Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

I'm fascinated by the fundamental questions raised by the Double Slit Experiment at the quantum level. I found this "Dr Quantum" video clip which seems like a great explanation. But is it scientifically accurate?

share|improve this question
    
At the same time you can also copy paste the comments from area51 ... –  Cedric H. Nov 14 '10 at 18:10
    
@Cedric - not sure I want to copy / paste them since 1. duplication and 2. that info is changing but here's the link: area51.stackexchange.com/proposals/23848/theoretical-physics/… –  Wikis Nov 14 '10 at 18:12
1  
The video is qualitatively accurate but has a quantitative lapse, which, however, is also widely present in the diagrams of the general literature on the topic. Roughly, if the two single slit distributions are as cleanly separated as indicated in the video, then the double slit interference pattern would not be as 'broad' as indicated. Precisely, the double slit pattern must always lie between the square of the sum of the square roots of the single slit patterns and the square of the difference of the square roots of the single slit patterns. But that's asking a lot from a cartoon video! –  Gordon N. Fleming Nov 17 '10 at 23:07

4 Answers 4

up vote 13 down vote accepted

A bad thing about the video is how they explained the part where you try to observe which slit the electron goes through. They made it sound more mysterious than it really is.

What we have to ask ourselves is: what does it mean to observe an electron? What does it mean to observe anything? If we want to look at something, we need light. We see things because light is reflected off objects and our eyes collect this light which is then interpreted by our brains.

If we want to see which slit the electron goes through, we shine light upon it, but this fundamentally alters the experiment. Small particles are very sensitive to perturbations and shining light on an electron is a big perturbation. Now, and this is technical, the Heisenberg uncertainty relation tells you how much the electron's path will be perturbed by the light. The path is more perturbed as the energy of the photon is greater, but to determine the position of the electron accurately, you need high energy according to Heisenberg. High energy means perturbing the electron a lot and as a consequence destroy the interference pattern.

So, you might want to give up on accuracy to avoid perturbing the path of the electron too much, but if you do that, the Heisenberg relation will show you that you have to diminish the energy of the photon so much that you will not be able to locate the electron anymore. The interference pattern on the other hand will reappear.

More details can be found in the Feynman Lectures, Volume 3, Chapter 1.

share|improve this answer
    
very thorough and clear answer, thank you. –  Wikis Nov 14 '10 at 19:34
1  
And what about the delayed choice quantum eraser? That experiment cancels out any probability for the measurement devices to affect the result. It's only the choice of the observer - keep the slit data or not - which determines the result. In that experiment the detectors can be turned ON all the time, it's only the delayed choice eraser which gets turned on and off (observer's choice) AFTER the photon has already went through the slits (or one of them). –  Martin May 4 '12 at 19:45
1  
I don't get what observers have to do with it? Just make a more elaborate measurement device that makes the "choice" and the delayed choice quantum eraser results will stay the same. Or do you mean to say that the measurement device would then have to be counted as an observer? What is an observer and what isn't? It's a very shaky foundation to build fundamental physics on the existance of observers when we know that there have not always been observers in history. –  Raskolnikov May 5 '12 at 9:02
    
This answer is completely WRONG! As others have pointed out, how do you then explain delayed choice quantum erasure? The video is fairly accurate. I'm aghast that this incorrect answer has been upvoted so much. –  Dheeraj V.S. Jan 27 '13 at 4:31
    
Maybe you could explain to me what the delayed choice quantum erasure changesto the problem? Because just saying it makes a difference doesn't quite cut it. –  Raskolnikov Jan 27 '13 at 17:09

Note that the Dr. Quantum video is from the pseudoscience film "What the Bleep Do We Know?", which takes the following approach:

  1. Use examples of quantum physics to show the viewer that the universe is far more weird and complex than our basic human perception/intuition suggests.

  2. Attempt to convince the viewer that if quantum weirdness is real, then the weirdness of someone channeling a 35,000 year old warrior-spirit named Ramtha is also real.

  3. Profit.

Despite its flaws, the Dr. Quantum video on its own isn't terrible. But I think the source of scientific information should be taken into consideration when assessing its accuracy. There's a similar but better video here: http://www.youtube.com/watch?v=UMqtiFX_IQQ

share|improve this answer

I get this feeling that this is Jack Sarfatti as a digital avatar. If you have ever encountered him he has all sorts of strange ideas about things. Cedric’s comment about Area 51 sort of tipped me off, for Sarfatti has all sorts of UFO ideas.

