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17

Heisenberg's Uncertainty Principle is, in essence, a consequence of two basic facts: de Broglie's relation between a particle's momentum and its associated wavelength, and a mathematical fact known as the bandwidth theorem, which states an equivalent uncertainty relation between a wave's position and its wavelength. The fundamental physical leap here is of ...


9

You're distracted by fancy language. Instead of "photon emitter" think "light bulb." (I suppose you could think "laser pointer" instead, since that's an easier way to make all the light exactly the same color than a light bulb plus a prism.) Instead of "photon detector" think "eyeball" or "camera." You've probably noticed that you can't detect a photon ...


5

The general situation is the following one. There is a self-adjoint operator $H :D(H) \to \cal H$, with $D(H) \subset \cal H$ a dense linear subspace of the Hilbert space $\cal H$. (An elementary case is ${\cal H} = L^2(\mathbb R, dx)$, but what follows is valid in general for every complex Hilbert space $\cal H$ associated to a quantum physical system.) It ...


5

Here's an elegant way to show that any linear combination of (anti)symmetric states is always (anti)symmetric. We use Dirac notation here for the states, and we assume, for simplicity, that we are dealing with a two-component system so that states of the system are linear combinations of products $|\psi\rangle = |\psi_1\rangle|\psi_2\rangle$. First, we ...


4

What you are proposing is called a local hidden variable theory. Bell's theorem proves that any such local hidden variable theory is inconsistent with behavior predicted by quantum mechanics. Bell test experiments have been performed, which show that the predictions made by quantum mechanics are correct, in ways that cannot be explained by a local hidden ...


3

This question is really classical. If you model two-slit interference in Maxwell's electrodynamics, the same thing happens: opening the second slit causes the intensity at some points on the screen to decrease. It happens with water surface waves too, and any other kind of wave. Destructive interference at one point is always matched by constructive ...


2

The interpretation of the double slit experiment is very strange and i want to understand how they did it before I give up my concept of reality. The double slit experiment does not refute the idea that reality is objective. It does refute the idea that each photons goes through one slit or the other. Rather there are multiple versions of each photon ...


2

Yes, the two are intimately related. One way, as in QMechanic's answer, is via Wick rotations, but in general there is a lot more freedom once you allow integration contours to go over into the complex plane. In my area, strong field physics, the use of complex time to understand tunnelling problems is everyday bread and butter for many people, and it is the ...


2

Yes,the photoelectric effect can be explained without photons! One can read it in L. Mandel and E. Wolf, Optical Coherence and Quantum Optics, Cambridge University Press, 1995, a standard reference for quantum optics. Sections 9.1-9.5 show that the electron field responds to a classical external electromagnetic radiation field by emitting ...


2

To "see another World" would require doing a measurement that involves (partially) reversing the effect that led to the split. In practice this is impossible to realize because the observer is a macroscopic object itself and it will decohere very fast. Decoherence means that the system becomes correlated with the environment and that poses a big problem ...


2

Yes, that is correct. A more general form of the superposition of the stationary state would be $$a|0\rangle + b|1\rangle$$ where $a,b$ describes the probability of each state. The state : $$|0\rangle + |1\rangle$$ assumes that the state $|0\rangle$ and $|1\rangle$ are equally probable.


2

Close, but not quite. Since quantum mechanics deals in probabilities, it is necessary to "normalize" the state in order to use it in later calculations. The most general state for the two-state quantum system you're considering would be \begin{equation} |\psi\rangle = \alpha\,|0\rangle+\beta\,|1\rangle \end{equation} where the quantities $\alpha$ and ...


2

The general form, equivalent to eqn (5) but for outside the well, is $\Psi_o = A_o \dfrac{e^{-k_o r}}{\sqrt{r}} + B_o \dfrac{e^{k_o r}}{\sqrt{r}}$ But just like we know $\Psi_i$ is not infinite at the origin, we also know that $\Psi_o$ doesn't go to infinity at large values of r. So we know $B_o$ must be 0.


2

A superposition of (anti)symmetric states is always (anti)symmetric, but it is not necessarily decomposable. So, your $$\frac{1}{2}[(\chi(A)\psi(B)\pm\psi(A)\chi(B))+(\phi(A)\eta(B)\pm\eta(A)\phi(B))]$$ can be put in the form $$\frac{1}{\sqrt2}(f(A,B)\pm f(B,A)).$$ but in most cases there would exist no such functions $f_1$, $f_2$ that ...


