# Tag Info

16

What you describe in your question is the "Copenhagen interpretation" of quantum mechanics. There are more nuanced views of this nowadays that don't treat "measurements" quite so asymmetrically, see e.g. sources that talk about decoherence. I recommend watching the classic lecture "Quantum Mechanics in your face" by Sidney Coleman for a nice take on this ...

14

Well, I think you said the answer yourself when you used the words "projection operator." In the Heisenberg picture the operators get projected down to a subspace at the time of the collapse. In other words, the operator 'collapses' by picking up a projection piece that kills the unphysical part of the state. Forget about pictures for a second, the physical ...

12

Assuming wave-function collapse is correct (which can be a relatively hefty philosophical claim in some circles), then think of measurement this way: When you measure an observable on a system, you collapse the wave-function of the system into a Dirac delta function in the eigenbasis for that observable. If you measure position, you get a delta function in ...

12

Dear Jack, there is no physical phenomenon that could be called the collapse. The collapse of the wave function, as first emphasized by Werner Heisenberg and then many others, is just the event when we learn something about a physical property of a physical system. When we learn that Osama bin Laden is located in a building in Pakistan, his wave function - ...

12

It's tempting to think of the light as a little ball (the photon), and since little balls have a definite position the little ball has to be in a superposition of a state where it goes through one slit and a state where it goes through the other. However this is not a good description of what actually happens. The light is not a photon, and it's not a wave ...

10

Assuming that the incoming "first" particle is prepared in a pure state, interaction with another particle does seem necessary. Such an interaction might simply be the spontaneous emission of a photon or other particle by the original incoming particle, however. Most importantly, such an interaction is not itself sufficient. For a measurement event to ...

9

Aren't the particles this quantum state consists of interacting with each other? Why doesn't that cause the state to collapse? We have a mathematical model for the observations we can make of any system in the micro world. This model is quantum mechanics and its predictions have been verified experimentally over and over again. Observables are ...

8

The effect you are describing in your question is known as wave-particle duality and is a form of complementarity, it has been observed in various experiments. Realisations of Wheelers delayed choice thought experiment are what I find most interesting. In a delayed choice experiment the particles are not measured before they go through the slits but ...

8

There are two different issues. One of them is the sign of the momentum; the other one is whether the momentum is spread (it's not because of the unnatural boundary conditions). Concerning the first point, the standing wave (sine) is a real function and every real wave function has the same probability to carry momentum $+p$ and $-p$. So indeed, both of ...

8

There are currently two different accounts that give a larger picture of what happens when a quantum system is measured. One of them is the fact that many random interactions between the system (which might be a 1-body or N-body quantum system) and the environment (which is considered for most purposes a pseudo-classical system with infinite degrees of ...

8

Interactions merely involve a correlation developing. For example, if an electron is put through a Stern-Gerlach apparatus, a correlation develops between the distance travelled in the x direction and the distance deviated in the y direction. It is reversible. The measurement which occurs when the particle hits the photographic plate is irreversible. It ...

7

An electron, indeed any particle, is neither a particle nor a wave. Describing the electron as a particle is a mathematical model that works well in some circumstances while describing it as a wave is a different mathematical model that works well in other circumstances. When you choose to do some calculation of the electron's behaviour that treats it either ...

7

In the following answer I am going to refer to the unitary evolution of a quantum state vector (basically Schrodinger's Equation which provide the rate of change with respect to time of the quantum state or wave function) as $\mathbf{U}$. I am going to refer to the state vector reduction (collapse of the wave function) as $\mathbf{R}$. It is important to ...

7

Answer Rigorous adherence to the liturgical rituals of the "Church of the Larger Hilbert Space" is feasible in principle yet exponentially inefficient in practice. Exercise One way to answer this question is by reference to a feasible numerical computation. So fire-up MatLab; specify the dynamical system as (say) $n\sim 10$ interacting qubits; specify ...

7

If you place a camera you will not see any interference pattern. So, the answer is yes. The camera will cause the wavefunction to "collapse". But I don't like the term "wavefunction collapse", because wavefunction is not really any physical object. What the camera will basically do is cause an abrupt change in the state of the particle. Here is the ...

