# $E=hf$ true for all implies $E=pv$ true for all?

In the answer to this question: Can I apply $$E=hf$$ to a particle having mass? It was stated that $$E=hf$$ is true for all particles. If so, doesn't this imply that $$E=$$momentum x velocity is true for all particles? Because $$E=hf=hv/\lambda=(h/\lambda)v=pv~?$$ I'm representing frequency with $$f$$ and velocity with $$v$$ here.

• The $v$ in your expression is the phase velocity of the wave, however the velocity of the particle is associated with the wave's group velocity – By Symmetry Dec 26 '18 at 14:25

It's rather $$E = hf = hu/\lambda,$$ where $$u = c^2/v$$ is the phase velocity. This gives $$E = hc^2/(v\lambda) = (h/\lambda)c^2/v = pc^2/v.$$

Now, since $$p = mv$$ (with $$m$$ being the relativistic mass), we get $$E = mc^2 = \gamma m_0 c^2,$$ where $$m_0$$ is the invariant mass (often called rest mass) and $$\gamma = 1/\sqrt{1-v^2/c^2}.$$

I will try to explain this a bit differently.

First, please allow me to briefly mention two concepts here. A group velocity and a phase velocity. For an ordinary sinusoidal wave, if I ask you the velocity of that wave you will have no confusion on giving that velocity a meaning. However, consider a wave which is not sinusoidal but a mixture of two sinusoidal waves. You can now visualize two different types of velocities --one at which a specific "crest/trough" travels and one at which the "overall shape" of the wave itself travels. The former is called phase velocity and the latter is called the group velocity. I have linked the Wikipedia entries for both concepts in case you have never encountered them before and want to drill into them further. This diagram, also from those articles, illustrates this difference rather well:

Now you can see that $$v = f\lambda$$ actually applies to the phase velocity. So, why am I writing about all this? How is the group velocity relevant here?

Consider now, what we mean when we say particles behave as waves. We actually believe that for these entities that we thought were particles, there is a wavefunction (a mathematical function that when plotted on a graphing paper looks like a wave) which gives interesting insights into the particle's physical properties. For example, the wavefunction $$\psi(x,y,z)$$ when expressed in position-space (i.e., as a function of $$x$$, $$y$$, and $$z$$) can be used to determine the probability of "finding" the particle within a region of space (say between $$x$$ and $$x + \Delta x$$). Technically, this probability distribution function comes out to $$|\psi|^2$$, where I have used the mod sign because $$\psi$$, as you may know, can be complex valued. This is basically the amplitude of the wave. The higher the wave's amplitude, the higher the probability of locating the particle in that neighbourhood. And the speed with which this amplitude travels through space is tied to our classical notion of the particle's velocity.

To summarize, the $$v$$ that goes in your formula $$v = f\lambda$$ and the $$v$$ that goes into momentum formula are two different concepts altogether and cannot be compared in that way.

There is also another conceptual problem with comparing energy and momentum that way. In QM formulation you will find that the Hamiltonian operator (which is the operator you apply to a wave function to determine the particle's energy) does not always commute with the momentum operator (the operator you will apply to that wave function to determine the particle's momentum). All this means that you cannot always determine the energy and momentum exactly for the particle in question simulatenously. So the relation that $$E = pv$$ is not universal in that sense. One case where you can measure $$E$$ and $$p$$ simulatenously is that of a free particle (i.e., a particle not subject to any force field). In that case $$E = p^2/2m$$ --non-relativistic case.

No, $$E^2=(pc)^2+(mc^2)^2$$.

At low speeds, $$p \approx mv$$, but at high speed the full relativistic formula for momentum must be used. For a particle with non-zero mass, that's $$p=mv\gamma$$, where $$\left(\frac 1{\gamma}\right)^2=1-\left(\frac v c \right)^2$$;
for massless particles we can use $$p=E/c$$.

• Ok? Did I use p=mv somewhere without realising it? – Brain Stroke Patient Dec 26 '18 at 16:26
• @AbrarFaiyaz No, you didn't. The main point of this answer was to show the relativistic formula relating energy, momentum and mass. I added that info about momentum for the benefit of other readers who may not be familiar with special relativity. I almost posted this answer as a comment, but the mods don't like it when we do that. ;) – PM 2Ring Dec 26 '18 at 16:53

No this is not the case. For a massive particle $$f \neq v/\lambda$$ do the derivation does not hold in ge eral.