$\lambda=\frac{2h}{p}$?

Question

I am studying quantum physics and there is something I don't understand: I know that for any particle $E=hf$ (Einstein relation) and $v=\lambda f$ ($v$ is the speed of the particle). 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 ?

Answer 1

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, p.172, and this Phys.SE post.)

II) So the answer to OP's question(v1) 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.

Answer 2

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.

Answer 3

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.

Answer 4

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.