Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Consider a particle free to translate in one dimension. The classical Hamiltonian is \(H=p^{2} / 2 m\). a. Determine \(q_{\text {classical}}\) for this system. To what quantum system should you compare it in order to determine the equivalence of the classical and quantum statistical mechanical treatments? b. Derive \(q_{\text {classical}}\) for a system with translational motion in three dimensions for which \(H=\left(p_{x}^{2}+p_{y}^{2}+p_{z}^{2}\right) / 2 m\).

Short Answer

Expert verified
The classical partition function \(q_{\text{classical}}\) for a one-dimensional system with Hamiltonian \(H = \frac{p^2}{2m}\) is \(\infty\). For a three-dimensional system with Hamiltonian \(H = \frac{p_x^2 + p_y^2 + p_z^2}{2m}\), the classical partition function is given by \(\infty^3\times\left(\frac{2m\pi}{\beta}\right)^{\frac{3}{2}}\), which is not finite. The classical and quantum mechanical treatments are not equivalent in these cases.

Step by step solution

01

Recall the classical partition function formula

The classical partition function, q_classical, is given by the following integral: \(q_{classical} = \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} e^{-\beta H(q, p)} dq dp\) where \(\beta = 1/k_B T\). Here, the Hamiltonian H is a function of position 'q' and momentum 'p'; 'k_B' is the Boltzmann constant, and 'T' is the temperature.
02

Substitute the Hamiltonian for a 1D system

The one-dimensional Hamiltonian is given as: \(H = \frac{p^2}{2m}\) Substituting this into the partition function formula, we get: \(q_{classical} = \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} e^{-\beta \frac{p^2}{2m}} dq dp\)
03

Perform integration and determine q_classical

The above integral can be split into two separate integrals as follows: \(q_{classical} = \int_{-\infty}^{\infty} e^{-\beta \frac{p^2}{2m}}dp \int_{-\infty}^{\infty} dq\) The integral over 'q' is a trivial one, and it should evaluate to infinity because there are no limits on the particle's position in one dimension: \(\int_{-\infty}^{\infty} dq = \infty\) The integral over 'p' is a Gaussian integral, which evaluates to: \(\int_{-\infty}^{\infty} e^{-\beta \frac{p^2}{2m}}dp = \sqrt{\frac{2m\pi}{\beta}}\) Multiplying both integrals, we get: \(q_{classical} = \sqrt{\frac{2m\pi}{\beta}} \times \infty = \infty\) Comparing this with quantum systems, such as the infinite square well, the classical partition function is not finite; thus, classical and quantum mechanical treatments will not be equivalent in this case. #Part_b#
04

Substitute the Hamiltonian for a 3D system

The three-dimensional Hamiltonian is given as: \(H = \frac{p_x^2 + p_y^2 + p_z^2}{2m}\) Substituting this into the partition function formula, we get: \(q_{classical} = \int e^{-\beta \frac{(p_x^2 + p_y^2 + p_z^2)}{2m}} dq_x dq_y dq_z dp_x dp_y dp_z\)
05

Perform integrations

Like before, we can separate the integrations. This time there will be three position integrals and three momentum integrals. Each position integral is unbounded and will evaluate to infinity. Each momentum integral is again a Gaussian integral: \(q_{classical} = \left(\int_{-\infty}^{\infty} dq_x \right) \left( \int_{-\infty}^{\infty} dq_y \right) \left( \int_{-\infty}^{\infty} dq_z \right) \left( \int_{-\infty}^{\infty} e^{-\beta \frac{p_x^2}{2m}}dp_x \right) \left( \int_{-\infty}^{\infty} e^{-\beta \frac{p_y^2}{2m}}dp_y \right) \left( \int_{-\infty}^{\infty} e^{-\beta \frac{p_z^2}{2m}}dp_z \right)\) Thus, we get: \(q_{classical} = (\infty) (\infty) (\infty) \left(\sqrt{\frac{2m\pi}{\beta}}\right) \left(\sqrt{\frac{2m\pi}{\beta}}\right) \left(\sqrt{\frac{2m\pi}{\beta}}\right)\) Simplifying we now receive: \(q_{classical} = \infty^3 \times \left(\frac{2m\pi}{\beta}\right)^{\frac{3}{2}}\) The classical partition function for the given three-dimensional system is thus proportional with cubic infinity times the factor \(\left(\frac{2m\pi}{\beta}\right)^{\frac{3}{2}}\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Statistical Mechanics
Statistical mechanics is a branch of theoretical physics that uses probability theory to study the average behavior of a mechanical system where the state of the system is uncertain. This discipline is particularly important for explaining the thermodynamic behavior of large systems. In the context of our exercise, statistical mechanics helps us to compute the partition function, which plays a crucial role in determining the macroscopic properties of the system from its microscopic constituents.

