Can particles at rest have wave nature?

Question

Can particles have wave nature even when they are at rest? I think this is possible due to the formation of standing waves

Answer 1

In quantum mechanics, particles cannot be at absolute rest due to Heisenberg's uncertainly principle. A quantum mechanical particle is neither a classical wave nor a classical particle. The question should be whether it manifests its particle nature or wave nature and that depends on how you probe it.

Answer 2

I think the previous answers transpose the classical concept of "particle at rest" much too literally onto the quantum mechanical domain.

If we conceive of a "quantum mechanical particle at rest" as one that

a) is conceptually equivalent to a classical point particle and therefore is entirely localized at a single point in space, $\langle x\rangle = x_0$ and $\langle \Delta x^2 \rangle = 0$,

and

b) has precisely null momentum relative to the considered frame, $\langle p \rangle = 0$ and $\langle \Delta p^2 \rangle = 0$,

then we obviously run into immediate conflict with the Uncertainty Principle, as the other answers already noticed ($\langle \Delta x^2 \rangle \langle \Delta p^2 \rangle = 0 < \hbar$).

If however we dispense with the "point particle" specification and remove the 1st statement as completely inapplicable in quantum context, we can relax the 2nd statement above to the following acceptable definition in terms of averages:

"A free particle is said to be at rest in an inertial frame where its average relative momentum is null".

In the non-relativistic limit this means that the average position does not change (free particle!), since $\langle {\vec p} \rangle = 0$ implies $$ \frac{d \langle {\vec x} \rangle}{dt} = \frac{\langle {\vec p} \rangle}{m} = 0 $$ as expected of a particle "at rest". But now the zero momentum "rest state" is by no means unique. The particle can be in any state with null average momentum, and it always complies with the Uncertainty Principle. If the state is a momentum eigenstate, $|\phi_{\vec p = 0}\rangle$, ${\vec p} |\phi_{\vec p = 0}\rangle = 0$, then the particle is completely delocalized, and the probability to locate it at any given position is uniform throughout space. If the state is an arbitrary wave packet of null average momentum, then there will be an uncertainty both in momentum and in position such that $\langle \Delta x^2 \rangle \langle \Delta p^2 \rangle \ge \hbar$. The wave packet is partly localized, but generally spreads out in time, although the average position remains the same.

Note: For relativistic particles defining a well-behaved position operator becomes a problem, but defining a "rest frame" in terms of null average momentum works fine.

Answer 3

Classicly, particles at "rest" have a rest energy due to their rest mass, E = mc^2, and you can associate a frequency f to this energy by E = (hbar)*f as well as an associated wavelength, known as Compton wavelength.

However, Quantum mechanicaly, the notion of a particle "beeing at rest" is almost meaningless, by uncertainity principle. In more detail, a particle's-at-rest wavefunction would be a Dirac delta function, which, by Fourier analysis, is an integral (or a superposition, if you like) of momentum eigen-states, each one contributing with the same amplitude and the integral is over the whole momentum space.

So, yes, theoreticaly particles at rest do have a wave nature, as a superposistion of equaly-contributing momentum eigenstates (exp(ikx)) over the whole range of momentum. But you see that such a case introduces an infinite uncertainity to the momentum of the particle.

Answer 4

In quantum mechanics, we encounter the concept of a particle "at rest" in the study of the Dirac equation (when we transform to the electron's rest frame). For simplicity, let's consider a single spatial dimension. The wavefunction of a particle "at rest" may be given by $\Psi(x,t)=Ae^{-\frac{iEt}{\hbar}}$. The particle has zero velocity (or, more aptly, zero momentum): $\hat p\Psi = -i\hbar\frac{\partial\Psi}{\partial x}=-i\hbar(0)=0$. It has a total energy equal to its rest energy $E=mc^2$. The particle is located at position $x$: $\hat x\Psi=xAe^{-\frac{iEt}{\hbar}}=x\Psi$, where $-\infty < x < +\infty$.

The probability of finding the particle at any position is uniform. Clearly, the wavefunction is not normalizable: $\int_{-\infty}^{\infty}|\Psi(x)|^2dx=A^2\int_{-\infty}^{\infty}dx \rightarrow \infty$. Therefore, it appears, at first glance, that we cannot, with certainty, even construct a wavefunction which would correspond to a single particle "at rest" (more generally, this would be a free particle with any constant momentum $p$). However, this is not true. In this particular case, the apparent failure has do with the fact that we have chosen coordinates (Cartesian) with infinite extent in one spatial direction.

We can do better by wrapping the configuration space around a finite ring or circle $S^1$ of radius $R$. For example, a free particle in a ring of radius $R$ in the ground state $n=0$ has a wavefunction $\Psi(\theta,t)=\frac{1}{\sqrt{2\pi}}e^{-i\omega t}$. This by contrast is normalizable: $\int_{0}^{2\pi}|\Psi(\theta)|^2d\theta=\left. \frac{\theta}{2\pi} \right|^{2\pi}_0=1$. Note that the particle is "at rest": $\hat L_z\Psi=-i\hbar\frac{\partial\Psi}{\partial\theta}=-i\hbar(0)=0$. Finding the particle at any position $\theta$ in the ring is equiprobable.

We qualify a particle is "at rest" by saying that it is located at a position $q$ with a velocity (or momentum) $p=0$. In quantum mechanics, such statements tend to raise alarm bells, as operators associated with the position and momentum observables do not commute $[\hat p,\hat q]\neq 0$, hence, these quantities cannot be measured simultaneously with arbitrary precision.

However, the fact that we do not know the precise position of the particle prior to a measurement does not mean that we cannot say it is "at rest." For such a system (as above), every measurement in which we locate the particle at position $q$ will correspond to a state of definite momentum $p=0$. It's just that the statistical distribution of position measurements will be evenly distributed, corresponding exactly to the norm-squared of the wavefunction.