4.3 The Hydrogen Atom

This section examines the criti­cally important case of the hydrogen atom. The hydrogen atom consists of a nucleus which is just a single proton, and an electron encircling that nucleus. The nucleus, being much heavier than the electron, can be assumed to be at rest, and only the motion of the electron is of concern.

The energy levels of the electron determine the photons that the atom will absorb or emit, allowing the powerful scientific tool of spectral analysis. The electro­nic structure is also essential for under­standing the properties of the other elements and of chemical bonds.

4.3.1 The Hamiltonian

The first step is to find the Hamiltonian of the electron. The electron experi­ences an electro­static Coulomb attraction to the oppositely charged nucleus. The corre­sponding potential energy is

$$V = -\frac{e^2}{4\pi\epsilon_0 r}$$ (4.28)

with $r$ the distance from the nucleus. The constant

$$e=1.6\;10^{-19} \mbox{ C}$$ (4.29)

is the magnitude of the electric charges of the electron and proton, and the constant

$$\epsilon_0=8.85\;10^{-12} \mbox{ C$^2$/J m}$$ (4.30)

is called the “permit­tivity of space.”

Unlike for the harmonic oscillator discussed earlier, this potential energy cannot be split into separate parts for Cartesian coordinates $x$, $y$, and $z$. To do the analysis for the hydrogen atom, you must put the nucleus at the origin of the coordinate system and use spherical coordinates $r$ (the distance from the nucleus), $\theta$ (the angle from an arbitrarily chosen $z$-axis), and $\phi$ (the angle around the $z$-axis); see figure 4.7. In terms of spherical coordinates, the potential energy above depends on just the single coordinate $r$.

To get the Hamiltonian, you need to add to this potential energy the kinetic energy operator ${\widehat T}$. Chapter 3.3 gave this operator as

$${\widehat T}= - \frac{\hbar^2}{2m} \nabla^2$$

where $\nabla^2$ is the Laplacian. The Laplacian in spherical coordinates is given in the notations, (N.5). Then the Hamiltonian is found to be:

$$H = - \frac{\hbar^2}{2m_{\rm e}r^2} \left\{ \frac{\p... ...l \phi^2} \right\} - \frac{e^2}{4\pi\epsilon_0}\frac1r %$$ (4.31)

where

$$m_{\rm e}=9.109\;10^{-31} \mbox{ kg}$$ (4.32)

is the mass of the electron.

It may be noted that the small proton motion can be corrected for by slightly adjusting the mass of the electron to be an effective 9.104,4 10$\POW9,{-31}$ kg, {A.5}. This makes the solution exact, except for extremely small errors due to relativistic effects. (These are discussed in addendum {A.38}.)

Key Points

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To analyze the hydrogen atom, you must use spherical coordinates.

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The Hamiltonian in spherical coordinates has been written down. It is (4.31).

4.3.2 Solution using separation of variables

This subsection describes in general lines how the eigenvalue problem for the electron of the hydrogen atom is solved. The basic ideas are like those used to solve the particle in a pipe and the harmonic oscillator, but in this case, they are used in spherical coordinates rather than Cartesian ones. Without getting too much caught up in the mathematical details, do not miss the opportunity of learning where the hydrogen energy eigen­functions and eigen­values come from. This is the crown jewel of quantum mechanics; brilliant, almost flawless, criti­cally important; one of the greatest works of physical analysis ever.

The eigenvalue problem for the Hamiltonian, as formulated in the previous subsection, can be solved by searching for solutions $\psi$ that take the form of a product of functions of each of the three coordinates: $\psi$ $\vphantom0\raisebox{1.5pt}{$=$}$ $R(r)\Theta(\theta)\Phi(\phi)$. More concisely, $\psi$ $\vphantom0\raisebox{1.5pt}{$=$}$ $R\Theta\Phi$. The problem now is to find separate equations for the individual functions $R$, $\Theta$, and $\Phi$ from which they can then be identified. The arguments are similar as for the harmonic oscillator, but messier, since the coordinates are more entangled. First, substituting $\psi$ $\vphantom0\raisebox{1.5pt}{$=$}$ $R\Theta\Phi$ into the Hamiltonian eigenvalue problem $H\psi$ $\vphantom0\raisebox{1.5pt}{$=$}$ $E\psi$, with the Hamiltonian $H$ as given in the previous subsection and $E$ the energy eigenvalue, produces:

$$\begin{array}{r} \displaystyle \left[ - \frac{\hbar^... ...\Phi\; \\ [15pt] \displaystyle= E R\Theta\Phi \end{array}$$

