Solved Problem on Wave Function
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Given the Gaussian distribution
\[
\begin{gather}
\rho(x)=A\operatorname{e}^{-\lambda(x-a)^{2}}
\end{gather}
\]
where A, a and λ are constants.a) Determine the constant A;
b) Determine \( \langle x\rangle \), \( \langle x^{2}\rangle \) e σ;
c) Sketch the graph of ρ(x).
Solution
a) The value of the constant A is calculated by the integral of the normalization of the function ρ(x)
\[
\begin{gather}
\bbox[#99CCFF,10px]
{\int_{-\infty}^{\infty}\rho(x)\;dx=1}
\end{gather}
\]
\[
\begin{gather}
\int_{-\infty}^{\infty}A\operatorname{e}^{-\lambda(x-a)^{2}}\;dx=1 \tag{I}
\end{gather}
\]
Integral of
\( \displaystyle \int_{-\infty}^{\infty}A\operatorname{e}^{-\lambda (x-a)^{2}}\;dx \)
Squaring the integral
\( \displaystyle x=r\cos \theta +a \)
\( \displaystyle \frac{\partial x}{\partial r}=\frac{\partial (r\cos \theta +a)}{\partial r}=\cos \theta \frac{\partial r}{\partial r}+\frac{\partial a}{\partial r}=\cos \theta .1+0=\cos \theta \)
\( \displaystyle \frac{\partial x}{\partial \theta}=\frac{\partial (r\cos \theta+a)}{\partial \theta}=r\;\frac{\partial (\cos \theta )}{\partial\theta}+\frac{\partial a}{\partial \theta}=r(-\operatorname{sen}\theta )+0=-r\operatorname{sen}\theta \)
\( \displaystyle y=r\operatorname{sen}\theta +a \)
\( \displaystyle \frac{\partial y}{\partial r}=\frac{\partial(r\operatorname{sen}\theta+a)}{\partial r}=\operatorname{sen}\theta \;\frac{\partial r}{\partial r}+\frac{\partial a}{\partial r}=\operatorname{sen}\theta.1+0=\operatorname{sen}\theta \)
\( \displaystyle \frac{\partial y}{\partial \theta}=\frac{\partial(r\operatorname{sen}\theta +a)}{\partial \theta}=r\;\frac{\partial(\operatorname{sen}\theta )}{\partial \theta}+\frac{\partial a}{\partial \theta}=r\cos \theta +0=r\cos \theta \)
Figure 1
Changing of variable
for r = 0
we have u = −λ.02 = 0
for r = ∞
we have u = −λ.∞2 = ∞
Squaring the integral
\[
\begin{gather}
I^{2}=\left(\int_{-\infty}^{\infty}A\operatorname{e}^{-\lambda (x-a)^{2}}\;dx\right)^{2}\\[5pt]
I^{2}=\int_{-\infty}^{\infty}A\operatorname{e}^{-\lambda(x-a)^{2}}\;dx\;\int_{-\infty}^{\infty}A\operatorname{e}^{-\lambda (x-a)^{2}}\;dx
\end{gather}
\]
changing to the second variable, x = y
\[
\begin{gather}
I^{2}=\int_{-\infty}^{\infty}A\operatorname{e}^{-\lambda (x-a)^{2}}\;dx\int_{-\infty}^{\infty}A\operatorname{e}^{-\lambda (y-a)^{2}}\;dy\\[5pt]
I^{2}=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}A\operatorname{e}^{-\lambda(x-a)^{2}}A\operatorname{e}^{-\lambda(y-a)^{2}}\;dx\;dy\\[5pt]
I^{2}=A^{2}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\operatorname{e}^{-\lambda[(x-a)^{2}+(y-a)^{2}]}\;dx\;dy \tag{II}
\end{gather}
\]
changing of variables from Cartesian coordinates to polar coordinates
\[
\begin{gather}
\left\{
\begin{matrix}
x-a=r\cos \theta \\[5pt]
y-a=r\operatorname{sen}\theta
\end{matrix}
\right. \tag{III}
\end{gather}
\]
The area element in Cartesian coordinates is
\[
\begin{gather}
dA=dx\;dy
\end{gather}
\]
to obtain the area element in polar coordinates, we calculate the Jacobian given by the determinant
