How Do Function Widths and Uncertainty Principles Relate in Quantum Mechanics?

[ognik]

Homework Statement

$\phi(k_x) = \begin{cases}\phantom{-} \sqrt{2 \pi},\; \bar{k_x} - \frac{\delta}{2} \le k_x \le \bar{k_x} + \frac{\delta}{2} \\ - \sqrt{2 \pi},\; \bar{k_x} - \delta \le k_x \le \bar{k_x} - \frac{\delta}{2} \:AND \: \bar{k_x} + \frac{\delta}{2} \le k_x \le \bar{k_x} + \delta \end{cases}$
Calculate $\psi(x, 0)$ and find $\Delta x \Delta k_x$

Homework Equations

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

The Attempt at a Solution

Firstly, $\Delta k_x = 2 \delta$

$\psi(x, 0) = -\int^{k_x - \frac{\delta}{2}}_{k_x - \delta} e^{i k_x x} dk_x +\int^{k_x + \frac{\delta}{2}}_{k_x - \frac{\delta}{2}} e^{i k_x x} dk_x -\int^{k_x + \delta}_{k_x + \frac{\delta}{2}} e^{i k_x x} dk_x$

$= \frac{1}{i x} [ -e^{i k_x x (\bar{k_x} - \frac{\delta}{2})} + e^{i k_x x (\bar{k_x} - \delta)} + e^{i k_x x (\bar{k_x} +\frac{\delta}{2})} - e^{i k_x x (\bar{k_x} - \frac{\delta}{2})} - e^{i k_x x (\bar{k_x} +\delta)} + e^{i k_x x (\bar{k_x} + \frac{\delta}{2})} ]$

$= \frac{1}{i x} e^{i x \bar{k_x}} [ -2 (e^{i x \frac{\delta}{2}} - e^{-i x \frac{\delta}{2}}) - (e^{i x \delta} - e^{-i x \delta})$

$= \frac{1}{i x} e^{i x \bar{k_x}} [4 sin(x \frac{\delta}{2}) - 2sin(x \delta) ]$

$= \frac{1}{i x} e^{i x \bar{k_x}} [ 2 \delta sinc (\frac{x \delta}{2}) - \delta sinc (x \delta )]$

Taylor to 2 terms...
$[...] = 2 \delta (1 - \frac{(x \delta)^2}{6} ) - \delta (1 - \frac{(x \delta)^2}{6} )$

$= 2 \delta - \frac{x^2 \delta^3}{12} - \delta + \frac{x^2 \delta^3}{6}$

$= \delta (1 + \frac{x^2 \delta^2}{12} )$

$\therefore \frac{1}{2} |\psi |^2 = \frac{\delta^2}{2} (1 + \frac{x^2 \delta^2}{12} ) ^2$

I hope this is right so far, please check? When I plot this in Wolfram I get a positive parabola, so no maximum, so I must have done something wrong but can't find it ...

 

[RUber]

Homework Statement

$\phi(k_x) = \begin{cases}\phantom{-} \sqrt{2 \pi},\; \bar{k_x} - \frac{\delta}{2} \le k_x \le \bar{k_x} + \frac{\delta}{2} \\ - \sqrt{2 \pi},\; \bar{k_x} - \delta \le k_x \le \bar{k_x} - \frac{\delta}{2} \:AND \: \bar{k_x} + \frac{\delta}{2} \le k_x \le \bar{k_x} + \delta \end{cases}$
Calculate $\psi(x, 0)$ and find $\Delta x \Delta k_x$

Homework Equations

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

The Attempt at a Solution

Firstly, $\Delta k_x = 2 \delta$

$\psi(x, 0) = -\int^{k_x - \frac{\delta}{2}}_{k_x - \delta} e^{i k_x x} dk_x +\int^{k_x + \frac{\delta}{2}}_{k_x - \frac{\delta}{2}} e^{i k_x x} dk_x -\int^{k_x + \delta}_{k_x + \frac{\delta}{2}} e^{i k_x x} dk_x$

$= \frac{1}{i x} [ -e^{i k_x x (\bar{k_x} - \frac{\delta}{2})} + e^{i k_x x (\bar{k_x} - \delta)} + e^{i k_x x (\bar{k_x} +\frac{\delta}{2})} - e^{i k_x x (\bar{k_x} - \frac{\delta}{2})} - e^{i k_x x (\bar{k_x} +\delta)} + e^{i k_x x (\bar{k_x} + \frac{\delta}{2})} ]$
...

