New answers tagged

0

Is momentum of a particle "random" because it is uncertain, or is it uncertain in addition to being random? In quantum mechanics systems are represented using wave-functions (wave-vectors). The momentum of a particle is completely uncertain if it's position is certain (a localized particle) . But it is also possible to create wave-functions that have a ...


0

A particle is a wave. It's wave function (consider non relativistic Quantum Mechanics), when absolute valued squared, is a probability density function. The particle's momentum is a multiple of the gradient of the wave function, with h, Planck's constant, one of the proportionality constants. That is then the probability density function for momentum. It is ...


0

If we setup the camera to record like above but NEVER EVER EVER look at the result of what was recorded. Does the wave function still collapse? The answer is that we just don't know. We can tell that the wave function has collapsed (in Copenhagen terms) only when we humans look at the system -- in the canonical experiment that means looking at the ...


7

Experimental determination of $c_i$ values starts with preparing multiple identical systems, then making measurements. From all the measurements, one determines the probabilities, which are the $|c_i|^2$. The square root of the probabilities will tell you the $c_i$ to within a phase factor of the form $e^{i\beta}$, where $\beta$ is real, and may or may not ...


4

Once a measurement ( observation ) is made on a quantum system the system will be in an eigenstate of that property, so if the energy of an electron is measured the electron will afterwards remain in an energy eigenstate ( until some other measurement or interaction occurs ), but if the angular momentum is measured the electron system will afterwards remain ...


0

I think you are probably misinterpreting the context here. If you read the previous line carefully it says "there is always an undetermined interaction between observer and observed; there is nothing we can do to avoid the interaction or to allow for it ahead of time. And later he just says due to the fact that photon can be scattered within the 2θ' angle ...


0

S., yes, you are correct, once you measure the energy, the state collapses to one of the two, with the associated probability that you have correctly identified. Now, energy eigenkets are sometimes called ''stationary states", because they do not evolve in time, being eigenstates of the Hamiltonian (time evolution generator). Secondly yes, in the case ...


2

As everyone points out commutation is not a shorthand for equivalence, as, given your relations, the Jacobi identity, [A,[B,C]]+[C,[A,B]]+[B,[C,A]]=0 dictates that when the first two terms vanish, the third must too, so that B must commute with [C,A], non vanishing in general, as remarked repeatedly. Lie algebra commutators do, nevertheless, parameterize ...


3

The oldest work on this preceeds quantum mechanics by more than 100 years. it was done by Malus in 1809 about experiments with polarized light. See http://www.mat.univie.ac.at/~neum/papers/physpapers.html#CQlightslides


3

Both are correct, actually. If you measure an observable for that wave function you'll either find the eigenvalue corresponding to state 1 with probability |c1|^2 (similarly for state 2), subject to the condition |c1|^2 + |c2|^2 = 1. Edit: What Griffiths is saying is that before you perform the measurement, the particle is neither in state 1 or 2, but in a ...


1

I will address the title, ignoring the content of the question . If I repeated a quantum measurement, would it be the same? This is the method of gathering data in particle physics in order to check with as good an accuracy as possible the quantum mechanical predictions for the interactions. For example one sets up a beam of identical particles ...


2

As you said, the result of the measurment depends on probability. Each eigenvalue (i.e. result of a measurment) has a certain probability of coming out when a some characteristic of the system is measured. Think of this in an easier example. Suppose that you have a pair of dice. Both are exactly the same, so, for both dice, the probability of each number to ...


2

Take a large ensemble of particles and preselect state $\left|\psi\right\rangle$ Then take a large sub-ensemble of this ensemble and postselect state $\left|\phi\right\rangle$. In other words, set up the experiment, with the ensemble starting in state $\left|\psi\right\rangle$. Make a weak measurement of the observable $\hat{A}$, then make a strong ...


0

The wave function you've provided is a linear superposition of two distinct energy eigenfunctions, $\psi_1(x)$ and $\psi_2(x)$, that are assumed to have distinct energy eigenvalues, $E_1$ and $E_2$ respectively. It is not possible to predict with absolute certainty the outcome of a measurement of the energy observable.


0

As I stated in the comments the post by Arnold Neumaier in reply to this question answer mine too. In particular the papers he links (arXiv papers cond-mat/0102428 and cond-mat/0203460). In my mind the two papers are essentially a continuation of what Leggett and Caldeira showed (that the density matrix becomes diagonal when coupling to certain statistical ...


1

In all these discussions about entanglement, all the measured observables of Alice always commute with those of Bob. Their degrees of freedom describe two factors ${\mathcal H}_A$ and ${\mathcal H}_B$ of the overall Hilbert space of possibilities which is the tensor product $$ {\mathcal H} = {\mathcal H}_A\otimes {\mathcal H}_B $$ This factorization of the ...


0

I am not satisfied of the published replies, so I will try my own, as a metrologist (expert in measurement units, but not in theoretical physics). The question clearly is referring to the experimental frame while all the answers are referring only to the theoretical frame, so they do not talk nor understand with each other. Here we are dealing with two ...


0

The short answer is No. Because there is no definite energy state of this particle. It is in a superposition of two energy states $E_0$ and $E_1$.So, all we know is that any measurement will result in the collapse of wavefunction in one of the energy eigenstates. But where will it collapse is completely random. All we know is that there is equal probability ...


1

The statement is correct. Let's be really specific. Is there a way to, in one measurement, determine whether you have: $$ | \psi_1 \rangle = \frac{1}{\sqrt{2}} \left( | 0 \rangle + | 1 \rangle \right) $$ or $$ | \psi_2 \rangle = | 1 \rangle $$ ? You should probably suspect that this is not possible. Whatever measurement you would get on the $| \psi_2 ...


3

There is no randomness in quantum mechanics, there is only uncertainty. , as stated, whoever may have said it. Mathematical definition of randomness: The fields of mathematics, probability, and statistics use formal definitions of randomness. In statistics, a random variable is an assignment of a numerical value to each possible outcome of an event ...


-1

I had a little research on it. First of all, Quantum Mechanics is all based on probabilities. What is a probability. It gives you the chance for a specific event to occur from a list of possible events. Now what is the probability of a probability. Suppose if you say that there is 80% probability of finding a particle, there is a most possible chance for ...


0

The core of your question is subtle, so I'll be careful in how I set up my answer. In my understanding of quantum mechanics, wave function collapse is the closest a physical process can be to the mathematical idealization of a random variable. However, before the collapse, a complicated many-body process, the wave function evolution of the system is ...


-2

Randomness is a behavior that is unpredictable and there can be no mathematical pattern to it. However, uncertainty does have a mathematical pattern, thus proving randomness as false. For instance, you can scribble all over a piece of paper and you can eventually find some mathematical pattern even though what you've done seemed like it was pure ...


3

Actually, the outcome of the experimental apparatus is an interval $(x_0-\delta, x_0+\delta)$. $\delta>0$ stays for the accuracy of the instrument which can be made smaller and smaller but cannot be removed. It is therefore assumed (Luders-von Neumann's axiom) that, if the state immediately before the measurement was determined by the wavefunction ...



Top 50 recent answers are included