\[
\begin{gather}
J=\left|
\begin{matrix}
\;\dfrac{\partial x}{\partial r}&\dfrac{\partial x}{\partial \theta}\;\\[5pt]
\;\dfrac{\partial y}{\partial r}&\dfrac{\partial y}{\partial \theta}\;
\end{matrix}
\right|
\end{gather}
\]
calculation of the partial derivatives of the functions x and y given in (III)
\( \displaystyle x=r\cos \theta +a \)
\( \displaystyle \frac{\partial x}{\partial r}=\frac{\partial (r\cos \theta +a)}{\partial r}=\cos \theta \frac{\partial r}{\partial r}+\frac{\partial a}{\partial r}=\cos \theta .1+0=\cos \theta \)
\[ \displaystyle \frac{\partial x}{\partial r}=\frac{\partial (r\cos \theta +a)}{\partial r}=\cos \theta \frac{\partial r}{\partial r}+\frac{\partial a}{\partial r}=\cos \theta .1+0=\cos \theta \]
\( \displaystyle \frac{\partial x}{\partial \theta}=\frac{\partial (r\cos \theta+a)}{\partial \theta}=r\;\frac{\partial (\cos \theta )}{\partial\theta}+\frac{\partial a}{\partial \theta}=r(-\operatorname{sen}\theta )+0=-r\operatorname{sen}\theta \)
\[ \displaystyle \frac{\partial x}{\partial \theta}=\frac{\partial (r\cos \theta+a)}{\partial \theta}=r\;\frac{\partial (\cos \theta )}{\partial\theta}+\frac{\partial a}{\partial \theta}=r(-\operatorname{sen}\theta )+0=-r\operatorname{sen}\theta \]
\( \displaystyle y=r\operatorname{sen}\theta +a \)
\( \displaystyle \frac{\partial y}{\partial r}=\frac{\partial(r\operatorname{sen}\theta+a)}{\partial r}=\operatorname{sen}\theta \;\frac{\partial r}{\partial r}+\frac{\partial a}{\partial r}=\operatorname{sen}\theta.1+0=\operatorname{sen}\theta \)
\[ \displaystyle \frac{\partial y}{\partial r}=\frac{\partial(r\operatorname{sen}\theta+a)}{\partial r}=\operatorname{sen}\theta \;\frac{\partial r}{\partial r}+\frac{\partial a}{\partial r}=\operatorname{sen}\theta.1+0=\operatorname{sen}\theta \]
\( \displaystyle \frac{\partial y}{\partial \theta}=\frac{\partial(r\operatorname{sen}\theta +a)}{\partial \theta}=r\;\frac{\partial(\operatorname{sen}\theta )}{\partial \theta}+\frac{\partial a}{\partial \theta}=r\cos \theta +0=r\cos \theta \)
\[ \displaystyle \frac{\partial y}{\partial \theta}=\frac{\partial(r\operatorname{sen}\theta +a)}{\partial \theta}=r\;\frac{\partial(\operatorname{sen}\theta )}{\partial \theta}+\frac{\partial a}{\partial \theta}=r\cos \theta +0=r\cos \theta \]
\[
\begin{gather}
dA=dx\;dy=J\;dr\;d\theta
\end{gather}
\]
\[
\begin{gather}
J=\left|
\begin{matrix}
\;\cos \theta&-r\operatorname{sen}\theta \;\\[5pt]
\;\operatorname{sen}\theta &r\cos\theta \end{matrix}\right|\\[5pt]
J=\cos \theta .r\cos \theta-(-r\operatorname{sen}\theta .\operatorname{sen}\theta )\\[5pt]
J=r\cos^{2}\theta +r\operatorname{sen}^{2}\theta \\[5pt]
J=r(\underbrace{\cos^{2}\theta +\operatorname{sen}^{2}\theta}_{1})\\[5pt]
J=r
\end{gather}
\]
\[
\begin{gather}
dA=r\;dr\;d\theta
\end{gather}
\]
In Cartesian coordinates, the limits of integration are −∞ and +∞ in dx
and dy to cover the entire plane (Figure 1), and in polar coordinates to cover the whole
plane the limits are 0 and +∞ in dr and 0 and 2&pi in dθ.
