Let a (free) particle move in $[0,a]$ with cyclic boundary condition $\psi(0)=\psi(a)$.

The solution of the Schrödinger-equation can be put in the form of a plane wave. In this state the standard deviation of momentum is $0$, but $\sigma_x$ must be finite. So we find that $\sigma_x\sigma_p=0$. Is something wrong with the uncertainty principle?

# [Physics] How to solve this quantum mechanical “paradox”

heisenberg-uncertainty-principlemomentumquantum mechanicswavefunction

#### Related Solutions

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" .

The precise, mathematical statement of the uncertainty principle is $\sigma^2_x \sigma^2_k \geq 1/4$. The use of deltas is just an informal way of talking about it. Nevertheless, it's pretty common to say, for instance, that the width of a peak is either the standard deviation or some quantity proportional to it--see, for example, full width at half maximum, which ends up being about $2.35\sigma$.

I'm not really sure what a slit would look like in 1 dimension. It's easier for me to consider a particle in a 1d infinite square well. Note that the infinite well absolutely forbids any leakage of the particle into the forbidden region, just like the classical case. In this case, the variances depend on the energy of the particle. For a particle in one of the $n$th energy eigenstate of an infinite well with width $L$, the variances are (per wikipedia)

$$\sigma^2_x = \frac{L^2}{12} \Bigg ( 1 - \frac{6}{n^2 \pi^2} \Bigg), \quad \sigma_k^2 = \frac{n^2 \pi^2}{L^2}$$

The product of the variances is then $\sigma_x^2 \sigma_k^2 = (n^2 \pi^2/12 - 1/2)$. For $n=1$, this is about $.322 \geq .25$, as required.

You can't really see what the uncertainties will be by inspection, by the geometry of the problem. These are the uncertainties for energy eigenstates, and there's no reason to expect that a particle will be in an eigenstate (which would then make the computation more complicated).

Really, one just calculates the variances of the wavefunction with respect to $x$ and $k$. You *might* be able to get a rough idea from the quantities in the problem (for instance, the standard deviation with respect to $x$ is indeed proportional to $L$, but only proportional, not exactly $L$), but that's all.

You ask if a particle has nonzero probability of existing everywhere. To be pedantic, a particle has zero probability of existing at any specific point, but it typically has a nonzero probability of existing in a region of any finite size. This infinite square well is an exception, as the infinite potential around the box absolutely forbids particles.

Uncertainty really is just a loose, loose word to use. It almost always really means standard deviation of the wavefunction.

## Best Answer

This is what happens if one cares not for the subtlety that quantum mechanical operators are typically only defined on subspaces of the full Hilbert space.

Let's set $a=1$ for convenience. The operator $p =-\mathrm{i}\hbar\partial_x$ acting on wavefunctions with periodic boundary conditions defined on $D(p) = \{\psi\in L^2([0,1])\mid \psi(0)=\psi(1)\land \psi'\in L^2([0,1])\}$ is self-adjoint, that is, on the domain of definition of $p$, we have $p=p^\dagger$, and $p^\dagger$ admits the same domain of definition. The self-adjointness of $p$ follows from the periodic boundary conditions killing the surface terms that appear in the $L^2$ inner product $$\langle \phi,p\psi\rangle - \langle p^\dagger \phi,\psi\rangle = \int\overline{\phi(x)}\mathrm{i}\hbar\partial_x\psi(x) - \overline{\mathrm{i}\hbar\partial_x\phi(x)}\psi(x) = 0$$ for every $\psi\in D(p)$ and every $\phi\in D(p^\dagger) = D(p)$, but not for $\phi$ with $\phi(0)\neq\phi(1)$.

Now, for the question of the commutator: the multiplication operator $x$ is defined on the entire Hilbert space, since for $\psi\in L^2([0,1])$ $x\psi$ is also square-integrable. For the product of two operators $A,B$, we have the rule $$ D(AB) = \{\psi\in D(B)\mid B\psi\in D(A)\}$$ and $$ D(A+B) = D(A)\cap D(B)$$ so we obtain \begin{align} D(px) & = \{\psi\in L^2([0,1])\mid x\psi\in D(p)\} \\ D(xp) & = D(p) \end{align} and $x\psi\in D(p)$ means $0\cdot \psi(0) = 1\cdot\psi(1)$, that is, $\psi(1) = 0$. Hence we have $$ D(px) = \{\psi\in L^2([0,1])\mid \psi'\in L^2([0,1]) \land \psi(1) = 0\}$$ and finally $$ D([x,p]) = D(xp)\cap D(px) = \{\psi\in L^2([0,1])\mid \psi'\in L^2([0,1])\land \psi(0)=\psi(1) = 0\}$$ meaning the plane waves $\psi_{p_0}$ do

notbelong to the domain of definition of the commutator $[x,p]$ and you cannot apply the naive uncertainty principle to them. However, for self-adjoint operators $A,B$, you may rewrite the uncertainty principle as $$ \sigma_\psi(A)\sigma_\psi(B)\geq \frac{1}{2} \lvert \langle \psi,\mathrm{i}[A,B]\rangle\psi\rvert = \frac{1}{2}\lvert\mathrm{i}\left(\langle A\psi,B\psi\rangle - \langle B\psi,A\psi\rangle\right)\rvert$$ where the r.h.s. and l.h.s. are now both defined on $D(A)\cap D(B)$. Applying this version to the plane waves yields no contradiction.