Proof of DeBroglie's Wavelength

May 11th, 2019 by Aziz Lokhandwala

Assumptions: Let us consider a material particle such as electron as a standing wave system, $${f(x) = f(x0) * sin 2πft}$$ It's value at any instant 't' is given as above. Now assume that particle is given velocity 'v' along X- Axis. In order to represent variation in f(x) under this new condition we have to make use of transformation equations from special theory of relativity which are independent from frame of reference.

$$t = {{t' - {v * x \over c^2}} \over {\sqrt {1 - {v^2 \over c^2}}}}$$ Therefore, $${f(x) = f(x0) * sin 2π*f*{{t' - {v * x \over c^2}} \over {\sqrt {1 - {v^2 \over c^2}}}}}$$

Now, comparing above equation with $$y= {A * sin 2π*f'*({t - {x\over u}})}$$ In this 'u' is the velocity of wave along X-direction Therefore, $$ u = {c^2 \over v}$$ $$ f' = {f \over{\sqrt {1 - {v^2 \over c^2}}}}$$ Now from Einstein's mass energy equations we have, $$hf = m*c^2$$ $$f' = {{m*c^2}\over h*{\sqrt {1 - {v^2 \over c^2}}}}$$ Now, since $$m'= {{m} \over {\sqrt {1 - {v^2 \over c^2}}}}$$ Therefore, $$f' = {{m'*c^2} \over h}$$ Now λ for matter wave is \(u \over f'\) Therefore, $$λ = {{c^2 \over v}\over {m'*c^2 \over h}}$$ Therefore, $$λ = {h\over{m'*v}}$$ Therefore, $$λ = {h\over p}$$ Hence Proved