2

Look it like this, a ket is a vector of a (fancy) vector space and as such it has different components with different weights. Now, going to a more familiar vector space of the $R^3$, you can choose for example the cartesian coordinate system to describe a vector in space, but you can also choose any other set of three vectors (provided they're not ...


2

Unitarity of quantum mechanics prohibits information destruction. On the other hand, the second law of thermodynamics claims entropy to be increasing. If entropy is to be thought of as a measure of information content, how can these two principles be compatible? I don't think there's anything inherently quantum-mechanical about this paradox. The same ...


1

A physical state is defined by the density matrix, so, if you define the density matrix by : $\rho = \frac{|\psi\rangle \langle \psi|}{\langle \psi| \psi\rangle}$ it is easy to see that any multiplication by a complex number does not change the density matrix, so does not change the physical state. It is probably what Dirac means, while I have not the ...


1

The wave function isn't something that can be "deformed" in the way that you are thinking. The possible states of an electron in the vicinity of a proton can be found by solving Schrodinger's equation. This gives a discrete set of bound state solutions (energy < 0), labelled by the quantum numbers n and l (and also s, j, etc. once various spin effects ...


1

My understanding is that the Aspect experiment shows that your understanding is wrong. http://en.wikipedia.org/wiki/Aspect_experiment


1

You can say that QCD is the opposite of QED. I don't know how much you know of Renormalization group, but QCD has an asymptotic freedom, that mean that when you go to higher energies, you can use perturbative theory to do cross section calculation. That because the coupling constant is running with energy, due to the Beta-function. The goodness of your ...


1

Obviously I cannot know what Dirac is thinking, but I think it is just that his direction does not correspond exactly to your direction. We "know", just as Dirac does, that quantum states are members of some Hilbert space $\mathcal{H}$. We also know that scalar multiplication should not change the state, so $\lvert \psi \rangle$ and $\lambda \lvert \psi ...


1

Stated in a simpler way, kets are the generalisation of vectors to complex and potentially continuous/infinite dimension space (Hilbert spaces). Yet you can keep in mind the image of a vector to begin with. When you multiply a vector by a nonzero real number, its direction does not change. The same is valid for vectors from a Hilbert space. If you multiply ...


1

Mathematically you have a superposition of two Eigenstates and you use it because you don't know all parameters of the source. And yes, you use a probabilistic source, otherwise you will not do any experiment with it. Using your special source you get always the next result: Measuring one of the photons you know immediately the Eigenstates of the twisted ...


1

Think about it this way: When you measure some observable $O$, you get some measured value. If you measure many identically prepared systems, you may measure different values corresponding to the operator $O$. In the limit where you do this infinitely many times, you are able to recover exactly which values are possible, and with which probability. Basic ...


1

Evidently this is a homework problem. One way to approach the integral is to look at the symmetry of the integrand. To get started: the integrand has two factors. one of them, $y$, is odd under inversion of the axis. Once you have the symmetry of the integrand, you can make quick progress with the integral. Make a sketch of the integrand. That will ...


1

First of all, not every Hamiltonian admits a basis of eigenvectors. An operator has to be either compact or with compact resolvent to admit a basis of eigenvectors. With those type of operators, eigenvalues can accumulate (i.e. being distinct, but getting arbitrarily close) only at zero (compact) or infinity (compact resolvent). A discrete eigenvalue can ...


1

Disproof of Joy Christian’s “Disproof of Bell’s theorem” Florin Moldoveanu Committee for Philosophy and the Sciences, University of Maryland, College Park, MD 20742 http://arxiv.org/pdf/1109.0535v3.pdf


1

Hints to the question (v2): First note that the operator norm $||A||=||UA||=||AU||$ of an operator $A$ is invariant if we compose with an unitary operator $U$ from either left or right. Therefore $\dot{\rho}(t)$ is not the zero-operator: $|| \dot{\rho}(t) || = || [H, \rho(0) || \neq 0. $


1

A free quark wavefunction does not exist. Instead, inside nucleons quarks are relativistic and asymptotically free, which means that they only behave like individual particles for sufficiently large momentum/energy transfer. Imagine a classical solid state analog: if you apply a very small, slowly acting force to a single atom of a crystal, the force will ...


1

If you open up a new illuminated slit the interference pattern that results is different than it would be without the slit. If you open a new slit and keep the intensity of the light illuminating the screen the same energy that would formerly have been absorbed by the screen now arrives at the detectors. So the amount of energy that arrives at the detectors ...



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