5.2 The Hydrogen Molecule

This section uses similar approxi­mations as for the hydrogen molecular ion of chapter 4.6 to examine the neutral H$_2$ hydrogen molecule. This molecule has two electrons circling two protons. It will turn out that in the ground state, the protons share the two electrons, rather than each being assigned one. This is typical of covalent bonds.

Of course, “share” is a vague term, but the discussion will show what it really means in terms of the six-di­mensional electron wave function.

5.2.1 The Hamiltonian

Just like for the hydrogen molecular ion of chapter 4.6, for the neutral molecule the Born-Oppenheimer approxi­mation will be made that the protons are at given fixed points. So the problem simplifies to just finding the wave function of the two electrons, $\Psi({\skew0\vec r}_1,{\skew0\vec r}_2)$, where ${\skew0\vec r}_1$ and ${\skew0\vec r}_2$ are the positions of the two electrons 1 and 2. In terms of scalar arguments, the wave function can be written out further as $\Psi(x_1,y_1,z_1,x_2,y_2,z_2)$.

In the Hamiltonian, following the Newtonian analogy the kinetic and potential energy operators simply add:

$$H = - \frac{\hbar^2}{2m_{\rm e}} \left( \nabla_1^2 +... ...1{\vert{\skew0\vec r}_1 - {\skew0\vec r}_2\vert} \right) %$$ (5.3)

In this expression, the Laplacians of the first two, kinetic energy, terms are with respect to the position coordinates of the two electrons:

$$\nabla^2_1 = \frac{\partial^2}{\partial x_1^2} + \frac... ...ial^2}{\partial y_2^2} + \frac{\partial^2}{\partial z_z^2}.$$

The next four terms in the Hamiltonian (5.3) are the attractive potentials between the electrons and the protons, with $r_{1{\rm {l}}}$, $r_{2{\rm {l}}}$, $r_{1{\rm {r}}}$, and $r_{2{\rm {r}}}$ being the distances between electrons 1 and 2 and the left, respectively right proton. The final term represents the repulsive potential between the two electrons.

Key Points

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The Hamiltonian for the 6-di­mensional electron wave function has been written down.

5.2.1 Review Questions

1. Verify that the repulsive potential between the electrons is infinitely large when the electrons are at the same position.

   Note: You might therefore think that the wave function needs to be zero at the locations in six-di­mensional space where ${\skew0\vec r}_1$ $\vphantom0\raisebox{1.5pt}{$=$}$ ${\skew0\vec r}_2$. Some authors refer to that as a “Coulomb hole.” But the truth is that in quantum mechanics, electrons are smeared out due to uncertainty. That causes electron 1 to “see electron 2 at all sides”, and vice-versa, and they do therefore not encounter any unusually large potential when the wave function is non­zero at ${\skew0\vec r}_1$ $\vphantom0\raisebox{1.5pt}{$=$}$ ${\skew0\vec r}_2$. In general, it is just not worth the trouble for the electrons to stay away from the same position: that would reduce their uncertainty in position, increasing their uncertainty-demanded kinetic energy.

   Solution hmola-a

2. Note that the total kinetic energy term is simply a multiple of the six-di­mensional Laplacian operator. It treats all Cartesian position coordinates exactly the same, regardless of which direction or which electron it is. Is this still the case if other particles are involved?

   Solution hmola-b

5.2.2 Initial approximation to the lowest energy state

The next step is to identify an approximate ground state for the hydrogen molecule. Following the same approach as in chapter 4.6, it will first be assumed that the protons are relatively far apart. One obvious approximate solution is then that of two neutral atoms, say the one in which electron 1 is around the left proton in its ground state and electron 2 is around the right one.

To formulate the wave function for that, the shorthand notation $\psi_{\rm {l}}$ will again be used for the wave function of a single electron that in the ground state around the left proton and $\psi_{\rm {r}}$ for one that is in the ground state around the right hand one:

$$\psi_{\rm {l}}({\skew0\vec r}) \equiv \psi_{100}(\vert{\sk... ...si_{100}(\vert{\skew0\vec r}- {\skew0\vec r}_{\rm {rp}}\vert)$$

where $\psi_{100}$ is the hydrogen atom ground state (4.40), and ${\skew0\vec r}_{\rm {lp}}$ and ${\skew0\vec r}_{\rm {rp}}$ are the positions of the left and right protons.