The partition function, denoted as \( q_{classical} \), is a sum over all possible states of the system, each weighted by the Boltzmann factor \( e^{-\beta H(q, p)} \), where \(\beta = 1/k_B T\) is the inverse temperature multiplied by the Boltzmann constant, \(H(q, p)\) is the Hamiltonian representing the total energy of a state, 'q' represents coordinates, and 'p' represents momenta. For a classical system where positions and momenta can take any continuous values, this sum turns into an integral over all possible values of \((q, p)\).
Hamiltonian Mechanics
Hamiltonian mechanics is a reformulation of classical mechanics that predicts the same outcomes as Newtonian mechanics but works with different mathematical tools. It is especially useful for problems in statistical mechanics.

In Hamiltonian mechanics, the Hamiltonian function \(H\) represents the total energy of the system, which is the sum of kinetic and potential energies. For a particle in free space, like in our exercise, the Hamiltonian depends only on the kinetic energy since there is no potential. The kinetic energy for one-dimensional motion is given by \( \frac{p^2}{2m} \) for momentum \(p\) and mass \(m\), and it's straightforwardly extended for three-dimensional motion to \( \frac{p_x^2 + p_y^2 + p_z^2}{2m} \).

The classical partition function involves integrating over this Hamiltonian function multiplied by the Boltzmann factor, and in the absence of potential energy, it simplifies the problem significantly. However, we see from the exercise that this simplification results in an infinite value when the system's degrees of freedom are not confined.
1D and 3D Translational Motion
Translational motion refers to the movement of a particle or system of particles in space. In our exercise, we are considering a particle free to move in either one dimension (1D) or three dimensions (3D).

In the case of 1D motion, the particle is free to move along a single axis, and the partition function reflects this unconstrained motion as it integrates over all possible positions and momenta along that axis. The result, as indicated by the exercise, is an infinite partition function due to the integration over all possible positions.

Similarly, for 3D motion, the particle can move in three mutually perpendicular directions, and the partition function calculation requires integrating over all possible positions and momenta in three-dimensional space. Again, the contributions from the spatial integrals are infinite for free particles. Only the momentum integrals give finite Gaussian results, but when multiplied by the infinite spatial integrals, the overall partition function becomes infinite.

The infinite values in the classical partition function for free translational motion in both 1D and 3D hint at the limitations of classical physics in describing quantum systems. Quantum mechanics introduces a discrete nature to energy states, which often results in a finite partition function, whereas the classical approach leads to divergences when systems are not confined.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Evaluate the vibrational partition function for \(\mathrm{H}_{2} \mathrm{O}\) at \(2000 .\) K where the vibrational frequencies are 1615,3694 and \(3802 \mathrm{cm}^{-1}\)

In using statistical mechanics to describe the thermodynamic properties of molecules, high-frequency vibrations are generally not of importance under standard thermodynamic conditions since they are not populated to a significant extent. For example, for many hydrocarbons the \(\mathrm{C}-\mathrm{H}\) stretch vibrational degrees of freedom are neglected. Using cyclohexane as an example, the IR-absorption spectrum reveals that the \(C-H\) stretch transition are located at \(\sim 2850 \mathrm{cm}^{-1}\). a. What is the value of the vibrational partition function for a mode of this frequency at \(298 \mathrm{K} ?\) b. At what temperature will this partition function reach a value of \(1.1 ?\)

What is the symmetry number for the following molecules? a. \(^{35} \mathrm{Cl}^{37} \mathrm{Cl}\) b. \(^{35} \mathrm{Cl}_{2}\) \(\mathbf{c} .^{16} \mathrm{O}_{2}\) d. \(\mathrm{C}_{6} \mathrm{H}_{6}\) \(\mathbf{e} . \mathrm{CH}_{2} \mathrm{Cl}_{2}\)

Determine the total molecular partition function for gaseous \(\mathrm{H}_{2} \mathrm{O}\) at \(1000 .\) K confined to a volume of \(1.00 \mathrm{cm}^{3}\) The rotational constants for water are \(B_{A}=27.8 \mathrm{cm}^{-1}\) \(B_{B}=14.5 \mathrm{cm}^{-1},\) and \(B_{C}=9.95 \mathrm{cm}^{-1} .\) The vibrational frequencies are \(1615,3694,\) and \(3802 \mathrm{cm}^{-1}\). The ground electronic state is nondegenerate.

A general expression for the classical Hamiltonian is \(H=\alpha p_{i}^{2}+H^{\prime}\) where \(p_{i}\) is the momentum along one dimension for particle \(i, \alpha\) is a constant, and \(H^{\prime}\) are the remaining terms in the Hamiltonian. Substituting this into the equipartition theorem yields \(q=\frac{1}{h^{3 N}} \iint e^{-\beta\left(\alpha p_{i}^{2}+H^{\prime}\right)} d p^{3 N} d x^{3 N}\) a. Starting with this expression, isolate the term involving \(p_{i}\) and determine its contribution to \(q\) b. Given that the average energy \(\langle\varepsilon\rangle\) is related to the partition function as follows, \(\langle\varepsilon\rangle=\frac{-1}{q}\left(\frac{\delta q}{\delta \beta}\right)\) evaluate the expression the contribution from \(p_{i} .\) Is your result consistent with the equipartiton theorem?

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free