Why do tunneling photons outrace their non-tunneling counterparts in vacuum?

Question

If we describe a photons with a wave packet, moving towards a potential barrier and E smaller than V, there is a finite chance that it will tunnel to the other side. In this process it is likely that it will arrive before a photon that does not tunnel, not because it exceeds the the speed of light, but because the front of a wave packet will contribute most to tunneling. I would understand this if it was concerned with PARTICLES as follows: because each of the different waves that make a wave packet has a different momentum, the faster moving, higher energy, waves will move to the front of the wave packet when it disperses. According to the formula for tunneling probability indeed the higher energy parts will tunnel more often.

BUT, why would a photon wave packet also have the higher energy elements more in front of it's wave packet, since all light-frequencies move at the same speed in vacuum right? Or does the described system indeed not work in vacuum.

At page 14 of this link the mechanism I describe is presented:

http://www.physics.umass.edu/sites/physics/files/admupld/Tunneling-UMass-12Feb10.pdf

Answer 1

In my opinion the presentation you linked is a weird description that conflates different phenomena in a confusing way.

The following is a totally wrong explanation of the the results of that referenced paper (non-paywall link): "The front part of the pulse has the higher-energy part of the light, which is more likely to tunnel." Both parts of this sentence are wrong:

(1) The frequency profile at the front and back of the pulse is essentially the same in this experiment (the light pulse is not substantially "chirped");

(2) It is not true that higher-energy (higher-frequency) light tunnels with higher probability through this particular structure. (See Fig. 1 in the paper; the photon is around 702nm, where the transmission curve is almost flat. Technically, the transmission minimum is at 692nm, so the lower-frequency parts are transmitted with very slightly higher probability. However, this slight difference is not important for the effect--see the theoretical curves in Fig. 1.)

It is certainly true that the front part of the pulse is reduced a little bit while the back part is reduced a lot, which moves the center of the pulse forward. But the reason for this differential reduction is not that sentence above. I think it's a more subtle kind of wave-interference effect, which is not necessarily easy to explain intuitively.

Answer 2

A wave packet cannot represent a photon whereas electromagnetic wave packets are made up with photons.

The "wave" concept to quantum mechanical entities/particles, as a photon is a probability wave.

The waves in the classical electrodynamics wave packets are energy distribution waves composed of higher and lower h*nu photons.

A single photon, due to the uncertainty principle, when measured at a point with uncertainty delta(x) will have its momentum/ energy bounded by the Heisenberg Uncertainty Principle ( HUP).

Even the de Broglie hypothesis used on particles does not describe a variation in the mass/energy of the particle. The solutions of the Schrodinger equation which obey the principle are describing the probability of finding the particle ( whole) with an energy at the high limit of the HUP. The particle is not split in any sense.

So far for single photons.

Why do tunneling photons outrace their non tunneling counterparts in vacuum?

They do not.

A wave packet made up of photons of various frequencies as in your illustration: if it is in vacuum the group velocity and the phase velocity are the same, so it will not disperse, it will have the energy versus space distribution it was given by construction.

Since you propose a barrier then different frequencies may propagate differently, a barrier is a medium.

The phase velocity of an electromagnetic wave, when traveling through a medium, can routinely exceed c, the vacuum velocity of light. For example, this occurs in most glasses at X-ray frequencies.[11] However, the phase velocity of a wave corresponds to the propagation speed of a theoretical single-frequency (purely monochromatic) component of the wave at that frequency. Such a wave component must be infinite in extent and of constant amplitude (otherwise it is not truly monochromatic), and so cannot convey any information.[12] Thus a phase velocity above c does not imply the propagation of signals with a velocity above c.[13]

The "first" part will be hard to measure, though I did find an experiment setup to measure group , phase and signal velocities.

With a prism we transform energy ordering to angles and that is why we can detect the difference in transmission.

Answer 3

all frequencies move at the same speed in vacuum right?

No, not for matter waves. According to the de Broglie hypothesis (which Schrödinger's equation obeys in free space), the dispersion relation for waves is $$\omega(\mathbf{k})=\frac{\hbar|\mathbf{k}|^2}{2m}.$$ Thus the group velocity becomes $$\mathbf{v}_\text{g}=\frac{\partial\omega}{\partial\mathbf{k}}=\frac{\hbar\mathbf{k}}{m}.$$ This means that the high-energy components travel faster, and move to the front.