# Fluorescent screen and quantum measurement

In the well known, two slit gedanken experiment, used in most textbooks to demonstrate wave like properties of an electron, a fluorescent screen or photographic plate are used to detect interference patterns. It is assumed, that whenever an electron hits the screen, it leaves a point mark (a "spot") on it. But what exactly do we mean by saying, that an electron hits the plate, if according to the principle of uncertainty it does not have well defined position?

Let us consider a single electron described by localized gaussian wavepacket, which is heading more or less (to be consistent with the uncertainity principle) towards the planar fluorescent screen. Can we predict when it will be recorded by the screen, or what is the probabiliy distribution, that it will be detected at a certain moment of time? Let us assume, that the screen is divided into pixels of finite size and thus has limited spatial resolution.

• Schlosshauer's book is all about this sort of thing. – DanielSank Jun 12 '17 at 16:59

You are asking about the quantum mechanical motion of the dispersive wave packet, probably the second easiest problem you considered in your QM course—I’ll stick to Pauli’s Wave Mechanics for the sake of specificity.

If your freely moving electron did not disperse, and your Gaussian wave packet just translated forward, then, taking the x-axis as the direction of motion, the probability density of your screen detecting the electron would be a Gaussian in time, reflecting the Gaussian you started with, as in your picture, and the probability density of a hit at a distance off axis of the wave packet center would be identical to the transverse features of your original Gaussian.

But this is not what happens for a free wave-packet: the Gaussian spreads, with the celebrated width increase linear in the time. Let us first monitor, specifically, the motion of the center of the wave packet; so solve first the free one-dimensional Schroedinger equation of a Gaussian wavepacket ψ, starting from x=0 in your picture, and being detected at x=X, the location of your screen.

So your initial condition is $$\psi (x,0)= N e^{-\frac{x^2}{2a} +ikx} =N e^{-ak^2/2} e^{-(x-iak)^2/2a}$$ where $N^2\sqrt{\pi a }=1$ and $k=mv/\hbar$ for the center of your wave packet. For a typical electron leaving the electron gun in an antique CRT TV (not flat screen!), v is about $10^{7}m/s$, so 4% of c, whence $k\sim 10^{11}/m$.

Now free propagation is specified by $$i\hbar\frac{\partial \psi}{\partial t} = -\frac{\hbar^2}{2m} \frac{\partial^2 \psi}{\partial x^2}~~,$$ readily yielding for all times and distances $$\psi (x,t)= \frac{N}{\sqrt{1+it\hbar/ma}}~ e^{-ak^2/2} e^{-(x-it\hbar/ma)^2/(2a(1+ik\hbar/ma))},$$ so the resulting probability density is just $$\psi^* \psi = P(x,t)=\frac{1}{\sqrt{\pi a (1+(t\hbar/ma)^2)}} e^{-\frac{(x-t\hbar k/ma)^2}{a(1+(t\hbar/ma)^2)}}$$ The electron wavepacket has spread: from the initial narrow width $\sqrt a$, so $\Delta x = \sqrt{a/2}$ to $\Delta x \sim t\hbar/m\sqrt{2a}$ for large t, driving Heisenberg’s inequality far, far from saturation!

On the screen, x=X, then, the time probability profile for the electron being there (time of observation) thus is not a plain Gaussian in time: $$P(X,t)= \frac{1}{\sqrt{\pi a (1+(t\hbar/ma)^2)}} e^{-\frac{\hbar^2 k^2(t-Xma/\hbar k)^2}{m^2a^3(1+(t\hbar/ma)^2)}}.$$

We may now address the y and z components. The full wave function factorizes, and the three separate components do not influence each other. The y and z wave functions are unaffected by this motion, so then, each is basically the above solution, except with k=0, $$P(y,t)=\frac{1}{\sqrt{\pi a (1+t^2\hbar^2/m^2a^2)}} e^{-\frac{y^2}{a(1+t^2\hbar^2/m^2a^2)}},$$ $$P(z,t)=\frac{1}{\sqrt{\pi a (1+t^2\hbar^2/m^2a^2)}} e^{-\frac{z^2}{a(1+t^2\hbar^2/m^2a^2)}}.$$ The surface probability profile for observation on your high-resolution screen is $P(X,t)P(y,t)P(z,t)$.

The probability density in the y and z directions is a 2d Gaussian, unlike the time distribution of observation events. Now, as a consequence of the uncertainty principle, the smaller the initial width of the wave packet $\sqrt a$, the larger the width on the screen, given the spreading above. For a very slowly changing width, e.g., if X=0.1 m, you must take $\sqrt{a}=10^{-6} m$, far above atomic dimensions.