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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. In this case an EM wave will be treated exactly the same as a de Broglie wave except an EM wave has power, force, and energy density associated with it so its interaction with a atom is more direct than a de Broglie wave. Quantum field collapse, where a quantum field collapses in order to create a quantum event is not understood in the Standard Model. This animation is not meant to provide a better understanding of collapse. Instead it attempts to provide an intuitive vision of how a quantum field might collapse and the results of that collapse. There are two requirements:

1. Enough EM energy was be transferred to the atom's electron to equal the ionization energy.

2. The 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 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 and 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 `dU*dt>=ℏ/2` we can easily show that the length can be reduced to `dt=ℏ/(2dU)` where dU is now the ionization energy. Since the total energy of the single photon wave is also the ionization energy, we know that `dU=hf` and dt becomes: `dtgtℏ/(2hf)=1/(4pi*f)`. Then, since the photon moves at the speed of light, the spatial length of the pulse`c*dt>=c/(4pi*f)=lambda/(4pi)`

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 `U=(epsilon_0/2)E*E` the power per unit area is `c*epsilon_0/2(E*E)` so that`E=sqrt(8pi(fdU)/(c*A*epsilon_0))`

With parameters `A=10^(-19)` `m^2`, `dU=10^(-19)` J, `c=3*10^8` m/sec, `f=10^(14)` Hz, and `epsilon_0=8.9*10^-(12)` we obtain `E=0.97*10^9` V/m. Let's compare this result with the electric field needed to straitghtaway pry an alectron away from a nucleus when its distance is `r=sqrt(10^(-19)/pi)` meters. The equation is simply`E=e/(4*pi*epsilon_0*r^2) = (1.6*10^(-19))/(4*pi*8.9*10^(-12)10^(-19))`

and the result is `44.9*10^9` V/m which is about 50 times larger than the minimum uncertainty result. The graphics show the progress in the reduction in size of the beam waist of the collapsing photon where the color scale is gaussian with red depicting the most intense field and purple depicting the weakest field. The axial E field varies inversely with the radius of the beam and is plotted in black. Note that the field is significant for only about 1/2 cycle of the EM wave as should be expected for axial compression of the wave. Also shown is a nucleus and electron where the latter is ejected when the E field peaks. Obviously the beam should be extinguished after interaction with the rotating electron but I chose not to show this. Click Here for Discussion