I am studying quantum physics and there is something I don't understand:

  1. I know that for any particle $E=hf$ (Einstein relation) and

  2. $v=\lambda f$ ($v$ is the speed of the particle).

  3. I also know that the kinetic energy is $E_k=\frac{mv^2}{2}$.

Solving those 3 equations for $\lambda$: $$h\frac{v}{\lambda}=\frac{mv^2}{2},$$ I finally find $$\lambda=\frac{2h}{mv}=\frac{2h}{p},$$ which is not consistent with the De Broglie relation $$\lambda=\frac{h}{p}.$$ Where am I wrong in my development?

  • 1
    $\begingroup$ De broglie used the relativistic version of mass-rest energy $ E=mc^{2} $ and then simply generalized to $ c=v $ this is why you get an extra term $ 2 $ $\endgroup$ Aug 15, 2012 at 9:18

4 Answers 4


I) One must distinguish between the group velocity

$$v_g~=~\frac{\partial E}{\partial p}~=~v,$$

and the phase velocity

$$v_p~=~\frac{E}{p}~=~\left\{ \begin{array}{cl} \frac{v}{2} & \text{in non-rel. QM (Schr. eq.) where}~ E~=~ \frac{p^2}{2m}, \cr \frac{c^2}{v}& \text{in rel. QM, QFT (Dirac eq., KG. eq.) where}~ E~=~ \sqrt{(pc)^2+(m_0 c^2)^2}. \end{array}\right. $$

of a matter wave. (See also the Wikipedia page about the de Broglie relations. The phase velocity $v_p$ is sensitive to where one puts the zero of the energy scale. In non-relativistic theories, one usually works with the kinetic energy (=total energy minus rest-energy). For a reduction from Klein-Gordon eq. to Schrödinger eq., see e.g. A. Zee, QFT in a Nutshell, Chap. III.5, and this Phys.SE post.)

II) So the answer to OP's question (v4) is that his first relation is correct, his second relation should read

$$v_p~=~\lambda f,$$

and in the relativistic case, the kinetic energy in his third relation should be replaced with the total energy.

  • $\begingroup$ In the case of a single particle isn't phase velocity equal to group velocity ? I learned that group velocity is defined for wave paquets. $\endgroup$
    – snickers
    Aug 15, 2012 at 12:56
  • 2
    $\begingroup$ No, the phase velocity and the group velocity are different, cf. e.g. Wikipedia. $\endgroup$
    – Qmechanic
    Aug 15, 2012 at 13:19
  • $\begingroup$ So in one case $v_p < c$, in the other $v_p > c$. Isn't it contradictory? $\endgroup$
    – marmistrz
    Jan 28, 2016 at 7:47
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    $\begingroup$ @ marmistrz: Good observation. This (superficial) discrepancy becomes most apparent in the non-rel. limit $v\to 0$ where the two cases predict $v_p \to 0$ and $v_p \to \infty$, respectively! The resolution is that the energy $E$ in the phase velocity formula $v_p=\frac{E}{p}$ depends on where we put the zero on the energy scale: In manifestly non-rel. theories we have implicitly subtracted the rest energy $m_0c^2$ in the definition of energy $E$. Consult e.g. the mentioned book by Zee to see how this works out consistently. $\endgroup$
    – Qmechanic
    Jan 28, 2016 at 16:06

This story actually starts with Einstein's paper on the photoelectric effect. Einstein proposed that for light waves, $E \propto f$, with a proportionality constant that eventually became known as $h$. Using the relation $E = pc$ from special relativity, you can derive that $pc = hf$, and with $\lambda f = c$ you get $\lambda = \frac{h}{p}$. Remember, though, so far this only applies to light. de Broglie's insight was to use the same relation to define the wavelength of a particle as a function of its momentum.

So where does your derivation go wrong? The key step is $v = \lambda f$, which applies to a wave, not a particle. As Qmechanic says, the wave velocity is not the same as the particle velocity. (The former is the phase velocity and the latter is the group velocity.)

Even though $\lambda = \frac{h}{p}$ was originally taken as an assumption, you can work backwards (or forwards, depending on your view) and derive it from a more general quantum theory. For example, suppose you start with the Schroedinger equation in free space,

$$i\hbar\frac{\partial \Psi(t,x)}{\partial t} = -\frac{\hbar^2}{2m}\frac{\partial^2 \Psi(t,x)}{\partial x^2}$$

Solutions to this equation take the form

$$\Psi(t,x) = \sum_n C_n\exp\biggl(-\frac{i}{\hbar}\bigl(E_nt \pm x\sqrt{2mE_n}\bigr)\biggr) = \sum_n C_n\exp\biggl[-i\biggl(\omega_nt \pm k_nx\biggr)\biggr]$$

This is a wave with multiple individual components, each having angular frequency $$\omega_n = E_n/\hbar$$ and wavenumber $$k_n = \frac{\sqrt{2mE_n}}{\hbar}$$ or equivalently, frequency $$f_n = E_n/h$$ and wavelength $$\lambda_n = \frac{h}{\sqrt{2mE_n}}$$ To come up with de Broglie's relation, you need to find an expression for the momentum carried by the wave. This is done using the momentum operator $\hat{p} = -i\hbar\frac{\partial}{\partial x}$ in $\hat{p}\Psi = p\Psi$. The thing is, it only works for a wavefunction with one component. So if (and only if) all the $C_n$ are zero except one, you can get

$$p_n = \mp\hbar k_n = \mp\sqrt{2m E_n}$$

and if you put that together with the definition of $\lambda_n$, you get $\lambda_n = \frac{h}{p_n}$.

