# Tag Info

41

The other answers here, while technically correct, might not be presented at a level appropriate to your apparent background. When the electron interacts with any other system in such a way that the other system's behavior depends on the electron's (e.g., it records one thing if the electron went left and another if it went right), then the electron no ...

19

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 ...

19

It is true that the reactions $$e + \gamma \to e, \quad e \to e + \gamma$$ cannot occur without violating energy or momentum conservation. But that doesn't mean that electrons can't interact with anything! For example, scattering $$e + \gamma \to e + \gamma$$ is perfectly allowed. And a classical electromagnetic field is built out of many photons, so the ...

18

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 ...

18

I think most arguments in the literature can be boiled down to the point that decoherence does in no way touch the linearity of the Schrödinger equation, and thus cannot make an "or" from an "and". This is complicated in the literature by very technical discussions, which I would like to avoid. Let me explain the basic point in more details. A widely ...

18

Good question. The textbook formalism in Quantum Mechanics & QFT just doesn't deal with this problem (as well as a few others). It deals with cases where there is a well-defined moment of measurement, and a variable with a corresponding hermitian operator $x, p, H$, etc is measured. However there are questions which can be asked, like this one, which ...

17

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 ...

17

What you describe is the process known as decoherence: any interaction of a quantum system with its environment (e.g. with photons or other particles passing by, and, yes, most likely interacting through gravity, although we don't have a theory to fully describe this yet) has the potential to destroy its genuinely quantum nature, turning quantum ...

16

The system doesn't "know" anything. The only uncontroversial statement one can make about the (strong) measurement of a quantum system is that you will make the correct predictions if you assume that the state after the measurement was the eigenstate corresponding to the measured value of the observable (so, for position, indeed a $\delta$-function, if we ...

15

There are many interpretations, and while there are good arguments in favor of one or another, they are currently not distinguished experimentally. Therefore it is often considered prudent to leave the question of which interpretation is best to the field of philosophy, and focus in a physics course on the falsifiable aspects of the theory. The field of ...

13

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 ...

13

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 -...

13

Unless the wavefunction collapses to an eigenstate of the Hamiltonian, the subsequent time-evolution will produce a superposition. The postulates clearly state that, if you measure the observable $\Lambda$ and obtain the outcome $\lambda$ (assumed non-degenerate for simplicity), then the state collapses to the eigenstate $\vert\psi_{\lambda}\rangle$ of $\... 13 The way I like to understand this is the following: suppose you have one observable$A$with spectrum$\sigma(A) = \{ a_n : n \in \mathbb{N}\}$which we will assume discrete and non-degenerate for simplicity. In constructing the theory you would like to have states where the value of$A$is indeed certain. Those states, the postulates of QM tell you to be ... 12 The answer by Craig Gidney is quite adequate for the question, but I want to address the word "collapse" in the title, since search engines will be homing in on it. From webster.com 1: to fall or shrink together abruptly and completely : fall into a jumbled or flattened mass through the force of external pressure <a blood vessel that collapsed&... 11 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 ... 11 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 quantities ... 11 So we are taught collapse at school, although there is no experimental evidence of collapse. You don't think this is good. You suggest that we are taught many-worlds instead, although there is no experimental evidence of many worlds. Why is it better? I would say there is no generally accepted interpretation of quantum theory (the mere existence of ... 11 A theorem of von Neumann says that it doesn't make a bit of difference whether you model the cat (or anything else along the causal chain between closing the box and opening it to observe the cat) as capable of collapsing the wave function. You'll make exactly the same testable predictions no matter where along the way you place the collapse. (The argument ... 10 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 \}$is an orthonormal ... 10 I agree in full with Marty Green except the explanations of chemistry in which I was unable to follow so well (that doesn't say that I disagree with them). But, let me put the things in short. The collapse is a phenomenon that is supposed to occur when a quantum object comes in contact with a quantum system. For instance, a quantum particle falls on a beam-... 10 No, the detector is not always collapsing the state. When the particle is in an undecayed state its wave function is physically localised with a vanishingly small amplitude in the region of the detector, so the detector doesn't interact with it and isn't 'always' measuring it. It is only when the particle's state evolves to the point at which it has a ... 9 Both are correct, actually. If you measure an observable for that wave function you'll either find the eigenvalue corresponding to state 1 with probability$|c_1|^2$(similarly for state 2), subject to the condition$|c_1|^2 + |c_2|^2 = 1\$. Edit: What Griffiths is saying is that before you perform the measurement, the particle is neither in state 1 or 2, ...

9

Quantum mechanics was developed in order to match experimental data. The seemingly very weird idea that some observables do not have a definite value before their measurement is not something physicists have been actively promoting, it is something that theoretical considerations followed by many actual experiments have forced them to admit. I don't think ...

9

Welcome to SE -- good question! Decoherence does not mean that there won't be a wavefunction anymore, it just means that if the electron becomes coupled to the surrounding environment, its state will be described by a probabilistic mixture of orbital wavefunctions rather than a (coherent) superposition thereof. The electron in an atom doesn't have some "non-...

9

My take on this is that in the original thought experiment, you don't get to monitor the detector. When the detector detects, it kills the cat. But it doesn't tell you then. You only find out when you open the box. If it tells you immediately, then you know immediately. And then there's the question whether the detector detects 100%. If the Geiger counter ...

9

I just finished a thesis on this subject and I'm happy to share. None of the linked papers are my own. The time of arrival in quantum mechanics is actually a subject of ongoing research. It is certainly a question which begs for an answer, as experiments have been able to measure the distribution of arrival times for decades (see for example Fig. 3 of this ...

8

''No elementary quantum phenomenon is a phenomenon until it is a registered ('observed', 'indelibly recorded') phenomenon, brought to a close' by 'an irreversible act of amplification'.'' (W.A. Miller and J.A. Wheeler, 1983, http://www.worldscientific.com/doi/abs/10.1142/9789812819895_0008 ) A measurement is an influence of a system on a measurement device ...

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