How Do Widths of Functions Relate to Uncertainty Principles?

[ognik]

(corrections edited in)
1. Homework Statement
Assume ## \psi(x, 0) = e^{-\lambda |x|} \: for \: -\infty < x < +\infty ##. Calculate ## \phi(k_x) ## and show that the widths of ## \phi, \psi ##, reasonably defined, satisfy ##\Delta x \Delta k_x \approx 1 ##

Homework Equations

## \phi(k_x) = \frac{1}{\sqrt{2 \pi}} \int \psi(x) e^{-ik_x x} dx ##

The Attempt at a Solution

## \phi(k_x) = \frac{1}{\sqrt{2 \pi}} [ \frac{1}{(\lambda - i k_x)} e^{(\lambda - i k_x)x} |^0_{-\infty} -
\frac{1}{(\lambda + i k_x)} e^{-(\lambda + i k_x)x} |_0^{\infty} ] ##

## = \frac{1}{\sqrt{2 \pi}} [\frac{1}{(\lambda - i k_x)} + \frac{1}{(\lambda + i k_x)} ] ##

## = \frac{1}{\sqrt{2 \pi}} \frac{2\lambda}{\lambda^2 + k_x^2} ##

What does the width of ##\phi, \psi## mean here? They are ## \Delta k_x, \Delta x ##?

 

[blue_leaf77]

I believe, the wavefunction in position should be ##\psi(x,0) = e^{-\lambda |x|}##.

## = \frac{1}{\sqrt{2 \pi}} [\frac{1}{(\lambda - i k_x)} + \frac{1}{(\lambda + i k_x)} ] ##

## = \frac{1}{\sqrt{2 \pi}} \frac{\lambda}{\lambda^2 + k_x^2} ##

Check again the last line whether a factor of two should appear or not.

What does the width of ##\phi, \psi## mean here? They are ## \Delta k_x, \Delta x ##?

There are several ways to define the width of a function. The two most common ways are using full-width at half maximum (FWHM) and the standard deviation.

 

[ognik]

I believe, the wavefunction in position should be ##\psi(x,0) = e^{-\lambda |x|}##.

Sorry, yes.

Check again the last line whether a factor of two should appear or not.

Again sorry, yes

There are several ways to define the width of a function. The two most common ways are using full-width at half maximum (FWHM) and the standard deviation.

Reading between the lines they are thinking of FWHM - do I take half of ## |\psi \psi^*| ## etc.?

 

[blue_leaf77]

Reading between the lines they are thinking of FWHM - do I take half of |ψψ∗||ψψ∗| |\psi \psi^*| etc.?

Yes, try that. Actually the difference in the form of the uncertainty relation ##\Delta x \Delta k_x## for different definition of the width is on the constant on the RHS, and they do not differ more than one order of magnitude.

 

[ognik]

Hi - off course that is the magnitude not the maximum, please critique the following:

##\psi(x,0) = e^{-\lambda |x|}## is even around 0, ie it has a max. at ##x = 0, \therefore \psi_{max} = 1, \therefore \frac{1}{2}\psi_{max} = \frac{1}{2} ##
Therefore at half max, ## x = \pm \frac{1}{\lambda}ln2, \therefore FWHM \Delta x = \frac{2}{\lambda}ln2 ## ?

Similarly ## \phi(k_x) = \frac{2 \lambda}{\sqrt{2 \pi}} \frac{1}{\lambda^2 + k_x^2} ## is even around 0, therefore max at ## k_x = 0, \therefore \phi_{max}= \frac{2}{\sqrt{2 \pi}} \frac{1}{\lambda}, \therefore\frac{1}{2}\phi_{max} = \frac{1}{\lambda \sqrt{2 \pi}} = \frac{2 \lambda}{\sqrt{2 \pi}} \frac{1}{\lambda^2 + k_x^2}, \therefore \Delta k_x = 2\lambda ##

Something is not quite right above, 'cos I expect ##\Delta x \Delta k_x \approx 1## but I get ## \frac{2 ln 2}{\lambda} 2 \lambda = 4 ln 2 ##?

(Side point, I assume I could also have found the maxes by setting the 1st differential to 0?)

 

[blue_leaf77]

You should calculate the width of the probability functions ##|\psi(x)|^2## and ##|\phi(k_x)|^2##, not the wavefunctions.

(Side point, I assume I could also have found the maxes by setting the 1st differential to 0?)

Not always, sometimes the probability function has more than one maxima, or worse no first derivative at all like ##\psi(x)## in your case.

