Consider the wave function
${\psi }_{(x,t)}=A{e}^{-\lambda \left|x\right|}{e}^{-i\omega t}$

whereA, λ, and ω are positive real constants. (We’ll see in Chapter for what potential (V) this wave function satisfies the Schrödingerequation.)

(a) Normalize$\psi$ .

(b) Determine the expectation values of$x$ and $x^2$.

(c) Find the standard deviation of . Sketch the graph of${\left|\mathbf{\Psi }\right|}^{\mathbf{2}}$ , as a function ofx, and mark the points $(⟨x⟩+\sigma )$and $(⟨x⟩-\sigma )$, to illustrate the sense in whichσ represents the “spread” inx. What is the probability that the particle would be found outside this range?

Given wave function is,${\psi }_{(x,t)}=A{e}^{-\lambda \left|x\right|}{e}^{-i\omega t}$

Where the real constants are A, λ and ω.

a)

The condition for normalization of wave equation is:

${\int }_{-\infty }^{+x}{\psi }^{*}\psi dx=1$

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

${\psi }^{*}\left(x\right)=A·e^{-\lambda |·|}{e}^{iet}$

The normalization condition can be written:

$|A{|}^2{\int }_{-\infty }^{+\infty }{e}^{-2\lambda \left|x\right|}·dx=1$

For being even function, the equation can be written as

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

So, the normalization of the wave equation is $\sqrt{\lambda }{e}^{-\lambda \mid x}{e}^{i\omega t}$.

(b)

Theaverage value of x is $⟨x⟩={\int }_{-\infty }^{+\infty }{\psi }^{*}x\psi dx$.

As$\psi ={\psi }^{*}$ . So,

$\begin{array}{rcl}⟨x⟩& =& {\int }_{-\infty }^{+\infty }x|y{|}^2·dx\\ & =& |A{|}^2{\int }_{-\infty }^{+\infty }x·e^{-2\lambda \left|x\right|}dx\\ & & \end{array}$

Being $x·e^{-2\lambda \left|x\right|}$an odd function,the value of $⟨x⟩=0$.

So, the value of $⟨x^2⟩$ is as follows:

According to the gamma function, the value will be ${\int }_0^{\infty }{x}^{n-1}{e}^{-ax}·dx=\frac{\Gamma \left(n\right)}{a^n}$.

So,

The expectation value of x is 0 and .

(c)

The standard deviation is given by,

After substitution the value of and $⟨x⟩$ is:

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

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

$\begin{array}{rcl}{\left|\psi (\pm \sigma )\right|}^2& =& |A{|}^2{e}^{-2\lambda \sigma }\\ & =& \lambda ·e^{-2\lambda \left(\frac{1}{\sqrt{2}}\right)\lambda }\\ & =& \lambda ·e^{-\sqrt{2}}\\ & =& \lambda e^{-1.414}\\ & =& \frac{\lambda }{e^{1.414}}\\ & =& 0.2431\lambda \\ & & \end{array}$

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

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

$\begin{array}{rcl}P& =& {\int }_{-\infty }^{-\sigma }|\psi {|}^{2}dx+{\int }_a^{+\infty }|\psi {|}^{2}dx\\ & =& 2{\int }_{\sigma }^{\infty }|\psi {|}^{2}dx\\ & =& 2|A{|}^2{\int }_0^{\infty }{e}^{-2\lambda x}dx\\ & =& 2\lambda {\left(\frac{e^{-2\lambda x}}{-2\lambda }\right)}_0^{\infty }\\ & & \end{array}$

Further solving above equation as,

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

So,the value of P is $0.2431$.