pettingyg0

2022-04-27

De Broglie's Matter wave equation dividing by zero

I was just thinking about De Broglie's matter wave equation: $\lambda =\frac{h}{p}$ where $p$ is the momentum of the object. But what if the object is at rest? Won't we be dividing by zero? What if we take the limit as momentum tends to zero, won't we start to get noticeable waves? Can someone please explain to me where I went wrong?

I was just thinking about De Broglie's matter wave equation: $\lambda =\frac{h}{p}$ where $p$ is the momentum of the object. But what if the object is at rest? Won't we be dividing by zero? What if we take the limit as momentum tends to zero, won't we start to get noticeable waves? Can someone please explain to me where I went wrong?

luminarc24lry

Beginner2022-04-26Added 1 answers

Strictly speaking, you must use the relativistic "momentum". For most objects, it's not simply $mv$. Rather,

$\lambda =h/p=hc/pc=hc/\sqrt{(}{T}^{2}+2Tm{c}^{2})$

For an electron, say, even at quite low energy (e.g. 1 eV) the term $2Tm{c}^{2}$ is quite high, and so the wavelength still ends up being quite low (~ angstrom scale). In this limit of ${T}^{2}<<2Tm{c}^{2}$, $\lambda \approx h/p$. But it's true, slower moving particles will have longer wavelengths, and those wavelengths can become observable if you have precise enough control of your system and those speeds.

This is also relatable to the Heisenberg Uncertainty principle. If you knew that $p=0$ very precisely, for $\mathrm{\Delta}x\mathrm{\Delta}p\ge \hslash /2$ we'd need larger and larger uncertainties in x to compensate. This would correspond to the particles wavelength becoming larger. In the limit toward $\mathrm{\Delta}p=0$ (which is unphysical) the wavelength of and uncertainty of the particle's position would have to become infinite to compensate.

$\lambda =h/p=hc/pc=hc/\sqrt{(}{T}^{2}+2Tm{c}^{2})$

For an electron, say, even at quite low energy (e.g. 1 eV) the term $2Tm{c}^{2}$ is quite high, and so the wavelength still ends up being quite low (~ angstrom scale). In this limit of ${T}^{2}<<2Tm{c}^{2}$, $\lambda \approx h/p$. But it's true, slower moving particles will have longer wavelengths, and those wavelengths can become observable if you have precise enough control of your system and those speeds.

This is also relatable to the Heisenberg Uncertainty principle. If you knew that $p=0$ very precisely, for $\mathrm{\Delta}x\mathrm{\Delta}p\ge \hslash /2$ we'd need larger and larger uncertainties in x to compensate. This would correspond to the particles wavelength becoming larger. In the limit toward $\mathrm{\Delta}p=0$ (which is unphysical) the wavelength of and uncertainty of the particle's position would have to become infinite to compensate.

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