1.7: Introduction to diffusion

An Ideal Gas of Monatomic Distinguishable Particles

The energy levels for an ideal gas in a three-dimensional cubic infinite square well of volume \(V\) are

\[E_j (N, V) = E_{\textbf{n}_1 \cdots \textbf{n}_N} (N,V) = \dfrac{\hbar^2 \pi^2}{2mV^{2/3}} \sum_{i=1}^N \left| \textbf{n}_i \right|^2 \label{7.1}\]

Since the particles do not interact, this sum, like the single particle in a three-dimensional box, is just a sum of \(3N\) independent terms:

\[\begin{align} E_j (N, V) &= E_{n_{x_1}, n_{y_1}, n_{z_1}, n_{x_2}, n_{y_2}, n_{z_2}, \cdots, n_{x_N}, n_{y_N}, n_{z_N}} \\ &= \dfrac{\hbar^2 \pi^2}{2mV^{2/3}} \left( n_{x_1}^2 + n_{y_1}^2 + n_{z_1}^2 + n_{x_2}^2 + n_{y_2}^2 + n_{z_2}^2 + \cdots + n_{x_N}^2 + n_{y_N}^2 + n_{z_N}^2 \right) \end{align} \label{7.2}\]

Since the partition function for a single particle in a cubic box (here denoted \(q(V, T)\)) could be written as

\[\begin{align} q(V, T) &= \sum_{n_{x=1}}^\infty \sum_{n_{y=1}}^\infty \sum_{n_{z=1}}^\infty e^{-\beta \hbar^2 \pi^2 \left( n_x^2 + n_y^2 + n_z^2 \right) / 2mV^{2/3}} \\ &= \left( \sum_{n=1}^\infty e^{-\beta \hbar^2 \pi^2 n^2 / 2mV^{2/3}} \right)^3 \\ &= \left( \int_0^\infty e^{-\beta \hbar^2 \pi^2 n^2 / 2mV^{2/3}} \: dn \right)^3 \\ &= \left( \dfrac{m}{2 \pi \beta \hbar^2} \right)^{3/2}V \end{align} \label{7.3}\]

The partition function for \(N\) particles in a cubic box is just

\[\begin{align} Q(N, V, T) &= \left( \sum_{n_{x=1}}^\infty \sum_{n_{y=1}}^\infty \sum_{n_{z=1}}^\infty e^{-\beta \hbar^2 \pi^2 \left( n_x^2 + n_y^2 + n_z^2 \right) / 2mV^{2/3}} \right)^N \\ &= \left[ q(V, T) \right]^N \\ &= \left[ \dfrac{m}{2 \pi \beta \hbar^2} \right]^{3/2} V^N \end{align} \label{7.4}\]

The total energy of the gas is then given by

\[E = -\dfrac{\partial}{\partial \beta} \: \text{ln} \: Q = \dfrac{3}{2} N k_B T = \dfrac{3}{2} nRT \label{7.5}\]

The pressure is given by

\[P = k_B T \left( \dfrac{\partial \: \text{ln} \: Q}{\partial V} \right) = \dfrac{N k_B T}{V} \label{7.6}\]

which can be rearranged to read

\[PV = N k_B T = nRT \label{7.7}\]

which is the equation of state of an ideal gas. Thus, starting from a microscopic description of the system in terms of energy levels, we are able to predict the correct thermodynamics for an ideal gas.

An Ideal Gas of Monatomic Indistinguishable particles

Boltzmann particles

In a monatomic ideal gas, the quantum spin degree of freedom is the only truly quantum property of a particle with no classical analog. If we neglect the spin degree of freedom of particles in a monatomic gas, then there is essentially no difference between a classical and a quantum monatomic ideal gas. This means that the calculation of the partition function will overcount the number of microscopic states by \(N!\) and Equation \(\ref{7.4A}\) needs to be divided by \(N!\) to generate the correct counting for indistinguishable particles:

\[Q(N, V, T) = \dfrac{\left[ q(V, T) \right]^N}{N!} \label{7.8}\]

From Equation \(\ref{7.4A}\), we see that the partition function for indistinguishable monatomic Boltzmann ideal-gas particles is

\[Q(N, V, T) = \dfrac{1}{N!} \left( \dfrac{m}{2 \pi \beta \hbar^2} \right)^{3N/2} V^N \label{7.9}\]

Note, however, that this partition function gives the same energy and pressure as in Equations \(\ref{7.5}\) and \(\ref{7.6}\), respectively, for distinguishable particles.