The first thing I see is that you still have k_x in the exponent, even though that was your variable of integration.
$= \frac{1}{i x} [ -e^{i x (\bar{k_x} - \frac{\delta}{2})} + e^{i x (\bar{k_x} - \delta)} + e^{ix (\bar{k_x} +\frac{\delta}{2})} - e^{ix (\bar{k_x} - \frac{\delta}{2})} - e^{ix (\bar{k_x} +\delta)} + e^{ix (\bar{k_x} + \frac{\delta}{2})} ]$

$= \frac{1}{i x} e^{i x \bar{k_x}} [ -2 (e^{i x \frac{\delta}{2}} - e^{-i x \frac{\delta}{2}}) - (e^{i x \delta} - e^{-i x \delta})$

$= \frac{1}{i x} e^{i x \bar{k_x}} [4 sin(x \frac{\delta}{2}) - 2sin(x \delta) ]$
...

Don't forget the i in the denominator of the sine function.
$sin(x) = \frac{e^{ix}-e^{-ix}}{2i}$ so what you have should be $4i\sin(x \frac{\delta}{2})- 2i sin(x \delta)$
From there, it seems like a double angle or half angle formula might be appropriate.

$= \frac{1}{i x} e^{i x \bar{k_x}} [ 2 \delta sinc (\frac{x \delta}{2}) - \delta sinc (x \delta )]$

Taylor to 2 terms...
$[...] = 2 \delta (1 - \frac{(x \delta)^2}{6} ) - \delta (1 - \frac{(x \delta)^2}{6} )$

$= 2 \delta - \frac{x^2 \delta^3}{12} - \delta + \frac{x^2 \delta^3}{6}$

$= \delta (1 + \frac{x^2 \delta^2}{12} )$

$\therefore \frac{1}{2} |\psi |^2 = \frac{\delta^2}{2} (1 + \frac{x^2 \delta^2}{12} ) ^2$

I hope this is right so far, please check? When I plot this in Wolfram I get a positive parabola, so no maximum, so I must have done something wrong but can't find it ...

 

[ognik]

Sorry, I was cutting & pasting and didn't notice I had left the $k_x, i$ in, had it right on paper ...

So I think my solution is correct up to this:
$\psi(x,0) = \frac{1}{x} e^{i x \bar{k_x}} [ 4 sin (\frac{x \delta}{2}) - 2 sin (x \delta)]$; - The exponent will vanish when I find $| \psi |^2$

I couldn't simplify this using dbl angle identities, I got for example $4 Sinx\delta (1 - Cos x\delta)$ I couldn't use identities like sin A - sin B because of the different coefficients?

Taylor expanding (more carefully) I get $\psi = e^{i x \bar k_x}.\frac{x^2 \delta^3}{4}$, but doesn't look right?

 

[RUber]

Why doesn't that look right? What characteristics are you expecting to see? If you aren't seeing them, try to keep more terms in your Taylor expansion.
Looking at the functional form
$\psi(x,0) = e^{ixk_x}\frac{4\sin(x\delta) (1 - \cos(x\delta) ) }{x}$ makes it pretty clear that this function is bounded.

 

[ognik]

If I plot $x^4$ I get an upside down parabola, centre 0, instead of being periodic as I would expect - so I suppose that would need a few additional terms in the expansion?

May I assume my solution of $\psi(x,0) = \frac{1}{x} e^{i x \bar{k_x}} [ 4 sin (\frac{x \delta}{2}) - 2 sin (x \delta)]$ is correct? When I plot the square of this it is periodic, but each cycle is not quite symmetric. The max appears to be close to 27 from the plot, but I can't see how to find this and $\Delta x$ analytically?
Normally textbook exercises simplify readily, which makes me feel my solution must be wrong somehow? So, I'd appreciate a bit more help - I think once I see how to do this, the rest of the section will follow.

 

[ognik]

Hi - I am effectively self-taught with this forum my only source of interaction and assistance, so would really, really appreciate any and all help making progress please ...

 

[mfb]

The taylor expansion to second order is always a parabola. A taylor expansion to finite order cannot be periodic. Why do you want to use a taylor expansion of anything?

The sum of two sinc functions is right as answer, but I would expect to see them shifted. Do you have the corrected version of your integrals somewhere?

Firstly, $\Delta k_x = 2 \delta$

How did you get this?

 

[ognik]

The taylor expansion to second order is always a parabola. A taylor expansion to finite order cannot be periodic. Why do you want to use a taylor expansion of anything?

The sum of two sinc functions is right as answer, but I would expect to see them shifted. Do you have the corrected version of your integrals somewhere?

