Quantum Chemistry/Example 24

Write a question and its solution showing the probability of a particle being located in the bottom left quadrant of the ground state of the particle in a square.

Particle In a Square (2D)
The particle in a square model is where a particle in confined to a restricted

region inside a 2D box. For this model we consider a particle which is free to move

anywhere inside the x,y dimensions but cannot exist outside of the 2D box where it would

have infinite potential energy.

The conditions for the potential energy are $$\nu= \begin{Bmatrix} \infty, & x<0,y<0 \\ 0, & 0\leq x, y\leq L \\\infty, & x,y>L \end{Bmatrix}$$ meaning that the particle spends 100% of its time within the walls of the box.

Since the particle in a square is a 2D model the wave function must take into account the x and y direction ; $$(\Psi(x,y)) $$

$$

The states the quantum system is in are defined by the quantum numbers $$n_x, n_y $$

In quantum energy levels only discrete energy levels are possible. In this model the particle cannot be motionless meaning the lowest possible energy level must be non-zero (n=1). Since the particle is always moving even in the ground state the particle has a a non- zero energy level. Therefore the ground state of a particle in a box is $$n_x=1, n_y=1 $$. In quantum mechanical systems there is a presence of nodes where in states above the ground state for the 2D model there are nodal lines that divide the square. As the quantum numbers increase so does the amount of nodal lines as $$N nodal lines =n_x,_y -1 $$. The sign of the wavefunction changes between the nodes and has a value of zero on the nodes.

Probability Distributions
The probability (P) of a particle being found in an interval can be found by integrating the probability distribution (P(x))

The probability of finding a particle in the range $$\left [ a_1,a_2 \right ] $$,$$\left [b_1,b_2 \right ] $$ is given by; $$P =\int\limits_{a_1}^{a_2} \int\limits_{b_1}^{b_2 } P(x,y)dxdy$$

The probability distribution of a a particle is the wave function squared.

For one particle in 2D (particle in a box):

$$P(x,y)=\left \vert \Psi(x,y)^2 \right \vert= \Psi(x,y)^* \cdot \Psi(x,y) $$ ,where * is the complex conjugate of the wave-function

Therefore the probability of finding a particle over a range can be re-written as:

$$P = \int\limits_{a_1}^{a_2} \int\limits_{b_1}^{b_2 } \Psi^* \bigl(x,y)\Psi \bigl(x,y)dxdy$$

Solution
If the particle is in the bottom left quadrant it is in the interval; $$x=\left [ 0, \frac{ L}{2} \right ] $$ $$y=\left [ 0, \frac{ L}{2} \right ] $$

Therefore the probability density is:

$$P =\int\limits_{0}^{ \tfrac{L}{2} } \int\limits_{ 0}^{ \tfrac{L} {2} } \Psi^* \bigl(x,y)\Psi \bigl(x,y)dxdy$$

The wave-function for a particle in a square is:

$$\Psi(x,y)=\frac{2}{L} sin\left ( \frac{n_x\pi}{L} x\right )\left ( \frac{n_y\pi}{L} y\right )\cdot\frac{2}{L} sin\left ( \frac{n_x\pi}{L} x\right )\left ( \frac{n_y\pi}{L} y\right )$$

In the ground state $$n_x =1$$ $$n_y =1$$, therefore the probability distribution can be written as:

$$P = \int\limits_{ 0}^{ \frac{L}{2}} \int\limits_{ 0}^{ \frac{L} {2}} \biggl ({2 \over L}\biggr) sin \left ( \frac{\pi}{L }x \right ) sin \left ( \frac{\pi}{L }y \right )\centerdot\biggl ({2 \over L}\biggr) sin \left ( \frac{\pi}{L }x \right ) sin \left ( \frac{\pi}{L }y \right )dx dy$$

Then solving the integral,

$$P= \left ( \frac{4}{L^2} \right )\int_{0}^{\frac{L}{2}} \int_{0}^{\frac{L}{2}}sin^2 \left ( \frac{ \pi}{ L} x\right )sin^2 \left ( \frac{\pi}{ L} y\right ) dx dy$$

$$P= \left ( \frac{4}{L^2} \right )\int_{0}^{\frac{L}{2}} sin^2 \left ( \frac{ \pi}{ L} x\right )dx \cdot \int_{0}^{\frac{L}{2}} sin^2 \left ( \frac{\pi}{ L} y\right ) dy$$

$$

$$P= \left ( \frac{4}{L^2} \right )\left [ \frac{ y}{2} - \frac{sin\bigl(2 \frac{\pi}{L}y \bigr)}{4\left ( \frac{\pi}{L} \right )} \right ]_0^L \left [ \frac{ x}{2} - \frac{sin\bigl(2 \frac{\pi}{L}x \bigr)}{4\left ( \frac{\pi}{L}  \right )} \right ]_0^L $$

$$P= \left ( \frac{4}{L^2} \right )\Biggl( \frac{ \frac{L}{2}}{2} - \frac{Lsin\bigl(2 \frac{\pi}{L}\frac{L}{2} \bigr)}{4\left ( \pi \right )} \Biggr) \Biggl( \frac{ \frac{L}{2}}{2} - \frac{Lsin\bigl(2 \frac{\pi}{L}\frac{L}{2} \bigr)}{4\left ( \pi  \right )} \Biggr) $$

$$P= \left ( \frac{4}{L^2} \right ) \Biggl( \frac{L}{4} - \frac{Lsin\bigl(\pi \bigr)}{4\left ( \pi \right )} \Biggr) \Biggl(  \frac{L}{4} - \frac{Lsin\bigl(\pi \bigr)}{4\left ( \pi  \right )} \Biggr) $$

Since $$sin\pi $$ is zero the whole term goes to zero

$$P= \left ( \frac{4}{L^2} \right )\Biggl(\frac{L}{4}-0\biggr) \Biggl(\frac{L}{4}-0\Biggr) $$

$$P= \frac{4}{8} $$

$$P= \frac{1}{4}$$

Therefore the the probability of finding a particle in the bottom left quadrant of a square in the ground state is 0.25 of 25%.