# Frustrated total internal reflection

1. Apr 28, 2007

### Sonden

While making calculations of FTIR/tunneling isn't that hard, I don't understand the phenomenon, and therefore there are some predictions I'm unable to make. So I have some questions (or it's probably just one).

If you have a prism with refractive index n in vacuum, total internal reflection occurs if the angle of incidence is greater than arcsin(n). But if you have a second prism at some distance from the first one, the wave will not be completely reflected, a part of it will be transmitted to the second prism. If the wavelength is large (eg microwaves) you'll get substantial transmission even if the distance between the prisms is "macroscopic".

This is probably "thinking classically" when I should be thinking quantum physically, but how does the wave "know" there's a second prism at some distance away? If the distance is d, and the wave is partially reflected in the first prism at time t, when is the wave transmitted into the second prism? It should be at T=t+d/c?

What if d is large (which should allow transmission if the wavelength is large) and there's no second prism at this distance at t, the second prism is being put there at, for example, t_2=t+d/(3c). Will the wave be transmitted to the second prism as if the second prism had been there all along (that is, at time T)? Or not at all? What if you measure the part of the wave that was reflected in the first prism (at some time, before t_2, between t_2 and T, or after T, that shouldn't matter, if it's just a short wave packet), then you could predict (if you make the measure before t_2) that someone will, in the future, put a second prism at some distance to the first! Which must be nonsense.

On the other hand, if the second prism is there from the start, and you measure the reflected wave, the you can draw the conclusion that the wave "knew", at t, that there's a second prism at distance d! And that no one will remove it prior to T! Which, to me, sounds equally ridiculous.

So, what's wrong here? I don't think it should have anything to do with relativity. The uncertainty principle - but how? Something else - what?

I'd be very grateful to get an answer (not too advanced, this is my first course in quantum physics), as this is making me more frustrated than the Frustrated total internal reflection...

2. Apr 29, 2007

### Meir Achuz

The wave is infinite in extent, so the mathematics knows everything.
You would have the same conceptual problem in any interference, classical or quantum.

3. Apr 29, 2007

### Sonden

That's a totally unclear answer, doesn't help me (and in my example it's a very short wave packet). I still don't know which of the two alternatives (the nonsensical or the ridiculous...) that's right (if any).

(I don't see the problem arising in interference, but even if it were so, a second problem doesn't solve the first one.)

Last edited: Apr 29, 2007
4. Apr 29, 2007

### akhmeteli

Although your question is short and crystal clear, and your style is extremely welcoming, I'll try to reply.

I think I have two things to say.

First, as I wrote here some time ago, contrary to what some textbooks may say, quantum tunneling does have classical analogs. Let me quote myself:"I suspect theoretically you can jump over a classical barrier having lesser kinetic energy than the potential energy your mass would have at the top of the barrier. In fact, in the course of a high jump you can bend over the barrier in such a way that your center of gravity will be outside of your body and pass under the barrier."

But I guess it would be more convenient to consider another (although similar) example to make sense of your application. Imagine a long train of mass m moving on rails by inertia (no engine running) and having kinetic energy W. If the rails rise on a hill of height h somewhere, W does not have to exceed mgh (the potential energy of mass m at height h) for the train to get over the hill, because, if the train is longer than the hill, it will never get to height h with all its carriages being there simultaneously. You can thus model your situation. The case where there is only one prism corresponds to the situation where the hill is infinite in one direction, so the train will be "reflected" by the hill. The case of two prisms corresponds to the situation where the hill has finite length, so the train can get over it. Of course, every analogy has its weak point. In our case, the train, unlike a wave, cannot divide in two parts (that is, it can, in principle , but you know what I mean).

Second, either "the wavelength is large", or "it's a very short wave packet". You cannot have it both (I guess that's why you were told that "The wave is infinite in extent". If it's a very short wave packet, its Fourier expansion will have short-wavelength components. Each Fourier component reaches beyond the plane of total internal reflection approximately by its wavelength (its amplitude decreases exponentially). The front of your wave packet contributes to shorter-wavelength Fourier components, so initially your packet will propagate beyond the plane with speed c (in general, remember that the very front of any wave propagates with speed c). However, the total length to which the packet will propagate beyond the plane will be limited by the maximum wavelength of its Fourier components.

5. Apr 29, 2007

### Sonden

Nice, that made it clearer. If it made it clear enough... I'll ponder a bit on that.

6. May 11, 2007