# Tag Info

## New answers tagged uncertainty-principle

0

Your situation is a little confusing since the velocity of the wave packet is not related to the momentum. If you want to speak about the uncertainty principle for classical waves, the 'momentum' is proportional to the inverse wavelength $\lambda^{-1}$. In this situation the speed of the wave does not depend on the wavelength, only things like the density of ...

4

I think you have misunderstood the meaning of the equation $$\Delta X \Delta P \geq \hbar / 2 \, .$$ This is not surprising given that the notation used here is really, really misleading. It should be written like this $$\sigma_X \sigma_P \geq \hbar / 2 \, .$$ To understand this we have to explain what $\sigma_X$ and $\sigma_P$ mean. Suppose you have a ...

1

Given two observables $A$ and $B$ such that $[A,B] = iC$, the most general form of the uncertainty principle is $$\Delta_\omega(A)\Delta_\omega(B)\geq\frac12|\omega(C)|,$$ where $\omega$ is any state of the algebra of observables. By the Riesz-Markov theorem, there is a regular probability measure such that $$\omega(f(A)) = \int_{\sigma(A)}f(\lambda)\ \text ... 0 With the uncertainty priciple, you cannot know the position and momentum of a particle (or anything for that matter) beyond a maximum precision. You can know where and how fast a particle is moving, just not with infinite precision in both variables. It's not just a measurement issue, but a fundamental fact. So a particle accelerator can still focus a ... 1 To answer your first point, as far as I know, a test beam is used to calibrate the position and accuracy of the system, before the main beam is sent, to avoid, as you point out, any damage to the equipment. An indication of how few actual collisions there are is given by the fact that in 1978, a scientist did get a high energy beam right in the face, not at ... 0 If the pencil were at absolute zero it would necessarily assume its lowest energy state, which is not the vertical state. If the pencil were modeled as a quantum rotor with an infinite potential barrier covering half its solid angle space (i.e. the table) then there are certainly excited but stable states where the pencil remains in a more or less vertical ... 0 Here \rho is the density matrix for a pure state, but it's nonetheless a density matrix (simply one with von Neumann entropy of nought, but that doesn't matter). Recall that for any Hermitian operator \hat{M}, the n^{th} moment of the probability distribution of the measurement made by \hat{M} is \mathrm{tr}(\rho\,\hat{M}^n). See, for example, the ... 0 GIVE IT A SPIN, with the max possible rpm for such massive object. Do the experiment at the ISS, well above Earth surface. Unconstrain the problem from the 'surface' issue. Remove the 'surface' below the atom tip -- where is bellow without gravity? -- or approximate the surface as much as you can but without contact (the atom's electronic cloud prevent this, ... 0 If your measurement would give you an exact upper and/or lower bound, but no more information (i.e. a probability distribution), the state would collapse into the projection onto the subspace of possible values, so it would still be a superposition. More generally and realistically, we would measure a value, and assign decreasing (classical) probabilities ... 2 If black holes have mass but no size, does that imply zero uncertainty in position? If so, what does that imply for uncertainty in momentum? I mean to say that the particles which were originally separate have theoretically come to occupy the same point in space. Does the uncertainty principle apply to this phenomenon? Zero size doesn't ... 4 General relativity is a classical theory. The Heisenberg uncertainty principle does not apply to it. The research frontier in physics now exists in quantizing gravity and unifying it with the other three forces (strong , weak, electromagnetic). Once that is done the solution for the black hole will become a probability distribution and the Heisenberg ... 1 Did you cover the uncertainty principle? In quantum mechanics there is and uncertainty between energy and time:$$ \Delta E \Delta t > \frac{h}{4\pi}$$this means that if you try to measure Energy with perfect accuracy you will have a great uncertainly in time (actually an infinity uncertainty). I guess this is what the professor was referring to, and ... 1 First of all I think I should clarify the confusion of the terminology "uncertainty in x". What we mean by uncertainty is simply the standard deviation of the observable defined by:$$\Delta x= \sqrt{ \left< x^2 \right> - \left< x \right>^2} where $<A>$ denotes the expectation value of the operator $A$ is some state $\psi$. You should ...

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