7

I'll start with the second one. $\int\phi^\ast\psi\,\mathrm{d}x$ is, as Chris says in the comments, the scalar (or dot) product of $\phi$ and $\psi$. In the Dirac notation, it is written as $\langle\phi|\psi\rangle$ and it gives the overlap of the two wavefunctions. In other words, it gives the probability amplitude (i.e., what you call square root of ...

6

I've never seen a single prediction based upon MWI. I've also never heard of the Cophenhagen interpretation called an approximation. If that were the case, then the Copenhagen interpretation must fail in at least one limit. Does Max provide such limits? Both of these statements seem to lean towards sensationalism than towards mathematical rigor.

6

I think you've misconstrued the Schrodinger picture. The Schrodinger and Heisenberg pictures are physical theories that make testable predictions, are strictly mathematically equivalent to each other, and are unitary. Neither theory says anything about wavefunction collapse. Collapse (of the wavefunction or of an operator) is a feature of a particular ...

6

In a true measurement procedure of position, the outcome is an interval $(a-\delta,a+\delta)$, $\delta>0$ being the precision of the instrument. In view of Luders-von Neumann's postulate on the reduction of the state, if the state before the measurement was described by the normalized vector $\psi \in L^2(R)$, immediately after the measurement the state ...

6

An observation is an act by which one finds some information – the value of a physical observable (quantity). Observables are associated with linear Hermitian operators. The previous sentences tautologically imply that an observation is what "collapses" the wave function. The "collapse" of the wave function isn't a material process in any classical sense ...

6

The collapse of the wavefunction is generally attributed to decoherence. This is time asymmetric in the same way the second law of thermodynamics is time asymmetric. I suppose it's theoretically possible for a wavefunction to uncollapse, but this is like saying it's theoretically possible for a broken egg to reassemble itself.

6

When position is measured, the uncertainty of the resulting delta spike's position is 0 This notion is the root of the problem. Quantum states which are actually eigenstates of the position operator are mathematically pathological and also completely unphysical. Some math tools Consider a one dimensional system. Suppose $\{|x\rangle \}$ is an ...

5

As far as I understand it, your claim is not per se about measurement but more like "if I prepare a particle in an eigenstate of either position or momentum, doesn't it mean that I have product of uncertainties $0.\infty$?" isn't it? One first problem with this claim as it is, is that the corresponding "wave functions" of these states are not ...

5

Let me take a slightly more "pop science" approach to this than Luboš, though I'm basically saying the same thing. Suppose you have some system in a superposition of states: a spin in a mix of up/down states is probably the simplest example. If we "measure" the spin by allowing some other particle to interact with it we end up with our original spin and the ...

5

''How isolated must a system be for it's wave function to be considered not collapsed?'' Experimentally, a system whose collapse is observable must be so small that one can prepare it in a well-defined pure state. If this is not the case, one can only speculate about what happened, leaving much room to imagination. This means that even when the carrier of ...

5

With imprecise measurements you need to use the more general statistical quantum mechanics rather than simple pure-state quantum mechanics. Specifically, rather than ending up with a pure quantum state, after an imprecise measurement you instead have a "dirty" state (mixed quantum state) which is blurred by uncertainty in the classical sense. This mixture ...

5

Leaving aside the quantum measurement problem (i.e. whether or not there is a "collapse" of quantum state to an eigenstate of an observable on measurement) and talking wholly about quantum state between "measurements" and its unitary evolution, I'd say that the transition is definitely a smooth shifting from one "eigenstate" to another, so that the ...

5

You already have some very erudite and good answers. I will give an experimentalist's point of view: I'm wondering, when an electron changes state, does it move from one state to another over some (very small) time period? Or does it change from one state to another in no time? An electron is par excellence an elementary particle and it is quantum ...

5

Your question contains a false statement: If I am not mistaken, by "observed" we mean interaction with any other particle You are mistaken. In different interpretations of quantum mechanics the definition of "measurement" is different. But I think it would be enough if I give just five of which you can choose yourself. In Copenhagen/von Neuman ...

5

1) "In what situations is it okay to separate the complete wavefunction like that?" As a simple product, only for distinguishable non-interacting particles. If you add antisymmetry, as Ruslan notes, you get a Slater determinant wavefunction which is exact for indistinguishable non-interacting fermions. Since electrons interact through the Coulomb operator, ...

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