 

[ognik]

Thanks for being patient, I have a few gaps to find as my degree was over 20 years back ...
I think there is a derivative for ## \psi(x,0), \frac{-x e^{-|x|}}{|x|} ## ? (checked on Wolfram)

Why the width of the probability functions? My text seems to look at the function widths, Ex: " ...it follows that the widths of ## \psi ## and ## \phi ## are effectively related by ## \Delta x \Delta k_x \approx 1 ##

Worked those out anyway, got ## |\psi|^2 = e^{\lambda^2 x^2}, \Delta x = \frac{2}{\lambda} ln \frac{1}{2};... |\phi|^2 = \frac{4 \lambda^2}{(\lambda^2 + k_x^2)^2}, \Delta k_x = 2 \lambda (\sqrt{2} - 1) ## but then ## \Delta x \Delta k_x ## gives a negative value (-1.148) ?

I think ## \Delta x , \Delta k_x ## should always be positive real?
Thanks
-------------------------------------
I hope I am not breaking any rules below ...
Understanding all this clearly will help me with my other current posts, especially https://www.physicsforums.com/threads/fourier-analysis-of-wave-packet.855706/.
https://www.physicsforums.com/threads/wave-packet-expansion.855999/ is slightly different, would appreciate some help with my last query please?
Finally I would appreciate some more help with https://www.physicsforums.com/threads/non-spreading-wave-packet.856281/

Much appreciated

 

[blue_leaf77]

I think there is a derivative for ψ(x,0),−xe−|x||x|ψ(x,0),−xe−|x||x| \psi(x,0), \frac{-x e^{-|x|}}{|x|} ? (checked on Wolfram)

It doesn't have a first derivative at ##x=0##.

Why the width of the probability functions?

Because the quantity which is directly related to the measurement is the probability density, not the wavefunction. Remember that the uncertainty principle tells you the uncertainties in the measurement outcomes.

##|\psi|^2 = e^{\lambda^2 x^2}##

That's not correct, remember you are squaring ##\psi##, which means
$$
|\psi(x)|^2 = e^{-\lambda |x|}\cdot e^{-\lambda |x|} = ?
$$
I am assuming ##\lambda## is real.

##|\phi|^2 = \frac{4 \lambda^2}{(\lambda^2 + k_x^2)^2}

I am sure you are forgetting some constants.

 

[ognik]

It doesn't have a first derivative at ##x=0##.

Just exploring this a bit further - it is multivalued at x=0, so there is a maximum (and a minimum) at x=0?

That's not correct, remember you are squaring ##\psi##, which means
$$
|\psi(x)|^2 = e^{-\lambda |x|}\cdot e^{-\lambda |x|} = ?
$$
I am assuming ##\lambda## is real.

Oops ## e^{-\lambda |x|} \cdot e^{-\lambda |x|} = e^{-2\lambda |x|} ## ?
but ##\frac{1}{2} |\psi|^2 = \frac{1}{2} ## still, then I get ## \Delta x = \frac{ln 0.5}{- \lambda} ## ? Is that right so far?

 

[blue_leaf77]

Just exploring this a bit further - it is multivalued at x=0, so there is a maximum (and a minimum) at x=0?

The wavefunction itself is single valued at the origin, so there can be either only maximum or minimum, to decide which one, one can check the limit of the derivatives when approached from left and from right. For example if it turns out that
$$
\lim_{x\rightarrow 0^-} \frac{d}{dx}f(x) > 0 \\
\lim_{x\rightarrow 0^+} \frac{d}{dx}f(x) < 0
$$
then ##f(x)## must have a maximum at ##x=0##, otherwise it's a minimum.

Oops ##e^{−λ|x|}⋅e^{−λ|x|}=e^{−2λ|x|}##?

Yes.

but ##\frac{1}{2} |\psi|^2 = \frac{1}{2} ## still, then I get ## \Delta x = \frac{ln 0.5}{- \lambda} ## ? Is that right so far?

That's right and also note that ## \Delta x = \frac{\ln 0.5}{- \lambda} > 0##.

 

[ognik]

Yay! Feeling a bit stoopid at this point...

So ##| \phi |^2 = \frac{1}{2 \pi} \frac{4 \lambda^2}{(\lambda^2 + k_x^2)^2}, ##
## \therefore \frac{1}{2}| \phi |^2_{max} = \frac{1}{\pi \lambda^2} = \frac{1}{2 \pi} \frac{4 \lambda^2}{(\lambda^2 + k_x^2)^2} ##

## \therefore (\lambda^2 + k_x^2)^2 = 2 \lambda^4, ##
## \therefore k_x = \pm \lambda (\sqrt{2} - 1)^{\frac{1}{2}}, ## (at half width)
## \therefore \Delta k_x =2 \lambda (\sqrt{2} - 1)^{\frac{1}{2}} ## ?

## \therefore \Delta x \Delta k_x = \frac{ln .5}{- \lambda} 2\lambda (\sqrt{2} - 1) ^{\frac{1}{2}} = 0.89 ##?

A bit lower than 1, but > 0.5 so I think it's ok?

 

[blue_leaf77]

A bit lower than 1, but > 0.5 so I think it's ok?

You are asked to show that ##\Delta x \Delta k_x \approx 1##, therefore as long as your result does not deviate by more than one order of magnitude, that should be fine.