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When we measure an electron's position we know that the wave function $\psi$ peaks at the measured position and the wave function as a function of momentum is a harmonic function.

When it makes the transition from the state $\psi$ that it was in before the measurement to the state $\psi'$ which is a peak after the measurement, does it transition in a smooth (but rapid) way or does it make the transition instantaneously?

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The transition is due to the interaction of your small system that you want to experiment on with the environment (including the measurement apparatus). If the interaction is brief and the environment part of the system macroscopic, then what you see is the transition from the initial total wave function $\psi(t_0) = \psi_{\mathrm{in}} = \psi_{\mathrm{in,sys}} \otimes \psi_{\mathrm{in,env}}$ to $$\psi(t_1) = U(t_1,t_0) \, \psi_{\mathrm{in}} = \psi_{\mathrm{out}} \approx \psi_{\mathrm{out,sys}} \otimes \psi_{\mathrm{out,env}}$$ The fact that the total wave function is initially a product is an assumption you have to make (but physically that just means there is no entanglement between the small system and the environment). The outgoing state is just the time-evolved state with respect to the total Hamiltonian $$H(t) = H_{\mathrm{sys}} \otimes \mathbb{1} + \mathbb{1} \otimes H_{\mathrm{env}} + \varepsilon \, W(t),$$ and I have denoted the associated time evolution from $t_0$ to $t_1$ with $U(t_1,t_0)$. If the interaction $W(t)$ between system and environment is weak, i. e. $\varepsilon$ is small, then in specific situations you can show that the outgoing state is still an approximate product state. That means the evolution of the total wave function is still perfectly continuous and it satisfies the Schrödinger equation, $$\mathrm{i} \partial_t \psi(t) = H(t) \psi(t), \qquad \qquad \psi(t_0) = \psi_{\mathrm{in}},$$ but the evolution of the factor $\psi_{\mathrm{in}} \to \psi_{\mathrm{out}}$ need not be because you are tracing out the environment.

As an application, think of a hydrogen atom interacting with the quantized radiation field (i. e. photons): absorbing or emitting photons can cause level jumps, and if you were to look at the electronic wave function in the hydrogen atom, it would seem as if the evolution is sudden and discontinuous (indeed, it seems as if energy is lost or gained). But of course you must not forget about the photons that facilitated the transition — and the “missing” energy is carried away or provided by these photons.

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The following is a bit oversimplified but I think it conveys the basic idea of what's going on. Let's say the state of the system and the measuring apparatus before the measurement is $$ (|a\rangle_e+|b\rangle_e)|0\rangle_M, $$ where the $e$ subscript refers to the state space of the electron, and $M$ is the apparatus state space.

To measure the state of the electron in the basis in which it is written above you have to induce an evolution that has the following effect: $$ |i\rangle_e|0\rangle_M\to|i\rangle_e|i\rangle_M, $$ and the evolution operator is $$ U_m = \sum_i |i\rangle_e|i\rangle_{M}\langle i|_M\langle 0| +\dots $$ The dots refer to states in which the measuring apparatus is not in the $|0\rangle_M$ at the start, so we can ignore them. This can't happen instantly, so the evolution operator is something like $$ U(t) = a(t)I+b(t)U_m $$ where $a(0)=1,b(0)=0,a(\tau)=0,b(\tau) =1$, where $\tau$ is the time taken for the measurement.

When the measurement is concluded the state is now $$ |a\rangle_e|a\rangle_M+|b\rangle_e|b\rangle_M. $$ At this point it is customary to say that the state somehow becomes one or the other of the terms in this superposition. But this is entirely unnecessary. As long as the measurement is not undone, the two terms can't interfere, so if you are in one of the terms you can't interact with the other term. See the literature on decoherence for more details and more complicated models. You can sometimes tell that the other term must exist even after a measurement: this is part of the explanation for the EPR experiment, see also

http://arxiv.org/abs/1109.6223.

The account I gave is commonly called the many worlds interpretation of quantum mechanics and for reasons that are not very clear it is controversial. Many physicists want to modify quantum mechanics to get rid of all but one of the terms, but this ruins many explanations such as the explanation of the EPR experiment. Other physicists obfuscate the issue of whether the processes described happen in reality. This latter strategy does no good. If the above process doesn't describe what's really happening, then quantum theory will have to be replaced by a theory that does give a description of what's actually happening. If the above process does describe reality, then it also does no good to waffle obscurely about it not being real. So the other "interpretations" either contradict the existing theory by wanting to change the equations of motion to get rid of the other terms, or they waffle vaguely about non-reality of the only available explanation. In either case they are not interpretations of quantum mechanics, they are either alternate theories or vague philosophical muttering. So a better description of the account I gave is "following the implications of quantum mechanics in a consistent manner rather than fudging".

To summarise, there is a continuous transition from a state in which the electron is peaked in a couple of different places and could undergo interference, to a state in which there are two non-interacting versions of the electron and the measuring apparatus, where each version has a record of the electron in one of those places.

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  • $\begingroup$ Nice answer, but I would leave out "Single particle theories don't really work for reasons discussed in lots of books on quantum field theory. A very clear discussion is given in "Quantum Field Theory for the Gifted Amateur" by Lancaster and Blundell Chapter 8." Also I think the "Me" in your evolution operator equation should just be "e". $\endgroup$ – Virgo Jul 11 '15 at 14:52
  • $\begingroup$ "for reasons that are not very clear it is controversial." some reasons: quantum formalism is applied to macroscopic measurement apparatus as well and it is not clear how this could be consistent with experimental practice, where results of measurement are obtained and are definite. The attempts to solve this problem by many worlds hypothesis is a red flag to many. $\endgroup$ – Ján Lalinský Oct 25 '15 at 10:47
  • $\begingroup$ Quantum mechanics predicts that any particular version of you will see only one outcome, so that's consistent with experimental results. And saying something is a red flag without explaining why doesn't make the issue clearer. $\endgroup$ – alanf Oct 26 '15 at 20:31
  • $\begingroup$ "Quantum mechanics predicts that any particular version of you will see only one outcome" I do not think equations of quantum theory imply that, because there are no preferred states and no hint of possible existence of "versions of observers" in the t-d Schr. eq. Also, there is no experience of different simultaneous outcomes of one experiment. $\endgroup$ – Ján Lalinský Nov 24 '15 at 21:05
  • $\begingroup$ Many different outcomes is just an interpretation of the superposition, to make sense of "a+b" when a,b are considered to be physical states. It is particularly bad one, since decomposition of any $\psi$ function into components of some orthogonal system is completely arbitrary - it makes no difference to value of $\psi$ and probability densities in configuration space it gives. $\endgroup$ – Ján Lalinský Nov 24 '15 at 21:07

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