Answer
(a) $C=4$
(b) $P\left( {\ell = 2;n = 3;X \le \frac{1}{4};Y \le \frac{1}{8}} \right) \simeq 0.0379$
Work Step by Step
We have the joint probability density function:
$p\left( {x,y} \right) = \left\{ {\begin{array}{*{20}{c}}
{C{{\sin }^2}\left( {2\pi \ell x} \right){{\sin }^2}\left( {2\pi ny} \right)}&{{\rm{if}}{\ }\left( {x,y} \right) \in {\cal R}}\\
0&{{\rm{otherwise}}}
\end{array}} \right.$
where ${\cal R} = \left[ {0,1} \right] \times \left[ {0,1} \right]$.
(a) Recall the conditions that a joint probability density function must satisfy:
1. First condition: $p\left( {x,y} \right) \ge 0$ for all $x$ and $y$, since probabilities cannot be negative.
Since the square of sine functions are nonnegative, we require that $C \ge 0$.
2. Second condition: the normalization condition must hold, namely Eq. (5) must be satisfied:
(5) ${\ \ \ \ \ }$ $\mathop \smallint \limits_{ - \infty }^\infty \mathop \smallint \limits_{ - \infty }^\infty p\left( {x,y} \right){\rm{d}}y{\rm{d}}x = 1$
From the definition of $p\left( {x,y} \right)$ we obtain the domain ${\cal D}$, where $p$ is nonzero. So, the description of ${\cal D}$ is
${\cal D} = \left\{ {\left( {x,y} \right)|0 \le x \le 1,0 \le y \le 1} \right\}$
Then, using Eq. (5) we evaluate:
$\mathop \smallint \limits_{ - \infty }^\infty \mathop \smallint \limits_{ - \infty }^\infty p\left( {x,y} \right){\rm{d}}y{\rm{d}}x = C\mathop \smallint \limits_{x = 0}^1 \mathop \smallint \limits_{y = 0}^1 {\sin ^2}\left( {2\pi \ell x} \right){\sin ^2}\left( {2\pi ny} \right){\rm{d}}y{\rm{d}}x = 1$
$C\left( {\mathop \smallint \limits_{x = 0}^1 {{\sin }^2}\left( {2\pi \ell x} \right){\rm{d}}x} \right)\left( {\mathop \smallint \limits_{y = 0}^1 {{\sin }^2}\left( {2\pi ny} \right){\rm{d}}y} \right) = 1$
Using the Double-angle formulas in Section 1.4:
${\sin ^2}x = \frac{1}{2}\left( {1 - \cos 2x} \right)$
we get
$\frac{C}{4}\left( {\mathop \smallint \limits_{x = 0}^1 \left( {1 - \cos \left( {4\pi \ell x} \right)} \right){\rm{d}}x} \right)\left( {\mathop \smallint \limits_{y = 0}^1 \left( {1 - \cos \left( {4\pi ny} \right)} \right){\rm{d}}x} \right) = 1$
$\frac{C}{4}\left( {\left( {x - \frac{1}{{4\pi \ell }}\sin \left( {4\pi \ell x} \right)} \right)|_0^1} \right)\left( {\left( {y - \frac{1}{{4\pi n}}\sin \left( {4\pi ny} \right)} \right)|_0^1} \right) = 1$
$\frac{C}{4}\left( {1 - \frac{1}{{4\pi \ell }}\sin \left( {4\pi \ell } \right)} \right)\left( {1 - \frac{1}{{4\pi n}}\sin \left( {4\pi n} \right)} \right) = 1$
Since $\ell $ and $n$ are integers, we have $\sin \left( {4\pi \ell } \right) = 0$ and $\sin \left( {4\pi n} \right) = 0$. Thus,
$\frac{C}{4} = 1$
So, $C=4$.
(b) From part (a) we obtain:
$p\left( {x,y} \right) = \left\{ {\begin{array}{*{20}{c}}
{4{{\sin }^2}\left( {2\pi \ell x} \right){{\sin }^2}\left( {2\pi ny} \right)}&{{\rm{if}}{\ }\left( {x,y} \right) \in {\cal R}}\\
0&{{\rm{otherwise}}}
\end{array}} \right.$
where ${\cal R} = \left[ {0,1} \right] \times \left[ {0,1} \right]$.
Evaluate the probability that a particle with $\ell = 2$, $n=3$ lies in the region $\left[ {0,\frac{1}{4}} \right] \times \left[ {0,\frac{1}{8}} \right]$:
$P\left( {\ell = 2;n = 3;X \le \frac{1}{4};Y \le \frac{1}{8}} \right) = 4\mathop \smallint \limits_{x = 0}^{1/4} \mathop \smallint \limits_{y = 0}^{1/8} {\sin ^2}\left( {4\pi x} \right){\sin ^2}\left( {6\pi y} \right){\rm{d}}y{\rm{d}}x$
$ = 4\left( {\mathop \smallint \limits_{x = 0}^{1/4} {{\sin }^2}\left( {4\pi x} \right){\rm{d}}x} \right)\left( {\mathop \smallint \limits_{y = 0}^{1/8} {{\sin }^2}\left( {6\pi y} \right){\rm{d}}y} \right)$
Using the Double-angle formulas in Section 1.4:
${\sin ^2}x = \frac{1}{2}\left( {1 - \cos 2x} \right)$
we get
$P\left( {\ell = 2;n = 3;X \le \frac{1}{4};Y \le \frac{1}{8}} \right) = \left( {\mathop \smallint \limits_{x = 0}^{1/4} \left( {1 - \cos \left( {8\pi x} \right)} \right){\rm{d}}x} \right)\left( {\mathop \smallint \limits_{y = 0}^{1/8} \left( {1 - \cos \left( {12\pi y} \right)} \right){\rm{d}}x} \right)$
$ = \left( {\left( {x - \frac{1}{{8\pi }}\sin \left( {8\pi x} \right)} \right)|_0^{1/4}} \right)\left( {\left( {y - \frac{1}{{12\pi }}\sin \left( {12\pi y} \right)} \right)|_0^{1/8}} \right)$
$ = \left( {\frac{1}{4} - \frac{1}{{8\pi }}\sin \left( {2\pi } \right)} \right)\left( {\frac{1}{8} - \frac{1}{{12\pi }}\sin \left( {\frac{3}{2}\pi } \right)} \right)$
$ = \left( {\frac{1}{4}} \right)\left( {\frac{1}{8} + \frac{1}{{12\pi }}} \right) \simeq 0.0379$
So, $P\left( {\ell = 2;n = 3;X \le \frac{1}{4};Y \le \frac{1}{8}} \right) \simeq 0.0379$.
