The gaussian wave packet. A free particle has the initial wave function

$\mathbf{Y}\mathbf{(}\mathit{x}\mathbf{,}\mathbf{0}\mathbf{)}\mathbf{=}\mathit{A}{\mathit{e}}^{\mathbf{-}\mathbf{a}{\mathbf{x}}^{\mathbf{2}}}$

where$\mathit{A}$and are constants ( is real and positive).

(a) Normalize$\mathit{Y}\mathbf{(}\mathit{x}\mathbf{,}\mathbf{0}\mathbf{)}$

(b) Find $\mathbf{Y}\mathbf{(}\mathit{x}\mathbf{,}\mathit{t}\mathbf{)}$. Hint: Integrals of the form

${\mathbf{\int }}_{\mathbf{-}\mathbf{\infty }}^{\mathbf{+}\mathbf{\infty }}{\mathit{e}}^{\mathbf{-}\left(a{x}^2+bx\right)}\mathit{d}\mathit{x}$

Can be handled by “completing the square”: Let$\mathit{y}\mathbf{=}\sqrt{\mathbf{a}}\left[x+\left(bl2a\right)\right]$, and note that$\left(a{x}^2+bx\right)\mathbf{=}{\mathit{y}}^{\mathbf{2}}\mathbf{-}\left(b^{2}l4a\right)$. Answer:

localid="1658297483210" $\mathit{Y}\mathbf{(}\mathit{x}\mathbf{,}\mathit{t}\mathbf{)}\mathbf{=}{\mathbf{\left(}\frac{\mathbf{2}\mathbf{a}}{\mathbf{\pi }}\mathbf{\right)}}^{\mathbf{1}\mathbf{/}\mathbf{4}}\frac{{\mathbf{e}}^{\mathbf{-}\mathbf{e}{\mathbf{x}}^{\mathbf{2}}\mathbf{l}\mathbf{\left[}\mathbf{1}\mathbf{+}\mathbf{(}\mathbf{2}\mathbf{i}\mathbf{h}\mathbf{a}\mathbf{t}\mathbf{}\mathbf{l}\mathbf{m}\mathbf{)}\mathbf{\right]}}}{\sqrt{\mathbf{1}\mathbf{+}\mathbf{(}\mathbf{2}\mathbf{i}\sout{\mathbf{h}}\mathbf{a}\mathbf{t}\mathbf{}\mathbf{l}\mathbf{}\mathbf{m}\mathbf{)}}}$

(c) Find . Express your answer in terms of the quantity

localid="1658297497509" $\mathbf{\omega }\mathbf{=}\sqrt{\frac{\mathbf{a}}{\mathbf{1}\mathbf{+}{\mathbf{(}\mathbf{2}\mathbf{i}\mathbf{}\sout{\mathbf{h}}\mathbf{at}\mathbf{}\mathbf{l}\mathbf{}\mathbf{m}\mathbf{)}}^{\mathbf{2}}}}$

Sketchlocalid="1658124147567" ${\left|Y\right|}^{\mathbf{2}}$(as a function of $\mathit{x}$) at $\mathbf{t}\mathbf{=}\mathbf{0}$, and again for some very large $\mathbf{t}$. Qualitatively, what happens to ${\left|Y\right|}^{\mathbf{2}}$, as time goes on?

(d) Find $<x>\mathbf{,}<p>\mathbf{,}<x^2>\mathbf{,}<p^2>\mathbf{,}{\mathit{\sigma }}_{\mathbf{x}}$and ${\mathit{\sigma }}_{\mathbf{P}}$. Partial answer:localid="1658297458579" $\mathbf{<}{\mathbf{p}}^{\mathbf{2}}\mathbf{>}\mathbf{=}\mathit{a}{\sout{\mathbf{h}}}^{\mathbf{2}}$, but it may take some algebra to reduce it to this simple form.

(e) Does the uncertainty principle hold? At what time $\mathbf{t}$does the system come

closest to the uncertainty limit?

