5.3 Two-State Systems

Two-state systems are systems in which only two quantum states are of importance. That makes such systems the simplest non­trivial quantum systems. A lot of quali­tative under­standing can be obtained from them. Among others, this section will shed some light on the reason why chemical bonds tend to involve pairs of electrons.

As seen in chapter chapter 4.6, the protons in the H$_2^+$ hydrogen molecular ion are held together by a single shared electron. However, in the H$_2$ neutral hydrogen molecule of the previous section, they are held together by a shared pair of electrons. In both cases a stable bond was formed. So why are chemical bonds involving a single electron relatively rare, while bonds involving pairs of shared electrons are common?

The unifying concept relating the two bonds is that of two-state systems. Such systems involve two intuitive basic states $\psi_1$ and $\psi_2$.

For the hydrogen molecular ion, one state, $\psi_1$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\psi_{\rm {l}}$, described that the electron was in the ground state around the left proton. A physi­cally equivalent state, $\psi_2$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\psi_{\rm {r}}$, had the electron in the ground state around the right proton. For the hydrogen molecule, $\psi_1$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\psi_{\rm {l}}\psi_{\rm {r}}$ had electron 1 around the left proton and electron 2 around the right one. The other state $\psi_2$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\psi_{\rm {r}}\psi_{\rm {l}}$ was physi­cally the same, but it had the electrons reversed.

There are many other physical situations that may be described as two state systems. Covalent chemical bonds involving atoms other than hydrogen would be an obvious example. Just substitute a positive ion for one or both protons.

As another example of a two-state system, consider the C$_6$H$_6$ “benzene molecular ring.” This molecule consists of a hexagon of 6 carbon atoms that are held together by 9 covalent bonds. The logical way that 9 bonds can be arranged between the atoms of a 6 atom ring is to make every second bond a double one. However, that still leaves two possi­bilities; the locations of the single and double bonds can be swapped. So there are once again two different but equivalent states $\psi_1$ and $\psi_2$.

The NH$_3$ “ammonia molecule” consists of an nitrogen atom bonded to three hydrogen atoms. By symmetry, the logical place for the nitrogen atom to sit would surely be in the center of the triangle formed by the three hydrogen atoms. But it does not sit there. If it was in the center of the triangle, the angles between the hydrogen atoms, measured from the nitrogen nucleus, should be 120$\POW9,{\circ}$ each. However, as discussed later in chapter 5.11.3, valence bond theory requires that the angles should be about 90$\POW9,{\circ}$, not 120$\POW9,{\circ}$. (The actual angles are about 108$\POW9,{\circ}$ because of reasons similar to those for water as discussed in chapter 5.11.3.) The key point here is that the nitrogen must sit to the side of the triangle, and there are two sides, producing once again two different but equivalent physical states $\psi_1$ and $\psi_2$.

In each case described above, there are two intuitive physical states $\psi_1$ and $\psi_2$. The peculiarities of the quantum mechanics of two-state systems arise from states that are combin­ations of these two states, as in

$$\psi=c_1\psi_1+c_2\psi_2$$

Note that according to the ideas of quantum mechanics, the square magnitude of the first coefficient of the combined state, $\vert c_1\vert^2$, represents the proba­bility of being in state $\psi_1$ and $\vert c_2\vert^2$ the proba­bility of being in state $\psi_2$. Of course, the total proba­bility of being in one of the states should be one:

$$\vert c_1\vert^2+\vert c_2\vert^2=1$$

(This is only true if the $\psi_1$ and $\psi_2$ states are ortho­normal. In the hydrogen molecule cases, ortho­normalizing the basic states would change them a bit, but their physical nature would remain much the same, especially if the protons are not too close.)

The key question is now what combin­ation of states has the lowest energy. That will be the ground state $\psi_{\rm {gs}}$ of the two-state system. The expec­tation value of energy is

$$\big\langle E\big\rangle = \langle c_1\psi_1+c_2\psi_2\vert H \vert c_1\psi_1+c_2\psi_2\rangle$$

This can be multi­plied out, taking into account that numerical factors come out of the left of an inner product as complex conjugates. The result is

$$\big\langle E\big\rangle = \vert c_1\vert^2 \langle{E}_... ...12} + c_2^*c_1H_{21} + \vert c_2\vert^2 \langle{E}_2\rangle$$

using the shorthand notation

$$\langle{E}_1\rangle = \langle\psi_1\vert H\psi_1\rangle, \... ...uad \langle{E}_2\rangle = \langle\psi_2\vert H\psi_2\rangle$$

Note that $\langle{E}_1\rangle$ and $\langle{E}_2\rangle$ are real, (2.16). They are the expec­tation energies of the states $\psi_1$ and $\psi_2$. The states will be ordered so that $\langle{E}_1\rangle$ is less or equal to $\langle{E}_2\rangle$. (In all the examples mentioned so far, $\langle{E}_1\rangle$ and $\langle{E}_2\rangle$ are equal because the two states are physi­cally equivalent.) Normally, $H_{12}$ and $H_{21}$ are not real but complex conjugates, (2.16). However, you can always change the definition of, say, $\psi_1$ by a complex factor of magnitude one to make $H_{12}$ equal to a real and negative number, and then $H_{21}$ will be that same negative number.

