Uncertainty of a Gaussian wavepacket

  • #1
Isaac0427
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Hi,

I know that a Gaussian wavepacket has minimum uncertainty. The issue is, some sources are telling me that σxσp=ħ and others are telling me that σxσp=ħ/2. I am really confused. I think the latter is correct due to what I have been taught about the uncertainty principle, but then I don't understand what sources like the following are telling me:
http://oer.physics.manchester.ac.uk/QM/Notes/jsmath/Notesse39.html (see equation 10.17)

However, my own math (to my understanding) gives me a result similar to the link above. Working this out on my own, this is how I understand it:
A normalized Gaussian, where ##\left<x\right>=0## with a standard deviation ##\sigma_x## is
$$\psi (x) = \frac{1}{\sigma_x \sqrt{\pi}}e^{-x^2/2\sigma_x^2}$$
It's Fourier transform is
$$\tilde{\psi} (k) = \frac{\sigma_x}{\sqrt{\pi}}e^{-k^2\sigma_x^2/2}=\frac{1}{\sigma_k\sqrt{\pi}}e^{-k^2/2\sigma_k^2}$$
where the uncertainty in k is
$$\sigma_k=\frac{1}{\sigma_x}$$

This seems to yield
$$\sigma_x\sigma_k=1$$
or
$$\sigma_x\sigma_p=\hbar$$
Since the Gaussian minimizes uncertainty, the uncertainty principle would thus be
$$\sigma_x\sigma_p\geq \hbar$$
as the linked article suggests. To my knowledge, however, the uncertainty principle is
$$\sigma_x\sigma_p\geq\frac{\hbar}{2}$$
What is going on here?
 
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  • #2
Isaac0427 said:
I think the latter is correct
True
Isaac0427 said:
but then I don't understand what sources like the following are telling me:
http://oer.physics.manchester.ac.uk/QM/Notes/jsmath/Notesse39.html (see equation 10.17)
Equation 10.13 is correct, but 10.17 cannot be (they contradict!). I don't think that site is very reliable and accurate. I spotted other typos there too. May be try doing the math instead of trusting it.
 
  • #3
Stavros Kiri said:
May be try doing the math instead of trusting it.
Right-- I edited that into the original post. Could you tell me what I did wrong there?
 
  • #4
Looking back at the link, I think I found the mistake. Equation 10.21 is the key and is correct (yields Δp = ħ/σ√2), but equations 10.17 and 10.22 are both wrong (typo).

Do you still want me to check your Fourier transform method?
 
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  • #5
Stavros Kiri said:
Do you still want me to check your Fourier tranforms?
If you could, as I still don't know what I did wrong there.
 
  • #6
Ok. I think that although the original gaussian is correct and normalized, the Fourier transform isn't normalized. When you normalize it you should get the same form as the original (just x and k switched, I think) and the correct result ...
 
  • #7
Stavros Kiri said:
Ok. I think that although the original gaussian is correct and normalized, the Fourier transform isn't normalized. When you normalize it you should get the same form as the original (just x and k switched, I think) and the correct result ...
What would the correct result be (i.e. the correct transform)?
 
  • #8
Isaac0427 said:
A normalized Gaussian, where ##\left<x\right>=0## with a standard deviation ##\sigma_x## is
$$\psi (x) = \frac{1}{\sigma_x \sqrt{\pi}}e^{-x^2/2\sigma_x^2}$$
It's Fourier transform is
Just switch x and k. Just noticed you have it there:
Isaac0427 said:
It's Fourier transform is
$$\tilde{\psi} (k) = \frac{\sigma_x}{\sqrt{\pi}}e^{-k^2\sigma_x^2/2}=\frac{1}{\sigma_k\sqrt{\pi}}e^{-k^2/2\sigma_k^2}$$

But the first part of this last double equation seems to be wrong. How did you get it? Is that equation normalized? The second part is.
 
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  • #9
Hang on. I am fixing the quotes.
 
  • #10
Isaac0427 said:
A normalized Gaussian, where ##\left<x\right>=0## with a standard deviation ##\sigma_x## is
$$\psi (x) = \frac{1}{\sigma_x \sqrt{\pi}}e^{-x^2/2\sigma_x^2}$$
It's Fourier transform is
$$\tilde{\psi} (k) = \frac{\sigma_x}{\sqrt{\pi}}e^{-k^2\sigma_x^2/2}=\frac{1}{\sigma_k\sqrt{\pi}}e^{-k^2/2\sigma_k^2}$$
 
  • #11
All fixed. Just see and reply to #8 above.
 
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