Accounting for spin: Fermions and bosons

The Boltzmann particles considered in the last subsection do not actually exist in nature. In reality, all particles carry the quantum spin degree of freedom. Neglect of spin is just an approximation, which is, nevertheless, often an accurate one. However, in many cases, we cannot neglect spin, and when spin must be included, the calculation of the partition function becomes significantly more difficult. We will not carry out the calculation in detail, but we will explore what it is that makes the problem difficult.

In nature, the spin of a particle can have either an integer or half-integer value. Particles with integer spin are called bosons (after the Indian physicist Satyendra Bose), while those with half-integer spin are called fermions (after the Italian physicist Enrico Fermi). Fermions are governed by the Pauli exclusion principle, which states that no two fermions can occupy the same quantum state. Bosons, on the other hand, have no such restriction. It is, therefore, clear that state occupancy is the key distinguishing feature between the two types of particles, and therefore, we need to express the energy of a system of fermions or bosons in terms of these state occupancies.

Recall that the energy for one particle in a cubic box is

\[\mathcal{E}_\textbf{n} = \dfrac{\hbar^2 \pi^2}{2mV^{2/3}} | \textbf{n} | \label{7.10}\]

The energy \(\mathcal{E}_\textbf{n}\) is called a single-particle energy level. Now if there are \(f_\textbf{n}\) particles occupying this energy level, then the contribution of this energy level to the total energy must be \(\mathcal{E}_\textbf{n} f_\textbf{n}\). The number \(f_\textbf{n}\) is called an occupation number for the state with energy \(\mathcal{E}_\textbf{n}\). The total energy of a system is now obtained simply by summing over all of the single-particle levels. The total energy is now characterized by the complete set of occupation numbers, which we will denote simply as \(f\). Thus,

\[E_f = \sum_\textbf{n} \mathcal{E}_\textbf{n} f_\textbf{n} \label{7.11}\]

Note, however, that there is a restriction on the choice of occupation numbers, namely, that the occupation numbers must sum to the total number of particles \(N\):

\[\sum_\textbf{n} f_\textbf{n} = N \label{7.12}\]

For fermions, the only allowable values for any one occupation number \(f_\textbf{n}\) is \(0\) or \(1\), consistent with the Pauli exclusion principle. However, for bosons, \(f_\textbf{n}\) can take on any value between \(0\) and \(N\). In principle, the partition function can be computed by summing over all possible values of the occupation numbers as follows:

\[Q(N, V, T) = \sum_f e^{-\beta E_f} = \sum_f e^{-\beta \sum_\textbf{n} \mathcal{E}_\textbf{n} f_\textbf{n}} \label{7.13}\]

However, the sum over \(f\) must be carried out subject to the restriction in Equation \(\ref{7.12}\), and this makes the calculation of partition functions for fermions and bosons highly nontrivial. It also makes the thermodynamics of fermionic and bosonic ideal gases nontrivial as well. In fact, there is no exact solution to either problem. At very high temperatures, it can be shown that these gases behave approximately as classical ideal gases, which means that at high temperatures, the effect of the spin degree of freedom can be safely neglected. For many systems ``high temperature'' can simply mean room temperature! However, at low temperatures, where quantum effects become extremely important, the deviation from classical ideal gas behavior is significant.

At low temperatures, an ideal electron gas (electrons are spin-1/2 particles) behaves somewhat like the quasi-free conduction electrons in a metal. At low temperatures, an ideal boson gas exhibits a phenomenon known as Bose-Einstein condensation, in which all of the particles "condense" into the ground state. Bose-Einstein condensation is at the heart of phenomena such as superconductivity and superfluidity. If you have an interest in learning more about quantum ideal gases, see Lecture 18.

An Ideal Gas of Indistinguishable Diatomic Molecules

We will now consider an ideal gas of diatomic molecules. Each atom in a molecule consists of two nuclei and a distribution of electrons. Let \(r_\text{eq}\) be the equilibrium bond length of the molecule. We will assume that the spin of the nuclei can be safely neglected. In addition, we will assume that the Born-Oppenheimer approximation holds so that there is an approximate separation of nuclear and electronic energy levels, and the nuclear energy levels must be computed on each electronic Born-Oppenheimer surface. This is illustrated in the figure below:

FIG 7.1: Illustration of two bonding Born-Oppenheimer electronic surfaces with nuclear levels. Rotational levels in blue, vibrational levels in red.