It may seem like a problem that this procedure only works for single-component waves. It's okay, though, because the wave doesn't actually have a single well-defined wavelength anyway unless it consists of only one component. This is a key point: whenever you talk about the wavelength of a particle, or more precisely the wavelength of the matter wave associated with a particle, you're implicitly assuming that the matter wave has only a single frequency component. This is generally a useful approximation for real particles, but it's never exactly true.


The energy $E$ in this case is the total energy i.e. the kinetic energy plus the rest mass energy. For a photon there is no contribution from the rest mass because the rest mass is zero, but for a massive particle you need to consider both terms. The energy $E$ is given by (this is the relativistic expression so it applies both to photons and massive particles):

$$E^2 = p^2c^2 + m_0^2c^4 $$

We can also write $E$ as the sum of the kinetic energy and the rest mass energy:

$$E = KE + m_0c^2$$

Setting the two expressions for the energy equal gives:

$$KE^2 + 2KEm_0c^2 + m_0^2c^4 = p^2c^2 + m_0^2c^4$$

or with a quick rearrangement:

$$p^2 = \frac{KE}{c^2} (KE + 2m_0c^2)$$

Note that we get very different behaviours for the photon and massive particles. For a photon $m_0$ is zero so the expression reduces to:

$$p^2 = \frac{KE^2}{c^2}$$

and setting $KE = hc/\lambda$ immediately gives $p = h/\lambda$ or $\lambda = h/p$. However for a massive particle the kinetic energy is normally tiny compared to the rest mass, so this time our expression simplifies to:

$$p^2 \approx \frac{\space KE}{c^2}2\space m_0c^2 \approx 2\space KE \space m_0$$

The last step is to go back to the de Broglie relation:

$$ \lambda = \frac{h}{p} $$

and substitute for $p$ using the equation above to get:

$$ \lambda \approx \frac{h}{\sqrt{2 \space KE \space m_0}} $$

Finally use the low energy approximation $KE = 1/2m_0v^2$ to substitute for KE and this gives:

$$ \lambda \approx \frac{h}{\sqrt{2 \space \frac{1}{2}m_0v^2 \space m_0}} \approx \frac{h}{m_0v}$$

and of course $m_0v$ is just the non-relativistic momentum.

  • 1
    $\begingroup$ Thanks for the explanation about the proton. For the massive particles, there is something strange about your reasoning. You use the de Broglie relation before you prove it (you proved it for the photon-like particles but not for the massive particles). There must be another way to prove it. $\endgroup$
    – snickers
    Aug 15, 2012 at 13:20
  • 1
    $\begingroup$ This answer is not relevant, this is a classic phase-velocity group-velocity confusion. For nonrelativistic particles, the phase velocity is twice the group velocity. For relativistic particles, the phase velocity is always superluminal. This answer is just doing formal manipulations without regard to OP's question. $\endgroup$
    – Ron Maimon
    Aug 16, 2012 at 3:34
  • $\begingroup$ de Broglie's relation is a pretty fundamental postulate and cannot in good faith be rigorously derived from "first principles". It's not formally a postulate of QM but it is equivalent to the operator identity $\hat{p}=-i\hbar\frac{\partial}{\partial x}$ and therefore contained in the canonical commutation relation $[\hat{x},\hat{p}]=i\hbar$. $\endgroup$ Aug 16, 2012 at 3:37
  • $\begingroup$ Yes, I was hoping to get an answer without having to consider the group and phase velocities, but I think I got lost halfway through and ended up with a circular argument. $\endgroup$ Aug 16, 2012 at 6:27

Your mistake is in identification of symbols. The phase velocity is $v_p={E \over p}={\omega \over k}={\nu \lambda}$, while the group velocity is $v_g={\partial E\over \partial p}={\partial \omega\over \partial k}$

$$E=\hbar \omega$$ and $$p=\hbar k$$

In Galilean relativity $E={p^2 \over 2m}$

$$ v_g= {\partial\omega \over \partial k } = {\hbar k\over m} = {p\over m}=2{\omega \over k}=2v_p $$

The group velocity is twice the phase velocity.

However, in special relativity:

$$ v_g= {\partial\omega \over \partial k } = {\partial \sqrt{(mc^2)^2+(pc)^2} \over \partial p }= {pc^2\over \sqrt{(mc^2)^2+(pc)^2}}={c^2 \over v_p} $$

The group velocity goes with the inverse of the phase velocity.

  • 1
    $\begingroup$ This answer was based on a previous deleted answer by Ron Maimon which contained some mistakes that did not affect its main argument. This answer corrected these mistakes and added the special relativistic case that was not contained in any answer at that moment. $\endgroup$ Aug 16, 2012 at 22:03

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