To reduce this problem, premul­tiply by $2{m_{\rm e}}r^2$$\raisebox{.5pt}{$/$}$$R\Theta\Phi$ and then separate the various terms:

$$\begin{array}[b]{r} \displaystyle - \frac{\hbar^2}{R} ... ...}{4\pi\epsilon_0}\frac1r = 2m_{\rm e}r^2 E \end{array} %$$ (4.33)

Next identify the terms involving the angular derivatives and name them $E_{\theta\phi}$. They are:

$$\frac{1}{\Theta\Phi} \left[ -\frac{\hbar^2}{\sin\theta... ...2}{\partial \phi^2} \right] \Theta\Phi = E_{\theta\phi}$$

By this definition, $E_{\theta\phi}$ only depends on $\theta$ and $\phi$, not $r$. But it cannot depend on $\theta$ or $\phi$ either, since none of the other terms in the original equation (4.33) depends on them. So $E_{\theta\phi}$ must be a constant, independent of all three coordinates. Then multi­plying the angular terms above by $\Theta\Phi$ produces a reduced eigenvalue problem involving $\Theta\Phi$ only, with eigenvalue $E_{\theta\phi}$:

$$\left[ -\frac{\hbar^2}{\sin\theta} \frac{\partial}{\pa... ...hi^2} \right] \Theta\Phi = E_{\theta\phi} \Theta\Phi %$$ (4.34)

Repeat the game with this reduced eigenvalue problem. Multiply by $\sin^2\theta$$\raisebox{.5pt}{$/$}$$\Theta\Phi$, and name the only $\phi$-dependent term $E_\phi$. It is:

$$- \frac{1}{\Phi} \hbar^2 \left( \frac{\partial^2}{\partial\phi^2} \right) \Phi = E_\phi$$

By definition $E_\phi$ only depends on $\phi$, but since the other two terms in the equation it came from did not depend on $\phi$, $E_\phi$ cannot either, so it must be another constant. What is left is a simple eigenvalue problem just involving $\Phi$:

$$- \hbar^2 \left( \frac{\partial^2}{\partial\phi^2} \right) \Phi = E_\phi \Phi$$

And that is readily solvable.

In fact, the solution to this final problem has already been given, since the operator involved is just the square of the angular momentum operator $\L _z$ of section 4.2.2:

$$- \hbar^2 \left( \frac{\partial^2}{\partial\phi^2} \... ...\partial}{\partial\phi} \right)^2 \Phi = \L _z^2 \Phi$$

So this equation must have the same eigen­functions as the operator $\L _z$,

$$\Phi_m=e^{{\rm i}m\phi}$$

and must have the square eigen­values

$$E_\phi=(m\hbar)^2$$

(each appli­cation of $\L _z$ multi­plies the eigen­function by $m\hbar$). It may be recalled that the magnetic quantum number $m$ must be an integer.

The eigenvalue problem (4.34) for $\Theta\Phi$ is even easier; it is exactly the one for the square angular momentum $L^2$ of section 4.2.3. (So, no, there was not really a need to solve for $\Phi$ separately.) Its eigen­functions are therefore the spherical harmonics,

$$\Theta\Phi = Y_l^m(\theta,\phi)$$

and its eigen­values are

$$E_{\theta\phi} = l (l+1) \hbar^2$$

It may be recalled that the azimuthal quantum number $l$ must be an integer greater than or equal to $\vert m\vert$.

Returning now to the solution of the original eigenvalue problem (4.33), replacement of the angular terms by $E_{\theta\phi}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $l(l+1)\hbar^2$ turns it into an ordinary differen­tial equation problem for the radial factor $R(r)$ in the energy eigen­function. As usual, this problem is a pain to solve, so that is again shoved away in a note, {D.15}.

