# Quantum Chemistry/Probability and Statistics

Probability distributions describe the likelihood of a variable taking on a given range of values. The probability (P) of an arbitrary event happening can be calculated using the following formula, where M represents the number of times the event occurred over N number of trials.
    Definition of Probability $P=\lim _{N\to \infty }{M \over N}$
The value of P can range between 0 (event is impossible) and 1 (event is guaranteed). Suppose an event has a range of possible outcomes. By repeatedly calculating the frequency of each individual outcome, one can derive a probability distribution over all outcomes, describing the likelihood of any possible outcome occurring in a single trial.
Several key properties characterize these distributions. The probability of any event or outcome must be a non-negative value, as negative probabilities are physically meaningless. Second, the total probability must be normalized, meaning the sum of probabilities for all possible outcomes must equal one. Additionally, probability distributions can either be discrete or continuous. Discrete distributions apply when a variable can take on a limited number of distinct values, allowing the probability of a specific outcome to be determined. In contrast, continuous distributions apply for variables that can assume any value within a range, such as the position of an electron in space. This is common in quantum mechanics, where probabilities are associated with continuous variables, like the x-axis. In such cases, calculating the probability of finding a particle at an exact point (e.g., x = 0.5000) is practically meaningless, as the probability at any single point is effectively zero. Instead, it is more useful to consider the probability of locating the particle within a small interval along the x-axis. The probability (P) can be calculated by integrating the probability distribution (P(x)) over an interval $a_{1}\leq x\leq a_{2}$.
    Probability of Finding a Value within an Interval $P=\int \limits _{a_{1}}^{a_{2}}P(x)dx$
In quantum mechanics, probability and statistics play an essential role in interpreting and predicting the behavior of particles. The system is defined by a wavefunction, 𝜓, which is used to derive various properties of the system, such as position, momentum and energy. The probability distribution for a quantum mechanical particle is the square of the wavefunction. The probability distribution can then be used to determine the probability of a particle being in a certain interval or the average value of that property, via integration.
    The probability distribution of a quantum mechanical particle in 3D $|\psi (x,y,z)|^{2}=\psi (x,y,z)^{*}\cdot \psi (x,y,z)=P(x,y,z)$
The simplest quantum system that can be described is called “particle in a 1D Box”. The particle has a probability of existing anywhere along a 1D box ($P=1$ when $0<x<L$ ) but can never exist outside of the box ($P=0$ when $x<0$ or $x>L$ ). The probability of the particle existing within the bounds of the box is one, since the total probability of it being located somewhere in the allowed region is 100%. Using probability distributions, one can determine the likelihood of a particle being in any specific range along the x-axis.

## Example

Calculate the probability of the particle being located between $x=0$ and $x=L/2$ in the first quantum state (n=1) for a particle in a 1D box.
Solution: The wavefunction for a particle in a 1D box is,
$\psi (x)={\sqrt {\frac {2}{L}}}\sin \left({\frac {n\pi x}{L}}\right)$
The probability density is the square of the wavefunction:
$|\psi (x)|^{2}={\frac {2}{L}}\sin ^{2}\left({\frac {n\pi x}{L}}\right)$
For the interval $0<x<L/2$, the integral becomes,
$P=\int \limits _{a_{1}}^{a_{2}}P(x)dx=\int \limits _{0}^{L/2}{\frac {2}{L}}\sin ^{2}\left({\frac {n\pi x}{L}}\right)dx={\frac {2}{L}}\int \limits _{0}^{L/2}\sin ^{2}\left({\frac {\pi x}{L}}\right)dx$
Substituting $a=\pi /L$,
$P={\frac {2}{L}}\int \limits _{0}^{L/2}\sin ^{2}\left({\frac {\pi x}{L}}\right)dx={\frac {2}{L}}\int \limits _{0}^{L/2}\left[{\frac {x}{2}}-{\frac {\sin(2ax)}{4a}}\right]dx$
$P={\frac {2}{L}}\left[\left({\frac {L}{4}}\right)-{\frac {L\sin \left({\frac {2\pi {\frac {L}{2}}}{L}}\right)}{4\pi }}\right]={\frac {2}{L}}\left[{\frac {L}{4}}-{\frac {L\sin(\pi )}{4\pi }}\right]={\frac {2}{L}}\cdot {\frac {L}{4}}={\frac {1}{2}}$
Therefore, the probability of finding a particle in the first half of the 1D box is $1/2$, or $50\%$.