How to interpret the velocity in de Broglie's equation?

Question

Just wondering if anyone can help me understand the basic principle of quantum theory.

De Broglie's equation allows one calculate the wave length of the physical object, following the fundamental wave-particle duality of quantum theory.

$\lambda = h/mv$

Since velocity $v$ is always relative to the reference frame of observer, does it imply that the wave property is not inherent but displays itself differently to different observers?

Thanks for the answer. Another related question, since momentum/velocity has a direction, how is that direction related to the direction of the wave? — wang1908, Jan 28 '18 at 16:15
the wave has no direction, it is a probability wave :the probability of finding a particle at (x,y,z,t) shows wave properties. Look at the double slit experiments single particle at a time https://en.wikipedia.org/wiki/Double-slit_experiment#Interference_of_individual_particles . That is where the wave like properties reside — anna v, Jan 28 '18 at 16:22
@ anna v - But hasn't a plane de Broglie wave $$\psi (\vec r)=\psi_0 \exp (i\vec k \vec r-i\omega t)$$ clearly the direction of the wave vector $\vec k$? — freecharly, Jan 28 '18 at 16:54
Although the wave is scalar-valued, $\mathbf{k}$ is the wave's direction of propagation (provided $\omega>0$). — J.G., Jan 28 '18 at 17:22
About de Broglie relations, what exactly is $E$? Its energy of what? — Frobenius, Jan 28 '18 at 17:25

score 1 · Answer 1 · answered Jan 28 '18 at 16:20

1

If you want to think about how the wavelike behaviour varies by reference frame, it's more helpful to work with the wavevector $\mathbf{k}=\hbar^{-1}\mathbf{p}$ where $\hbar=h/(2\pi),\,k=2\pi/\lambda$. Extending from 3-vectors to special relativity's 4-vectors, you can even write $k^\mu=\hbar^{-1}p^\mu$, showing light is also affected. (Changing reference frame alters a photon's momentum and energy, albeit not its speed, so also changes the wavevector and frequency). General relativity shows gravity can also have this effect (see here).

answered Jan 28 '18 at 16:20

J.G.

24,837

the de broglie wave length has to do with the quantum mechanical probability distribution describing the particle, so it has no vector properties, it is just the speed. – anna v Jan 28 '18 at 16:24
Photon is difficult to understand since, with zero mass, it's so different from everyday objects. But is it correct to say as an observer, the part of the world that is motionless relative to me will behave wave-like while the part move fast behaves classically? If so why can a person, assuming standing perfectly still, see the wave-like behavior of a building in front of him/her? Although the mass is huge here but we can control the velocity/momentum to really close to zero, say a magnitude of $10^-40$. – wang1908 Jan 28 '18 at 17:06
@Wang-X-Y The problem with plugging a very small momentum into the de Broglie formula is it neglects the uncertainty in the momentum. – J.G. Jan 28 '18 at 18:16

How to interpret the velocity in de Broglie's equation?

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