It turns out that the solutions of the radial problem can be numbered using a third quantum number, $n$, called the “principal quantum number”. It is larger than the azimuthal quantum number $l$, which in turn must be at least as large as the absolute value of the magnetic quantum number:

$$\fbox{$\displaystyle n > l \mathrel{\raisebox{-1pt}{$\geqslant$}}\vert m\vert $}$$ (4.35)

so the principal quantum number must be at least 1. And if $n$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1, then $l$ $\vphantom0\raisebox{1.5pt}{$=$}$ $m$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0.

In terms of these three quantum numbers, the final energy eigen­functions of the hydrogen atom are of the general form:

$$\fbox{$\displaystyle \psi_{nlm} = R_{nl}(r) Y_l^m(\theta,\phi) $} %$$ (4.36)

where the spherical harmonics $Y_l^m$ were described in section 4.2.3. The brand new radial wave functions $R_{nl}$ can be found written out in table 4.4 for small values of $n$ and $l$, or in derivation {D.15}, (D.8), for any $n$ and $l$. They are usually written in terms of a scaled radial distance from the nucleus $\rho$ $\vphantom0\raisebox{1.5pt}{$=$}$ $r$$\raisebox{.5pt}{$/$}$$a_0$, where the length $a_0$ is called the “Bohr radius” and has the value

$$\fbox{$\displaystyle a_0=\frac{4\pi\epsilon_0\hbar^2}{m_{\rm e}e^2} \approx 0.529{,}177\;10^{-10}\mbox{ m} $} %$$ (4.37)

or about half an Ångstrom. The Bohr radius is a really good length scale to describe atoms in terms of. The Ångstrom itself is a good choice too, it is 10$\POW9,{-10}$ m, or one tenth of a nanometer.

Table 4.4: The first few radial wave functions for hydrogen.

The energy eigen­values are much simpler and more inter­esting than the eigen­functions; they are

$$\fbox{$\displaystyle E_n = - \frac{\hbar^2}{2 m_{\rm e... ...{\hbar^2}{2 m_{\rm e}a_0^2} = - 13.605{,}7 \mbox{ eV} $} %$$ (4.38)

where eV stands for electron volt, a unit of energy equal to 1.602,18 10$\POW9,{-19}$ J. It is the energy that an electron picks up during a 1 volt change in electric potential.

You may wonder why the energy only depends on the principal quantum number $n$, and not also on the azimuthal quantum number $l$ and the magnetic quantum number $m$. Well, the choice of $z$-axis was arbitrary, so it should not seem that strange that the physics would not depend on the angular momentum in that direction. But that the energy does not depend on $l$ is non­trivial: if you solve the simpler problem of a particle stuck inside an impenetrable spherical container, using procedures from {A.6}, the energy values depend on both $n$ and $l$. So, that is just the way it is. (It stops being true anyway if you include relativistic effects in the Hamiltonian.)

Since the lowest possible value of the principal quantum number $n$ is one, the ground state of lowest energy $E_1$ is eigen­function $\psi_{100}$.

Key Points

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Skipping a lot of math, energy eigen­functions $\psi_{nlm}$ and their energy eigen­values $E_n$ have been found.

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There is one eigen­function for each set of three integer quantum numbers $n$, $l$, and $m$ satisfying $n$ $\raisebox{.3pt}{$>$}$ $l$ $\raisebox{-.5pt}{$\geqslant$}$ $\vert m\vert$. The number $n$ is called the principal quantum number.

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The typical length scale in the solution is called the Bohr radius $a_0$, which is about half an Ångstrom.

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The derived eigen­functions $\psi_{nlm}$ are eigen­functions of

• $z$ angular momentum, with eigenvalue $L_z$ $\vphantom0\raisebox{1.5pt}{$=$}$ $m\hbar$;
• square angular momentum, with eigenvalue $L^2$ $\vphantom0\raisebox{1.5pt}{$=$}$ $l(l+1)\hbar^2$;
• energy, with eigenvalue $E_n$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\vphantom0\raisebox{1.5pt}{$-$}$$\hbar^2$$\raisebox{.5pt}{$/$}$$2{m_{\rm e}}a_0^2n^2$.

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The energy values only depend on the principal quantum number $n$.

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The ground state is $\psi_{100}$.

