get the energy change in that case, you need to compute the second order term, which is a pain. And that term will be much smaller still than even $e{\cal E}_{\rm {ext}}a_0$ for reasonable field strengths.

Now first suppose that you ignore the warnings on good eigen­functions, and just compute the energy changes using the inner product $\langle\psi_{nlm}{\updownarrow}\vert H_1\psi_{nlm}{\updownarrow}\rangle$. You will then find that this inner product is zero for whatever energy eigen­function you take:

$$\langle\psi_{nlm}{\updownarrow}\vert e{\cal E}_{\rm ext}z\... ...ownarrow}\rangle=0 \mbox{ for all $n$, $l$, $m$, and $m_s$}$$

The reason is that negative $z$ values integrate away against positive ones. (The inner products are integrals of $z$ times $\vert\psi_{nlm}\vert^2$, and $\vert\psi_{nlm}\vert^2$ is the same at opposite sides of the nucleus while $z$ changes sign, so the contributions of opposite sides to the inner product pairwise cancel.)

So, since all first order energy changes that you compute are zero, you would naturally conclude that to first order approxi­mation none of the energy levels of a hydrogen atom changes due to the electric field. But that conclusion is wrong for anything but the ground state energy. And the reason it is wrong is because the good eigen­functions have not been used.

Consider the operators $\L ^2$, $\L _z$, and $S_z$ that make the energy eigen­functions $\psi_{nlm}{\updownarrow}$ unique. If $H_1$ $\vphantom0\raisebox{1.5pt}{$=$}$ $e{\cal E}_{\rm {ext}}z$ commuted with them all, the $\psi_{nlm}{\updownarrow}$ would be good eigen­functions. Unfortunately, while $z$ commutes with $\L _z$ and $S_z$, it does not commute with $\L ^2$, see chapter 4.5.4. The quantum number $l$ is bad.

Still, the two states $\psi_{100}{\updownarrow}$ with the ground state energy are good states, because there are no states with the same energy and a different value of the bad quantum number $l$. Really, spin has nothing to do with the Stark problem. If you want, you can find the purely spatial energy eigen­functions first, then for every spatial eigen­function, there will be one like that with spin up and one with spin down. In any case, since the two eigen­functions $\psi_{100}{\updownarrow}$ are both good, the ground state energy does indeed not change to first order.

But now consider the eight-fold degenerate $n$ $\vphantom0\raisebox{1.5pt}{$=$}$ 2 energy level. Each of the four eigen­functions $\psi_{211}{\updownarrow}$ and $\psi_{21{-1}}{\updownarrow}$ is a good one because for each of them, there is no other $n$ $\vphantom0\raisebox{1.5pt}{$=$}$ 2 eigen­function with a different value of the bad quantum number $l$. The energies corre­sponding to these good eigen­functions too do indeed not change to first order.

However, the remaining two $n$ $\vphantom0\raisebox{1.5pt}{$=$}$ 2 spin-up states $\psi_{200}{\uparrow}$ and $\psi_{210}{\uparrow}$ have different values for the bad quantum number $l$, and they have the same values $m$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0 and $m_s$ $\vphantom0\raisebox{1.5pt}{$=$}$ ${\textstyle\frac{1}{2}}$ for the good quantum numbers of orbital and spin $z$-momentum. These eigen­functions are bad and will have to be combined to produce good ones. And similarly the remaining two spin-down states $\psi_{200}{\downarrow}$ and $\psi_{210}{\downarrow}$ are bad and will have to be combined.

It suffices to just analyze the spin up states, because the spin down ones go exactly the same way. The coefficients $c_1$ and $c_2$ of the good combin­ations $c_1\psi_{200}{\uparrow}+c_2\psi_{210}{\uparrow}$ must be eigen­vectors of the matrix

$$\left( \begin{array}{cc} \langle\psi_{200}{\uparrow}\v... ...angle \end{array} \right) \qquad H_1=e{\cal E}_{\rm ext}z$$

The “diagonal” elements of this matrix (top left corner and bottom right corner) are zero because of cancell­ation of negative and positive $z$ values as discussed above. And the top right and bottom left elements are complex conjugates, (2.16), so only one of them needs to be actually computed. And the spin part of the inner product produces one and can therefore be ignored. What is left is a matter of finding the two spatial eigen­functions involved according to (4.36), looking up the spherical harmonics in table 4.2 and the radial functions in table 4.4, and integrating it all against $e{\cal E}_{\rm {ext}}z$. The resulting matrix is

$$\left( \begin{array}{cc} 0 & - 3 e{\cal E}_{\rm ext}... ... - 3 e{\cal E}_{\rm ext}a_0 & 0 \end{array} \right)$$

The eigen­vectors of this matrix are simple enough to guess; they have either equal or opposite coefficients $c_1$ and $c_2$:

$$\begin{eqnarray*} & \displaystyle \left( \begin{array}{cc} 0 & - 3 e{\... ...{array}{c} \sqrt{\frac12} \\ -\sqrt{\frac12} \end{array}\right) \end{eqnarray*}$$

