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We know that there is uncertainty principle, so question: can we ever measure wavefunction of particles? I do not think this is possible, but I am not sure. I guess that everything is probabilistic. (that's why, I believe, we have sigma level when we say we discovered some particles...)

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Duplicate of physics.stackexchange.com/questions/10240/…? –  dmckee May 18 '13 at 23:49
    
The wavefunction itself is not physical, its modulus squared is (as a probability density function). So we cannot measure the wavefunction directly. And to be clear: the level of confidence we have in the discovery of new particles is not related to the uncertainty principle or the wavefunction of the particle, it is a purely statistical notion about the experiment. –  Wouter May 19 '13 at 0:05
    
For a photon, the wavefunction is simply the electromagnetic field, which is measurable. –  Ben Crowell May 19 '13 at 2:29
    
There has been some interesting recent developments in this question, which are summarized in this article: physicsworld.com/cws/article/news/2011/jun/15/… To be clear, in this experiment you still need many copies of the system to gather statistics, but the observables directly correspond to the real and imaginary parts of the wavefunction. –  Rococo May 19 '13 at 3:04

2 Answers 2

If you are able to produce multiple copies of the same pure quantum state, then it is possible to reconstruct the wavefunction. In that case, you need a relatively precise experiment, as just measuring the position and building a histogram will only give you the mod-squared of the wavefunction. To get some information of the phase, you might try measuring the momentum distribution, which tells you about the mod-squared of the wavefunction's Fourier transform, but that comes up a bit short; to get the full quantum state you need to perform a state tomography protocol.

If you only have a single particle, getting the quantum state is impossible, because it contains more information than you can get in any one measurement and that one measurement will destroy the state. Even for a qubit, where only two states (and coherent superpositions of them) are allowed, any tomography protocol needs at least two measurements, with a large number of repetitions determined by the required precision.

Now, as to whether you can "measure" the "wavefunction", it is still a complicated open problem in quantum foundations. Operationally, in a tomographical protocol what you're doing is essentially diagnosing your preparation procedure, and there's no one right now that even pretends to know for sure whether the quantum state is in fact a physical quantity, or whether it is an epistemic thing related to our statistical knowledge of quantum systems.

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"If you are able to produce multiple copies of the same pure quantum state..." How do you do that without violating the no-cloning theorem? –  Ben Crowell May 19 '13 at 2:09
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The no-cloning theorem prohibits the copying of an unknown quantum state. Making multiple copies of the same state is required for many experimental systems. If I want a spin-up particle, and I build a system that always generates spin-up particles, then I have made multiple copies of the same state. –  emarti May 19 '13 at 6:20
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... which is why, operationally, tomography should really be seen as a diagnosis of the preparation procedure you use to prepare multiple particles in the same state. –  Emilio Pisanty May 19 '13 at 12:42
    
Another example: You could start out with N copies of a known state and time-evolve each under the same unknown Hamiltonian. That would give you N copies of the same unknown state. You could then proceed to measure the heck out of that ensemble and try to estimate the Hamiltonian. –  Siva May 20 '13 at 19:46

You can measure the amplitude squared of the wave function if you have many copies of the system. You can then make a histogram of the recorded observations of the systems. This histogram will tend to the amplitude squared of the wave function as the number of copies tends to infinity.

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The downvote wasn't mine, but this only says how to measure the square of the wavefunction, not the wavefunction itself. –  Ben Crowell May 19 '13 at 3:46
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It is still a measurement of the wave function, it is just not a complete measurement as the phase is lost. –  physicsphile May 19 '13 at 9:58

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