The terminology of collapse of the wavefunction is an unfortunate one .
Take an oscillating AC line and use a scope to measure it and display it. Is the AC 50 herz wavefunction collapsed because we observe it on the scope?
The AC wave function is just a mathematical description of the voltage and current on the line and allows us to calculate the amplitude and time dependance of the energy it carries.
An equally unfortunate concept is the matter wave. The particle is not a continuous soup distributing its matter in space and time the way of an AC voltage or other classical wave. You will never find 1/28th of a particle, it is either there in your measuring instruments or it is not, and it is governed by a probability wave mathematical description, not a "matter wave"
Even more so, the wavefunction manifestation of a particle does not collapse when we measure it the way a balloon collapses when pierced by a pin, because it is just a mathematical description of the probability to find a particle in a particular (x,y,z) with a particular (p_x,p_y,p_z) within the constraints of the Heisenber Uncertainty Principle.
When wavefunction collapse, doesn't σx become 0?, as we will know the location of the particle. Or does standard deviation just become smaller?
We know the location of the particle at that specific coordinate where we had our measuring instrument with the specifice momentum that our insturments measured, within the instrument errors. The probability of finding it there after the fact is 1. It is the nature of all probability distributions that after the detection they become one. example: the probability I will die in the next ten years is 50%. At the instant of my death the probability is one that I am dead.
σx is not a standard deviation in the error sense.
σxσp≥ℏ2 says that: if I want to know the location of my particle within a region about the x point with uncertainty/accuracy σx , the σp I can measure simultaneously is constrained to be within an uncertainty that follows the constraint σxσp≥ℏ2.
Does the particle resurrect into a wavefunction form?
The particle keeps it dual nature of particle or probability wave according to the momentum it still carries and will be appropriately detected as a particle or a probability wave by the next experimenter. It is not a balloon to have been destroyed by the measurement.
What can be an observer that triggers wavefunction collapse? (electron wavefunction does not collapse when meeting with electrons; but some macroscopic objects seem to become observers....)
In principle, any interaction of a particle that changes its momentum and position is an observer except that some interactions are quantum mechanical because of the HUP and the nature of the interaction and some are macroscopic manifestations in our instruments of the passage of a particle or probability wave of a particle. We usually call observers the classical macroscopic detectors, be they people or instruments. At the microcosm quantum level we have interactions governed by the probability wave functions.
What happens to the energy of a particle/wave packet after the collapse?
Energy and momentum are conserved absolutely, so it will depend on what sort of detection of the particle took place. Some will be carried off by the particle if it has not been absorbed into the detector, as for example these particles in this bubble chamber photograph which continually interact with the transparent liquid of the bubble chamber. In this case a tiny bit of the energy is taken by kicked off electrons (first detector atom of liquid, final detector photographic plate) which show by the ionisation the passage of the particle, which is certainly not idiotically "collapsing" .
Nick,
Don't be surprised that this is confusing. There are a lot of concepts intermixed in the discussion of the uncertainty principle that are frequently not clearly understood and are intertwined unintentionally.
Although one often sees that these are stated in statistical terms, the standard deviation does not directly require multiple observations of a sample to understand. Traditional statistics does rely upon repeated sampling in order to develop a standard deviation, however in quantum mechanics the idea is more closely associated with properties associated with the Fourier transform.
To understand the Fourier transform one must first understand what a Fourier series is. The hyperlink will take you to a discussion about the Fourier series as it relates to sound. Starting at about minute two you see a representation of a saw-tooth like wave form. When they show you in the video how the saw-tooth like wave has many components, those components are determined by performing a Fourier transform. In many cases, they transform time series functions into frequency functions (which is directly proportional to energy) but the transform is also applicable to situations where one is transforming position into momentum.
Essentially what happens, is that if one wants to have complete certainty in the value of momentum (or energy), one must look at the entire position (or time) spectrum. In other words, a definite position, when transformed into the momentum domain, requires the entire momentum domain. If one allows a little uncertainty in the position, one does not require the entire momentum domain.
This relationship can be well defined as it relates to Fourier Transforms. This is the real source of the uncertainty principle, and does not require a statistical interpretation to understand.
Best Answer
The argument that one can only detect positions precisely using high-energy photons is very often used to justify the uncertainty principle but it its not its ultimate justification. The argument's place in physics is curious: it essentially states that, even if quantum mechanics were wrong, these kind of observer effects would also prevent us from measuring position and momentum simultaneously.
The real ground the uncertainty principle stands on is de Broglie's postulate: particles of matter are also waves (of some sort) and these waves have wavelength $$\lambda=\frac hp.$$ If you substitute this expression for the particle's momentum into the uncertainty principle, you get it in the form $\Delta x \cdot\Delta\frac1\lambda\geq\frac1{2\pi}$, or in terms of the wavenumber $k=2\pi/\lambda$, $$\Delta x \cdot\Delta k\geq1.$$ This now expresses a fundamental fact of wave physics: you can have a localized wavepacket, but you can only localize it to - roughly - a region the size of the wavelength. More precisely, the more you want to localize a wavepacket, the more waves with different wavelengths you will need to superpose to create it.