\[
\begin{align}
\int_{-\infty}^{\infty}\int_{-\infty}^{\infty} & \operatorname{e}^{-\lambda[(x-a)^{2}+(y-a)^{2}]}\;dx\;dy\Rightarrow \int_{{0}}^{\infty}\int_{{0}}^{{2\pi}}\operatorname{e}^{-\lambda [(r\cos \theta)^{2}+(r\operatorname{sen}\theta)^{2}]}r\;dr\;d\theta \Rightarrow\\[5pt]
& \Rightarrow \int_{{0}}^{\infty}\int_{{0}}^{{2\pi}}\operatorname{e}^{-\lambda r^{2}[\underbrace{\cos ^{2}\theta +\operatorname{sen}^{\;2}\theta}_{1}]}r\;dr\;d\theta \Rightarrow \\[5pt]
& \Rightarrow \int_{{0}}^{\infty}\int_{{0}}^{{2\pi}}\operatorname{e}^{-\lambda r^{2}}r\;dr\;d\theta \Rightarrow \\[5pt]
& \Rightarrow \int_{{0}}^{\infty}r\operatorname{e}^{-\lambda r^{2}}\;dr\underbrace{\int_{{0}}^{{2\pi}}d\theta}_{2\pi} \tag{IV}
\end{align}
\]
Integral of
\( \displaystyle \int_{{0}}^{\infty}r\operatorname{e}^{-\lambda r^{2}}\;dr \)
Changing of variable
\[
\begin{array}{l}
u=-\lambda r^{2}\\[5pt]
du=-2\lambda r\;dr\Rightarrow d r=-{\dfrac{du}{2\lambda r}}
\end{array}
\]
changing the limits of integration
for r = 0
we have u = −λ.02 = 0
for r = ∞
we have u = −λ.∞2 = ∞
\[
\begin{align}
\int_{{0}}^{\infty}r\operatorname{e}^{-\lambda r^{2}}\;dr & \Rightarrow \int_{{0}}^{-\infty}r\operatorname{e}^{u}\left(-{\frac{du}{2\lambda r}}\right)\Rightarrow-\frac{1}{2\lambda}\int_{{0}}^{-\infty}\operatorname{e}^{u}\;du\Rightarrow \\[5pt]
& \Rightarrow -\frac{1}{2\lambda}\left(\left.\operatorname{e}^{u}\right|_{\;0}^{\;-\infty}\right)\Rightarrow -\frac{1}{2\lambda}\left(\operatorname{e}^{-\infty}-\operatorname{e}^{0}\right)\Rightarrow\\[5pt]
& \Rightarrow -\frac{1}{2\lambda}\left(\frac{1}{\operatorname{e}^{\infty}}-1\right)\Rightarrow-\frac{1}{2\lambda}\left(\frac{1}{\infty}-1\right)\Rightarrow \\[5pt]
& \Rightarrow -\frac{1}{2\lambda}\left(0-1\right)\Rightarrow \frac{1}{2\lambda}
\end{align}
\]
substituting into the expression (IV)
\[
\begin{gather}
\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\operatorname{e}^{-\lambda[(x-a)^{2}+(y-a)^{2}]}\;dx\;dy=\frac{1}{2\lambda}.2\pi =\frac{\pi}{\lambda}
\end{gather}
\]
Using equations (I) and (II) and substituting the result above
\[
\begin{gather}
I^{2}=A^{2}\frac{\pi}{\lambda}=1\\[5pt]
A^{2}=\frac{\lambda}{\pi}
\end{gather}
\]
\[
\begin{gather}
\bbox[#FFCCCC,10px]
{A=\sqrt{\frac{\lambda}{\pi}}}
\end{gather}
\]
b) The expectation value of x is calculated by the integral
\[
\begin{gather}
\bbox[#99CCFF,10px]
{\langle x\rangle =\int_{-\infty}^{\infty}x\rho(x)\;dx}
\end{gather}
\]
\[
\begin{gather}
\langle x\rangle =\int_{-\infty}^{\infty}xA\operatorname{e}^{-\lambda (x-a)^{2}}\;dx\\[5pt]
\langle x\rangle =A\int_{-\infty}^{\infty}x\operatorname{e}^{-\lambda(x-a)^{2}}\;dx \tag{V}
\end{gather}
\]
Integral of
\( \displaystyle \int_{-\infty}^{\infty}x\operatorname{e}^{-\lambda (x-a)^{2}}\;dx \)
Changing of variable
for x = −∞
we have u = −∞−a = −∞
for x = ∞
we have u = ∞−a = ∞
1st method
Function \( f(u)=u \) is an odd function, and function \( g(u)=\operatorname{e}^{-\lambda u^{2}} \) is an even function, an odd function multiplied by an even function equals an odd function, which integrated into a symmetric interval (from −∞ to ∞) equals zero.
