User:Mikayla Clarke

Find 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.

The conditions for the potential energy are $$\nu= \begin{Bmatrix} \infty, & x<0,y<0 \\ 0, & 0\leq x, y\leq L \\\infty, & x,y>L \end{Bmatrix}$$

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 wave function for a particle in a square
$$\Psi(x,y)=\frac{2}{L} sin\left ( \frac{n_x\pi}{L} x\right )\left ( \frac{n_y\pi}{L} y\right )\cdot\frac{2}{L} sin\left ( \frac{n_x\pi}{L} x\right )\left ( \frac{n_y\pi}{L} y\right )$$

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

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

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,

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

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

Using the integral- from table

$$\int sin^2(ax)dx=\frac{x}{2}- \frac{sin(2ax)}{4a} $$

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

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

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

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%.