Why do we need wave packet? [closed]

Question

My question is that why we need wave packet to describe the free particle states? Will this packet decompose after propagating for a period of time?

Answer 1

If you simply consider the math problem represented by the free Schrödinger equation $$ -\frac{\hbar^2}{2m} \frac{\partial^2}{\partial x^2} \Psi(x,t) = i \hbar \frac{\partial}{\partial t} \Psi(x,t) \;,$$ then the simplest solution is $$ \Psi_\text{math}(x,t) = A \exp \left[ i \left( kx - \omega t \right) \right]\;, $$ for $A$ some amplitude still to be determined and $\omega$ and $k$ arbitrary constants. However, if this is to represent a particle of mass $m$ and (kinetic) energy $E$, we must also have $\omega = E/\hbar$ and $k = p/\hbar = \sqrt{2mE}/\hbar = \sqrt{2m\omega/\hbar}$. We can re-write the above as $$ \Psi_\text{pw}(x,t;\omega) = A \exp \left[ i \left( \sqrt{\frac{2m\omega}{\hbar}}x - \omega t \right) \right] \;, \tag{1} $$ where "pw" is intended to indicate "plane wave".

However, physical wave-functions must meet some requirements on continuity (which this does) and normalizability. The normalization requirement is $$ \int_\text{all space} \Psi^* \Psi \, \mathrm{d}x = 1 \;, $$ which (1) does not respect.

So that "simplest" solution to the math problem does not represent a physical state of affairs.

This is where the notion of a wave-packet enters. If we define some new states $$ \Psi_\text{wp}(x,t) = \int_\text{all frequencies} A(\omega) \Psi_\text{pw}(x,t;\omega) \,\mathrm{d}\omega \tag{2}$$ (in which "wp" stands for "wave-packet") for a suitable choice(s) of envelope function $A(\omega)$ we find that (2) can be normalized (and can therefore represent a physical state).

BTW, we know that (2) solves the Schrödinger equation without testing it explicitly because it is a linear combination of solutions.

That said, it is a combination of states with different $\omega$ and therefore with different kinetic energy. But they all have the same mass, which means that the contributions to the wave-packet move a different speeds and as a result the wave-packet spreads out over time.

So, to finally answer your questions

1. A wave-packet is necessary because a plane wave is not normalizable.

   (Or to frame it in terms of the Heisenberg principle: knowing the momentum exactly precludes you from have any information about the position).

2. The wave-packet does not "decompose" in the sense of ceasing to exist, but it does spread out in space meaning that however good your original knowledge of the particle's position was it grows worse in time.

   (Framed in Heisenberg terms: because you start knowing something about it's position you no longer have perfect information about momemtum and thus can't predict it's future position with precision.)

Answer 2

Let $\phi(x) = \langle x | \phi\rangle = e^{-ikx}$ be the representation of a solution of the free particle Schrödinger equation $$ i\hbar \frac{\partial}{\partial t}|\phi\rangle = \hat{p}^2|\psi\rangle. $$ Since the above equation is linear, one can show that the expression $$ \psi(x) = \langle x | \psi\rangle = \int dk\, a(k)\,e^{-ikx} $$ is also a solution; in particular the latter is normalisable in the sense of $L_2$ norm on the real line, while the former is not (although it is in the dual space): in the literature the latter state is referred to as wave packet, the former as plain wave. As such the wave packets naturally arise simply as (normalisable) solution to the Schrödinger equation: compared to the plane wave, they are the actual states that describe physical particles. Very often in the literature all mathematical properties are proven for plain waves only for the sake of simplicity: the argument that authors use is that whichever property holds for the plane wave, it must hold the same for a linear superposition thereof, the equation of motion of quantum mechanics being linear in $|\psi\rangle$. It is worth noticing en passant that wave packets possess a few good properties, in particular their classical limits approximation for expected value of position and momentum obey Newton's-like equations of motion.

Now equipped with the wave packet state $|\psi\rangle$ the dispersion on its position is given, by definition, by the expectation value of the standard deviation of the position operator upon such a state, namely $$ (\Delta x)^2 = \langle \hat{x}^2 \rangle - (\langle \hat{x} \rangle)^2 $$ where the expectation value is intended to be taken against the state represented by $|\psi\rangle$, namely $\langle \hat{A} \rangle = \langle \psi | \hat{A} | \psi\rangle$. One can show (a result that undergoes the name of Ehrenfest theorem) that the expectation value of an operator must satisfy the Schrödinger-like equation of motion $$ i\hbar \frac{\partial}{\partial t}\langle \hat{A}\rangle = \langle [\hat{A}, \hat{H}]\rangle + i\hbar \langle \frac{\partial \hat{A}}{\partial t}\rangle $$ Plugging the expression of $(\Delta x)^2$ into the above one can show that it assumes the form of a second order differential equation$^1$, for general states; as such, the dispersion $(\Delta x)^2$ depends on the time. In general it happens to be the case that $$(\Delta x)^2(t)\geq (\Delta x)^2(t=0)$$ and one refers to said behaviour as to spreading of the wave packet.

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$^1$ explicit calculations can be found in standard quantum mechanics textbooks, like the ones by Messiah or Cohen-Tannoudji.

Answer 3

In quantum field theory, there are posited particle fields for each elementary particle entering the theory. An electron field etc as in the table. These fields are represented by the plane wave function of the solution of the corresponding quantum mechanical equation, and as plane waves they are not localized in space. On these fields quantum mechanical creation and annithilation operators act, and thus a particle propagates and interactions can be calculated using Feynman diagrams.

For real particles interacting with the apparatus one has to model them with a wave packet around a central wavelength, in order for the Heisenberg uncertainty to be fulfilled. The spreading of wavepackets should be answered by a theorist but afaik the relativistic case can have non spreading wavepackets.