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While reading a text book about basics of Quantum Mechanics, I came across a situation in which it is said that

$E=\hbar\omega$ and also

$E = \frac12mv^2=p^2/2m$

where

$h$ Planck's constant

$\hbar=\frac{h}{2\pi}$ Planck's reduced constant

$\omega=2\pi f$ angular frequency

$m$ mass

$v$ velocity

$p$ momentum

But if I take the first definition,E=(h/2pi)*w,then

E=(h/2pi)*2*pi*f (because w=2*pi*f)

 = h*f

 = h*(v/λ)       (because v=fλ)

 = p*v           (de-Broglie's wave-particle duality p=h/λ )

 = mv*v          (because p=m*v, the momentum)

E = m*v^2

This is not same as definition $E=\frac12mv^2$.

What am I missing in the derivation above?

Qmechanic
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gmax
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    While the question may be based on a mistake, it is asked in good faith and is certainly answerable, so I don't see why it is off topic. –  Nov 04 '14 at 20:16
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    Essentially a duplicate of http://physics.stackexchange.com/q/34214/2451 – Qmechanic Nov 06 '14 at 15:42

1 Answers1

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The relationship $$ v=\lambda\nu=\frac{\omega}{k} $$ describes the phase velocity ($v_p\neq v$) and not the group velocity ($v_g=v$), so it should be $$ v_p=\frac{\omega}{k}=\frac{\hbar\omega}{\hbar k}=\frac{E}{p}=\frac{p}{2m}=\frac{v}{2}\tag{1} $$ which does follow from the de Broglie relation ($p=h/\lambda=\hbar k$). Inserting (1) into your 3rd line gives $$ E=pv_p=mvv_p=mv\left(\frac12v\right)=\frac12mv^2 $$

Kyle Kanos
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  • in the above derivation I used the definition"The wavelength λ of a sinusoidal waveform traveling at constant speed v is given by λ=v/f", while substituting for frequency 'f'. So 'v' is velocity not frequency.Then used the definition, momentum p = mass(m)velocity(v). Which resulted E=mv^2(and of course a big confusion in mind :)). – gmax Nov 06 '14 at 05:03
  • @GaneshHegde: You are correct. I've found your error though and have updated my answer. – Kyle Kanos Nov 06 '14 at 14:49