Integral of \( \displaystyle \int_{-\infty}^{\infty}\operatorname{e}^{-\lambda u^{2}}\;du \)
The integral \( I=\int_{-\infty}^{\infty}\operatorname{e}^{-\lambda u^{2}}\;du \), is of the same type as the integral solved in part (a) \( I=\int_{-\infty}^{\infty}A\operatorname{e}^{-\lambda(x-a)^{2}}\;dx \) so its result is the same
Changing of variable
\[
\begin{array}[l]
u=x-a\Rightarrow x=u+a\\[5pt]
du=dr
\end{array}
\]
changing the limits of integration
for x = −∞
we have u = −∞−a = −∞
for x = ∞
we have u = ∞−a = ∞
\[
\begin{align}
\int_{-\infty}^{\infty} & x\operatorname{e}^{-\lambda (x-a)^{2}}\;dx\Rightarrow \int_{-\infty}^{\infty}(u+a)\operatorname{e}^{-\lambda u^{2}}\;du\Rightarrow\\[5pt]
& \Rightarrow \int_{-\infty}^{\infty}u\operatorname{e}^{-\lambda u^{2}}\;du+\int_{-\infty}^{\infty}a\operatorname{e}^{-\lambda u^{2}}\;du\Rightarrow\\[5pt]
& \Rightarrow \int_{-\infty}^{\infty}u\operatorname{e}^{-\lambda u^{2}}\;du+a\int_{-\infty}^{\infty}\operatorname{e}^{-\lambda u^{2}}\;du \tag{VI}
\end{align}
\]
Integral of
\( \displaystyle \int_{-\infty}^{\infty}u\;\operatorname{e}^{-\lambda \;u^{\;2}}\;du \)
1st method
\[
\begin{align}
\int_{-\infty}^{\infty}u\operatorname{e}^{-\lambda u^{2}}\;du & \Rightarrow \left.-{\frac{1}{2\lambda}}\operatorname{e}^{-\lambda u^{2}}\;\right|_{\;-\infty}^{\;\infty}\Rightarrow -\frac{1}{2\lambda}\left(\operatorname{e}^{-\lambda (\infty)^{2}}-\operatorname{e}^{-\lambda (-\infty)^{2}}\right)\Rightarrow\\[5pt]
& \Rightarrow -\frac{1}{2\lambda}\left(\operatorname{e}^{-\infty}-\operatorname{e}^{-\infty}\right)\Rightarrow -\frac{1}{2\lambda}\left(\frac{1}{\operatorname{e}^{\infty}}-\frac{1}{\operatorname{e}^{\infty}}\right)\Rightarrow\\[5pt]
& \Rightarrow -\frac{1}{2\lambda}\left(\frac{1}{\infty}-\frac{1}{\infty}\right)\Rightarrow -\frac{1}{2\lambda}\left(0-0\right)\Rightarrow 0
\end{align}
\]
2nd method
Function \( f(u)=u \) is an odd function, and function \( g(u)=\operatorname{e}^{-\lambda u^{2}} \) is an even function, an odd function multiplied by an even function equals an odd function, which integrated into a symmetric interval (from −∞ to ∞) equals zero.
Integral of \( \displaystyle \int_{-\infty}^{\infty}\operatorname{e}^{-\lambda u^{2}}\;du \)
The integral \( I=\int_{-\infty}^{\infty}\operatorname{e}^{-\lambda u^{2}}\;du \), is of the same type as the integral solved in part (a) \( I=\int_{-\infty}^{\infty}A\operatorname{e}^{-\lambda(x-a)^{2}}\;dx \) so its result is the same
\[
\begin{gather}
I^{2}=\frac{\pi}{\lambda}\\[5pt]
I=\sqrt{\frac{\pi}{\lambda}\;}
\end{gather}
\]
Note: The difference between the two functions is that the function ρ(x)
is centered at point a on the abscissa axis, (x−a) in the exponent, and the function
in the integrand above, is centered at the origin, (u−0). Substituting the results above into
expression (VI)
\[
\begin{gather}
\int_{-\infty}^{\infty}x\operatorname{e}^{-\lambda(x-a)^{2}}\;dx=0+a\;\sqrt{\frac{\pi}{\lambda}\;}=a\;\sqrt{\frac{\pi}{\lambda}\;}
