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2

Griffiths' formulation makes it explicit that operators which commute are not restricted by the uncertainty principle. Your boxed expression obscures this physical and mathematical insight.


2

The most correct relation is the following general relation, that actually contains both terms. If you omit the inequality $(|z|^2\geq(Im(z))^2)$ from the derivation, the next steps toward the uncertainty relation would be: $$\sigma_A^2 \sigma_B^2\geq|\langle f\lvert g\rangle|^2$$ $$|\langle f\lvert g\rangle|^2=\langle f\lvert g\rangle\langle g\lvert ...


2

Yes, you're right ! The uncertainty principle tells us that the thickness of the energy state $\Delta E$ is linked to the typical decay time of this energy level. If $\Delta E$ is large then $\Delta t \sim \tau$ (decay constant of the energy level) is small and then this energy state is very unstable.


2

Usually, when you want to probe the lifetime $\tau$ of a particle (or a quasi-particle), what you basically do is looking for the way the associated wave-function $\psi$ is significantly decreasing : $$\psi\sim\psi_0\,e^{-\left(\frac{1}{\tau}+\,i\frac{E}{\hbar}\right)t}$$ Let us consider a free particle with a given energy $E$. As $\langle E\rangle=E$ is ...


0

Let us take an electron's track in a bubble chamber where there is also a magnetic field. We can measure the momentum of the electron, the change due to ionisation, and its position as it goes through the spiral and finally know its final (x,y,z) at rest, and 0 momentum. Even though we are dealing with an elementary particle we are still, with our ...


1

Intuitively, if the potential energy is a function only of the position, if you measure the position precisely, you can just calculate the potential energy using that precise measurement. More formally, if $V(\hat x)$ is any function of $\hat x$, the position operator $$[V(\hat x), \hat x] = 0$$ which really is just that any operator commutes with itself.


2

When you have two operators $\hat{A}, \hat{B}$ satisfying equation $[\hat{A}, \hat{B}] = \imath \hat{C}$, you can prove with Schwarz inequality that $\sigma_{\hat{A}, \psi} \sigma_{\hat{B}, \psi} \geq \frac{1}{2} | \hat{C} |$. Unless there would be stronger inequality that can be used in calculations, it gives us the lower bound of uncertainty principle. ...


0

The particle is described by a wave that is spread out in position and momentum space. It is not possible to produce a wavepacket that does not respect the HUP. You can make such wavefunctions undergo interference and when it's undergoing interference the square amplitudes don't respect the calculus of probability, for an example see ...


1

Classical mechanics assumes that the state of a particle can be described, with arbitrary precision, by its position and momentum. That is in classical mechanics the state of the particle is, mathematically speaking, a phase vector. (And, the physical laws governing the particle's motion can be thought of as a trajectory in this vector space.) But quantum ...


0

From my understanding of your questions, you are confusing the "scientific method" and the "uncertainty principle. The scientific method says that "given the same starting conditions, within a controlled environment, etc., the "results" should be the same (ei. repeatable within some degree of accuracy). The uncertainty principle "deals" with an entirely ...


2

In this article electrons seen in a bubble chamber are shown. The spiral is an electron knocked off from an atom of hydrogen , a bubble chamber is filled with supercoole liquid hydrogen in this case. Th accuracy of measuring the tracks is of order of microns. The momentum of the electon can be found if one knows the magnetic field and the curvature. The ...


2

As you said if two operators commute they share eigenvectors. Physically this means that you can have a definite value for both. For example in the hydrogen atom the Hamiltonian $H$, which is the energy, and $J^2$, the magnitude of angular momentum, commute. A hydrogen atom can be in a state of definite energy and definite angular momentum. However, the ...


3

1.Uncertainty in position and in momentum If you've taken a class of statistics, you should recall that for some variable $y$, \begin{align} \sigma_y^2 = \sum (y_i-\bar y_i)^2 = \sum y_i^2 - \left(\sum y_i\right)^2 = \langle y^2\rangle - \langle y\rangle^2. \end{align} In other words, the expectation value of the position of the harmonic oscillator ...


1

Different trajectory each time due to probabilistic nature inherent in quantum mechanics and the uncertainty principle. The uncertainty principle is not "removable", it is not a constraint based on practical limitations in an experiment, but inherent to quantum mechanics calculations. There was an interesting thought experiment, given absolute control over ...


5

The problem with this question is that, even if you perfectly controlled the conditions at the macroscopic level -- including somehow releasing the leaf in exactly the same way every time (nearly impossible), and using leaves in identical starting state (the same leaf may have lost some water by the time you repeat the experiment, and suppressing all ...


5

No. The Uncertainity Principle states the following: The position and momentum of a particle cannot be simultaneously measured with arbitrarily high precision. There is a minimum for the product of the uncertainties of these two measurements. There is likewise a minimum for the product of the uncertainties of the energy and time. $$\Delta x \Delta p \geq ...


15

You are misunderstanding the Uncertainty Principle. The Uncertainty Principle says that a particle cannot simultaneously have a definite momentum and a definite position. This is not due to our incomplete knowledge of parameters. This is a fundamental law of the universe and arises from the fact that the momentum and position operators do not commute in ...


3

In the case of a leaf's trajectory, yes, it'd be same in both cases if the environment is exactly the same. And you can generalize this to any macroscopic event. As for the formation of Earth simulating the universe since its beginning, it's complex because it's not a macroscopic event. In the quantum world, everything is probability driven. In a heap of ...


0

Lubos's answer is excellent, but a simpler answer is that after you measure the particles, they're not entangled anymore. If the first is known (after a measurement) to be in some state A and the second is known to be in some state B, then (quite regardless of whether you measured the same or different observables), the pair is in state $A\otimes B$, which ...


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Consider two entangled quantum systems which are described by three properties, A, B and C each which can take a value of up or down. The two systems can be entangled such that if you measure the same property on both systems then you will get the same result 100% of the time, and each case occurs 50% of the time. That is to say both systems give up 50% of ...


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You are confused about what is called the Observer Effect. Read about it from here Observer Effect. Also, read some parts of this. The Uncertainty Principle just says that there is a fundamental limit to which the position and momentum of an object can be measured. Also, as PhotonicBoom said, the Observer Effect for Quantum Mechanics is not experienced at a ...


0

A clock is a collection of particles that behaves classically, i.e its a combination of all the wavefunctions of its particles, and due to that "coupling" it does not behave probabilistically anymore. You are refering to the phenomenon of wavefunction collapse due to a conscious observer, but in reality decoherence occurs. Wavefunction collapse happens in ...


0

"△x△p>h/2" is a simple consequence of the fundamental principle of using wavefunctions ("Amplitudes") to determine the probability of finding a particle. A plane wave is evenly spread over all space and is the eigenfunction of one precisely known value of p. In order to get anything other than such complete indeterminacy of position x, one must add several ...


1

It is an intrinsic property of our universe. There were some alternative interpretations, like the "hidden variables" (there are a swarm of deterministic random things going on that we don't know or cannot know about that cause the quantum randomness) but they have been experimentally disproven (Bell's theorem). You have a nice list of the experiments here. ...


2

There is nothing like bubble chamber photos to clarify such issues. Here is an anti Lamda: The part relevant to your question is that we can measure with an accuracy of microns the vertex of the decaying antilamda, and with an accuracy of an MeV/c the momenta of the particles it decays into, an antiproton and a pion. The antiproton identifiable by ...



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