The slide called "QUANTA" here says that "One Quantum has a definite mass" and the picture shows a wave. So, What is meant by the mass of a wave?


First of all, the quanta in the slide are not "waves", they are "wave packets".

Secondly one has to distinguish between wave packets in the classical world and wave packets in quantum mechanics which rules in the microworld of particles. In the macroscopic world described by classical waves and wave packets, the waves are variations in density of a specific medium, for sound, and water waves.

Electromagnetic waves and wave packets straddle the divide between classical and quantum definitions, as described classically they are propagating variations in the electromagnetic field. As described quantum mechanically they have a dual nature, a quantum of light with an energy and momentum four vector, and a probability wave which describes the probability of finding the wave packet as a photon in a specific space time point: $(x,y,z,t)$. The classical electromagnetic wave blends with the quantum one, but still the interpretation of the wave packet is in terms of probabilities.

In the particle world all particles, including photons, are described as quanta of either probability waves or actual particles, in a dual manifestation depending on the experiment observing them.

The mass of the probability wave packet is the mass given by the four vector representation of its energy and momentum as a particle: $m^2=E^2-p^2$. In the case of the photon that mass is 0.

The mass in a classical water or sound wave has to be defined because it was never a useful concept for the study of classical waves.

  • $\begingroup$ I'm confused about your remark that electromagnetic wave packets can be a "quantum of light with mass." Did you mean energy? $\endgroup$ – kleingordon Sep 21 '12 at 5:15
  • $\begingroup$ @kleingordon the yes was for your observation that I misrepresented the photon. I have since edited. $\endgroup$ – anna v Sep 21 '12 at 6:28
  • $\begingroup$ anna v wrote: "the quanta in the slide are not "waves", they are "wave packets"" and wave packets are not waves? $\endgroup$ – Zeynel Sep 26 '12 at 23:55
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    $\begingroup$ @Zeynel Wave packets are composed by a superposition of wave functions, but are more than waves. In a similar way that this message is composed with letters but is more than letters. $\endgroup$ – anna v Sep 27 '12 at 2:57
  • $\begingroup$ The way I see this picture hyperphysics.phy-astr.gsu.edu/hbase/waves/imgwav/wpac5.gif waves do not combine like words. Words combine to make sentences, waves combine to make waves. If this were not true, there would not be Fourier analysis. How do you justify that a wave is not a wave? $\endgroup$ – Zeynel Sep 28 '12 at 0:54

It's not that the wave has a mass, really; it's that a quantum of the wave has a mass. A quantum of the wave is one of those little wave packets shown in the slide, and the mass is the minimum amount of energy that one of those wave packets can have. The mass of a quantum is set by a fundamental property of the field that the wave propagates in.

You can find a more detailed explanation of what he's talking about on Prof. Strassler's website, in particular starting at section 4 of his series on quantum fields and particles. The gist of it is as follows: you can infer in various ways that a wave has to satisfy the differential equation

$$\frac{\partial^2\phi}{\partial t^2} - c^2 \frac{\partial^2\phi}{\partial x^2} = -\mu^2c^4(\phi - \phi_0)$$

where $\mu$ is a property of the field and $c$ is the universal invariant speed (a.k.a. the speed of light). Equations like this come out of quantum field theory, for example. The solution to this formula is a linear combination of plane waves of the form

$$\phi(x,t) = \phi_0 + e^{i(kx - \omega t)}$$

where $\omega$ and $k$ satisfy a particular equation called the dispersion relation, in this case

$$\omega^2 = k^2 c^2 + \mu^2 c^4$$

The wavenumber $k$ is inversely related to wavelength and thus can take on any real value, but assuming $\mu$ is real, the frequency $\omega$ is restricted to values greater than or equal to $\mu c^2$.

Now, if you multiply through by $\hbar^2$, you get

$$\begin{align}(\hbar\omega)^2 &= (\hbar k)^2 c^2 + (\hbar\mu)^2 c^4\\ E^2 &= p^2c^2 + m^2c^4\end{align}$$

where $E = \hbar\omega$, $p = \hbar k$, and $m = \hbar\mu$. So this property of the field (or wave), $\mu$, is related to what we perceive as the mass $m$ of the corresponding quantum (particle).

  • $\begingroup$ Great answer, thanks. You say that "the frequency w is restricted to values greater than or equal to mu". Is it given that mu has dimensions of frequency? Otherwise how can we compare frequency and mu (a property of the field). $\endgroup$ – Zeynel Sep 30 '12 at 11:55
  • $\begingroup$ Also, in the definition of m written as m=hbar*mu, does mass have units of kg? Thanks. $\endgroup$ – Zeynel Sep 30 '12 at 12:00
  • $\begingroup$ You can work out the units from the equations. Though when I say $\omega\ge\mu$, that's in natural units; I'm leaving implicit the necessary powers of $c$. $\endgroup$ – David Z Sep 30 '12 at 12:03
  • $\begingroup$ When you say "this property of the field (or wave), mu, is related to what we perceive as mass m..." do you mean that mu is the property of both field and the wave or do you mean that the field and wave are the same thing? $\endgroup$ – Zeynel Sep 30 '12 at 12:18
  • $\begingroup$ It's probably more correct to consider it a property of the field. The sense in which it is a property of the wave is only that the wave "inherits" the properties of the field it propagates in. (also: on second thought I went back and edited in the conversion factor from $\mu$ to $\omega$, it's clearer that way.) $\endgroup$ – David Z Sep 30 '12 at 12:29

Very short theoretical answer:

The mass is a term $m$ which appears in the Klein-Gordon, Dirac, and even Schrödinger equations.

For example, for scalar particles, it is the term $m$ from the Klein-Gordon equation:

$$\frac{1}{c^2}\frac{\partial^2}{\partial t^2}\psi - \nabla^2\psi + \frac{m^2 c^2}{\hbar^2}\psi = 0$$

How is that? The answer is that particles, in quantum theory, are understood as vectors in a Hilbert space. These Hilbert spaces associated to particles are obtained from the unitary representations of the Poincaré group. In fact, all this boils down to the representations of the $SL(2,\mathbb C)$ group, which is the double cover of the (proper orthochronous) Lorentz group. $SL(2,\mathbb C)$ has the same representations as those of its the maximal compact subgroup, which is the group $SU(2)$. These representations are classified by spin and mass (the Casimir invariants). The spin tells which equation to use, so that its solutions are such vectors. The mass is the coefficient $m$ from the equation.

In practice, as anna v said, you can determine the mass from the momentum and energy, so the particle has to be in an eigenstate of the momentum. But the proper mass makes sense even if the particle is not an eigenstate of the momentum.

If you want a classical picture of how the mass is associated to a particle, originally, Schrodinger interpreted the wavefunction classically. He thought that the mass density is proportional to the squared amplitude of the wave: $\psi(x)\psi^*(x)$. He had to abandon this picture when he realized that for many particles would not hold. Instead, $\psi(x)\psi^*(x)$ was interpreted by Born as density of probability.

  • $\begingroup$ "describing one particle..." in physics the word "particle" has dozens of meanings, can you specify which meaning of particle you are using here? $\endgroup$ – Zeynel Sep 28 '12 at 0:14

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