The wave function that describes that electron 1 is in the ground state around the left proton and electron 2 around the right one will be approximated to be the product of the single electron states:

$$\psi({\skew0\vec r}_1,{\skew0\vec r}_2) = \psi_{\rm {l}}({\skew0\vec r}_1) \psi_{\rm {r}}({\skew0\vec r}_2)$$

Taking the combined wave function as a product of single electron states is really equivalent to an assumption that the two electrons are independent. Indeed, for the product state, the proba­bility of finding electron 1 at position ${\skew0\vec r}_1$ and electron 2 at ${\skew0\vec r}_2$ is:

$$\vert\psi_{\rm {l}}({\skew0\vec r}_1)\vert^2 {\,\rm d}^3 {... ...{\rm {r}}({\skew0\vec r}_2)\vert {\,\rm d}^3 {\skew0\vec r}_2$$

or in words:

$$\begin{array}{l} \mbox{[probability of finding 1 at ${\s... ... ${\skew0\vec r}_2$\ unaffected by where 1 is]} \end{array}$$

Such product proba­bilities are charac­teristic of statisti­cally independent quantities. As a simple example, the chances of getting a three in the first throw of a die and a five in the second throw are $\frac16\times\frac16$ or 1 in 36. Throwing the three does not affect the chances of getting a five in the second throw.

Key Points

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When the protons are well apart, an approximate ground state is that of two neutral atoms.

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Single electron wave functions for that case are $\psi_{\rm {l}}$ and $\psi_{\rm {r}}$.

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The complete wave function for that case is $\psi_{\rm {l}}({\skew0\vec r}_1)\psi_{\rm {r}}({\skew0\vec r}_2)$, assuming that electron 1 is around the left proton and electron 2 around the right one.

5.2.2 Review Questions

1. If electron 2 does not affect where electron 1 is likely to be, how would a grey-scale picture of the proba­bility of finding electron 1 look?

   Solution hmolb-a

2. When the protons are close to each other, the electrons do affect each other, and the wave function above is no longer valid. But suppose you were given the true wave function, and you were once again asked to draw the blob showing the proba­bility of finding electron 1 (using a plotting package, say). What would the big problem be?

   Solution hmolb-b

5.2.3 The probability density

For multiple-particle systems like the electrons of the hydrogen molecule, showing the magnitude of the wave function as grey tones no longer works since it is a function in six-di­mensional space. You cannot visualize six-di­mensional space. However, at every spatial position ${\skew0\vec r}$ in normal space, you can instead show the “proba­bility density” $n({\skew0\vec r})$, which is the proba­bility per unit volume of finding either electron in a vicinity ${\rm d}^3{\skew0\vec r}$ of the point. This proba­bility is found as

$$n({\skew0\vec r}) = \int \vert\Psi({\skew0\vec r}, {\ske... ...0\vec r}_1, {\skew0\vec r})\vert^2{\,\rm d}^3{\skew0\vec r}_1$$ (5.4)

since the first integral gives the proba­bility of finding electron 1 at ${\skew0\vec r}$ regardless of where electron 2 is, (i.e. integrated over all possible positions for electron 2), and the second gives the proba­bility of finding 2 at ${\skew0\vec r}$ regardless of where 1 is. Since ${\rm d}^3{\skew0\vec r}$ is vanishingly small, the chances of finding both particles in it at the same time are zero.

The proba­bility density $n({\skew0\vec r})$ for state $\psi_{\rm {l}}({\skew0\vec r}_1)\psi_{\rm {r}}({\skew0\vec r}_2)$ with electron 1 around the left proton and electron 2 around the right one is shown in figure 5.1. Of course the proba­bility density for the state $\psi_{\rm {r}}({\skew0\vec r}_1)\psi_{\rm {l}}({\skew0\vec r}_2)$ with the electrons exchanged would look exactly the same.

