# Can we measure “wavefunction” of quantum particles?

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...)

• Duplicate of physics.stackexchange.com/questions/10240/…? – dmckee --- ex-moderator kitten 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. – user4552 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

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.

• "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? – user4552 May 19 '13 at 2:09
• 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
• ... 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

According to this paper, an experiment was performed that measured the single electron's physical wave function by causing it to interfere with itself. The interference pattern matched the predictions of the Schrodinger equation. So, apparently this was a direct measurement of an electron's wave. Hydrogen Atoms under Magnification: Direct Observation of the Nodal Structure of Stark States A. S. Stodolna, A. Rouzée, F. Lépine, S. Cohen, F. Robicheaux, A. Gijsbertsen, J. H. Jungmann, C. Bordas, and M. J. J. Vrakking Phys. Rev. Lett. 110, 213001 (2013) Published May 20, 2013

I also believe that Bose Einstein Condensates are examples of macroscopic quantum wave functions. Since it occurs when the wave functions of individual atoms "overlap" thus forming a single matter wave which evolves according to Schrodinger's equation. However, I believe this begs the question as to what is actually occurring. If the wave function is physical real but only collapses when there is some "decoherence", then physics needs to explain this mechanism of wave function collapse. Furthermore, tests of BEC's show that the matter waves behave both classically and quantum mechanically based on whether all of the atoms in the condensates are accounted for by the experiment.(http://www.lkb.ens.fr/IMG/pdf/Toulouse-Leggett_April2012-2.pdf)

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.

• The downvote wasn't mine, but this only says how to measure the square of the wavefunction, not the wavefunction itself. – user4552 May 19 '13 at 3:46
• It is still a measurement of the wave function, it is just not a complete measurement as the phase is lost. – Virgo May 19 '13 at 9:58
• Just to add, by measuring in a different basis you can also measure the phase. However, of course with one measurement you can determine almost nothing about ψ at all. – doublefelix Jun 21 '17 at 0:37