Tag Info

New answers tagged

0

For a connection between Schr. eq. and complex Klein-Gordon eq., see e.g. A. Zee, QFT in a Nutshell, Chap. III.5, and this Phys.SE post plus links therein. The complex Klein-Gordon eq. in QFT describes both particle and anti-particle excitations. When scaling to the appropriate non-relativistic one-particle sector to derive the Schr. eq. for the complex ...


1

Klein-Gordon (and actually also Dirac) equation is usually considered a classical field equation. To obtain a quantum field theory, you have to quantize it to become an operator on the symmetric (for bosons) Fock space. Then you have again a Schrödinger-type equation for the wavefunction, with an Hamiltonian that embodies the Klein-Gordon dynamics. For a ...


1

The Heisenberg inequalities reads : $$\Delta x \Delta p_x \geq \frac{\hbar}{2},$$ where $$\hbar = \frac{h}{2\pi}. $$ Therefore, for a free particle, your add the expressions $$ p_x=mv_x$$ so you get : $$\Delta x \Delta v_x \geq \frac {\hbar}{2m}$$ you have the uncertainty on the velocity : $\Delta v_x \geq \frac{\hbar}{2m\Delta x}$ the Heisenberg ...


1

What makes you think an electron reflects photons as you have drawn? Electrons scatter photons in any direction, although not uniformly. (Examples: Thomson scattering, Rayleigh scattering, X-ray crystallography) The electron may absorb the photon for an arbitrary period of time, changing momentum and thus position, then release a photon of a different ...


0

It has been said that your second experiment is impossible. For me it is possible but can't be used to predict the position of the electron and its momentum. First consider that your "classic experiment" is true. In this experiment, the important fact is that one can't obtain the momentum of the electron when the photon angle is measured. Now, consider ...


8

First of all, the uncertainty principle is more than just disturbance of observation. From the Wikipedia article "Uncertainty principle": Historically, the uncertainty principle has been confused with a somewhat similar effect in physics, called the observer effect, which notes that measurements of certain systems cannot be made without affecting ...


2

The problem with your set up is that you have ignored the quantum mechanical nature of the photons. The photons are subject to the uncertainty principle as well as the electron, and so there is no way to send in 2 photons with precisely the same momentum at precisely the same time, and we can't guarantee they will scatter of in precisely the same way. The ...


0

Your classic experiment 1) is quantum mechanical. Electrons and photons are elementary particles and interact individually as quantum mechanical entities. The plot shows the elastic scattering of photon on electron, a computable process quantum mechanically thus the angular distribution is known. As we are talking quantum mechanics there is a probability ...


0

Yes - but not really. A single photon has a single value for it's related wavelength - so it's perfectly monochromatic, in a formal sense. But, in a physical sense, the concept of a spectrum (like: monochromatic) basically does not apply to a single photon. In practice, you do not have these "single photons, with some specific wavelength" to actively ...


5

Any measurement of the photon's energy (i.e. frequency, or free-space wavelength though making a direct identification of particle properties to wave properties is a little sketchy) will return a single value. Every time. But ... you can't fool Heisenberg and if you have confined the position of the photons---say by insisting that it hit the detector---then ...


2

This is a partial answer based on a quick read of the paper. If somebody would like to post a fuller analysis I'll delete this. Anyhow, the experiment is not defying the uncertainty principle. Instead it's effectively moving the uncertainty around. The uncertainty attached to the whole system is unchanged, but it's possible to measure one aspect of the ...



Top 50 recent answers are included