I was trying the expansion based on advice from a similar problem, that using sync wouldn't allow an analytic solution - see https://www.physicsforums.com/threads/fourier-analysis-of-wave-packet.855706/

$\psi(x, 0) = -\int^{k_x - \frac{\delta}{2}}_{k_x - \delta} e^{i k_x x} dk_x +\int^{k_x + \frac{\delta}{2}}_{k_x - \frac{\delta}{2}} e^{i k_x x} dk_x -\int^{k_x + \delta}_{k_x + \frac{\delta}{2}} e^{i k_x x} dk_x$

$= \frac{1}{i x} [ -e^{i x (\bar{k_x} - \frac{\delta}{2})} + e^{i x (\bar{k_x} - \delta)} + e^{i x (\bar{k_x} +\frac{\delta}{2})} - e^{i x (\bar{k_x} - \frac{\delta}{2})} - e^{i x (\bar{k_x} +\delta)} + e^{i x (\bar{k_x} + \frac{\delta}{2})} ]$

$= \frac{1}{i x} e^{i x \bar{k_x}} [ 2 (e^{i x \frac{\delta}{2}} - e^{-i x \frac{\delta}{2}}) - (e^{i x \delta} - e^{-i x \delta})$

$= \frac{1}{x} e^{i x \bar{k_x}} [4 sin(x \frac{\delta}{2}) - 2sin(x \delta) ]$

$\therefore \psi(x,0) = e^{i x \bar{k_x}} 2 \delta [ sinc (\frac{x \delta}{2}) - sinc (x \delta )]$

I am stuck here because I want $\frac{1}{2} | \psi |^2_{max}$ but the max of Sinc anything = 1 so Sinc - Sinc = 0?

Firstly, $\Delta k_x = 2 \delta$
How did you get this?

From the function definition, its a square wave from $\bar{k_x} - \delta$ to $\bar{k_x} + \delta$
Also the book says it will be the same width as I found in a similar problem - again see https://www.physicsforums.com/threads/fourier-analysis-of-wave-packet.855706/

 

[mfb]

Why do you want to find the maximum amplitude of your wave function, and why do you expect it at zero?
There is a formula for Δx that did not appear in this thread yet. Start with that formula before you calculate things that don't help.

The square wave has a width of 2δ, but that is not Δk. Well, if you don't need the precise prefactors it is okay.

 

[vela]

Have you plotted the function? As you noted, you can't calculate $\langle x^2 \rangle$, but you can come up with a reasonable estimate for the width by other methods. You don't have to use FWHM.

 

[ognik]

Why do you want to find the maximum amplitude of your wave function, and why do you expect it at zero?

I want to find FWHM and thus the values of $k_x$ at 1/2 max of $|psi|^2$ - its the method this section of the text seems to be focused on. I don't expect it at 0, but using the Sinc function it workes out to 0 - which I'm sure is wrong, but can't see my mistake.

There is a formula for Δx that did not appear in this thread yet. Start with that formula before you calculate things that don't help.

I don't think I know that formula?

The square wave has a width of 2δ, but that is not Δk. Well, if you don't need the precise prefactors it is okay.

Probably I don't, we're at the start of the book - but if you don't mind I'd like more awareness of how those prefactors would impact this please?

 

[ognik]

Have you plotted the function? As you noted, you can't calculate $\langle x^2 \rangle$, but you can come up with a reasonable estimate for the width by other methods. You don't have to use FWHM.

Ah, assumptions :-(, the expression is only 0 at x=0, looks like a capital M ...
For $\delta = 1$, the $max^2$ is $2 \times 0.664$, but I would need to know this in terms of $\delta$ to use FWHM.

slightly off-topic to help me get a feel for these things, what would be an approx. range for $\delta$ in this case, for (say) a free electron? An Angstrom maybe?

The book only mentions FWHM, and Taylor expansions don't seem good enough - could you walk me through an appropriate alternative please?

 

[mfb]

I don't think I know that formula?

Okay, fine.

Probably I don't, we're at the start of the book - but if you don't mind I'd like more awareness of how those prefactors would impact this please?

Well, your answer can be wrong by something like a factor of 2 if you just take something that looks like a width instead of the mathematical definition of the standard deviation.

For $\delta = 1$ , the $max^2$ is $2 \times 0.664$ , but I would need to know this in terms of $\delta$ to use FWHM.

The maximum scales nicely with δ.

slightly off-topic to help me get a feel for these things, what would be an approx. range for $\delta$ in this case, for (say) a free electron? An Angstrom maybe?

δ is an inverse length. x is the width of the pattern (if you imagine the setup as similar to a slit experiment), probably more of the order of micrometers, so δ is the inverse of that.