Figure 5.1: State with two neutral atoms.

Key Points

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The proba­bility density is the proba­bility per unit volume of finding an electron, whichever one, near a given point.

5.2.3 Review Questions

1. Suppose, given the wave function $\psi_{\rm {l}}({\skew0\vec r}_1)\psi_{\rm {r}}({\skew0\vec r}_2)$, that you found an electron near the left proton. What electron would it probably be? Suppose you found an electron at the point halfway in between the protons. What electron would that likely be?

   Solution hmolc-a

5.2.4 States that share the electrons

This section will examine the states where the protons share the two electrons.

The first thing is to shorten the notations a bit. So, the state $\psi_{\rm {l}}({\skew0\vec r}_1)\psi_{\rm {r}}({\skew0\vec r}_2)$ which describes that electron 1 is around the left proton and electron 2 around the right one will be indicated by $\psi_{\rm {l}}\psi_{\rm {r}}$, using the convention that the first factor refers to electron 1 and the second to electron 2. In this convention, the state where electron 1 is around the right proton and electron 2 around the left one is $\psi_{\rm {r}}\psi_{\rm {l}}$, shorthand for $\psi_{\rm {r}}({\skew0\vec r}_1)\psi_{\rm {l}}({\skew0\vec r}_2)$. It is of course physi­cally the same thing as $\psi_{\rm {l}}\psi_{\rm {r}}$; the two electrons are identical.

The “every possible combin­ation” idea of combining every possible state for electron 1 with every possible state for electron 2 would suggest that the states $\psi_{\rm {l}}\psi_{\rm {l}}$ and $\psi_{\rm {r}}\psi_{\rm {r}}$ should also be included. But these states have the electrons around the same proton, and that is not going to be energeti­cally favorable due to the mutual repulsion of the electrons. So they are not useful for finding a simple approximate ground state of lowest energy.

States where the electrons are no longer assigned to a particular proton can be found as linear combin­ations of $\psi_{\rm {l}}\psi_{\rm {r}}$ and $\psi_{\rm {r}}\psi_{\rm {l}}$:

$$\psi = a \psi_{\rm {l}}\psi_{\rm {r}} + b \psi_{\rm {r}}\psi_{\rm {l}}$$ (5.5)

In such a combin­ation each electron has a proba­bility of being found about either proton, but wherever it is found, the other electron will be around the other proton.

The eigen­function must be normalized, which noting that $\psi_{\rm {l}}$ and $\psi_{\rm {r}}$ are real and normalized produces

$$\langle \psi \vert \psi \rangle_6 = \langle a\psi_{\rm {... ...b \langle \psi_{\rm {l}} \vert \psi_{\rm {r}} \rangle^2 = 1$$ (5.6)

assuming that $a$ and $b$ are real. As a result, only the ratio $a$$\raisebox{.5pt}{$/$}$$b$ can be chosen freely. The proba­bility density of the combin­ation can be found to be:

$$n = \psi_{\rm {l}}^2+\psi_{\rm {r}}^2 + 2ab \langle \psi_{... ...m {r}} \rangle (\psi_{\rm {l}}^2+\psi_{\rm {r}}^2) \right\}$$ (5.7)

The most important combin­ation state is the one with $b$ $\vphantom0\raisebox{1.5pt}{$=$}$ $a$:

$$\psi({\skew0\vec r}_1,{\skew0\vec r}_2) = a \left[ \... ...}({\skew0\vec r}_1)\psi_{\rm {l}}({\skew0\vec r}_2) \right]$$ (5.8)