Note that a diatomic molecule, which has two atoms, has a total of \(3 \times 2 = 6\) degrees of freedom. Three of these will be overall translations of the molecule in the box, two will be rotations of the molecule about an arbitrarily chosen axis, and the remaining degree of freedom is motion along the bond axis of the molecule.

Consider first the classical energy of the two nuclei. If the nuclei have momenta \(\textbf{p}_1\) and \(\textbf{p}_2\) and positions \(\textbf{r}_1\) and \(\textbf{r}_2\), then the classical energy is

\[\mathcal{E}_\text{nuc} = \dfrac{\textbf{p}_1^2}{2m_1} + \dfrac{\textbf{p}_2^2}{2m_2} + u \left( | \textbf{r}_1 - \textbf{r}_2 | \right) \label{7.14}\]

From this expression, it would seem that the two atoms in each molecule are inextricably coupled because the potential \(u\) is a function of the distance \( | \textbf{r}_1 - \textbf{r}_2 | \) between them. However, the energy can, in fact, be separated into independent contributions by a simple change of variables. In particular, we transform to center-of-mass and relative coordinates by the following change of variables:

\[\begin{align} \textbf{R} &= \dfrac{m_1 \textbf{r}_1 + m_2 \textbf{r}_2}{M} \\ \textbf{P} &= \textbf{p}_1 + \textbf{p}_2 \\ \textbf{r} &= \textbf{r}_1 - \textbf{r}_2 \\ \textbf{p} &= \dfrac{m_2 \textbf{p}_1 - m_1 \textbf{p}_2}{M} \end{align} \label{7.15}\]

where \(M = m_1 + m_2\). Here \(\textbf{R}\) and \(\textbf{P}\) are the center of mass position and momentum, respectively, while \(\textbf{r}\) and \(\textbf{p}\) are the relative coordinate and momentum, respectively. The transformation can be inverted to give

\[\begin{align} \textbf{r}_1 &= \textbf{R} + \dfrac{m_2}{M} \textbf{r} \\ \textbf{p}_1 &= \dfrac{m_1}{M} \textbf{P} + \textbf{p} \\ \textbf{r}_2 &= \textbf{R} + \dfrac{m_1}{M} \textbf{r} \\ \textbf{p}_2 &= \dfrac{m_2}{M} \textbf{P} - \textbf{p} \end{align} \label{7.16}\]

Substituting these into the expression for \(\mathcal{E}_\text{nuc}\), we obtain

\[\begin{align} \mathcal{E}_\text{nuc} &= \dfrac{1}{2m_1} \left( \dfrac{m_1}{M} \textbf{P} + \textbf{p} \right)^2 + \dfrac{1}{2m_2} \left( \dfrac{m_2}{M} \textbf{P} - \textbf{p} \right)^2 + u(r) \\ &= \dfrac{1}{2m_1} \left( \dfrac{m_1^2}{M^2} \textbf{P}^2 + \dfrac{2m_1}{M} \textbf{P} \cdot \textbf{p} + \textbf{p}^2 \right) + \dfrac{1}{2m_2} \left( \dfrac{m_1^2}{M^2} \textbf{P}^2 - \dfrac{2m_1}{M} \textbf{P} \cdot \textbf{p} + \textbf{p}^2 \right) \\ &= \dfrac{m_1}{2M^2} \textbf{P}^2 + \dfrac{m_2}{2M^2} \textbf{P}^2 + \dfrac{1}{M} \textbf{P} \cdot \textbf{p} - \dfrac{1}{M} \textbf{P} \cdot \textbf{p} + \dfrac{\textbf{p}^2}{2m_1} + \dfrac{\textbf{p}^2}{2m_2} \\ &= \dfrac{m_1 + m_2}{2M^2} \textbf{P}^2 + \dfrac{1}{2} \dfrac{m_1 + m_2}{m_1 m_2} \textbf{p}^2 \end{align} \label{7.17}\]

We define

\[\mu = \dfrac{m_1 m_2}{m_1 + m_2} \label{7.18}\]

which is called the reduced mass. In terms of \(M\) and \(\mu\), the energy can be expressed as

\[\mathcal{E}_\text{nuc} = \dfrac{\textbf{P}^2}{2M} + \dfrac{\textbf{p}^2}{2 \mu} + u(r) \label{7.19}\]