4.3.2 Review Questions

1. Use the tables for the radial wave functions and the spherical harmonics to write down the wave function
   
   $$\psi_{nlm} = R_{nl}(r) Y_l^m(\theta ,\phi)$$
   
   
   for the case of the ground state $\psi_{100}$.

   Check that the state is normalized. Note: $\int_0^{\infty}e^{-2u}u^2{\,\rm d}{u}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\frac 14$.

   Solution hydb-a

2. Use the generic expression
   
   $$\psi_{nlm} = -\frac{2}{n^2} \sqrt{\frac{(n-l-1)!}{[(n+l)!a_0... ...+1}\left(\frac{2\rho}n\right) e^{-\rho /n} Y_l^m(\theta ,\phi)$$
   
   
   with $\rho$ $\vphantom0\raisebox{1.5pt}{$=$}$ $r$$\raisebox{.5pt}{$/$}$$a_0$ and $Y_l^m$ from the spherical harmonics table to find the ground state wave function $\psi_{100}$. Note: the Laguerre polynomial $L_1(x)$ $\vphantom0\raisebox{1.5pt}{$=$}$ $1-x$ and for any $p$, $L_1^p$ is just its $p$-th derivative.

   Solution hydb-b

3. Plug numbers into the generic expression for the energy eigen­values,
   
   $$E_n = - \frac{\hbar^2}{2m_{\rm e}a_0^2} \frac 1{n^2},$$
   
   
   where $a_0$ $\vphantom0\raisebox{1.5pt}{$=$}$ $4\pi\epsilon_0\hbar^2$$\raisebox{.5pt}{$/$}$${m_{\rm e}}e^2$, to find the ground state energy. Express in eV, where 1 eV equals 1.602,2 10$\POW9,{-19}$ J. Values for the physical constants can be found at the start of this section and in the notations section.

   Solution hydb-c

4.3.3 Discussion of the eigenvalues

The only energy values that the electron in the hydrogen atom can have are the “Bohr energies” derived in the previous subsection:

$$E_n = - \frac{\hbar^2}{2 m_{\rm e}a_0^2} \frac1{n^2} \qquad n = 1, 2, 3, \ldots$$

This subsection discusses the physical consequences of this result.

Figure 4.8: Spectrum of the hydrogen atom.

To aid the discussion, the allowed energies are plotted in the form of an energy spectrum in figure 4.8. To the right of the lowest three energy levels the values of the quantum numbers that give rise to those energy levels are listed.

The first thing that the energy spectrum illustrates is that the energy levels are all negative, unlike the ones of the harmonic oscillator, which were all positive. However, that does not mean much; it results from defining the potential energy of the harmonic oscillator to be zero at the nominal position of the particle, while the hydrogen potential is instead defined to be zero at large distance from the nucleus. (It will be shown later, chapter 7.2, that the average potential energy is twice the value of the total energy, and the average kinetic energy is minus the total energy, making the average kinetic energy positive as it should be.)

A more profound difference is that the energy levels of the hydrogen atom have a maximum value, namely zero, while those of the harmonic oscillator went all the way to infinity. It means physi­cally that while the particle can never escape in a harmonic oscillator, in a hydrogen atom, the electron escapes if its total energy is greater than zero. Such a loss of the electron is called “ionization” of the atom.

There is again a ground state of lowest energy; it has total energy

$$E_1=-13.6 \mbox{ eV} %$$ (4.39)

(an eV or “electron volt” is 1.6 10$\POW9,{-19}$ J). The ground state is the state in which the hydrogen atom will be at absolute zero temperature. In fact, it will still be in the ground state at room temperature, since even then the energy of heat motion is unlikely to raise the energy level of the electron to the next higher one, $E_2$.

The ionization energy of the hydrogen atom is 13.6 eV; this is the minimum amount of energy that must be added to raise the electron from the ground state to the state of a free electron.

If the electron is excited from the ground state to a higher but still bound energy level, (maybe by passing a spark through hydrogen gas), it will in time again transition back to a lower energy level. Discussion of the reasons and the time evolution of this process will have to wait until chapter 7. For now, it can be pointed out that different transitions are possible, as indicated by the arrows in figure 4.8. They are named by their final energy level to be Lyman, Balmer, or Paschen series transitions.

