Tag Info

New answers tagged

0

I'm not sure I can answer your questions about quantum mechanics honestly without equations, but I can tell you something about the details of generating entangled photons with BBO. First, there are two things you need to know about laser light: it has a definite polarization (orientation of the electric field) and definite energy (or, equivalently, a ...


0

In order to generate entanglement you need an interaction, by which I mean that the dynamics have a term that is a function of two different degrees of freedom that you intend to entangle$^1$. The type of nonlinearity in this case is what is known as spontaneous parametric downconversion or SPD, which is a nonlinear optical process. 1) How does this ...


1

Punk Physicist's answer is spot on. But I'd like to add a little to his/her last two paragraphs, in particular, a description of what it is that you see in an interference pattern. You cannot define a position observable, but you can of course define the state of the second quantized field. Moreover you can describe the probability amplitude for a photon to ...


2

There's an old argument by Newton and Wigner, that the photon as a massless particle can't have a position operator and therefore no position space wave function. The paper you're thinking of is T. Newton and E. Wigner, "Localized States for Elementary Systems," Rev. Mod. Phys. 21, 400–406 (1949) doi:10.1103/RevModPhys.21.400. Photons are ...


3

The de Broglie wavelength, $\lambda$, is given by $$\lambda = {h \over p}$$ where $h$ is planck's constant and $p$ is momentum. If we take the mass to be $160$ kg and speed to be $100$ ms$^{-1}$ then we get $\lambda = 4\times10^{-38}$m given a slow light spaceship and only one person (more people and speed would increase the momentum and decrease the value ...


1

Fundamentally, you can slow down double-slit experiments as much as you like but a version of the Heisenberg principle will not allow you to put definite "gaps" between the detection events or to be sure to produce a photon on demand, at the right frequency etc. It is commonly stated that the particle must be interacting with itself in a double slit ...


2

When we look at individual photons, the experiment is about how the individual photons interact with themselves. Since the photon is interfering with itself and not with other photons in the experiment, the interval between the photons does not matter. In a real experiment, however, you have to deal with the fact that your screen is not perfectly isolated ...


1

All measurements come from interactions. In our macroscopic world, our intuition is that we can observe things without affecting them, but this is not true at quantum scales. In order for that measurement at the slit to happen, the photon has to interact with something — an electron, say — which couples the state of the photon to the state of the ...


1

You are wrong, but understandably so given the ridiculous way in which quantum measurement is commonly discussed. A measurement is an interaction that makes a record of the value of some observable. If such an interaction happens during an interference experiment, the interference doesn't take place. What matters is whether there is a record, not whether ...


2

The second approach is wrong. You cannot use the relation $E_{total} = h\dfrac{c}{\lambda}$ because this is equivalent to saying that $E_{total} = pc$ which is true only for massless particles. In this case it should be $E_{total} = \sqrt{p^2c^2+m_0^2c^4}$.


3

If you shoot only one photon, you will get only one "spot" (pixel, if your detector is digital). One "spot" does not a pattern make. However, you can predict that there are regions on your detector where the photon will not hit, and no "spot" will appear. So there is a sense in which the rather poorly defined phrase "a photon interferes with itself" can ...


3

Short answer: no. The interference pattern is formed only after many electrons are shot through the slits. The "natural" conclusion of this experiment is that we cannot predict where an electron will be detected, only the probability of the electron being detected in particular locations, as indicated by the interference pattern. Another remarkable aspect ...


6

You can't predict where the electron will hit, but you can measure that it will hit at some discrete point. The probability distribution of final positions on the detector corresponds to the interference pattern. You will see the pattern only after shooting many electrons: https://physicsforme.files.wordpress.com/2012/04/slit.jpg


2

The second screen makes it two separate Fraunhofer single slits with their independent diffraction pattern, if the slits are narrow enough. On the vertical screen there will be spots uncorrelated, the same as would be found on the second screen if one closed off sequentially each slit. If the second screen were made partially transparent the degree of ...


4

The problem is that the dual slits are not infinitely narrow - the slit pattern is in fact the convolution of two infinitely narrow slits with a single wider slit. By the convolution theorem, the diffraction pattern, which is the Fourier Transform of the aperture function, is the product of the two Fourier transforms - so you see a series of equal intensity ...


2

Assuming you mean some special screen 2 that "lights up" when hit from either side, then no, there won't be a pattern, because the length (or rather:time) from hole A to some spot on the screen will be the same as the length/time from hole B. There will be no destructive interference "on" the screen 2.


0

To demonstrate the double slit experiment, you might consider using a thin sewing needle, two razor blades, two binder clamps and a laser pointer, as shown in the photo below. The sewing needle is held in a vertical position by sticking it into a piece of cardboard. The binder clamps hold the razor blades up. The razor blades are brought close to the ...


0

I think that for the demonstration, a laser and double slits are probably the best way to show the interference pattern. First showing with the 1st slit only, and then the 2nd. Finally show with the 2 slits opened. You probably want to emphasize one of the wonderful aspect of the interference-based experiments, roughly said: "You put light on light, and you ...


2

You can have a pool of water and place a cardboard or a piece of plastic with two slits in the middle. Just create a wave that goes through both slits and the waves should start interfering.


2

No. In fact, in general the collapse of the wavefunction is not consciousness-dependent. A good example would be using quarter-wave rotators. That sounds like a mouthful of jargon, but you need to understand that the light coming through the slits has some polarization which usually has to be modeled as having two "linearly independent" components; in this ...


0

It depends on the coherence length of the light source. If the wavefronts are curved, and it's difficult to ensure that light generated at the same time is impinging on the slits at the same time -- you need a relatively precise alignment in both the centre of the lens and the angle. If you collimate a beam, the wavefronts are planar, so it's less sensitive ...


4

I think your confusion is assuming that the energy of two waves add, when in reality there is an interference term. The short, physics-y answer is that it is not that any energy has disappeared, rather the interference has caused some of the energy that 'would have been there' to show up in a different place. The energy got shifted but not destroyed. This ...


2

What experiments should be done to test a theory depends on both the theory and its rivals. At some point, somebody may come up with a better theory. The crucial experiments will then depend on where quantum theory and its rival differ. If you're asking about quantum physics versus classical physics, there are many experiments where they would make ...



Top 50 recent answers are included