Uncertainty values for non-Gaussian functions

[PhyPsy]

Uncertainty values for non-Gaussian wave packet functions

Homework Statement

$\phi(k)= \left\{ \begin{array}{cc} \sqrt{\frac{3}{2a^3}}(a-|k|), & |k| \leq a \\ 0, & |k|>a \end{array} \right.$
$\psi(x)= \frac{4}{x^2}sin^2 (\frac{ax}{2})$
Calculate the uncertainties $\Delta x$ and $\Delta p$ and check whether they satisfy the uncertainty principle.

Homework Equations

$\Delta x\Delta p \geq h/2$

The Attempt at a Solution

The solution is worked out in the book, which is $\Delta k=a$ and $\Delta x=\pi /a$. I understand that for a Gaussian distribution, you can use the standard deviation as $\Delta x$ and $\Delta k$, and this leads to the lowest limit of the uncertainty relation, $h/2$. I don't see how I'm supposed to come up with $\Delta x$ and $\Delta k$ for non-Gaussian functions, though. The book seems to just pick $\Delta k=a$ and $\Delta x=\pi /a$ somewhat arbitrarily without explaining why these values were chosen. Could someone tell me if there is a method for figuring out what the uncertainties should be for non-Gaussian functions like this one?

 

[fzero]

If we are measuring the eigenvalue $a$ of an observable $A$, then we may characterize the uncertainty $\Delta a$ in $a$ using the statistical variance:

$$(\Delta a)^2 = \langle A^2 \rangle - \langle A\rangle^2.$$

This definition works for any state we use to compute the expectation values. In fact, this definition is the one used in the derivation of the generalized uncertainty principle.

 

[PhyPsy]

$\langle A\rangle=0$ for both functions, so $(\Delta a)^2= \int ^\infty _{-\infty}a^2 f(a) da$

$(\Delta k)^2= \int ^\infty _{-\infty} k^2 |\phi(k)|^2 dk = \frac{3}{2a^3} [\int ^0 _{-a} k^2 (a+k)^2 dk + \int ^a _0 k^2 (a-k)^2 dk]$
$=\frac{3}{2a^3} [(\frac{1}{3} a^2 k^3 + \frac{1}{2} a k^4 + \frac{1}{5} k^5)| ^0 _{-a} + (\frac{1}{3} a^2 k^3 - \frac{1}{2} a k^4 + \frac{1}{5} k^5)| ^a _0$
$=\frac{3}{2a^3} \frac{2}{30} a^5 = \frac{a^2}{10}$
$\Delta k = \frac{a}{\sqrt{10}}$
$\Delta p = \frac{a \hbar}{\sqrt{10}}$

$(\Delta x)^2= \int ^\infty _{-\infty} x^2 |\psi(x)|^2 dx = 16 \int ^\infty _{-\infty} \sin ^4 (\frac{ax}{2}) dx / x^2$
$=16[\frac{1}{2} a Si(ax) - \frac{1}{4} a Si(2ax) - \sin ^4 (\frac{ax}{2}) / x]|^\infty _{-\infty} = 16(\frac{a \pi}{2} - \frac{a \pi}{4}) = 4a \pi$
$\Delta x = 2 \sqrt{a \pi}$

$\Delta p \Delta x = \frac{2a \hbar \sqrt{a \pi}}{\sqrt{10}}$
So whether or not this is greater than $\hbar / 2$ depends on what a is. Did I do something wrong here?

 

[fzero]

$\langle A\rangle=0$ for both functions, so $(\Delta a)^2= \int ^\infty _{-\infty}a^2 f(a) da$

$(\Delta k)^2= \int ^\infty _{-\infty} k^2 |\phi(k)|^2 dk = \frac{3}{2a^3} [\int ^0 _{-a} k^2 (a+k)^2 dk + \int ^a _0 k^2 (a-k)^2 dk]$
$=\frac{3}{2a^3} [(\frac{1}{3} a^2 k^3 + \frac{1}{2} a k^4 + \frac{1}{5} k^5)| ^0 _{-a} + (\frac{1}{3} a^2 k^3 - \frac{1}{2} a k^4 + \frac{1}{5} k^5)| ^a _0$
$=\frac{3}{2a^3} \frac{2}{30} a^5 = \frac{a^2}{10}$
$\Delta k = \frac{a}{\sqrt{10}}$
$\Delta p = \frac{a \hbar}{\sqrt{10}}$

I agree with this, but we'd also want to calculate $\Delta p$ using $\hat{x} = -i\hbar d/dp$ in the momentum representation. Note that $\psi(x)$ is not the Fourier transform of $\phi(k)$, so we really have two problems to solve here.

$(\Delta x)^2= \int ^\infty _{-\infty} x^2 |\psi(x)|^2 dx = 16 \int ^\infty _{-\infty} \sin ^4 (\frac{ax}{2}) dx / x^2$
$=16[\frac{1}{2} a Si(ax) - \frac{1}{4} a Si(2ax) - \sin ^4 (\frac{ax}{2}) / x]|^\infty _{-\infty} = 16(\frac{a \pi}{2} - \frac{a \pi}{4}) = 4a \pi$
$\Delta x = 2 \sqrt{a \pi}$

$\Delta p \Delta x = \frac{2a \hbar \sqrt{a \pi}}{\sqrt{10}}$
So whether or not this is greater than $\hbar / 2$ depends on what a is. Did I do something wrong here?

Again, this is a completely different wavefunction, so we can't use the uncertainty in momentum that we computed from the other one. Use $\hat{p} = -i\hbar d/dx$ to compute it.

 

[PhyPsy]

Actually, $\psi (x)$ is the Fourier transform of $\phi (k)$, or at least it is supposed to be...

$\psi (x)= \frac{1}{\sqrt{2 \pi}} \int ^\infty _{-\infty} \phi(k) e^{ikx} dk$
$=\frac{\sqrt{3}}{2 \sqrt{\pi a^3}} [\int ^0 _{-a} (a+k)e^{ikx} dk + \int ^a _0 (a-k)e^{ikx} dk]$
$=\frac{\sqrt{3}}{2 \sqrt{\pi a^3}} [\int ^0 _{-a} ke^{ikx} dk - \int ^a _0 ke^{ikx} dk + a \int ^a _{-a} e^{ikx} dk]$
$=\frac{\sqrt{3}}{2 \sqrt{\pi a^3}} [\frac{a}{ix} e^{-iax} + (1 - e^{-iax} )/ x^2 - \frac{a}{ix} e^{iax} - (e^{iax} - 1)/ x^2 + 2a \sin (ax) /x]$

At this point, the book just says that after some calculations, the answer is $\psi (x) = 4 \sin ^2 (\frac{ax}{2}) / x^2$. I don't know how they came up with that; the answer I got is $\frac{\sqrt{3}}{x^2 \sqrt{\pi a^3}} [1 - \frac{ax}{2} \sin (ax) - \cos (ax)]$.