# Is wavelenth of a particle relative according to Wave Particle Duality?

De Broglie's equation regarding the wave particle duality of matter $$\lambda =\frac{h}{mv}$$ depends on velocity. Now, velocity is relative and depends on the frame of reference of the observer. But does this mean that even the wavelength is relative and will be different for different observers at different frames of reference? At first I almost readily agreed to this conclusion but then I realized that this means that two people at different speeds will perceive wave"length" at different amounts. This is what struck me as odd, as the only place where I've heard of length disparities among different observers is in special relativity

Can someone explain what does it really mean to have $$\lambda = \frac{h}{mv}$$ when velocity is relative?

The de Broglie relation is a forerunner of the quantum mechanical wave equations. We now know that the wavelength in the wave particle duality is not a wavelength in space. It is the wavelength of the wavefunction describing the quantum mechanical probability distribution of finding the particle at (x,y,z) at time t. This means that to measure the wavelength you need an accumulation of measurements with the same boundary conditions.

wavelength is relative and will be different for different observers at different frames of reference

Yes, distributions can differ if the measurement happens in different inertial frames. But an accumulation of measurements is always necessary to see the wave-particle duality.

To get an intuition of wave particle duality , see how the distribution of electrons from a double slit experiment show the wavelength for those particular boundary condition The wavelength referred for the probability distribution should be referred to the rest frame of the experiment.

For a consistent treatment you can have a look at the Klein-Gordon equation, which is a (spinless) relativistic version of the Schrödinger equation $$\left(\frac{1}{c^2}\frac{\partial^2}{\partial t^2}-\nabla^2+\frac{m^2c^2}{\hbar^2}\right)\psi(t,\mathbf x)=0$$ It has solutions of the form $$e^{-ip_\mu x^\mu/\hbar}$$ where $$p_\mu x^\mu=\hbar\omega t-\hbar\,\mathbf k\cdot \mathbf x$$. This last quantity is invariant when you change your frame of reference in a relativistic way.