# Tag Info

## New answers tagged measurement-problem

5

I don't intend to repeat the points already made in the other answers, consider this as a small addition to those, with the intention to give a more practical description (reminding some of the basic ideas) without getting into observer-philosophies. Quite clearly the act of observing, i.e. measuring a quantum system can be done via many different ...

1

I think you may have confused two separate things. By orbital we normally mean the distinct atomic orbitals like $1s$, $2s$, $2p$, etc. These actually only exist for hydrogen atoms (and single electron ions) so let's just consider hydrogen for now. In practice if we do an experiment with a sample of hydrogen its energy is well characterised and under normal ...

2

Quantum mechanics tells us the probability for an electron in a certain orbital to be in a certain position or momentum state. Inverting this to find the probability for the electron to be in a certain orbital given its position or momentum state requires some a priori idea of what orbitals are the most probable. This is stated succinctly by Bayes' theorem: ...

4

When does an interaction drop the system into an eigenstate? (i.e. when is a measurement=) This is an ill-posed question because, first of all, the system $S$ doesn't drop into any state but each observer $O$ has a state about it, as a state $\rho$ is nothing but the coding of past measurements (so it should be named with reference to being dependent on ...

0

An observable basically is something you can measure. However compared to just any general measurement, an observable has one specific property: If you repeat the measurement without the system changing in between, you get the same result (this is obviously not true for all possible measurement procedures). Such measurement have indeed been experimentally ...

0

Quantum optics demonstrates the existence of interaction-free measurements: the detection of objects without light—or anything else—ever hitting them. Paul Kwiat, Harald Weinfurter and Anton Zeilinger SciAm November 1996 http://www.arturekert.org/sandvox/quantum-seeing-in-the-dark.pdf

8

This is not a settled question. Just as it is still debated whether or not there is wavefunction collapse, so is it debated what exactly we should understand by a measurement. In the following, we will go through the ideas behind the von Neumann measurement scheme, which is one way to try and talk about measurement in quantum mechanics. An interaction ...

5

All measurements are based on interactions , but not all interactions are measurable. The "set of measurements" is a subset of the "set of interactions". The simplest mathematical way to see this is by Feynman diagrams. Feynman diagrams have one to one correspondence with integrals, and when they describe a measurement, the crossection or lifetime can be ...

0

take a system initially uncorrelated with the environment, with both being in pure states. It interacts with the environment, forming an entangled state. Taking a partial trace over the environment leads to a mixed state matrix for the system with nearly zero off-diagonal entries. The problem is we can reverse both the environment and the system so that the ...

1

#What is a wavefunction? A wave function is a mathematical solution of one of the basic quantum mechanical equations: Schrodinger, Klein Gordon, Dirac. By the postulates of quantum mechanics the square of this wavefunction gives the probability of finding the system under study when looking at (x,y,z,t) or (p_x, p_y, p_z,E) or similar four vector spaces. ...

1

I am going to answer this in a hurry because the question is on the edge of being closed. Quantum mechanics isn't just about "wavefunctions", it is also about "observables". An observable is something like: energy, position, momentum... i.e. it includes all the properties of classical physics. The wavefunction (or state vector or quantum state) is the ...

0

The question you ask is essentially: How to solve the measurement problem? As you can see from that article (although I wouldn't say it's a very good one), there are several approaches to either get a theory where no collapse occurs (thereby rendering the question futile) or to explain the collapse. So far, nothing has been so satisfactory as to be a ...

2

I agree in full with Marty Green except the explanations of chemistry in which I was unable to follow so well (that doesn't say that I disagree with them). But, let me put the things in short. The collapse is a phenomenon that is supposed to occur when a quantum object comes in contact with a quantum system. For instance, a quantum particle falls on a ...

-1

In theory, every example of wavefunction collapse should be explainable through a mechanism of normal time evolution of the wave function. The apparent collapse is only an illusion. People who agree with this idea like to talk about "decoherence", which is a fancy sounding word, but it doesn't really tell you anything. In fact, there is very little interest ...

4

When a particle passes through an ideal pair of slits, it's wave function can be written in the form $$\psi(x) = \alpha\, \psi_1(x) + \beta \,\psi_2(x)$$ where $\psi_{1,2}(x)$ are the wave-functions you would get by closing either slit 2 or slit 1, respectively. We say that the wavefunction is in a coherent superposition. What is coherent about this ...