This little video is correct, but lapses into mysteriousness at the end. The process of measurement of a system in a superposition is to replace that superposition with an entanglement. We might think of it as a process where the phase associated with the “waviness” of the superposition of a system is transferred to a nonlocal property of this system with another. Consider a two slit experiment where a photon wave function interacts with a screen. The wave vector is of the form $$ |\psi\rangle~=~e^{ikx}|1\rangle~+~e^{ik’x}|2\rangle $$ as a superposition of states for the slits labeled $1$ and $2$. The normalization is assumed. The state vector is normalized as $$ \langle\psi|\psi\rangle~=~1~=~\langle 1|1\rangle~+~\langle 2|2\rangle~+~ e^{i(k’~+~k)x}\langle 1|2\rangle~+~ e^{-i(k’~+~k)x}\langle 2|1\rangle $$ The overlaps $\langle 1|2\rangle$ and $\langle 2|1\rangle$ are multiplied by the oscillatory terms which are the interference probabilities one measures on the photoplate. We now consider the classic situation where one tries to measure which slit the photon traverses. We have a device with detects the photon at one of the slit openings. We consider another superposed quantum state. This is a spin space that is $$ |\phi\rangle~=~\frac{1}{\sqrt {2}}(|+\rangle~+~|-\rangle). $$ This photon quantum state becomes entangled with this spin state. So we have $$ |\psi,\phi\rangle~=~e^{ikx}|1\rangle|+\rangle~+~e^{ik’x}|2\rangle|-\rangle $$ which means if the photon passes through slit number 1 the spin is + and if it passes through slit 2 the spin is in the – state. Now consider the norm of this state vector $$ \langle\psi,\phi|\psi,\phi\rangle~=~\langle 1|1\rangle\langle +|+\rangle~+~\langle 2|2\rangle\langle-|-\rangle~+~ e^{i(k’~+~k)x}\langle 1|2\rangle\langle +|-\rangle~+~ e^{-i(k’~+~k)x}\langle 2|1\rangle\langle-|+\rangle. $$ The spin states $|+\rangle$ and $|-\rangle$ are orthogonal and thus $\langle +|-\rangle $ and $\langle-|+\rangle$ are zero. This means the overlap or interference terms are removed. In effect the superposition has been replaced by an entanglement.

This analysis does not tell us which state is actually measured, but it does tell us how the interference term is lost due to the entanglement of the system we measure with an instrument quantum state. So one does not need to invoke an outright collapse to illustrate how a superposition is lost.

How the actual state is obtained is some matter of debate. We might think of there being some other system which now measures this spin state. So with the $|\pm\rangle$ states we now entangle another system with two states. Yet it is clear this does not help much, for we could do this inductively “forever” and get presumably no closer to finding out which state obtains. However, maybe this third state could be a heftier spin, or an angular momentum in this case, say a spinning buckyball cooled to some low temperature. The buckyball could enter into an entanglement, as various quantum properties of these have been observed. What has this accomplished? The path integral for the entire entangled system is now narrowed closer to a classical path. We have a bit of quantum superposition properties here, but “barely.” Now we need to measure the buckyball’s rotational state. This carries us up to an even larger system and …, well we have the Schrodinger cat issue. However, some sort of asymmetry enters into the picture with the buckyball which puts the buckyball in a .7 to .3 probability ratio of being either angular momentum configuration. Further entangling this reduces the probability ratio further to .9 to .1 and the so forth. Somehow entanglement phase is being transferred completely out of the picture or into the environment (or demolished) , which then gives this state reduction in a measurement. From the perspective of a quantum path integral the set of paths are reduced to an ever narrow set of paths which corresponds to the outcome.

share|improve this answer

The video is horrifyingly bad. It shows a single-slit electron pattern over here, and then puts a second slit in, and shows the pattern from the second slit over there. Then it says: what if you have both slits open at the same time?

In fact, since the pattern from the first slit is separated from the pattern of the second slit, NOTHING DIFFERENT happens when you open both slits at the same time. There is no interference. But the video shows multiple bands. This is wrong.

You only get multiple bands when the INDIVIDUAL patterns of each slit occupy the same area on the screen. Then, when you open both slits at once, you get interference within that common area.

What the video shows is complete nonsense.

share|improve this answer

protected by Qmechanic Sep 15 '13 at 23:47

Thank you for your interest in this question. Because it has attracted low-quality answers, posting an answer now requires 10 reputation on this site.

Would you like to answer one of these unanswered questions instead?

Not the answer you're looking for? Browse other questions tagged or ask your own question.