The energy lost by the electron during a transition is emitted as a quantum of electro­magnetic radiation called a photon. The most energetic photons, in the ultra­violet range, are emitted by Lyman transitions. Balmer transitions emit visible light and Paschen ones infrared.

The photons emitted by isolated atoms at rest must have an energy very precisely equal to the difference in energy eigen­values; anything else would violate the requirement of the orthodox inter­pretation that only the eigen­values are observable. And according to the “Planck-Einstein relation,” the photon’s energy equals the angular frequency $\omega$ of its electro­magnetic vibration times $\hbar$:

$$E_{n_1}-E_{n_2} = \hbar \omega.$$

Thus the spectrum of the light emitted by hydrogen atoms is very distinctive and can be identified to great accuracy. Different elements have different spectra, and so do molecules. It all allows atoms and molecules to be correctly recognized in a lab or out in space.

(To be sure, the spectral frequencies are not truly mathemati­cally exact numbers. A slight “spectral broadening” is unavoidable because no atom is truly isolated as assumed here; there is always some radiation that perturbs it even in the most ideal empty space. In addition, thermal motion of the atom causes Doppler shifts. In short, only the energy eigen­values are observable, but exactly what those eigen­values are for a real-life atom can vary slightly.)

Atoms and molecules may also absorb electro­magnetic energy of the same frequencies that they can emit. That allows them to enter an excited state. The excited state will eventually emit the absorbed energy again in a different direction, and possibly at different frequencies by using different transitions. In this way, in astronomy atoms can remove specific frequencies from light that passes them on its way to earth, resulting in an absorption spectrum. Or instead atoms may scatter specific frequencies of light in our direction that was originally not headed to earth, producing an emission spectrum. Doppler shifts can provide infor­mation about the thermal and average motion of the atoms. Since hydrogen is so prevalent in the universe, its energy levels as derived here are particularly important in astronomy. Chapter 7 will address the mechanisms of emission and absorption in much greater detail.

Key Points

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The energy levels of the electron in a hydrogen atom have a highest value. This energy is by convention taken to be the zero level.

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The ground state has a energy 13.6 eV below this zero level.

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If the electron in the ground state is given an additional amount of energy that exceeds the 13.6 eV, it has enough energy to escape from the nucleus. This is called ionization of the atom.

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If the electron transitions from a bound energy state with a higher principal quantum number $n_1$ to a lower one $n_2$, it emits radiation with an angular frequency $\omega$ given by

$$\hbar \omega = E_{n_1}-E_{n_2}$$

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Similarly, atoms with energy $E_{n_2}$ may absorb electro­magnetic energy of such a frequency.

4.3.3 Review Questions

1. If there are infinitely many energy levels $E_1$, $E_2$, $E_3$, $E_4$, $E_5$, $E_6$, ..., where did they all go in the energy spectrum?

   Solution hydc-a

2. What is the value of energy level $E_2$? And $E_3$?

   Solution hydc-b

3. Based on the results of the previous question, what is the color of the light emitted in a Balmer transition from energy $E_3$ to $E_2$? The Planck-Einstein relation says that the angular frequency $\omega$ of the emitted photon is its energy divided by $\hbar$, and the wave length of light is $2{\pi}c$$\raisebox{.5pt}{$/$}$$\omega$ where $c$ is the speed of light. Typical wave lengths of visible light are: violet 400 nm, indigo 445 nm, blue 475 nm, green 510 nm, yellow 570 nm, orange 590 nm, red 650 nm.

   Solution hydc-c

4. What is the color of the light emitted in a Balmer transition from an energy level $E_n$ with a high value of $n$ to $E_2$?

   Solution hydc-d

4.3.4 Discussion of the eigenfunctions

The appearance of the energy eigen­states will be of great inter­est in under­standing the heavier elements and chemical bonds. This subsection describes the most important of them.

It may be recalled from subsection 4.3.2 that there is one eigen­function $\psi_{nlm}$ for each set of three integer quantum numbers. They are the principal quantum number $n$ (determining the energy of the state), the azimuthal quantum number $l$ (determining the square angular momentum), and the magnetic quantum number $m$ (determining the angular momentum in the chosen $z$-direction.) They must satisfy the requirements that

$$n > l \mathrel{\raisebox{-1pt}{$\geqslant$}}\vert m\vert$$

For the ground state, with the lowest energy $E_1$, $n$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1 and hence according to the conditions above both $l$ and $m$ must be zero. So the ground state eigen­function is $\psi_{100}$; it is unique.

