According to the De Broglie hypothesis, wavelength equals to h/mv(m-mass, v-velocity, h-Planck constant), so by moving at a slow speed, that is reducing the v(velocity factor) can human beings with such high masses(m) have a wavelength that is long enough to interfere?


If we assume that the human body can be treated as a single particle at the centre of mass, then we can tackle this problem. Optimal diffraction occurs when the wavelength is equal to the size of the aperture.

The average male shoulder width is .465m, so we can take the width of the aperture to be $.5$m. This would give a wavelength of $.5$m. So, $\lambda=0.5=\frac{h}{mv}$

The average male body weight is 81.9kg, which we can take as 80kg. Then, using $h=6.626 \times 10^{-34}$, we get:

$v=\frac{h}{m\lambda}=\frac{6.626 \times 10^{-34}}{40}$

This gives a velocity of $1.6565 \times 10^{-35}$ metres per second. If we then take a door to be $.3$m deep, it would take you $1.8 \times 10^{34}$ seconds, or $5.7 \times 10^{26}$ years. That's pretty slow.

Of course, all of this assumes that we could take the human body as a single particle - which we can't.

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  • $\begingroup$ physics.stackexchange.com/questions/57390/… But this explains that the de Broglie wavelength formula is valid to a non-fundamental (many body) object. $\endgroup$ – alst Sep 1 '16 at 7:47
  • $\begingroup$ If you look at John Rennie's answer, it states that all the particles are required to be coherent. Considering the complexity of the human body, and all the fluids, and the fact that it's almost impossible for a bullet, then the formula is not valid for a human. Though, if it were, then I have given the times involved. $\endgroup$ – Noah P Sep 1 '16 at 8:14
  • $\begingroup$ So is that means David Bar Moshe's answer is wrong? $\endgroup$ – alst Sep 1 '16 at 8:20
  • $\begingroup$ No, but it requires all the particles to be coherent, which would not be possible. $\endgroup$ – Noah P Sep 1 '16 at 8:25
  • $\begingroup$ I've read John Rennie's answer, he also said that we can prepare a bullet in a coherent state in principle, what does this mean? Can a human be in coherent state in principle too? $\endgroup$ – alst Sep 1 '16 at 9:07

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