If you want to check these expressions, note that the product of a matrix times a vector is found by taking dot products between the rows of the matrix and the vector. It follows that the good combin­ation $\sqrt{\frac12}\psi_{200}{\uparrow}+\sqrt{\frac12}\psi_{210}{\uparrow}$ has a first order energy change $-3e{\cal E}_{\rm {ext}}a_0$, and the good combin­ation $\sqrt{\frac12}\psi_{200}{\uparrow}-\sqrt{\frac12}\psi_{210}{\uparrow}$ has $+3e{\cal E}_{\rm {ext}}a_0$. The same applies for the spin down states. It follows that to first order the $n$ $\vphantom0\raisebox{1.5pt}{$=$}$ 2 level splits into three, with energies $E_2-3e{\cal E}_{\rm {ext}}a_0$, $E_2$, and $E_2+3e{\cal E}_{\rm {ext}}a_0$, where the value $E_2$ applies to the eigen­functions $\psi_{211}{\updownarrow}$ and $\psi_{21{-1}}{\updownarrow}$ that were already good. The conclusion, based on the wrong eigen­functions, that the energy levels do not change was all wrong.

Remarkably, the good combin­ations of $\psi_{200}$ and $\psi_{210}$ are the “sp” hybrids of carbon fame, as described in chapter 5.11.4. Note from figure 5.13 in that section that these hybrids do not have the same magnitude at opposite sides of the nucleus. They have an intrinsic “electric dipole moment,” with the charge shifted towards one side of the atom, and the electron then wants to align this dipole moment with the ambient electric field. That is much like in Zeeman splitting, where electron wants to align its orbital and spin magnetic dipole moments with the ambient magnetic field.

The crucial thing to take away from all this is: always, always, check whether the eigen­function is good before applying perturb­ation theory.

It is obviously somewhat disap­pointing that perturb­ation theory did not give any infor­mation about the energy change of the ground state beyond the fact that it is second order, i.e. very small compared to $e{\cal E}_{\rm {ext}}a_0$. You would like to know approximately what it is, not just that it is very small. Of course, now that it is established that $\psi_{100}{\uparrow}$ is a good state with $m$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0 and $m_s$ $\vphantom0\raisebox{1.5pt}{$=$}$ ${\textstyle\frac{1}{2}}$, you could think about evaluating the second order energy change (A.241), by integrating $\langle\psi_{100}{\uparrow}\vert e{\cal E}_{\rm {ext}}z\psi_{nl0}{\uparrow}\rangle$ for all values of $n$ and $l$. But after refreshing your memory about the analytical expression (D.8) for the $\psi_{nlm}$, you might think again.

It is however possible to find the perturb­ation in the wave function from the alternate approach (A.243), {D.81}. In that way the second order ground state energy is found to be

$$E_{100} = E_1 - \frac{3e{\cal E}_{\rm {ext}}a_0}{8\vert E_... ...rt} 3e{\cal E}_{\rm {ext}}a_0 \qquad E_1 = - 13.6\;{\rm eV}$$

Note that the atom likes an electric field: it lowers its ground state energy. Also note that the energy change is indeed second order; it is propor­tional to the square of the electric field strength. You can think of the attraction of the atom to the electric field as a two-stage process: first the electric field polarizes the atom by distorting its initially symmetric charge distribution. Then it inter­acts with this polarized atom in much the same way that it inter­acts with the sp hybrids. But since the polar­ization is now only propor­tional to the field strength, the net energy drop is propor­tional to the square of the field strength.

Finally, note that the typical value of 0.000,5 eV or so for $e{\cal E}_{\rm {ext}}a_0$ quoted earlier is very small compared to the about 100 eV for $8\vert E_1\vert$, making the fraction in the expression above very small. So, indeed the second order change in the ground state energy $E_1$ is much smaller than the first order energy changes $\pm3e{\cal E}_{\rm {ext}}a_0$ in the $E_2$ energy level.

A weird prediction of quantum mechanics is that the electron will eventually escape from the atom, leaving it ionized. The reason is that the potential is linear in $z$, so if the electron goes out far enough in the $z$-direction, it will eventually encounter potential energies that are lower than the one it has in the atom. Of course, to get at such large values of $z$, the electron must pass positions where the required energy far exceeds the $\vphantom0\raisebox{1.5pt}{$-$}$13.6 eV it has available, and that is impossible for a classical particle. However, in quantum mechanics the position of the electron is uncertain, and the electron does have some miniscule chance of “tunneling out” of the atom through the energy barrier, chapter 7.12.2. Realisti­cally, though, for even strong experi­mental fields like the one mentioned above, the “life time” of the electron in the atom before it has a decent chance of being found outside it far exceeds the age of the universe.