# wavefunction collapse and uncertainty principle

1. We all know that wavefunction collapse when it is observed. Uncertainty principle states that $\sigma_x \sigma_p \geq \frac {\hbar}{2}$. When wavefunction collapse, doesn't $\sigma_x$ become $0$?, as we will know the location of the particle. Or does standard deviation just become smaller?

2. After collapse occurs, what happens to the particle? Does the particle resurrect into a wavefunction form?

3. 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....)

4. What happens to the energy of a particle/wave packet after the collapse?

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" .

• What does this deviation in position show when the position of a particle is measured? I can't understand why this spread is observed. It is like the electron is measured many times but this can't be true. Thanks in advance. – Antonios Sarikas Dec 13 '19 at 21:57
• When one measures something, there is always an experimental error. If one makes many measurements one can have a distribution of experimental errors. The uncertainty principle says that there are variables coupled quantum mechanically, in this case momentum and position, that introduce and inherent deviation in the "error" curve. from the simple statistical errors. No matter how accurate one makes the instrument there will be Δx associate with it, dependent on the accuracy Δp of the momentum measurement. – anna v Dec 14 '19 at 4:43