The above expression for the expec­tation energy consists of two kinds of terms, which will be called:

$$\mbox{the averaged energy: } \vert c_1\vert^2 \langle{E}_1\rangle + \vert c_2\vert^2 \langle{E}_2\rangle$$ (5.11)

$$\mbox{the twilight terms: } (c_1^*c_2 + c_2^*c_1) H_{12}%$$ (5.12)

Each of those contributions will be discussed in turn.

The averaged energy is the energy that you would intuitively expect the combined wave function to have. It is a straight­forward sum of the expec­tation energies of the two component states $\psi_1$ and $\psi_2$ times the proba­bilities of being in those states. In particular, in the important case that the two states have the same energy, the averaged energy is that energy. What is more logical than that any mixture of two states with the same energy would have that energy too?

But the twilight terms throw a monkey wrench in this simplistic thinking. It can be seen that they will always make the ground state energy $E_{\rm {gs}}$ lower than the lowest energy of the component states $\langle{E}_1\rangle$. (To see that, just take $c_1$ and $c_2$ positive real numbers and $c_2$ small enough that $c_2^2$ can be neglected.) This lowering of the energy below the lowest component state comes out of the mathematics of combining states; absolutely no new physical forces are added to produce it. But if you try to describe it in terms of classical physics, it really looks like a mysterious new “twilight force” is in operation here. It is no new force; it is the weird mathematics of quantum mechanics.

So, what are these twilight terms physi­cally? If you mean, what are they in terms of classical physics, there is simply no answer. But if you mean, what are they in terms of normal language, rather than formulae, it is easy. Just have another look at the definition of the twilight terms; they are a measure of the inner product $\langle\psi_1\vert H\psi_2\rangle$. That is the energy you would get if nature was in state $\psi_1$ if nature was in state $\psi_2$. On quantum scales, nature can get really, really ethereal, where it moves beyond being describable by classical physics, and the result is very concrete, but weird, inter­actions. For, at these scales twilight is real, and classical physics is not.

For the twilight terms to be non­zero, there must be a region where the two states overlap, i.e. there must be a region where both $\psi_1$ and $\psi_2$ are non­zero. In the simplest case of the hydrogen molecular ion, if the atoms are far apart, the left and right wave functions do not overlap and the twilight terms will be zero. For the hydrogen molecule, it gets a bit less intuitive, since the overlap should really be visualized in the six-di­mensional space of those functions. But still, the terms are zero when the atoms are far apart.

The twilight terms are customarily referred to as “exchange terms,” but everybody seems to have a different idea of what that is supposed to mean. The reason may be that these terms pop up all over the place, in all sorts of very different settings. This book prefers to call them twilight terms, since that most clearly expresses what they really are. Nature is in a twilight zone of ambiguity.

The lowering of the energy by the twilight terms produces more stable chemical bonds than you would expect. Typically, the effect of the terms is greatest if the two basic states $\psi_1$ and $\psi_2$ are physi­cally equivalent, like for the mentioned examples. Then the two states have the same expec­tation energy, call it $\langle{E}\rangle_{1,2}$. For such symmetric systems, the ground state will occur for an equal mixture of the two states, $c_1$ $\vphantom0\raisebox{1.5pt}{$=$}$ $c_2$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\sqrt{\frac12}$, because then the twilight terms are most negative. (Complex coefficients do not really make a physical difference, so $c_1$ and $c_2$ can be assumed to be real numbers for convenience.) In the ground state, the lowest energy is then an amount $\left\vert H_{12}\right\vert$ below the energy of the component states:

$$\fbox{$\displaystyle \mbox{Symmetric 2-state systems:} ... ...rm{gs}} = \langle{E}\rangle_{1,2} - \vert H_{12}\vert $} %$$ (5.13)

On the other hand, if the lower energy state $\psi_1$ has significantly less energy than state $\psi_2$, then the minimum energy will occur near the lower energy state. That means that $\vert c_1\vert$ $\vphantom0\raisebox{1.1pt}{$\approx$}$ 1 and $\vert c_2\vert$ $\vphantom0\raisebox{1.1pt}{$\approx$}$ 0. (This assumes that the twilight terms are not big enough to dominate the energy.) In that case $c_1c_2$ $\vphantom0\raisebox{1.1pt}{$\approx$}$ 0 in the twilight terms (5.12), which pretty much takes the terms out of the picture completely.