This state is called “symmetric with respect to exchanging electron 1 with electron 2,” or more precisely, with respect to replacing ${\skew0\vec r}_1$ by ${\skew0\vec r}_2$ and vice-versa. Such an exchange does not change this wave function at all. If you change ${\skew0\vec r}_1$ into ${\skew0\vec r}_2$ and vice-versa, you still end up with the same wave function. In terms of the hydrogen ground state wave function, it may be written out fully as

$$\fbox{$\displaystyle \Psi \approx a \left[ \psi_{100... ...ew0\vec r}_2-{\skew0\vec r}_{\rm{lp}}\vert) \right] $} %$$ (5.9)

with $\psi_{100}(r)$ $\vphantom0\raisebox{1.5pt}{$\equiv$}$ ${e}^{-r/a_0}$$\raisebox{.5pt}{$/$}$$\sqrt{{\pi}a_0^3}$, where $a_0$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0.53 Å is the Bohr radius, and ${\skew0\vec r}_1$, ${\skew0\vec r}_2$, ${\skew0\vec r}_{\rm {lp}}$, and ${\skew0\vec r}_{\rm {rp}}$ are again the position vectors of the electrons and protons.

The proba­bility density of this wave function looks like figure 5.2. It has increased likelihood for electrons to be found in between the protons, compared to figure 5.1 in which each proton had its own electron.

Figure 5.2: Symmetric sharing of the electrons.

The state with $b$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\vphantom0\raisebox{1.5pt}{$-$}$$a$,

$$\psi({\skew0\vec r}_1,{\skew0\vec r}_2) = a \left[ \... ...}({\skew0\vec r}_1)\psi_{\rm {l}}({\skew0\vec r}_2) \right]$$ (5.10)

is called “anti­symmetric” with respect to exchanging electron 1 with electron 2: swapping ${\skew0\vec r}_1$ and ${\skew0\vec r}_2$ changes the sign of wave function, but leaves it further unchanged. As seen in figure 5.3, the anti­symmetric state has decreased likelihood for electrons to be found in between the protons.

Figure 5.3: Antisymmetric sharing of the electrons.

Key Points

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In state $\psi_{\rm {l}}\psi_{\rm {r}}$, the electron numbered 1 is around the left proton and 2 around the right one.

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In state $\psi_{\rm {r}}\psi_{\rm {l}}$, the electron numbered 1 is around the right proton and 2 around the left one.

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In the symmetric state $a(\psi_{\rm {l}}\psi_{\rm {r}}+\psi_{\rm {r}}\psi_{\rm {l}})$ the protons share the electrons equally; each electron has an equal chance of being found around either proton. In this state there is increased proba­bility of finding an electron somewhere in between the protons.

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In the anti­symmetric state $a(\psi_{\rm {l}}\psi_{\rm {r}}-\psi_{\rm {r}}\psi_{\rm {l}})$ the protons also share the electrons equally; each electron has again an equal chance of being found around either proton. But in this state there is decreased proba­bility of finding an electron somewhere in between the protons.

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So, like for the molecular ion, at large proton separations the weird trick of shuffling unobservable wave functions around does again produce different physical states with pretty much the same energy.

5.2.4 Review Questions

1. Obviously, the visual difference between the various states is minor. It may even seem counter-intuitive that there is any difference at all: the states $\psi_{\rm {l}}\psi_{\rm {r}}$ and $\psi_{\rm {r}}\psi_{\rm {l}}$ are exactly the same physi­cally, with one electron around each proton. So why would their combin­ations be any different?

   The quantum difference would be much more clear if you could see the full six-di­mensional wave function, but visualizing six-di­mensional space just does not work. However, if you restrict yourself to only looking on the $z$-axis through the nuclei, you get a drawable $z_1,z_2$-plane describing near what axial combin­ations of positions you are most likely to find the two electrons. In other words: what would be the chances of finding electron 1 near some axial position $z_1$ and electron 2 at the same time near some other axial position $z_2$?

   Try to guess these proba­bilities in the $z_1,z_2$-plane as grey tones, (darker if more likely), and then compare with the answer.

   Solution hmold-a

2. Based on the previous question, how would you think the proba­bility density $n(z)$ would look on the axis through the nuclei, again ignoring the existence of positions beyond the axis?

   Solution hmold-b

5.2.5 Variational approximation of the ground state

The purpose of this section is to find an approxi­mation to the ground state of the hydrogen molecule using the rough approxi­mation of the wave function described in the previous subsections.

