- #1
coccoinomane
- 19
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Localization vs De Broglie wavelength
First of all, I would like to say hello to everybody since this is my first post, even if it's been some time since I read Physics Forums.
Second, sorry for my bad English, I'm Italian :)
Third, the issue:
I read from my teacher's notes, that a particle (a proton in that case, but I don't think it matters) cannot be localized with precision superior to its De Broglie wavelength.
In my mind, if [tex]\Delta x[/tex] is the uncertainty on the particle x coordinate, and [tex]\lambda[/tex] is the particle's associated De Broglie wavelength, then my teacher's statement can be expressed as
[tex]\Delta x\geq\lambda[/tex]
at least in one dimension.
But all I get from the Uncertainty principle, assuming [tex]\Delta p=\frac{h}{\Delta\lambda}[/tex], is
[tex]\Delta x\Delta p=\Delta x\frac{h}{\Delta\lambda}\geq\frac{\hbar}{2}\Rightarrow\Delta x\geq\frac{\Delta\lambda}{4\pi}[/tex]
Where am I wrong? Maybe in considering [tex]\Delta p=\frac{h}{\Delta\lambda}[/tex]? Or am I missing some reasoning as a result of which [tex]\Delta\lambda\simeq\lambda[/tex]?
Thanks in advance for any answer.
Guido
First of all, I would like to say hello to everybody since this is my first post, even if it's been some time since I read Physics Forums.
Second, sorry for my bad English, I'm Italian :)
Third, the issue:
I read from my teacher's notes, that a particle (a proton in that case, but I don't think it matters) cannot be localized with precision superior to its De Broglie wavelength.
In my mind, if [tex]\Delta x[/tex] is the uncertainty on the particle x coordinate, and [tex]\lambda[/tex] is the particle's associated De Broglie wavelength, then my teacher's statement can be expressed as
[tex]\Delta x\geq\lambda[/tex]
at least in one dimension.
But all I get from the Uncertainty principle, assuming [tex]\Delta p=\frac{h}{\Delta\lambda}[/tex], is
[tex]\Delta x\Delta p=\Delta x\frac{h}{\Delta\lambda}\geq\frac{\hbar}{2}\Rightarrow\Delta x\geq\frac{\Delta\lambda}{4\pi}[/tex]
Where am I wrong? Maybe in considering [tex]\Delta p=\frac{h}{\Delta\lambda}[/tex]? Or am I missing some reasoning as a result of which [tex]\Delta\lambda\simeq\lambda[/tex]?
Thanks in advance for any answer.
Guido
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