\end{gather}
\]
Substituting the constant A determined in item (a) and the value of the integral calculated above, in equation (V)
\[
\begin{gather}
\langle x\rangle =\cancel{\sqrt{\frac{\lambda}{\pi}\;}}a\cancel{\;\sqrt{\frac{\pi}{\lambda}\;}}
\end{gather}
\]
\[
\begin{gather}
\bbox[#FFCCCC,10px]
{\langle x\rangle =a}
\end{gather}
\]
The expectation value of x2 is calculated by the integral
\[
\begin{gather}
\bbox[#99CCFF,10px]
{\langle x^{2}\rangle =\int_{-\infty}^{\infty}x^{\;2}\rho(x)\;dx}
\end{gather}
\]
\[
\begin{gather}
\langle x^{2}\rangle =\int_{-\infty}^{\infty}x^{2}A\operatorname{e}^{-\lambda (x-a)^{2}}\;dx\\[5pt]
\langle x^{\;2}\rangle =A\int_{-\infty}^{\infty}x^{\;2}\operatorname{e}^{-\lambda (x-a)^{2}}\;dx \tag{VII}
\end{gather}
\]
Integral of
\( \displaystyle \int_{-\infty}^{\infty}x^{2}\operatorname{e}^{-\lambda(x-a)^{2}}\;dx \)
Changing of variable
for x = −∞
we have u = −∞−a = −∞
for x = ∞
we have u = ∞−a = ∞
Using Integration by Parts \( \int fg'=fg-\int f'g \)
Figure 2
substituting expression (IX) into expression (VIII)
Changing of variable
\[
\begin{array}{l}
u=x-a\Rightarrow x=u+a\\[5pt]
du=dr
\end{array}
\]
changing the limits of integration
for x = −∞
we have u = −∞−a = −∞
for x = ∞
we have u = ∞−a = ∞
\[
\begin{align}
\int_{-\infty}^{\infty} & x^{\;2}\operatorname{e}^{-\lambda (x-a)^{2}}\;dx\Rightarrow \int_{-\infty}^{\infty}(u+a)^{2}\operatorname{e}^{-\lambda u^{2}}\;du\Rightarrow\\[5pt]
& \Rightarrow \int_{-\infty}^{\infty}(u^{2}+2ua+a^{2})\operatorname{e}^{-\lambda u^{2}}\;du\Rightarrow\\[5pt]
& \Rightarrow \int_{-\infty}^{\infty}u^{2}\operatorname{e}^{-\lambda u^{2}}+2ua\operatorname{e}^{-\lambda u^{2}}+a^{2}\operatorname{e}^{-\lambda u^{2}}\;du\Rightarrow \hfill\\[5pt]
& \Rightarrow \int_{-\infty}^{\infty}u^{2}\operatorname{e}^{-\lambda u^{2}}\;du+\int_{-\infty}^{\infty}2ua\operatorname{e}^{-\lambda u^{2}}\;du+\int_{-\infty}^{\infty}a^{2}\operatorname{e}^{-\lambda u^{2}}\;du\Rightarrow\\[5pt]
& \Rightarrow \int_{-\infty}^{\infty}u^{2}\operatorname{e}^{-\lambda u^{2}}\;du+2a\underbrace{\int_{-\infty}^{\infty}u\operatorname{e}^{-\lambda u^{2}}\;du}_{0}+a^{2}\underbrace{\int_{-\infty}^{\infty}\operatorname{e}^{-\lambda u^{2}}\;du}_{\sqrt{\frac{\pi}{\lambda}}} \tag{VIII}
\end{align}
\]
Integral of
\( \displaystyle \int_{-\infty}^{\infty}u^{2}\operatorname{e}^{-\lambda u^{2}}\;du=\int_{-\infty}^{\infty}u.u\operatorname{e}^{-\lambda u^{2}}\;du \)
\[ \displaystyle \int_{-\infty}^{\infty}u^{2}\operatorname{e}^{-\lambda u^{2}}\;du=\int_{-\infty}^{\infty}u.u\operatorname{e}^{-\lambda u^{2}}\;du \]
Using Integration by Parts \( \int fg'=fg-\int f'g \)
\[
\begin{gather}
\begin{array}{l}
f=u & \qquad g'=u\operatorname{e}^{-\lambda u^{2}}\\[5pt]
\begin{array}{l}
f'=1\\[5pt]
\phantom{{\frac{\phantom{{}}}{\phantom{{}}}}}\\[5pt]
\phantom{{\frac{\phantom{{}}}{\phantom{{}}}}}\\[5pt]
\phantom{{\frac{\phantom{{}}}{\phantom{{}}}}}
\end{array}
& \qquad
\begin{array}{l}