From this we see that the energy due to center-of-mass motion is purely kinetic. This term gives the energy of simple translations of the molecule throughout the box. The remaining term is purely internal motion of the molecule, called relative motion. Thus, we see that the energy is now separated into purely translational and relative contributions, and we write

\[\mathcal{E}_\text{nuc} = \mathcal{E}_\text{trans} + \mathcal{E}_\text{rel} \label{7.20}\]

Now for the relative motion, we resolve the vector \(\textbf{p}\) into a radial component \(\textbf{p}_r\) and a residual two-dimensional momentum in the two angular directions \((\theta, \phi)\), i.e., \(\textbf{p}_{(\theta, \phi)}\). The first component \(\textbf{p}_r\) is a purely radial momentum directed radial outward from the origin. The remaining two components describe rotational motion in the azimuthal (\(\phi\)) and polar ((\theta\)) directions. These two components are perpendicular so that \(\textbf{p}_r \cdot \textbf{p}_{(\theta, \phi)} = 0\). In this case, the relative energy becomes

\[\mathcal{E}_\text{rel} = \dfrac{\textbf{p}_r^2}{2 \mu} + \dfrac{\textbf{p}_{(\theta, \phi)}^2}{2 \mu} + u(r) \label{7.21}\]

The purely radial part contains the two terms \(\textbf{p}_r^2/2 \mu + u(r)\). This motion occurs entirely along the direction of the bond axis and constitutes the vibrational energy. The remaining purely kinetic term is the rotational motion of the molecule. Thus, we can write the relative energy as

\[\mathcal{E}_\text{rel} = \mathcal{E}_\text{vib} + \mathcal{E}_\text{rot} \label{7.22}\]

so that the total nuclear energy of the molecule becomes

\[\mathcal{E}_\text{nuc} = \mathcal{E}_\text{trans} + \mathcal{E}_\text{rot} + \mathcal{E}_\text{vib} \label{7.23}\]

Note that there is an approximation made here, namely, that rotations and vibrations are not coupled in any way. The approximation we are making here is that rotations are sufficiently slow compared to vibrations that we can consider the molecule as a rigid rotor of length \(r_\text{eq}\), where \(r_\text{eq}\) is the equilibrium bond length. If the vibrations are sufficiently fast, then on the time scale of rotations, the average distance between the atoms will be \(r_\text{eq}\), hence, the approximation is a reasonable one, but it must be kept in mind that this is what we are assuming.

Let us now construct the nuclear part of the partition function for a single molecule. The translational energy is just that of the center of mass, and hence, its energy levels are those of a particle of mass \(M\) in a box of volume \(V\), characterized by the usual quantum number \(\textbf{n} = (n_x, n_y, n_z)\):

\[\mathcal{E}^\text{trans}_\textbf{n} = \dfrac{\hbar^2 \pi^2}{2MV^{2/3}} \left| \textbf{n} \right|^2 \label{7.24}\]

For the rotational energy, we consider rotations of the molecule about an axis through the position \(\textbf{r}_1\) of atom \(1\). In this case, the moment of inertia introduced in the last lecture is just \(I = \mu r_\text{eq}^2\). The two quantum numbers we need for rotation are \(J\) and \(m_J\), where \(J = 0, 1, 2, \ldots\), and \(m_J = -J, -J+1, \ldots, J-1, J\). The energy levels are then given by

\[\mathcal{E}_\text{rot} = \dfrac{\hbar^2}{2I} J (J+1) \label{7.25}\]

Finally, there is the vibrational motion, the potential energy of which is \(u(r)\). Typically, \(u(r)\) resembles the plot shown in the figure below:

FIG 7.2: Typical shape of \(u(r)\) for a covalent bond, and the harmonic approximation to the curve.

If the temperature of the gas is sufficiently low, then only the lowest lying vibrational energy levels are important, and we can approximate the true vibrational potential with a harmonic potential (see figure) of natural frequency \(\omega\) and write

\[u(r) = \dfrac{1}{2} \mu \omega^2 \left( r - r_\text{eq} \right)^2 \label{7.26}\]

For a harmonic oscillator, the energy levels are given by

\[\mathcal{E}^\text{vib}_\nu = \left( \nu + \dfrac{1}{2} \right) \hbar \omega, \: \: \: \: \: \: \: \nu = 0, 1, 2, \ldots \label{7.27}\]

