Quantum Field Collapse

In this animation, I'll use the electromagnetic (EM) wave associated with a photon and show how it might collapse in order to eject an electron from an atom.

There are two requirements: 1. Enough EM energy has to be transferred to the atom's electron to equal the ionization energy. 2. The integrated force from the EM wave must equal the force needed to pry the electron away from the nucleus.

This ionization can occur by a couple of means: 1. A long EM wave can "phase lock" with the oscillating electron. On average, the wave is just as likely to decrease the kinetic energy of the electron as it is to increase its kinetic energy.

2. The second way is that the EM wave can be compressed to a very strong pulse so that its phase becomes unimportant. Only this second way will be addressed here.

Pulse Parameters

The parameters needed for the pulse will be calculated and illustrated here.

The length of the pulse must be less than a single wavelength of the original EM wave. From the energy-time uncertainty relation

$δ U δ t > \frac{ℏ}{2}$

(1.1)

where dU is energy uncertainty and dt is time uncertainty.

We can easily show that the pulse duration can be reduced to

$δ t = \frac{ℏ}{2 δ U}$

(1.2)

where dU is now the ionization energy. Since the total energy of the single photon wave is also the ionization energy, we know that $δ U = h f$ and then dt becomes:

$δ t > = \frac{ℏ}{2 h f} = \frac{1}{4 π f}$

(1.3)

Then, since the photon moves at the speed of light, the spatial length of the pulse

$δ x > = \frac{c}{4 π f} = \frac{λ}{4 π}$

(1.4)

which is much shorter than the wavelength, as expected.

The other item needed for the animation is the amplitude of this short pulse. Since the energy per unit volume of an EM wave field is

$\frac{d U}{d V} = \frac{ε_{0}}{2} E^{2}$

(1.5)

the power per unit area is

$\frac{d P}{d A} = \frac{c ε_{0}}{2} E^{2}$

(1.6)

The power of the original wave delivered to an atom of area A is then

$P = \frac{c A ε_{0}}{2} E^{2}$

(1.7)

The energy needed in time $δ t = \frac{1}{4 π f}$ is equal to dU. So what is the new pulse's value of E?

$δ U = P δ t = \frac{c A ε_{0}}{8 π f} E^{2}$

(1.8)

so that the amplitude of E in the pulse is:

$E = \sqrt{\frac{8 π f δ U}{c A ε_{0}}}$

(1.9)

For a typical atom

$\begin{array}{l} A = 10^{- 19} m^{2} \\ δ U = 10^{- 19} J \\ f = 10^{14} H z \\ c = 3 * 10^{8} \\ ε_{0} = 8.9 * 10^{- 12} \end{array}$

(1.10)

From this we arrive at a value of E for the pulse of:

$E = \sqrt{\frac{8 π 10^{14} 10^{- 19}}{3 * 10^{8} 10^{- 19} 8.9 * 10^{- 12}}} = 0.970 * 10^{9} V / m$

(1.11)

Let's compare this result with the electric field needed to pry a static electron away from a shielded nucleus (like the hydrogen nucleus) when its distance $r = \sqrt{\frac{1 e - 19}{π}}$ meters. The equation is simply

$E = \frac{e}{4 π ε_{0} r^{2}} = \frac{1.6 * 10^{- 19}}{4 * 8.9 * 10^{- 12} * 10^{- 19}} = 44.9 * 10^{9} V / m$

(1.12)