g=\int u\operatorname{e}^{-\lambda u^{2}}\;du=\int \operatorname{e}^{-\lambda u^{2}}u\;du\\[5pt]
k=-\lambda u^{2}\Rightarrow\dfrac{dk}{du}=-2\lambda u\Rightarrow u du=-{\dfrac{dk}{2\lambda}}\\[5pt]
g=-\int \operatorname{e}^{u}\;\dfrac{du}{2\lambda}=\dfrac{\operatorname{e}^{u}}{2\lambda}=\dfrac{\operatorname{e}^{-x^{2}}}{2\lambda}
\end{array}
\end{array}
\end{gather}
\]
\[
\begin{align}
\int_{-\infty}^{\infty} & u^{2}\operatorname{e}^{-\lambda u^{2}}\;du\Rightarrow\left.u\frac{\operatorname{e}^{-\lambda u^{2}}}{2\lambda}\;\right|_{\;-\infty}^{\;\infty}+\int_{-\infty}^{\infty}1.\frac{\operatorname{e}^{-\lambda u^{2}}}{2\lambda}\;du\Rightarrow\\[5pt]
& \Rightarrow\left.u\frac{\operatorname{e}^{-\lambda u^{2}}}{2\lambda}\;\right|_{\;-\infty}^{\;\infty}+\int_{-\infty}^{\infty}1.\frac{\operatorname{e}^{-\lambda u^{2}}}{2\lambda}\;du\Rightarrow\\[5pt]
& \Rightarrow\frac{1}{2\lambda}\left(\frac{\infty}{\operatorname{e}^{\lambda(\infty)^{2}}}-\frac{-\infty}{\operatorname{e}^{\lambda (-\infty)^{2}}}\right)+\frac{1}{2\lambda}\underbrace{\int_{-\infty}^{\infty}{\operatorname{e}^{-\lambda u^{2}}\;du}}_{\sqrt{\frac{\pi}{\lambda}}}\Rightarrow \frac{1}{2\lambda}\sqrt{\frac{\pi}{\lambda}} \tag{IX}
\end{align}
\]
Note: In the terms in parentheses
\( u=\pm \infty \)
tends to
\( \pm \infty \),
\( \operatorname{e}^{\lambda (\pm \infty )^{2}} \)
tends to
\( +\infty \),
but the fraction
\( \frac{1}{\operatorname{e}^{\lambda (\pm \infty )^{2}}} \)
tends to zero (Figure 2). As the fraction tends to zero faster than u tends to infinity, the
terms
\( \frac{\pm \infty}{\operatorname{e}^{\lambda (\pm \infty )^{2}}} \)
tend to zero.
\[
\begin{gather}
\int_{-\infty}^{\infty}x^{\;2}\operatorname{e}^{-\lambda(x-a)^{2}}\;dx\Rightarrow \frac{1}{2\lambda}\sqrt{\frac{\pi}{\lambda}}+a^{2}\sqrt{\frac{\pi}{\lambda}}\Rightarrow \left(\frac{1}{2\lambda}+a^{2}\right)\sqrt{\frac{\pi}{\lambda}}
\end{gather}
\]
Substituting the constant A determined in item (a) and the value of the integral calculated above, into equation (VII)
\[
\begin{gather}
\langle x^{\;2}\rangle =\cancel{\sqrt{\frac{\lambda}{\pi}}}\;\left(\frac{1}{2\lambda}+a^{2}\right)\cancel{\sqrt{\frac{\pi}{\lambda}}}
\end{gather}
\]
\[
\begin{gather}
\bbox[#FFCCCC,10px]
{\langle x^{2}\rangle =\frac{1}{2\lambda}+a^{2}}
\end{gather}
\]
The value of variance σ2 is given by
\[
\begin{gather}
\bbox[#99CCFF,10px]
{\sigma^{2}=\langle x^{2}\rangle -\langle x \rangle ^{2}}
\end{gather}
\]
substituting the values of
\( \langle x\rangle \)
and
\( \langle x^{2}\rangle \)
found above
\[
\begin{gather}
\sigma^{2}=\frac{1}{2\lambda}+a^{2}-a^{2}\\[5pt]
\sigma^{2}=\frac{1}{2\lambda}
\end{gather}
\]
the value of the standard deviation σ is
\[
\begin{gather}
\bbox[#FFCCCC,10px]
{\sigma=\frac{1}{\sqrt{2\lambda}\;}}
\end{gather}
\]
c) Graphic sketch (Figure 3)
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Fisicaexe - Physics Solved Problems by Elcio Brandani Mondadori is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License .