1

But the way I've seen it be explained, it seems that the reason things collapse is that upon interacting with macroscopic things, the wave function decoheres until there is only one component left. No. That's not the way decoherence works at all. A system $S$ interacts with the environment $E$ with a Hamiltonian that does the following: $$... 0 Decoherence is not a sufficient explanation for the collapse. Decoherence means that the phases between the components of the wave-function, (when the latter is expanded as a quantum superposition), are destroyed. But WHY in a measurement we get a certain ONE from these components, the decoherence doesn't explain. And what happens with the other components ... 1 A particle does not need to have a hard edge. It can for example, be a density function, which sort of fades to zero. One might note that waves can intersect each other and come out as if the other wave was not there. Particles with hard edges are more an artefact of our minds rather than 'what's really there', until we get some real evidence otherwise. ... 2 I like the way you're thinking, but I'd like to try to open your mind a bit further. A lot of our everyday concepts are a product of the net behaviour of billions of particles following the laws of physics. For example, consider a steel ball. It has a "hard edge". What does this mean? It means another ball can be anywhere outside of the hard edge, but ... 11 This is one of the key results of quantum field theory: particles are not single points, they are disturbances in quantum fields that are spread out over space. Typically the disturbance is not spread out very much, otherwise it looks more like what we know as a wave than a particle. The technical term for what you're calling a "smear" is a wavepacket. 7 In fact any object with non-zero physical extent is required to be deformable by relativity (but see below); otherwise pushing on it can transmit energy and information faster than light. And quantum mechanics rules at those scales so everything does have fuzzy edges. At the scale of the very small there are no sharply defined objects. A consequence of this ... 1 If we have two arbitrary quantum-mechanical operators \hat A and \hat B, then the corresponding observables have to satisfy the Robertson-Schrödinger uncertainty relation:$$\Delta A \Delta B \geq \frac12 \lvert\langle [\hat A,\hat B] \rangle\rvert This equation implies that it is impossible to measure both $A$ and $B$ to arbitrary precision at the ...

1

While I agree with Noah Steinberg's answer, there are some other points. Usually when quantizing a system, we label our states using quantum numbers, which are as close to classical parameters (also called c-numbers or commuting numbers) as you can get. The calculation is usually easiest when you can find the largest set of commuting observables, and ...

3

If two observables are compatible it means the eigenstate of one observable is also an eigenstate of the other and that the commutator of the two operators is 0. This means that if two observables are compatible one can make a measurement of the first observable and then measure the second observable without changing the state of the particle.

2

You speak of the as yet unresolved quantum measurement problem. Although it is unresolved, most physicists think that it has nothing to do with solipsism, which is a sensational explanation but there are many reasons to think that this is not the right one: Conscious observers themselves are quantum systems. Observation always comprises interaction ...

2

The electron does not move - it has no well-defined position in the orbital state, and hence no well-defined momentum. Neither does it "teleport" around - as long as it is not interacting with something that forces it to be at a definite position, its state is "smeared" all over the electron as an electron cloud. Yes, this is essentially the Bohr model, ...

1

I will answer this part In addition, we know that the Hamiltonian represents the sum of kinetic and potential energy in a system.However, I'm not quite sure why, intuitively, the time dependent version of the Schrodinger equation becomes Hψ=iℏ ∂/∂t ψ(r,t). Quantum mechanics was developed slowly, because experiments showed that light came in quanta ...

0

You ask a couple of different questions: 1) You say "by Heisenberg's uncertainty, we cannot measure the exact momentum or position of a particle/wave ever". No, Heisenberg's uncertainty principle doesn't say that. It says that IF you measure the position of a QUANTUM particle with precision Δx, i.e. you localize the particle within an interval Δx, then the ...

0

There are many confuse things in this question: 1) "The multiple paths the particle simultaneously travels, interfere with each other" It's not the paths that interfere, when these paths cross one another, i.e. the wave-packet on one path and the wave-packet on the other path pass through the same region at the SAME TIME. 2) "but as it is absorbed, it ...

0

Given that Bohmian mechanics (better named de Broglie-Bohm or dBB theory) contains the Schrödinger equation, it contains quantum theory completely, and therefore all things which make sense in quantum theory remain meaningful. Of course, for quasiparticles nobody plans to construct trajectories. And it is also not possible to simply "watch" Bohmian ...

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