User:
Example of free particles
Storyboard 
Once we have defined the method for counting states and estimating probabilities in situations of interest, we can delve into how a system of many free particles behaves.
ID:(435, 0)
Quantum Calculation of Number of States
Equation
Para poder calcular probabilidades debemos contabilizar las veces que una situación se da. En este caso podemos discretizar el espacio de fase en intervalos de largo
Cada una de estas celdas es de un 'volumen'
Por ello con se tiene que
| $\Delta p\Delta q \sim h$ |
Surge asi una red de puntos discretos en el espacio de fase. Cada sistema esta en un estado que corresponde a uno de estos puntos.
La probabilidad de que el sistema se encuentre en un punto en particular se determina contabilizando los sistemas del ensamble que cumplen esta condición dividido por todos los posibles estados.
ID:(527, 0)
Case Classical Mechanics
Description
In classical mechanics, a system is described by the coordinates $q_1, q_2, \ldots, q_f$ and momenta $p_1, p_2, \ldots, p_f$, where $f$ represents the number of degrees of freedom. The state of the system is represented as a point in phase space, given by $(q_1, q_2, \ldos, q_f, p_1, p_2, \ldos, p_f)$.
In the case of a system consisting of $N$ free particles, described using a total of $3N$ coordinates, the number of degrees of freedom is defined as $f = 3N$.
ID:(524, 0)
Case Quantum Mechanics
Description
In quantum mechanics, the state is described by the wave function $\psi$, which depends on the variables $q_1, q_2, \ldos, q_f$, where $f$ is the number of degrees of freedom of the system.
The wave function is a solution, in the non-relativistic case and for particles without spin, of the Schrödinger equation. Eigenvalues are associated with wave functions, which typically depend on integers. These integers represent possible states of the system that are bounded by the system's energy.
ID:(523, 0)
Case of a free particle gas, volume
Equation
In the case of free particles, there is no positional dependence, and when calculating the phase space, it is necessary to sum or integrate over all positions.
Therefore, using the notation of , we have:
 $V=\displaystyle\int_V d^3q$ |
ID:(522, 0)
Calculation of the number of states
Description
In classical mechanics, a system is described by the coordinates $q_1, q_2, \ldots, q_f$ and momenta $p_1, p_2, \ldots, p_f$, where $f$ represents the number of degrees of freedom. The state of the system is represented as a point in phase space, given by $(q_1, q_2, \ldots, q_f, p_1, p_2, \ldos, p_f)$.
In the case of a system consisting of $N$ free particles, described using a total of $3N$ coordinates, the number of degrees of freedom is defined as $f = 3N$.
ID:(10580, 0)
Case of a free particle gas
Equation
In the case of $N$ free particles in the classical approximation, we need to perform an integration in phase space with the constraint that describes the system.
Since it's a gas of free particles, the constraint is solely related to energy and doesn't depend on position. Therefore, we can express the energy as:
$E=\displaystyle\frac{1}{2m}\displaystyle\sum_i^N\vec{p}_i^2$
This expression can be simplified using mathematical notation as:
| $2mE=\displaystyle\sum_i^N\vec{p}_i^2$ |
This formula represents a "sphere of radius" $\sqrt{2mE}$ in a "phase space" of $3N$ dimensions.
ID:(528, 0)
Number of states of a gas of free particle
Equation
La integral sobre el espacio de estados se calcula con posición $m$ and volumen $m^3$ mediante
| $V=\displaystyle\int_V d^3q$ |
con la condición para la energía que con energía del sistema $J$, masa de la partícula $kg$, momento de la i-esima partícula $J$ and numero de Partículas $-$ es
| $2mE=\displaystyle\sum_i^N\vec{p}_i^2$ |
\\n\\nse deja integrar en forma simple en las coordenadas espaciales. Cada integral es igual al volumen
$3N-1\sim 3N$
Por ello se puede estimar con energía del sistema $J$, masa de la partícula $kg$, momento de la i-esima partícula $J$ and numero de Partículas $-$
| $\Omega(E,N)=\Omega_0\left(\displaystyle\frac{V}{\Delta q^3}\right)^N\left(\displaystyle\frac{2mE}{\Delta p^2}\right)^{3N/2}$ |
ID:(3433, 0)
Number of free particle states with cell
Equation
En el calculo del número de estados se obtiene el número de estados con energía del sistema $J$, factor de normalización $-$, incerteza en el momento $kg m/s$, incerteza en la posición $m$, masa de la partícula $kg$, numero de estados para energía y partículas dadas $-$, numero de Partículas $-$ and volumen $m^3$ son
| $\Omega(E,N)=\Omega_0\left(\displaystyle\frac{V}{\Delta q^3}\right)^N\left(\displaystyle\frac{2mE}{\Delta p^2}\right)^{3N/2}$ |
Como el elemento de volumen del espacio de fase es con incerteza en el momento $kg m/s$, incerteza en la posición $m$ and planck constant $J s$ igual a
| $\Delta p\Delta q \sim h$ |
por lo que el número de estados se deja simplificar con incerteza en el momento $kg m/s$, incerteza en la posición $m$ and planck constant $J s$ a
| $ \Omega = \Omega_0 \left(\displaystyle\frac{2 m }{ h ^2}\right)^{3 N /2} V ^ N E ^{3N/2}$ |
ID:(4805, 0)
0
Video
Video: Example of free particles