Quantum Numbers and Probability Theory: From Intractability to Prediction

Probability Theory in Many-Particle Systems

• Challenge with Mechanical Laws: For systems involving many particles (e.g., moles), applying classical mechanical laws to track the positions and velocities (or quantum numbers) of each individual particle becomes intractable. This complexity renders direct application of mechanical laws impractical for predicting system behavior.

• Solution: Probability Theory: To overcome the intractability, probability theory is employed as a powerful tool to describe the average behavior of many-particle systems.

  • Average of Discrete Measurements:

    • If $n$ represents one of the possible outcomes of a measurement, and $P<em>n$ is the probability of measuring outcome $n$, then the average (or expectation value) of many such measurements, denoted as $\langle n \rangle$, is calculated as a sum: $\langle n \rangle = \sum</em>n n P_n$

    • Example (Energy States): Consider a particle that can exist in one of three energy states:

      • State 1: Energy $E<em>1 = 1.0 \text{ eV}$ with probability $P</em>1 = 0.507$

      • State 2: Energy $E<em>2 = 0.01 \text{ eV}$ with probability $P</em>2 = 0.307$

      • State 3: Energy $E<em>3 = 0.1 \text{ eV}$ with probability $P</em>3 = 0.186$

      • The average energy ($\langle E \rangle$) is calculated as:
        $\langle E \rangle = (1.0 \text{ eV})(0.507) + (0.01 \text{ eV})(0.307) + (0.1 \text{ eV})(0.186) = 0.0321 \text{ eV}$

  • Average of Continuous Measurements:

    • When measuring continuous quantities (e.g., distance, speed), integrals are used instead of sums.

    • If $x$ is the outcome of a measurement, and the probability of outcome $x$ occurring within an infinitesimal range $dx$ is $P(x) dx$, then the average outcome ($\langle x \rangle$) is given by the integral:
      $\langle x \rangle = \int x P(x) dx$

    • Example (Bond Stretching/Compression): The probability of a particular bond being stretched or compressed by a distance $x$ from its equilibrium bond length can be described by the function: $P(x) = \frac{1}{\sqrt{\pi/\alpha}} e^{-\alpha x^2}$

      • The average distance the bond is compressed or stretched (its average displacement) would then be calculated as:
        $\langle x \rangle = \int x \frac{1}{\sqrt{\pi/\alpha}} e^{-\alpha x^2} dx$

Maxwell-Boltzmann Distribution

• Introduction: The Maxwell-Boltzmann distribution describes the distribution of molecular speeds in a perfect (ideal) gas at equilibrium.

• Starting Point: Boltzmann Distribution:

  • The fundamental Boltzmann distribution provides the likelihood that a system has a particular energy $E$:
    $P(E) = A e^{-E/(k_B T)}$

  • This distribution is universally applicable for any system at a constant temperature ($T$).

  • $k_B$ is the Boltzmann constant, with a value of approximately $1.380 \times 10^{-23} \text{ J/K}$. (The precise value is $1.380649 \times 10^{-23} \text{ J/K}$).

  • $T$ represents the absolute temperature.

  • For a perfect gas, it consists of non-interacting gas particles.

• Energy of One Gas Particle: The kinetic energy of a single gas particle at any instant, considering translational motion in three dimensions, is given by:
  $E = \frac{1}{2}m(v<em>x^2 + v</em>y^2 + v<em>z^2)$ where $m$ is the mass of the particle and $v</em>x, v<em>y, v</em>z$ are its velocity components.

• Probability Distribution of Velocities: Substituting the energy expression into the Boltzmann distribution, the probability of a particle having specific velocity components ($v<em>x, v</em>y, v<em>z$) is: $P(v</em>x, v<em>y, v</em>z) = A e^{-m(v<em>x^2 + v</em>y^2 + v<em>z^2)/(2k</em>B T)}$
  where $A$ is a normalization constant related to $(\frac{m}{2\pi k_B T})^{3/2}$.

• Transition from Velocities to Speeds:

  • In a perfect gas, the direction of molecular motion is random. Therefore, we are typically interested in the distribution of speeds ($v = \sqrt{v<em>x^2 + v</em>y^2 + v_z^2}$) rather than specific velocity components.

  • To obtain the speed distribution, we multiply the velocity distribution by the surface area of a sphere of radius $v$ in velocity space, which is $4\pi v^2$. This accounts for all possible directions for a given speed.

• The Maxwell-Boltzmann Speed Distribution Function: Combining these elements, the Maxwell-Boltzmann distribution function for molecular speeds, $f(v)$, is:
  $f(v) = 4\pi \left( \frac{m}{2\pi k<em>B T} \right)^{3/2} v^2 e^{-mv^2/(2k</em>B T)}$
  This function gives the proportion of molecules having a speed between $v$ and $v + dv$.

• Characteristic Speeds for a Gas: From the Maxwell-Boltzmann distribution, several characteristic speeds can be derived:

  • Most Probable Speed ($v<em>{mp}$): The speed at which the distribution function $f(v)$ has its maximum value. It represents the speed possessed by the largest number of molecules.
    $v</em>{mp} = \sqrt{\frac{2 k_B T}{m}}$

  • Average Speed ($\bar{v}$): The arithmetic mean of the speeds of all molecules.
    $\bar{v} = \sqrt{\frac{8 k_B T}{\pi m}}$

  • Root-Mean-Square Speed ($v<em>{rms}$): The square root of the average of the squares of the speeds. This is related to the average kinetic energy.
    $v</em>{rms} = \sqrt{\frac{3 k_B T}{m}}$

• Example Calculation: Nitrogen ($N_2$) Gas at $298 \text{ K}$

  • Temperature: $T = 298 \text{ K}$

  • Mass of an $N_2$ molecule: $m = 28.014 \text{ u} = 4.65 \times 10^{-26} \text{ kg}$

  • Boltzmann constant: $k_B = 1.38 \times 10^{-23} \text{ J/K}$

  • Calculated Characteristic Speeds:

    • Most probable speed ($v_{mp}$): $433 \text{ m/s}$

    • Average speed ($\bar{v}$): $488 \text{ m/s}$

    • Root-mean-square speed ($v_{rms}$): $529 \text{ m/s}$

  • Variance Related Quantity: The variance of speeds is typically calculated differently. However, a related quantity mentioned is:
    $\sigma^2 = \frac{3 k<em>B T}{m} - \frac{8 k</em>B T}{\pi m} = v<em>{rms}^2 - \bar{v}^2$ For the given $N</em>2$ example, this leads to an approximate standard deviation:
    $\sigma \approx 200 \text{ m/s}$