The given data is listed below as follows,

The initial wave function for a free particle is,$\mathrm{Y}(\mathrm{x},0)={\mathrm{Ae}}^{-2{\mathrm{x}}^2}$

The characteristics of a particular wave form is given by a function known as wave function. Also, wave equation is satisfied by this wave function. It is denoted by $\mathbf{\psi }$.

Normalize the given wave function in the following way,

$$\begin{gathered} 1-{\left|A\right|}^2{\int }_{-\infty }^{\infty }{e}^{-2a{x}^2}dx \\ 1-{\left|A\right|}^2\sqrt{\frac{\mathrm{p}}{2\mathrm{a}}} \\ A-{\left(\frac{2a}{\mathrm{p}}\right)}^{1/4} \end{gathered}$$

Thus, the normalize wave function is $A-{\left(\frac{2a}{\mathrm{p}}\right)}^{1/4}$.

Perform the integration.

$$\begin{gathered} {\int }_{-\infty }^{\infty }{e}^{-{\left(a{x}^2-bx\right)}^2}dx={\int }_{-\infty }^{\infty }{e}^{-y^2+(b^2/4a)}\frac{1}{\sqrt{a}}dy \\ =\frac{1}{\sqrt{a}}{e}^{b^2/4a}{\int }_{-\infty }^{\infty }{e}^{-y^2}dy \\ =\sqrt{\frac{p}{a}}{e}^{b^2/4a} \end{gathered}$$

Find the value of $\varphi \left(k\right)$.

role="math" localid="1658128927318" $$\begin{gathered} \varphi \left(k\right)=\frac{1}{\sqrt{2p}}A{\int }_{-\infty }^{\infty }{e}^{-a{x}^2}{e}^{-ikx}dx \\ =\frac{1}{\sqrt{2p}}{\left(\frac{2a}{\mathrm{\pi }}\right)}^{1/4}\sqrt{\frac{p}{a}}{e}^{-k^2/4a} \\ =\frac{1}{{(2pa)}^{1/4}}{e}^{-k^2/4a} \end{gathered}$$

Determine the value of .

$$\begin{gathered} Y(x,t)=\frac{1}{\sqrt{2p}}\frac{1}{{(2pa)}^{1/4}}{\int }_{-\infty }^{\infty }{e}^{-k^2/4a}{e}^{i\left(kx-ħk^2/2m\right)}dk \\ =\frac{1}{\sqrt{2p}}\frac{1}{{\left(2pa\right)}^{1/4}}\frac{\sqrt{p}}{\sqrt{{\displaystyle \frac{1}{4a}}}+{\displaystyle \frac{iħt}{2m}}}{e}^{-x^2/4(\frac{1}{4}+iħt/2m)} \\ ={\left(\frac{2a}{p}\right)}^{1/4}\frac{e^{-a{x}^{2}l\left(1+2iħt/m\right)}}{\sqrt{\left(1+2iħt/m\right)}} \end{gathered}$$

Thus, the value of $\mathrm{Y}\left(\mathrm{x},\mathrm{t}\right)$is ${\left(\frac{2\mathrm{a}}{\mathrm{p}}\right)}^{1/4}\frac{e^{-a{x}^2/\left(1+2iħ/m\right)}}{\sqrt{\left(1+2iħt/m\right)}}$.

Assume that$\theta =\frac{2ħat}{m}$

Substitute $\theta$for $\frac{2ħat}{m}$in $Y\left(x\right){\left(\frac{2a}{p}\right)}^{1/4}\frac{e^{-a{x}^2/(1+2iħt/m)}}{\sqrt{\left(1+2iħt/m\right)}}$.