The six quantum numbers \((n_x, n_y, n_z, J, m_J, \nu)\) correspond to the six degrees of freedom we identified at the beginning of this section. The complete nuclear energy levels can now be expressed as

\[\mathcal{E}_{n_x \: n_y \: n_z \: J \: m_J \: \nu} = \mathcal{E}_{ \textbf{n} \: J \: m_J \: \nu} \dfrac{\hbar^2 \pi^2}{2MV^{2/3}} \left| \textbf{n} \right|^2 + \dfrac{\hbar^2}{2I} J (J+1) + \left( \nu + \dfrac{1}{2} \right) \hbar \omega \label{7.28}\]

Given these energy levels, we can construct the corresponding energy levels for a system of \(N\) non-interacting diatomics as

\[E_{\textbf{n}_1 \: J_1 \: m_{J_1} \: \nu_1 \: \textbf{n}_2 \: J_2 \: m_{J_2} \: \nu_2 \cdots \textbf{n}_N \: J_N \: m_{J_N} \: \nu_N} = \sum_{i=1}^N \dfrac{\hbar^2 \pi^2}{2MV^{2/3}} \left| \textbf{n}_i \right|^2 + \dfrac{\hbar^2}{2I} J_i (J_i+1) + \left( \nu_i + \dfrac{1}{2} \right) \hbar \omega \label{7.29}\]

Since the energy levels are just a sum of single-molecule energy levels, Equation \(\ref{7.8}\) applies, and we can write the total partition function as

\[Q(N, V, T) = \dfrac{[q(V, T)]^N}{N!} \label{7.30}\]

where \(q(V, T)\) is the single-molecule partition function corresponding to the energy levels in Equation\(\ref{ 7.28}\).

The nuclear partition function for one molecule is

\[\begin{align} q_\text{nuc} (V, T) &= \sum_\textbf{n} \sum_{J=0}^\infty \sum_{m_J=-J}^J \sum_{\nu=0}^\infty \exp \left\{ -\beta \left[ \dfrac{\hbar^2 \pi^2}{2MV^{2/3}} \left| \textbf{n} \right|^2 + \dfrac{\hbar^2}{2I} J(J+1) + \left( \nu + \dfrac{1}{2} \right) \hbar \omega \right] \right\} \\ &= \left( \sum_\textbf{n} \exp \left[ -\dfrac{\beta \hbar^2 \pi^2}{2MV^{2/3}} \left| \textbf{n} \right|^2 \right] \right) \left( \sum_{J=0}^\infty \sum_{m_J=-J}^J \exp \left[ -\dfrac{\beta \hbar^2}{2I} J(J+1) \right] \right) \left( \sum_{\nu=0}^\infty \exp \left[ -\beta \left( \nu + \dfrac{1}{2} \right) \hbar \omega \right] \right) \\ &= q_\text{trans} (V, T) \: q_\text{rot} (T) \: q_\text{vib} (T) \end{align} \label{7.31}\]

Here

\[\sum_\textbf{n} = \sum_{n_x=1}^\infty \sum_{n_y=1}^\infty \sum_{n_z=1}^\infty \label{7.32}\]

Each of these partition functions were worked out in the previous lecture, so if we simply take over those results, we obtain the nuclear partition function as

\[q_\text{nuc} (V, T) = \left[ V \left( \dfrac{M}{2 \pi \beta \hbar^2} \right) \right] \left[ \dfrac{e^{-\beta \hbar \omega/2}}{1 - e^{-\beta \hbar \omega}} \right] \left[ \dfrac{1}{2 \pi \beta \hbar B} \right] \label{7.33}\]

where \(B = \hbar/(4 \pi I)\) is called the rotation constant. Remember that for the translational and rotational partition functions, we approximated the energy levels as a continuum so that the sums could be replaced by integrals.

The final energy term we need is the electronic energy. Recall that, within the Born-Oppenheimer approximation, each electronic energy surface, assuming that it has bonding rather than anti-bonding character, gives rise to a set of rotational and vibrational energy levels (translational energies are independent of the particular electronic surface). Let the global minima on these bonding surfaces be denoted \(\mathcal{E}_0^\text{(elec)}\), \(\mathcal{E}_1^\text{(elec)}\), \(\ldots\). As a reference energy, we choose the value of \(\mathcal{E}_0^\text{(elec)}\) and define each of the electronic energies relative to these, i.e., \(\mathcal{E}_{10}^\text{(elec)} = \mathcal{E}_1^\text{(elec)} - \mathcal{E}_0^\text{(elec)}\), \(\mathcal{E}_{20}^\text{(elec)} = \mathcal{E}_2^\text{(elec)} - \mathcal{E}_0^\text{(elec)}\), and so forth. If the degeneracies of these levels are \(g_0\), \(g_1\), \(g_2\),\(\ldots\), then the electronic contribution to the partition function is

\[q_\text{elec} (T) = g_0 + g_1 e^{-\beta \mathcal{E}_{10}^\text{(elec)}} + g_2 e^{-\beta \mathcal{E}_{20}^\text{(elec)}} + \cdots \label{7.34}\]

