If a slit is illuminated by light will the electron still preserve its interference pattern?

Question

Electrons in the double slit experiment show interference accordingly to their wave-particle duality but does illumination of those electrons by light change just their trajectory or also changes their behaveiour as wave-particles forcing them to lose their wave function?

Answer 1

The interference pattern will dissapear if the electron interacts with the photons (the illumination of the slit), because then there is a localized event happening and the trajectory of the electron, meaning which slit was taken, is clear.

If you send out only a few photons, the probability for interaction is small. No interaction in between means interference pattern. So it 's a question if something happens, and the probability for that will be proportional on the number of photons.

Answer 2

To find out where the electron goes, you have to accept first that the answer will be a probability distribution not a single point.

To find the probability for the electron to arrive at a point $P$ on the screen, you have to add two numbers, and then take the modulus square. These are complex numbers. Let's call them $a$ and $b$. Let $x$ be the location of the point on the screen where the electron is finally observed to arrive. Then the complex numbers (called quantum amplitudes) will depend on $x$, and the probability is $$ \mbox{Probability(electron arrives at $x$)} = | a(x) + b(x) |^2 $$ First let's suppose there is no light illuminating the slits. In this case the formula for $a(x)$ and $b(x)$ is fairly simple: you just take the distance from the relevant slit to the point at $x$ and divide by the de-Broglie wavelength of the electrons, and this tells you the phase (after multiplying by $2\pi$). It is just like a wave propagating, which is why $\psi(x) = a(x) + b(x)$ is often called a wavefunction. Anyway you find $$ a(x) = A e^{-i\phi(x)/2}, \;\;\;\;\; b(x) = B e^{i\phi(x)/2} $$ where $A$ and $B$ depend very little on $x$, but the phase $\phi(x)$ is quite sensitive: $$ \phi(x) = \frac{2 \pi x d }{\lambda L} $$ where $d$ is the separation of the slits and $L$ the distance from the slits to the screen, and we assumed $L \gg x,d$ so that the combination $x/L$ is the angle subtended at the slits by the various places $x$ on the screen.

Now the interference pattern that is observed (when no light illuminates the slits) is very much to do with the phase $\phi(x)$. Because when $A$ and $B$ are equal (which is a good approximation in practice) we have $$ \mbox{Probability} = |A|^2 |e^{-i\phi(x)/2} + e^{i\phi(x)/2}|^2 = 4|A|^2 \cos^2(\phi(x)/2). $$ That $\cos^2$ function is the interference pattern.

Ok so now at last we come to what happens when light illuminates the slits. Let's take the case where light illuminates just slit $B$. The effect of this is to introduce a change to $b(x)$. The interaction between the light at the electron gives a momentum change to the electron, so that it now propagates away from slit B in a new direction (another way to analyze invokes the idea of entanglement, but I won't adopt that approach). The direction taken by the electron after interacting with the photon is such as to conserve momentum, so it depends on the change in momentum of the photon. But in order to hit one slit and not the other, the light beam has to have a narrow focus and therefore the direction of travel of the photon is spread over a range (an example of Heisenberg uncertainty principle, here applied to the photons arriving at the slit). Consequently the direction of travel of the electron after interacting with the photon is also spread over a range. This range of angles is about $$ \Delta \theta \simeq \frac{p_{\rm photon}}{p_{\rm electron}} $$ where here $p$ refers to momentum, and we assume $p_{\rm photon} < p_{\rm electron}$. This is the formula because the electron gets a momentum kick of about $p_{\rm photon}$ so its direction of travel is turned to the side by about $\Delta \theta$. Now recall that $x/L$ is the angle (i.e. the angle away from the normal to the plane of the slits) of location $x$ on the screen. The contribution to the wavefunction from slit B is now steered off by $\Delta \theta$, which means it is steered off by $$ \Delta x = L \Delta \theta $$ So now we have for the probability for electron to arrive at $x$: $$ {\rm Probability} = 4 |A|^2 | e^{-i\phi(x)/2} + e^{i\phi(x + \Delta x)/2}|^2 $$ where we have to keep in mind that $\Delta x$ here is saying the amount by which $x$ may varying between one electron and the next as the interference pattern is built up. The maths may look complicated but the central idea is that the momentum-kick caused by the light shifts the interference pattern by a random shift that varies from one electron to the next. But when you add a bunch of randomly shifted interference patterns, the pattern washes out, because the bright regions of one pattern fill the dark regions of another.

Let's see how big this shift needs to be in order to wash out the pattern. It will require $$ \Delta x > \lambda L/d $$ (because that is the separation between the fringes). So it will require $$ \Delta \theta > \lambda / d $$ and therefore $$ p_{\rm photon} > p_{\rm electron} \lambda / d. $$ Now the de Broglie wavelength is related to the momentum by $\lambda = h / p$ where $h$ is Planck's constant, so we have $$ p_{\rm photon} > \frac{h}{d} $$ or in other words $$ p_{\rm photon} d > h. $$

The reason why I presented the mathematical details was really so as to make it clear that the statement in bold font above, about momentum kick, really does contain the physics here. There is no need to say "sometimes its a wave, sometimes its a particle" or anything like that. It is simply a case of one thing interacting with another, conservation of momentum, and the fact that the momentum kick involves a random direction taken either by the incoming or outgoing photon or both.

As I hinted above, the same result can also be calculated by keeping the quantum state of the photon in the calculation, and then you will get an entangled state and the issue becomes whether the possible final states of the photon are mutually orthogonal. If they are then the photon contains 'which path' information and the electron interference vanishes. This provides some nice further insight but the above calculation in terms of momentum kick is entirely equivalent.