This happens for the single-electron bond of the hydrogen molecular ion if the second proton is replaced by another ion, say a lithium ion. The energy in state $\psi_1$, where the electron is around the proton, will now be significantly less than that of state $\psi_2$, where it is around the lithium ion. For such asymmetrical single-electron bonds, the twilight terms are not likely to help much in forging a strong bond. While it turns out that the LiH$\POW9,{+}$ ion is stable, the binding energy is only 0.14 eV or so, compared to 2.8 eV for the H$_2^+$ ion. Also, the LiH$\POW9,{+}$ bond seems to be best described as a Van der Waals attraction, rather than a true chemical bond.

In contrast, for the two-electron bond of the neutral hydrogen molecule, if the second proton is replaced by a lithium ion, states $\psi_1$ and $\psi_2$ will still be the same: both states will have one electron around the proton and one around the lithium ion. The two states do have the electrons reversed, but the electrons are identical. Thus the twilight terms are still likely to be effective. Indeed neutral LiH lithium hydride exists as a stable molecule with a binding energy of about 2.5 eV at low pressures.

(It should be noted that the LiH bond is very ionic, with the “shared” electrons mostly at the hydrogen side, so the actual ground state is quite different from the covalent hydrogen model. But the model should be better when the nuclei are farther apart, so the analysis can at least justify the existence of a significant bond.)

For the ammonia molecule, the two states $\psi_1$ and $\psi_2$ differ only in the side of the hydrogen triangle that the nitrogen atom is at. Since these two states are physi­cally equivalent, there is again a significant lowering of the energy $E_{\rm {gs}}$ for the symmetric combin­ation $c_1$ $\vphantom0\raisebox{1.5pt}{$=$}$ $c_2$. Similarly, there is a significant raising of the energy $E_{\rm {as}}$ for the anti­symmetric combin­ation $c_1$ $\vphantom0\raisebox{1.5pt}{$=$}$ $-c_2$. Transitions between these two energy states produce photons of a single energy in the microwave range. It allows a maser (microwave-range laser) to be constructed. The first maser was in fact an ammonia one. It gave rise to the subsequent development of optical-range versions. These were initially called “optical masers,” but are now known as “lasers.” Masers are important for providing a single frequency reference, like in some atomic clocks. See chapter 7.7 for the operating principle of masers and lasers.

The ammonia molecule may well be the best example of how weird these twilight effects are. Consider, there are two common-sense states in which the nitrogen is at one side of the hydrogen triangle. What physical reason could there possibly be that there is a state of lower energy in which the atom is at both sides at the same time with a 50/50 proba­bility? Before you answer that, recall that it only works if you do the 50/50 case right. If you do it wrong, you end up raising the energy. And the only way to figure out whether you do it right is to look at the behavior of the sign of a physi­cally unobservable wave function.

It may finally be noted that in the context of chemical bonds, the raised-energy anti­symmetric state is often called an “anti­bonding” state.

Key Points

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In quantum mechanics, the energy of different but physi­cally equivalent states can be lowered by mixing them together.

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This lowering of energy does not come from new physical forces, but from the weird mathematics of the wave function.

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The effect tends to be much less when the original states are physi­cally very different.

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One important place where states are indeed physi­cally the same is in chemical bonds involving pairs of electrons. Here the equivalent states merely have the identical electrons inter­changed.

5.3 Review Questions

1. The effective­ness of mixing states was already shown by the hydrogen molecule and molecular ion examples. But the gener­alized story above restricts the “basis” states to be orthogonal, and the states used in the hydrogen examples were not.

   Show that if $\psi_1$ and $\psi_2$ are not orthogonal states, but are normalized and produce a real and positive value for $\langle\psi_1\vert\psi_2\rangle$, like in the hydrogen examples, then orthogonal states can be found in the form
   

   $$\bar\psi_1 = \alpha\left(\psi_1 - \varepsilon\psi_2\right) \qquad\bar\psi_2 = \alpha\left(\psi_2 - \varepsilon\psi_1\right).$$
   
   
   For normalized $\psi_1$ and $\psi_2$ the Cauchy-Schwartz inequality says that $\langle\psi_1\vert\psi_2\rangle$ will be less than one. If the states do not overlap much, it will be much less than one and $\varepsilon$ will be small.

   (If $\psi_1$ and $\psi_2$ do not meet the stated requirements, you can always redefine them by factors $ae^{{\rm i}{c}}$ and $be^{-{\rm i}{c}}$, with $a$, $b$, and $c$ real, to get states that do.)

   Solution 2state-a

2. Show that it does not have an effect on the solution whether or not the basic states $\psi_1$ and $\psi_2$ are normalized, like in the previous question, before the state of lowest energy is found.

   This requires no detailed analysis; just check that the same solution can be described using the non­orthogonal and orthogonal basis states. It is however an important observ­ation for various numerical solution procedures: your set of basis functions can be cleaned up and simplified without affecting the solution you get.

   Solution 2state-b