Generally, the electronic energy differences are so large at ordinary temperatures that the above sum converges after just a few terms. Often, we only need one or two terms in the above.

Putting everything together, we can write the partition function for the full \(N\)-particle ideal gas system as

\[Q(N, V, T) = \dfrac{1}{N!} \left[ V \left( \dfrac{M}{2 \pi \beta \hbar^2} \right) \right]^N \left[ \dfrac{e^{-\beta \hbar \omega/2}}{1 - e^{-\beta \hbar \omega}} \right]^N \left[ \dfrac{1}{2 \pi \beta \hbar B} \right]^N \left[ g_0 + g_1 e^{-\beta \mathcal{E}_{10}^\text{(elec)}} + g_2 e^{-\beta \mathcal{E}_{20}^\text{(elec)}} + \cdots \right]^N \label{7.35}\]

From this partition function, we can compute the thermodynamic properties of an ideal gas of indistinguishable diatomic molecules.

First the pressure is given by

\[\begin{align} P &= k_B T \left( \dfrac{\partial \: \text{ln} \: Q}{\partial V} \right) \\ &= \dfrac{N k_B T}{V} \end{align} \label{7.36}\]

which is equivalent to the ideal gas law \(PV = N k_B T = nRT\). Note that none of the internal properties of the molecules have any effect on the equation of state, which arises entirely from translational motion, this being the only volume-dependent contribution.

The energy is given by

\[\begin{align} E &= -\dfrac{\partial \: \text{ln} \: Q}{\partial \beta} \\ &= \dfrac{3}{2} N k_B T + \dfrac{1}{2} N \hbar \omega + \dfrac{N \hbar \omega}{e^{\beta \hbar \omega} - 1} + N k_B T + \dfrac{N \left( g_1 \mathcal{E}_{10}^\text{(elec)} e^{-\beta \mathcal{E}_{10}^\text{(elec)}} + \cdots \right)}{g_0 + g_1 e^{-\beta \mathcal{E}_{10}^\text{(elec)}} + \cdots} \\ &= \dfrac{5}{2} N k_B T + \dfrac{1}{2} N \hbar \omega + \dfrac{N \hbar \omega}{e^{\beta \hbar \omega} - 1} + \dfrac{N \left( g_1 \mathcal{E}_{10}^\text{(elec)} e^{-\beta \mathcal{E}_{10}^\text{(elec)}} + \cdots \right)}{g_0 + g_1 e^{-\beta \mathcal{E}_{10}^\text{(elec)}} + \cdots} \end{align} \label{7.37}\]

Calculating Partition Functions for Fully Interacting Systems using Feynman's Path Integral

The reason we can only discuss partition functions for the ideal gas is that it is nearly impossible to calculate energy levels for anything but the simplest systems, basically at the level of ideal gases. As we discussed in Lecture 4, the effort to solve the Schrödinger equation for systems with more than just a few atoms grows exponentially with the number of atoms.

As it happens, there is a way to compute the partition function and thermodynamic properties of systems containing thousands of atoms, which is often sufficiently large to approach the thermodynamic limit. This approach is based on a reformulation of quantum mechanics due to Feynman.