The expression for the wave function of the ground state is (from the results of subsection 4.3.2):

$$\psi_{100}(r) = \frac{1}{\sqrt{\pi a_0^3}} e^{-r/a_0} %$$ (4.40)

where $a_0$ is called the “Bohr radius”,

$$a_0 = \frac{4\pi\epsilon_0\hbar^2}{m_{\rm e}e^2} = 0.53 \times 10^{-10} \mbox{ m}$$ (4.41)

The square magnitude of the energy states will again be displayed as grey tones, darker regions corre­sponding to regions where the electron is more likely to be found. The ground state is shown this way in figure 4.9; the electron may be found within a blob size that is about thrice the Bohr radius, or roughly an Ångstrom, (10$\POW9,{-10}$ m), in diameter.

Figure 4.9: Ground state wave function of the hydrogen atom.

It is the quantum mechanical refusal of electrons to restrict themselves to a single location that gives atoms their size. If Planck's constant $\hbar$ would have been zero, so would have been the Bohr radius, and the electron would have been in the nucleus. It would have been a very different world.

The ground state proba­bility distribution is spheri­cally symmetric: the proba­bility of finding the electron at a point depends on the distance from the nucleus, but not on the angular orien­tation relative to it.

The excited energy levels $E_2$, $E_3$, ...are all degenerate; as the spectrum figure 4.8 indicated, there is more than one eigenstate producing each level. Let’s have a look at the states at energy level $E_2$ now.

Figure 4.10 shows energy eigen­function $\psi_{200}$. Like $\psi_{100}$, it is spheri­cally symmetric. In fact, all eigen­functions $\psi_{n00}$ are spheri­cally symmetric. However, the wave function has blown up a lot, and now separates into a small, more or less spherical region in the center, surrounded by a second region that forms a spherical shell. Separating the two is a radius at which there is zero proba­bility of finding the electron.

Figure 4.10: Eigen­function $\psi_{200}$.

The state $\psi_{200}$ is commonly referred to as the “2s” state. The 2 indicates that it is a state with energy $E_2$. The “s” indicates that the azimuthal quantum number is zero; just think “spheri­cally symmetric.” Similarly, the ground state $\psi_{100}$ is commonly indicated as “1s”, having the lowest energy $E_1$.

States which have azimuthal quantum number $l$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1 are called “p” states, for some historical reason. Histori­cally, physicists have always loved confusing and incon­sistent notations. In particular, the $\psi_{21m}$ states are called “2p” states. As first example of such a state, figure 4.11 shows $\psi_{210}$. This wave function squeezes itself close to the $z$-axis, which is plotted horizontally by convention. There is zero proba­bility of finding the electron at the vertical $x,y$ symmetry plane, and maximum proba­bility at two symmetric points on the $z$-axis.

Figure 4.11: Eigen­function $\psi_{210}$, or 2p$_z$.

Since the wave function squeezes close to the $z$-axis, this state is often more specifi­cally referred to as the “2p$_z$” state. Think “points along the $z$-axis.”

Figure 4.12 shows the other two “2p” states, $\psi_{211}$ and $\psi_{21-1}$. These two states look exactly the same as far as the proba­bility density is concerned. It is somewhat hard to see in the figure, but they really take the shape of a torus around the left-to-right $z$-axis.

Figure 4.12: Eigen­function $\psi_{211}$ (and $\psi_{21-1}$).

Eigen­functions $\psi_{200}$, $\psi_{210}$, $\psi_{211}$, and $\psi_{21-1}$ are degenerate: they all four have the same energy $E_2$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\vphantom0\raisebox{1.5pt}{$-$}$3.4 eV. The consequence is that they are not unique. Combin­ations of them can be formed that have the same energy. These combin­ation states may be more important physi­cally than the original eigen­functions.

