Suppose $\Psi (x,0)=\frac{A}{x^2+a^2}.(-\infty <x<\infty )$for constantsA and a.

(a) Determine A, by normalizing$\Psi (x,0)$

(b) Find, and(at time).

(c) Find the momentum space wave function$\mathit{\Phi }\mathbf{(}\mathit{p}\mathbf{,}\mathbf{0}\mathbf{)}$, and check that it is normalized.

(d) Use$\Phi (p,0)$to calculate, and${\sigma }_p$(at time$t=0$).

(e) Check the Heisenberg uncertainty principle for this state.

a)

It is easy to obtain the constant A by requiring wave function to be normalized, more formally as:

${\int }_{-\infty }^{\infty }\left|\psi \right(x){|}^{2}dx=1$

Which in this case turns into the following requirement:

${\int }_{-\infty }^{\infty }\frac{A^2}{{\left(x^2+a^2\right)}^2}dx=1$

Integrate using Mathematica but the integral can be calculated by the use of reduction formula to reduce the integral into several integrals with lower power in the denominator. The integral turns out to be:

$\begin{array}{rcl}{\int }_{-\infty }^{\infty }\frac{A^2}{{\left(x^2+a^2\right)}^2}dx& =& \frac{1}{2}\frac{A^2}{a^3}\pi \\ \frac{1}{2}\frac{A^2}{a^3}\pi & =& 1\\ A& =& \sqrt{\frac{2a^3}{\pi }}\\ & & \end{array}$

Thus, the value of A is$\sqrt{\frac{2a^3}{\pi }}$

b)

Let us calculate the required expectation values:

$\begin{array}{rcl}⟨x⟩& =& {\int }_{-\infty }^{\infty }x·\frac{A^2}{{\left(x^2+a^2\right)}^2}dx\\ & =& 0\\ & & \end{array}$

The integral is trivially zero, because of integrating an odd function on a symmetric interval. The next expectation value can be obtained as follows:

Again, this is not a simple integral, but can be analysed analytically by writing $x^2=x^2+a^2-a^2$to obtain two integrals:

$\begin{array}{rcl}{\int }_{-\infty }^{\infty }\frac{A^2{x}^2}{{\left(x^2+a^2\right)}^2}& =& A^2\left({\int }_{-\infty }^{\infty }\frac{x^2+a^2-a^2}{{\left(x^2+a^2\right)}^2}\right)\\ & =& A^2\left({\int }_{-\infty }^{\infty }\frac{x^2+a^2}{{\left(x^2+a^2\right)}^2}dx-{\int }_{-\infty }^{\infty }\frac{a^2}{{\left(x^2+a^2\right)}^2}dx\right)\\ & =& A^2\left({\int }_{-\infty }^{\infty }\frac{1}{x^2+a^2}dx-{\int }_{-\infty }^{\infty }\frac{a^2}{{\left(x^2+a^2\right)}^2}dx\right)\\ & & \end{array}$

The last two integrals are trivial, since the first one is a table integral and the second one was calculated in the previous part of the problem where the question asked to find the normalization constant. Therefore, the integrals are trivial and yield the following value:

Now, calculate the uncertainty, ${\sigma }_x$. It is defined as follows:

Thus, the value of$⟨x⟩$,, and${\sigma }_x$ are 0, $a^2$,andarespectively.

c)

To find the momentum space wave function,find the Fourier transform of the original coordinate space wave function, since it is a connection between these two different representations. Therefore:$\Phi \left(p\right)=\frac{1}{\sqrt{2\pi \hslash }}{\int }_{-\infty }^{\infty }\psi \left(x\right)e^{-ipx/\hslash }dx$

In this case this becomes the following:

$\begin{array}{rcl}\Phi \left(p\right)& =& \frac{1}{\sqrt{2\pi \hslash }}{\int }_{-\infty }^{\infty }\frac{A}{x^2+a^2}{e}^{-ipx/\hslash }dx\\ & =& \frac{1}{\sqrt{2\pi \hslash }}{\int }_{-\infty }^{\infty }\frac{A}{x^2+a^2}\left[\cos \right(px/\hslash )-i\sin (px/\hslash \left)\right]dx\\ & & \end{array}$

In the last stepuse the trigonometric expression for the exponential function. Now, the last integral separates into two integrals; one with sine and second with cosine function. The sine integral vanishes, because of integrating an odd function on a symmetric interval. Therefore, only need focus on the cosine integral:
$\begin{array}{rcl}\Phi \left(p\right)& =& \frac{1}{\sqrt{2\pi \hslash }}{\int }_{-\infty }^{\infty }\frac{A\cos (px/\hslash )}{x^2+a^2}dx\\ & =& \frac{2A}{\sqrt{2\pi \hslash }}\frac{\pi }{2a}{e}^{-\left|p\right|a/\hslash }\\ \Phi \left(p\right)& =& \sqrt{\frac{a}{\hslash }}{e}^{-\left|p\right|a/\hslash }\\ & & \end{array}$

Now, to check whether this function is normalized, weintegrate over all momentum space as follows:

$\begin{array}{rcl}{\int }_{-\infty }^{\infty }\left|\Phi \right(p){|}^{2}dp& =& {\int }_{-\infty }^{\infty }\frac{a}{\hslash }{e}^{-2\left|p\right|a/\hslash }dp\\ & =& 2·{\int }_0^{\infty }\frac{a}{\hslash }{e}^{-2pa/\hslash }dp\\ & =& 1\\ & & \end{array}$

In the last step use several properties, one being that the wave function is even and therefore integrate from to and multiply it by two, instead of integrating over the whole space. Integrate only in the positive domain of P , the absolute value in the exponential function turns out to be $\left|p\right|=p$, and the latter integral is then trivial.

d)

To find the desired expectation values, we can either work in the coordinate or momentum space. Since we already have the momentum space wave function, it is simpler to work in it (since the expectation values are just integrals of the wave function multiplied by the momentum P , same as expectation value of position in the coordinate space calculations). Then:

$\begin{array}{rcl}⟨p⟩& =& {\int }_{-\infty }^{\infty }p\Phi \left(p\right)dp\\ & =& {\int }_{-\infty }^{\infty }p\sqrt{\frac{a}{\hslash }}{e}^{-2\left|p\right|a/\hslash }dp\\ & =& 0\\ & & \end{array}$

This integral is again equal to 0, since we are integrating an odd function on a symmetric interval. Next, we need to find the expectation value of the squared momentum operator

Now, this integral can evaluated by integrating by parts twice. This is messy and takes time and there is no physics behind it, so I shall again evaluate the integral using Mathematica. If someone wishes to do it by hand, I have written the procedure required to do it. The expectation value turns out to be:

Therefore, the uncertainty is given by:

Thus, the value of , $⟨p⟩$, localid="1659011703261" and ${\sigma }_p$are 0, localid="1659011735931" $\frac{{\hslash }^2}{2a^2}$,and rlocalid="1659011745980" $\frac{\hslash }{\sqrt{2}a}$espectively.

e)

To check whether the uncertainty relation holds, we need to multiply the two uncertainties as follows:${\sigma }_x{\sigma }_p=a\frac{\hslash }{\sqrt{2}a}=\frac{\hslash }{\sqrt{2}}=\frac{\hslash }{2}·\sqrt{2}>\frac{\hslash }{2}$

Therefore, the uncertainty relation is satisfied.