$\left|Y\left(x,t\right)\right|=\sqrt{\frac{2a}{p}}\frac{e^{-a{x}^2/\left(1+i\theta \right)}{e}^{-a{x}^2/(1-i\theta )}}{\sqrt{\left(1+i\theta \right)(1-i\theta )}}$

Simplify the exponent.

$$\begin{gathered} -\frac{a{x}^2}{\left(1+i\theta \right)}-\frac{a{x}^2}{(1-i\theta )}=-a{x}^2\frac{\left(1-i\theta +1i\theta \right)}{\left(1-i\theta \right)(1+\theta )} \\ =\frac{-2a{x}^2}{1+{\theta }^2} \end{gathered}$$

Use the above value and find in ${\left|\mathrm{Y}(x,t)\right|}^2$.

${\left|Y(x,t)\right|}^2=\sqrt{\frac{2a}{p}}\frac{e^{-2a{x}^{2}l(1+{\theta }^2)}}{\sqrt{1+{\theta }^2}}$

It is known that $w=\sqrt{\frac{a}{1+{\theta }^2}}$so, ${\left|Y(x,t)\right|}^2=\sqrt{\frac{2}{p}}w{e}^{-2w^2{x}^2}$. Now plot the graph for ${\left|Y\left(x,t\right)\right|}^2$when $t=0$

It is known that as the value of $t$increases, the graph of ${\left|\mathrm{Y}\right|}^2$flattens out and broadens. So, plot the graph for ${\left|\mathrm{Y}(x,t)\right|}^2$when $t>0$.

Thus, the value of ${\left|\mathrm{Y}(x,t)\right|}^2$is $\sqrt{\frac{2a}{p}}\frac{e^{-2a{x}^2/(1-e^2)}}{\sqrt{1+{\theta }^2}}$.

Determine the value of in the following way,

It is the odd integrand.

Determine the value of in the following way,

Determine the value of in the following way,

Determine the value of in the following way,

Write the value of $\mathrm{y}$.

$y=B{e}^{-b{x}^2}$

Here, $B={\left(\frac{2a}{p}\right)}^{1/4}\frac{1}{\sqrt{1+i\theta }}$and $b=\frac{2}{1+i\theta }$.

Double differentiate $y=B{e}^{-b{x}^2}$with respect to $x$.

$\frac{d^{2}y}{d{x}^2}=-2bB(1-2b{x}^2)e^{-b{x}^2}$

Multiply $y*$with the above expression.

$y*\frac{d^{2}y}{d{x}^2}=-2b{\left|B\right|}^2(1-2b{x}^2)e^{-(b-b*)x^2}$

It is known that $b+b*=2w^2{\left|B\right|}^2=\sqrt{\frac{2}{p}}w$.

Substitute all the values in above expression.

$y*\frac{d^{2}y}{d{x}^2}=-2b\sqrt{\frac{2}{p}}w(1-2b{x}^2)e^{-2w^2{x}^2}$

Substitute the above value in .

It is known that $1-\frac{\mathrm{b}}{2{\mathrm{w}}^2}=\frac{\mathrm{a}}{2\mathrm{b}}$, so substitute it in above expression.

Determine the value of ${\sigma }_x$in the following way,

${\sigma }_x=\frac{1}{2w}$

Determine the value of ${\sigma }_p$in the following way

${\sigma }_p=ħ\sqrt{a}$

Thus, the value of is 0, is 0, is is ${ħ}^{2}a$, ${\sigma }_x$is ,$\frac{1}{2w}$ and ${\sigma }_p$is $ħ\sqrt{a}$

Find the condition that uncertainty principle holds or not.

$$\begin{gathered} {\sigma }_x{\sigma }_p=\frac{1}{2w}ħ\sqrt{a} \\ =\frac{ħ}{2}\sqrt{1+q^2} \\ =\frac{ħ}{2}\sqrt{1+{\left(2ħat/m\right)}^2} \\ \ge \frac{ħ}{2} \end{gathered}$$

Thus, the uncertainty principle holds and it is closest to the uncertainty limit at $t=0$, and at that time it is right at the uncertainty limit.