In order to introduce Feynman's approach, let us consider the expression for the partition function once again:

\[Q(N, V, T) = \sum_j e^{-\beta E_j (N, V)} \label{7.38}\]

Recall that the energy levels \(E_j (N, V)\) are eigenvalues of the Hamiltonian \(\hat{H}\). Once we know the eigenvalues of \(\hat{H}\), then \(\hat{H}\) can be represented as a diagonal matrix with the eigenvalues on the diagonal. Interestingly, if we take the Hamiltonian and form the object \(\exp( -\beta \hat{H})\), the quantities \(\exp( -\beta E_j (N, V))\) are its eigenvalues. Thus, when we compute the sum in Equation \(\ref{7.38}\), we are actually taking the trace of the matrix \(\exp( -\beta \hat{H})\). In general, the trace of any matrix \(A_{ij}\) is

\[\text{Tr} (A) = \sum_i A_{ii} \label{7.39}\]

That is, we just need to sum over the diagonal elements of the matrix. A key property of traces is that they are the same no matter how we represent the matrix \(A\). We do not need its eigenvalues, and we do not need \(A\) to be diagonal. Rather, we can use the positions of the quantum particles to compute the needed trace. Let \(X = \textbf{r}_1, \ldots, \textbf{r}_N\) represent the set of particle positions. Then, we can calculate the trace of \(\exp( -\beta \hat{h})\) as

\[Q(N, V, T) = \text{Tr} \left[ e^{-\beta \hat{H}} \right] = \int dX \: \left[ e^{-\beta \hat{H}} \right]_{XX} \label{7.40}\]

Note that because the positions are continuous rather than discrete variables, when we express the trace in terms of \(X\), we must do so as an integral rather than as a sum.

Feynman, who was born right here in Far Rockaway, devised a remarkably elegant way to calculate \([\exp( -\beta \hat{H})]_{XX}\). By multiplying and dividing the argument of the exponential by \(\hbar\), we obtain

\[Q(N, V, T) = \int dX \: \left[ e^{-\beta \hbar \hat{H}/\hbar} \right]_{XX} \label{7.41}\]

Note that the quantity \(\beta \hbar\) has units of time. Feynman recognized that this is no mere accident or coincidence. His key insight was to recognize that \(\exp[-\beta \hat{H}]_{XX}\) could be computed by considering some path, \(X(t)\), parameterized by time, that the system could take that starts at \(X\) at time \(t=0\), returns to \(X\) in time \(T= \beta \hbar\), but that can go anywhere in space between \(t=0\) and \(t=\beta \hbar\). These are known as cyclic paths. This is illustrated in the figure below:

FIG 7.3: Illustration of the Feynman path integral. Left: Cartoon of a path, beginning and ending at the point \(X\). Right: Path in the \(X-t\) plane.

Let this path be denoted \(X(t) = (\textbf{r}_1 (t), \ldots, \textbf{r}_N (t))\). Then Feynman showed that \(\exp[-\beta \hat{H}]_{XX}\) could be computed in terms of an integral of the energy, expressed as a function of positions and velocities, over the path:

\[\left[ e^{-\beta \hbar \hat{H}/\hbar} \right]_{XX} = \exp \left[ \dfrac{1}{\hbar} \int_0^{\beta \hbar} \left( \dfrac{1}{2} \sum_{i=1}^N m_i v_i^2 (t) + U(\textbf{r}_1 (t), \ldots, \textbf{r}_N (t)) \right) \: dt \right] \label{7.42}\]

This is why the technique is called a path integral. Let us denote the integral in the above expression as \(S\):

\[S = \int_0^{\beta \hbar} \left( \dfrac{1}{2} \sum_{i=1}^N m_i v_i^2 (t) + U (\textbf{r}_1 (t) , \ldots, \textbf{r}_N (t)) \right) \: dt \label{7.43}\]

Then, the partition function can be computed by summing over all possible paths that take the system from \(X\) back to \(X\) in time \(\beta \hbar\):

\[Q(N, V, T) = \sum_\text{paths} e^{-S/\hbar} \label{7.44}\]

The need to include all possible paths is a consequence of the uncertainty principle. If we have full knowledge of the positions \(X(t)\) along a path for all \(t\), then we have no knowledge of the momenta. However, classically, a definite path has specific values of positions and momenta simultaneously. Without knowledge of momenta, the particles could follow any path from \(X\) to \(X\), and because of the uncertainly, we must consider all of them.

The key to the use of Equation \(\ref{7.44}\) is to discretize the paths into \(P\) time intervals, each having a time step or interval \(\beta \hbar P\). Once these paths are discretized, the sum can be carried out on a computer with a computational cost no more than \(P\) times what it would cost to carry out the calculation using classical mechanics. The basic scheme is shown in the figure below for \(P=5\) and two interacting quantum particles. Notice that when the paths are discretized, neighboring points on each cyclic path are connected by harmonic springs.

FIG 7.4: Path integral for two particles discretized into \(P=5\) time intervals. Interactions are computed between points in the same time interval only.