In particular, the torus-shaped eigen­functions $\psi_{211}$ and $\psi_{21-1}$ are often not very useful for descriptions of heavier elements and chemical bonds. Two states that are more likely to be relevant here are called 2p$_x$ and 2p$_y$; they are the combin­ation states:

$$\mbox{2p$_x$: } \frac1{\sqrt2}\left(-\psi_{211}+\psi_{21-1... ...: } \frac{{\rm i}}{\sqrt2}\left(\psi_{211}+\psi_{21-1}\right)$$ (4.42)

These two states are shown in figure 4.13; they look exactly like the “pointer” state 2p$_z$ of figure 4.11, except that they squeeze along the $x$-axis, respectively the $y$-axis, instead of along the $z$-axis. (Since the $y$-axis is pointing towards you, 2p$_y$ looks rota­tionally symmetric. Seen from the side, it would look like p$_z$ in figure 4.11.)

Figure 4.13: Eigen­functions 2p$_x$, left, and 2p$_y$, right.

Note that unlike the two original states $\psi_{211}$ and $\psi_{21-1}$, the states 2p$_x$ and 2p$_y$ do not have a definite value of the $z$-component of angular momentum; the $z$-component has a 50/50 uncertainty of being either $+\hbar$ or $\vphantom0\raisebox{1.5pt}{$-$}$$\hbar$. But that is not important in most circumstances. What is important is that when multiple electrons occupy the p states, mutual repulsion effects tend to push them into the p$_x$, p$_y$, and p$_z$ states.

So, the four independent eigen­functions at energy level $E_2$ are best thought of as consisting of one spheri­cally symmetrical 2s state, and three direc­tional states, 2p$_x$, 2p$_y$, and 2p$_z$, pointing along the three coordinate axes.

But even that is not always ideal; as discussed in chapter 5.11.4, for many chemical bonds, especially those involving the important element carbon, still different combin­ation states called “hybrids” show up. They involve combin­ations of the 2s and the 2p states and therefore have uncertain square angular momentum as well.

Key Points

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The typical size of eigen­states is given by the Bohr radius, making the size of the atom of the order of an Å.

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The ground state $\psi_{100}$, or 1s state, is non­degenerate: no other set of quantum numbers $n,l,m$ produces energy $E_1$.

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All higher energy levels are degenerate, there is more than one eigenstate producing that energy.

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All states of the form $\psi_{n00}$, including the ground state, are spheri­cally symmetric, and are called s states. The ground state $\psi_{100}$ is the 1s state, $\psi_{200}$ is the 2s state, etcetera.

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States of the form $\psi_{n1m}$ are called p states. The basic 2p states are $\psi_{21-1}$, $\psi_{210}$, and $\psi_{211}$.

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The state $\psi_{210}$ is also called the 2p$_z$ state, since it squeezes itself around the $z$-axis.

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There are similar 2p$_x$ and 2p$_y$ states that squeeze around the $x$ and $y$ axes. Each is a combin­ation of $\psi_{21-1}$ and $\psi_{211}$.

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The four spatial states at the $E_2$ energy level can therefore be thought of as one spheri­cally symmetric 2s state and three 2p pointer states along the axes.

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However, since the $E_2$ energy level is degenerate, eigen­states of still different shapes are likely to show up in appli­cations.

4.3.4 Review Questions

1. At what distance $r$ from the nucleus does the square of the ground state wave function become less than one percent of its value at the nucleus? Express it both as a multiple of the Bohr radius $a_0$ and in Å.

   Solution hydd-a

2. Check from the conditions
   
   $$n > l \mathrel{\raisebox{-1pt}{$\geqslant$}}\vert m\vert$$
   
   
   that $\psi_{200}$, $\psi_{211}$, $\psi_{210}$, and $\psi_{21-1}$ are the only states of the form $\psi_{nlm}$ that have energy $E_2$. (Of course, all their combin­ations, like 2p$_x$ and 2p$_y$, have energy $E_2$ too, but they are not simply of the form $\psi_{nlm}$, but combin­ations of the “basic” solutions $\psi_{200}$, $\psi_{211}$, $\psi_{210}$, and $\psi_{21-1}$.)

   Solution hydd-b

3. Check that the states
   
   $$\mbox{2p$_x$}= \frac 1{\sqrt 2}\left(-\psi_{211}+\psi_{21-1}... ...$}= \frac{{\rm i}}{\sqrt 2}\left(\psi_{211}+\psi_{21-1}\right)$$
   
   
   are properly normalized.

   Solution hydd-c