Fourier analysis of wave packet

[ognik]

Homework Statement

Assume $\phi(k_x ) = \sqrt2 {\pi}$ for $\bar{k}_x - \delta \le k_x \le \bar{k}_x + \delta$, and $= 0$ for all other values of $k_x$. Calculate $\psi(x, 0)$, and show that $\Delta x \Delta k_x \approx 1$ holds if $\Delta x$ is taken as the width at half maximum.

Homework Equations

$\psi (x,0) = \frac{1}{\sqrt{2 \pi}} \int \phi (k_x) e^{i k_x x} dk_x$

The Attempt at a Solution

$\psi (x,0) = \frac{1}{\sqrt{2 \pi}} \int_{\bar{k}_x - \delta } ^{\bar{k}_x + \delta} \sqrt{2 \pi} e^{i k_x x} dk_x$

$= \frac{ e^{i k_x x}}{ix} |_{\bar{k}_x - \delta } ^{\bar{k}_x + \delta}$

$= \frac{ e^{i \bar{k}_x x}}{x} 2 sin(x\delta)$. If $\delta$ is small we can approximate to $e^{i \bar{k}_x x} 2 \delta$

Is this right please?
---------------
How do I argue that $\Delta x \Delta k_x \approx 1$ holds if $\Delta x$ is taken as the width at half maximum?
I think that $\psi_{max} = 2 \delta$ ?

 

[George Jones]

If $\delta$ is small we can approximate to

Don't make this approximation.

 

[ognik]

Don't make this approximation.

I expected that ... in the meantime I thought of using a taylor approx, making it

$= e^{i \bar{k}_x x} (2 - \frac{x^2}{3} + \frac{x^4}{60} )$ ?

 

[George Jones]

The sinc function is defined by

$$sinc(u) = \frac{sinu}{u}.$$

Play around on WolframAlpha to find the u that gives the half-maximum of sinc(u).

 

[ognik]

The sinc function is defined by
$$sinc(u) = \frac{sinu}{u}.$$

Sorry if I am being dense, but I have $\frac{sin(x \delta)}{x}$ ? Should I be thinking of $\frac{\delta sin(x \delta)}{x \delta}$

In which case, would the maximum be $|\psi^*| = (2 \delta sinc (\delta x))^2$ ?

 

[blue_leaf77]

Should I be thinking of $\frac{\delta sin(x \delta)}{x \delta}$

Yes, you can do that.

In which case, would the maximum be $|\psi^*| = (2 \delta sinc (\delta x))^2$ ?

I don't know what you actually want to calculate, if you want to calculate $\psi(x,0)$, then there you have it with the sinc function. If you want calculate its magnitude (not maximum), it will be $2\delta \textrm{sinc}(x\delta)$.

 

[ognik]

So now I have $\frac{1}{2} | \psi |^2 = 2 \delta^2 [sinc (x \delta)]^2$ but the maximum of $sinc^2$ is 1, so $\frac{1}{2} | \psi |^2 = 2 \delta^2 = 2 \delta^2 [sinc (x \delta)]^2$, but I have no idea how to solve this eqtn for x, and thence $\Delta x$?

Also the width of $\phi(k_x)$ is $2 \delta$, that is $\Delta k_x$?

 

[blue_leaf77]

$\frac{1}{2} | \psi |^2 = 2 \delta^2 = 2 \delta^2 [sinc (x \delta)]^2$

Shouldn't it be $\frac{1}{2} | \psi |^2 = \delta^2 = 2 \delta^2 [sinc (x \delta)]^2$?
Indeed it's not possible to solve for $x$ analytically, the best you can do without the help of computer is to approximate the sine with its Taylor expansion truncated after, e.g two terms. Doing this,
$$
1=2\left(\frac{x\delta - \frac{1}{6}(x\delta)^3}{x\delta}\right)^2 \\
\frac{x\delta}{\sqrt{2}} = x\delta - \frac{1}{6}(x\delta)^3
$$
Solving for $x$ should be easy.

Also the width of $\phi(k_x)$ is $2 \delta$, that is $\Delta k_x$?

Yes.

 

[ognik]

Thanks (divided one side by 2 then got distracted...).
I quite often see Taylor truncated to 2 terms, is that a reasonable rule of thumb for OM? Or just a convenience because of the complexity the 3rd term usually adds? Also, If you wouldn't mind checking my algebra here...my final answer seems a little high?

$\psi: 1 = 2(1- \frac{(x \delta)^2}{6})^2$
$\therefore \frac{(x \delta)^2}{3} = 1 - \frac{1}{\sqrt{2}}$
$\therefore x = \pm \frac{1}{\delta} [3(1 - \frac{1}{\sqrt{2}})]^{\frac{1}{2}}$
$\therefore \Delta x = \frac{2}{\delta} (0.54)$
$\therefore \Delta x \Delta k_x = 2 \delta \frac{1.08}{\delta} = 2.165$ ?

 

[blue_leaf77]

I quite often see Taylor truncated to 2 terms, is that a reasonable rule of thumb for OM?

The number of terms in the expansion to be retained depends on the desired accuracy of the calculation, if you want you can add the next term into the expansion above if you want it to be more accurate.

$\therefore \frac{(x \delta)^2}{3} = 1 - \frac{1}{\sqrt{2}}$

The denominator on the LHS is not correct.

my final answer seems a little high?

After taking into account the correction I pointed out above, the answer may even goes up slightly. So long as the numerical value is not greater than ten (or five to be safe), it's fine. In fact, the boxcar function is a wide localized function as compared to the other localized functions whose width can vary depending on the way it is defined.

 

[ognik]

Thanks, I think that 'feel' for the max will be useful, and I've learned a lot else through your help.
I get 5.3 after fixing that denominator (should have been 6, I mustn't take short cuts in my algebra)