91影视

Given wave function is,${蠄}_{(x,t)}=A{e}^{-位\left|x\right|}{e}^{-i蝇t}$

Where the real constants are A, 位 and 蝇.

a)

The condition for normalization of wave equation is:

${鈭�}_{-鈭�}^{+x}{蠄}^{*}蠄dx=1$

The value of ${蠄}^{*}\left(x\right)$ is $蠄\left(x\right)=A{e}^{-位\left|x\right|}{e}^{-iex}$ ,So

${蠄}^{*}\left(x\right)=A路e^{-位|路|}{e}^{iet}$

The normalization condition can be written:

$|A{|}^2{鈭�}_{-鈭�}^{+鈭�}{e}^{-2位\left|x\right|}路dx=1$

For being even function, the equation can be written as

$\begin{array}{rcl}2{\left|A\right|}^2{鈭�}_0^{+鈭�}{e}^{-2位x}dx& =& 1\\ \frac{2|A{|}^2}{-2位}\left(e^{-2位(*)}-e^{-2位\left(0\right)}\right)& =& 1\\ \frac{|A{|}^2}{-位}(0-1)& =& 1\\ A& =& \sqrt{位}\\ & & \end{array}$

So, the normalization of the wave equation is $\sqrt{位}{e}^{-位鈭�x}{e}^{i蝇t}$.

(b)

Theaverage value of x is $鉄�x鉄�={鈭�}_{-鈭�}^{+鈭�}{蠄}^{*}x蠄dx$.

As$蠄={蠄}^{*}$ . So,

$\begin{array}{rcl}鉄�x鉄�& =& {鈭�}_{-鈭�}^{+鈭�}x|y{|}^2路dx\\ & =& |A{|}^2{鈭�}_{-鈭�}^{+鈭�}x路e^{-2位\left|x\right|}dx\\ & & \end{array}$

Being $x路e^{-2位\left|x\right|}$an odd function,the value of $鉄�x鉄�=0$.

So, the value of $鉄�x^2鉄�$ is as follows:

According to the gamma function, the value will be ${鈭�}_0^{鈭�}{x}^{n-1}{e}^{-ax}路dx=\frac{螕\left(n\right)}{a^n}$.

So,

The expectation value of x is 0 and .

(c)

The standard deviation is given by,

After substitution the value of and $鉄�x鉄�$ is:

$\begin{array}{rcl}{蟽}^2& =& \frac{1}{2{位}^2}-0\\ & =& \frac{1}{2{位}^2}蟽\\ & =& \frac{1}{\left(\sqrt{2}\right)位}\\ & & \end{array}$

The expression $\left|蠄\right(卤蟽){|}^2$ have to be calculated to make the graph $|蠄{|}^2$and to mark the points. So,

$\begin{array}{rcl}{\left|蠄(卤蟽)\right|}^2& =& |A{|}^2{e}^{-2位蟽}\\ & =& 位路e^{-2位\left(\frac{1}{\sqrt{2}}\right)位}\\ & =& 位路e^{-\sqrt{2}}\\ & =& 位e^{-1.414}\\ & =& \frac{位}{e^{1.414}}\\ & =& 0.2431位\\ & & \end{array}$

According to the above value,the graph is shown below,

To determine the likelihood that the particle will be discovered outside of this range,

$\begin{array}{rcl}P& =& {鈭�}_{-鈭�}^{-蟽}|蠄{|}^{2}dx+{鈭�}_a^{+鈭�}|蠄{|}^{2}dx\\ & =& 2{鈭�}_{蟽}^{鈭�}|蠄{|}^{2}dx\\ & =& 2|A{|}^2{鈭�}_0^{鈭�}{e}^{-2位x}dx\\ & =& 2位{\left(\frac{e^{-2位x}}{-2位}\right)}_0^{鈭�}\\ & & \end{array}$

Further solving above equation as,

$\begin{array}{rcl}P& =& -\left(e^{-2位(鈭�)}-e^{-2位a}\right)\\ & =& -\left(\frac{1}{鈭�}-e^{-2位{\left(\frac{1}{\sqrt{2}}\right)}^{位}}\right)\\ & =& -\left(0-e^{-\sqrt{2}}\right)\\ & =& -(-0.2431)\\ & =& 0.2431\\ & & \end{array}$

So,the value of P is $0.2431$.