In the figures that follow, we show three examples of the path integral approach. The first is an example of a simple organic molecule, malonaldehyde, shown with an internal hydrogen bond. Because of the internal bond, a proton on one of the \(OH\) groups can transfer through the bond to the neighboring oxygen. The energetic barrier to this reaction is approximately \(14.23\) kJ/mol. However, recall that quantum mechanics allows for a phenomenon known as tunneling, which gives particles a non-zero probability to be located in classically forbidden regions. In malonaldehyde, tunneling largely dominates this internal proton transfer, thus causing a substantial reduction in the effective energy barrier, which we will learn is equivalent to the free energy barrier, from \(14.23\) kJ/mol to about \(6.7\) kJ/mol. The figure shows what a tunneling path looks like. We see that when the proton tunnels, it exists in a Schrödinger's cat-like state in which it is simultaneously bonded to both oxygens, and only an observation of the proton can localize it on one oxygen or the other.

FIG 7.5: Top: Illustration of the internal proton transfer reaction in malonaldehyde. Bottom: A tunneling path in malonaldehyde.

FIG 7.6: left: Model of the AT base pair and reaction free energy profile. right: Same for GC base pair.

FIG 7.7: left: Schematic of the hydronium diffusion mechanism showing concerted hydrogen bond breaking and forming events. right: A particular path showing the delocalization of the topological defect in the hydrogen bond network of the solution.

The second example concerns the occurrence of rare tautomers in DNA base pairs. The two strands of DNA are held together by hydrogen bonds formed between pairs of nucleotide bases. The so-called tautomeric shift problem occurs when the protons in the hydrogen bonds holding a pair of bases together transfer through the hydrogen bonds. It was posited that if the protons become stuck in their shifted positions at the time of DNA replication, the incorrect locations of the protons would cause mispairing when the DNA replicates itself, which could lead to diseases such as cancer. The implication of this supposition is that quantum mechanics, which allows the protons to tunnel into their shifted positions, is the evil force behind such afflictions. However, experiments were never able to see the spontaneous formation of such rare tautomers, and a satisfactory explanation as to why took several decades to emerge. In fact, if we study this tautomeric shift problem using the Feynman path integral, and plot the reaction free energy profiles, the answer becomes clear. In the figure, we see two curves in the reaction profiles. One treats the protons as classical particles, and the other uses a proper quantum treatment. The two reaction profiles are clearly quite different. In the classical profile, we see that it is energetically steeply uphill to go from the normal form to the rare tautomer, but once up and over the hump, there is another energetic barrier to return to the normal form. Thus, it would be possible for the system to become stuck as a rare tautomer. However, with a proper quantum treatment, we see that, while there is still a steep uphill clime to get to the tautomeric shifted form, there is absolutely no barrier to return to the normal form. Thus, if there is a rare fluctuation that takes the system to the tautomeric form, it will immediately return to the normal form, thus explaining why spontaneous formation of rare tautomers is not observed. Thus, quantum mechanics, rather than being the evil force that gives us cancer, is harnessed by nature to prevent this from occurring in the first place.

The last example is that of a bulk aqueous acid solution. We learn in freshman chemistry that acids contains excess protons in the form of hydronium (\(H_3O^+\)) ions. Hydronium, as it turns out, is a highly species in water. What makes it so mobile is the fact that it does not diffuse hydrodynamically as would water molecules in the solution. Rather, hydronium undergoes a process known as structural diffusion, in which the hydronium structure is transported as a topological defect in water's hydrogen bond network through a series of proton transfer reactions between hydronium and its first solvation-shell water molecules. This process is driven by a complex series of hydrogen bond breaking and forming events that allow the structure to migrate through the hydrogen bond network. The basic mechanism is illustrated in the figure. However, what makes the process especially fast is the fact that the hydronium ion is a highly quantum mechanical species that can become very delocalized over the hydrogen bond network. If we examine some of the quantum paths and follow the location of the hydronium around the path, we see something strange occurring. At each point of the path, we ask where the hydronium oxygen is, and we find that at different points along the path, the hydronium oxygen is at a different site in the solution. In fact, as the figure shows, the hydronium ion can become delocalized over as many as five hydrogen bonds! By allowing the paths to become such spatially extended objects, quantum mechanics allows the system to probe the "most promising" direction for the next structural diffusion event. If this were not possible, hydronium diffusion would not be so fast, and devices such as proton-exchange-membrane fuel cells would not work as well as they do.