Like for the hydrogen molecular ion of chapter 4.6.6, the idea is that since the true ground state is the state of lowest energy among all wave functions, the best among approximate wave functions is the one with the lowest energy. The approximate wave functions are here of the form $a\psi_{\rm {l}}\psi_{\rm {r}}+b\psi_{\rm {r}}\psi_{\rm {l}}$; in these the protons share the electrons, but in such a way that when one electron is around the left proton, the other is around the right one, and vice-versa.

A computer program is again needed to print out the expec­tation value of the energy for various values of the ratio of coefficients $a$$\raisebox{.5pt}{$/$}$$b$ and proton-proton distance $d$. And worse, the expec­tation value of energy for given $a$$\raisebox{.5pt}{$/$}$$b$ and $d$ is a six-di­mensional integral, and parts of it cannot be done analyti­cally; numerical integr­ation must be used. That makes it a much more messy problem, {D.23}.

You might just want to take it on faith that the binding energy, at the state of lowest energy found, turns out to be 3.2 eV, at a proton to proton spacing of 0.87 Å, and that it occurs for the symmetric state $a$ $\vphantom0\raisebox{1.5pt}{$=$}$ $b$.

Key Points

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An approximate ground state can be found for the hydrogen molecule using a varia­tional method much like that for the molecular ion.

5.2.6 Comparison with the exact ground state

The solution for the ground state of the hydrogen molecule obtained in the previous subsection is, like the one for the molecular ion, pretty good. The approximate binding energy, 3.2 eV, is not too much different from the experi­mental value of 4.52 eV. Similarly, the bond length of 0.87 Å is not too far from the experi­mental value of 0.74 Å.

Quali­tatively, the exact ground state wave function is real, positive and symmetric with respect to reflection around the symmetry plane and to rotations around the line connecting the protons, and so is the approximate one. The reasons for these properties are similar as for the molecular ion; {A.8,A.9}.

One very important new symmetry for the neutral molecule is the effect of exchanging the electrons, replacing ${\skew0\vec r}_1$ by ${\skew0\vec r}_2$ and vice-versa. The approximate wave function is symmetric (unchanged) under such an exchange, and so is the exact wave function. To understand why, note that the operation of exchanging the electrons commutes with the Hamiltonian, (exchanging identical electrons physi­cally does not do anything). So energy eigen­functions can be taken to be also eigen­functions of the “exchange operator.” Furthermore, the exchange operator is a Hermitian one, (taking it to the other side in inner products is equivalent to a simple name change of integr­ation variables,) so it has real eigen­values. And more specifi­cally, the eigen­values can only be plus or minus one, since swapping electrons does not change the magnitude of the wave function. So the energy eigen­functions, including the ground state, must be symmetric under electron exchange (eigenvalue one), or anti­symmetric (eigenvalue minus one.) Since the ground state must be everywhere positive, (or more precisely, of a single sign), a sign change due to swapping electrons is not possible. So only the symmetric possi­bility exists for the ground state.

One issue that does not occur for the molecular ion, but only for the neutral molecule is the mutual repulsion between the two electrons. This repulsion is reduced when the electron clouds start to merge, compared to what it would be if the clouds were more compact. (A similar effect is that the gravity force of the earth decreases when you go down below the surface. To be sure, the potential energy keeps going down, or up for electron clouds, but not as much as it would otherwise. Compare figure 13.7.) Since the nuclei are compact, it gives an advantage to nucleus-electron attraction over electron-electron repulsion. This increases the binding energy significantly; in the approximate model from about 1.8 eV to 3.2 eV. It also allows the protons to approach more closely; {D.23}.

The question has been asked whether there should not be an “activation energy” involved in creating the hydrogen molecule from the hydrogen atoms. The answer is no, hydrogen atoms are radicals, not stable molecules that need to be taken apart before recombining. In fact, the hydrogen atoms attract each other even at large distances due to Van der Waals attraction, chapter 10.1, an effect lost in the approximate wave functions used in this section. But hydrogen atoms that fly into each other also have enough energy to fly apart again; some of the excess energy must be absorbed elsewhere to form a stable molecule. According to web sources, hydrogen molecule formation in the universe is believed to typically occur on dust specks.

Key Points

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The approximate ground state is pretty good, considering its simplicity.