I know that relativistic DeBroglie wavelength is given by $λ = h/γmv$. And $γ ≥ 1$, so at higher speed $λ$ will get shorter and shorter, does this mean it will start behaving like a particle and wave picture would be destroyed?
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$\begingroup$ Note that the question you should be asking is "at high momentum", not at "high speed". Particles traveling at the speed of light have zero mass, so the de Broglie wavelength doesn't become zero. $\endgroup$– Abhimanyu Pallavi SudhirCommented Aug 6, 2018 at 6:27
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$\begingroup$ No, I have mentioned for "high speed" $\endgroup$– Chandra PrakashCommented Aug 6, 2018 at 6:29
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2 Answers
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Yes that is correct.
In general if the action functional associated with the object is much larger than the Planck quantum then the object will behave almost classical in this particular case like a particle.
In most case but not all a larger action functional is associated with a larger energy so in many cases a good rule of thumb is that the higher the energy is the more classical an object behaves. In this case the object behaves more like a classical particle than a wave. Note that high velocity is associated with high energy for your case.
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It depends on whether you're referring to the particle velocity, the group velocity, or the phase velocity. Looking at the equation you have, you are referring to the particle velocity. Which means that you are correct to assume the DeBroglie wavelength will behave more like a particle at high particle velocity.
In the DeBroglie wavelength:
E=mvc
E = Energy | m = mass of the particle | v = velocity of the particle | c = speed of light
The more particle velocity you have, the more energy you have, and the shorter the wavelength will be. Which means that at high particle velocity, DeBroglie wavelength will behave more like a particle than that of a wave.