Let's say I am given the dispersion relation for nearly-free electrons:

$$ E(k) = \frac{\hbar^2}{2m}(k^2+c\,k^4)$$

Where $c$ is a small constant of appropiate dimension.

How do I calculate the velocity of an electron given a fixed $k_1$ ?

Applying "classical" laws results in

$$v(k_1)=\sqrt{\frac{2E(k_1)}{m}} = \frac{\hbar k_1}{m}\sqrt{1+c k_1^2}$$

On the other hand, applying $$v(k_1) = \frac{\partial \omega(k)}{\partial k}\bigg|_{k=k_1} = \hbar^{-1} \frac{\partial E(k)}{\partial k}\bigg|_{k=k_1} = \frac{\hbar}{2m}(2k_1+4ck_1^3) = \frac{\hbar k_1}{m}(1+2c k_1^2)$$

Obviously both terms are not the same, so can anyone explain to me where is the difference ? I guess it has something to do with mixing up the velocity-concept of classical particles (1) and the group-velocity of electrons as waves (2).

  • $\begingroup$ If you make use of the effective (as distinct from the free) electron mass in any calculations, you are, in principle, using QM. $\endgroup$
    – jim
    Jun 23, 2016 at 19:55

1 Answer 1


For wave mechanics there is the phase velocity and group velocity. For the energy $E~=~\hbar\omega$ the phase velocity is $$ v_p~=~\frac{\omega}{k}~=~\frac{\hbar}{2m}(k~+~ck^3). $$ This is the velocity of a wave front, or where the phase of the wave is constant. There is also the group velocity that is $$ v_g~=~\frac{\partial\omega}{\partial k}~=~\frac{\hbar}{m}(k~+~2ck^3). $$

The classical idea suggested in this question is that $k^2~+~ck^4~=~k'^2$ so that $$ p'~=~\hbar k'~=~\hbar k\sqrt{1~+~ck^2} $$ This is a different definition of the momentum and thus velocity. I would say that a better approach is to write the Hamiltonian or energy according to $p~=~\hbar k$ $$ H~=~\frac{1}{2m}\left(p^2~+~\frac{c}{\hbar^2}p^4\right). $$ This Hamiltonian is an operator for $p~\rightarrow~-i\partial_x$. Now put the wave function $\psi(x,t)~=~Ae^{-ikx~+~i\omega t}$ in the Schrodinger equation to arrive at $$ H\psi~=~i\hbar\frac{\partial\psi}{\partial t} $$ to find $$ \hbar\omega~=~\frac{\hbar^2}{2m}\left(k^2~+~ck^4\right), $$ which agrees with the phase velocity above.

If you were to insist on doing a sort of classical form with the Hamiltonian above with $H~=~E$ you will arrive at a rather complicated equation $$ p^2~=~-\frac{\hbar^2}{2c}\left(1~-~\sqrt{1~+~\frac{8mc}{\hbar^2}E}\right). $$ This is from the quadratic equation and the choice with $p^2~=~0$ at $E~=~0$. If you let $E~=~\hbar\omega$ it is not hard to see this recovers the above result with the Schrodinger equation.

  • 1
    $\begingroup$ And which approach do I use? $\endgroup$
    – Christian
    Jun 25, 2016 at 6:55

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