Recent content by eaglelake

You can obtain the commutator relation between position and momentum by doing (again) the math. Here, the operators are \hat x = x and \hat p = - i\hbar \frac{\partial }{{\partial x}}, so that \left[ {\hat x,\hat p} \right]\psi (x) = \left[ {x, - i\hbar \frac{\partial }{{\partial x}}}...

You are missing the definition of uncertainty. The uncertainty in position is the root-mean-square deviation (the standard deviation in ordinary statistics): \Delta x = \sqrt {\left\langle {\hat x^2 } \right\rangle - \left\langle {\hat x} \right\rangle ^2 } . Likewise, the momentum uncertainty...

How does a person become a quantum object, a part of a quantum event? Which, incidently, requires a measurement result. (This is what Bohr called closure.) I see nothing in the postulates that require a conscious observer, or anything about a collapsing wavefunction. A quantum experiment does...

There is no redistribution of energy going on here! Here, we are discussing classical waves that carry energy and momentum as they propagate through a vibrating medium. The wavefunction \psi (\vec r,t) represents the displacement of the vibrating medium from its equilibrium position. In an...

When discussing the fundamental principles of quantum mechanics we always assume ideal experiments to purposely exclude effects caused by any limitations in the devices used in the experiments. It is understood, then, that any non-classical behavior is due to the quantum nature of the...

1) An interferometer experiment with waves is different than one with photons. The wave will split and travel both paths simultaneously. Likewise, both detectors will be triggered at the same time by a wave. But with photons the detectors are triggered one at a time, never simultaneously. The...

The difficulty here is that light itself is not either a particle or a wave. It depends on the experiment that we do. Some experiments with light exhibit wave properties. In the beginning this was always the case. We were all told that Young’s experiment proved conclusively that light was a...

John Wheeler said that consciousness has nothing to do with the results obtained in a quantum experiment. Most of us do not believe that we have the mental capacity to kill cats simply by looking at them, or that we can change the result of an experiment by choosing to ignore certain...

Welcome to the forum! You are correct! The photon is, by definition, a particle. When we detect a photon it is always localized as a particle. For example, on a detection screen, we see a dot where the photon hit. But, all quantum particles, including photons, have an associated wavefunction...

I meant that the textbook author was most likely trying to demonstrate the use of the uncertainty principle in order to give you a “feel” for it without doing any advanced math. He is giving you an introduction to the heart of quantum mechanics without all the hairy details, which will come...

Every observable in quantum mechanics is represented by a Hermitian operator. Spin ½ is the observable represented by the Pauli spin matrices, one for each component. If we do an experiment where we measure the z-component, say, of the spin, then the eigenvalues of the spin operator (the Pauli...

I agree that your English is fine! In order to calculate the uncertainty exactly you must know the corresponding wavefunction. Not knowing the wavefunction for a given experiment, we make an educated guess by assuming that the uncertainty in position will be comparable to the size of the...

This is true only for non-relativistic particles! You want v=c, which is relativistic. Most likely you are talking about light (electromagnetic radiation). In the relativistic case p=E/c, where c is the constant speed of light. For light the photons have zero mass and when you increase E the...

danR is correct. If \lambda decreases, then p increases. But, this is not a cause and effect thing. That aside, if v = c, then p \ne mv. Best wishes