How does an initial Gaussian wave packet of the form $\psi(x,t=0) = e^{-ax^2} \cdot e^{\imath k_{o}x}$ arise?

Question

The concept of a wave packet is universally introduced in introductory QM classics in order to "localize" a particle in free space (i.e. $V(x) = 0$ in the 1-Dimensional Schrodinger Eq.). Many textbooks use the example of the Gaussian wave packet most often written as $$\psi(x,t=0) = e^{-ax^2} \cdot e^{\imath k_{o} x} \tag{1} $$ as the initial wave packet at time $t=0$ with $a$ and $k_{o}$ being constants. I have omitted the normalization constant for brevity. However, if Equation (1) is placed into the Schrodinger Equation for a free particle $$-\frac{\hbar^2}{2m} \frac{d^2}{dx^2}\psi = E \psi$$ a real and imaginary part is obtained, after reducing and dividing by $e^{-ax^2} \cdot e^{-\imath k_{o}x}$ on both sides of the equation, with the resulting equations $$-\frac{\hbar^2}{2m} (4a^2x + 2a - k_{o}^2) = E \tag{2}$$ and $$\imath \cdot \frac{\hbar^2}{2m}(4axk_{o})=0. \tag{3}$$ In order for the Eq. 3 - the imaginary part - to be satisfied for all $x$, either $a=0$ or $k_{o}=0$. If it is the former, then there is no Gaussian envelope to shape the term $e^{\imath k_{o}x}$. On the other hand, if $k_{o}=0$, and keeping in mind that $a>0$ for all $x$, Eq.(2) mandates that $E<0$ for $x>0$ and $E<0$ for $x>0$ which seems odd. Furthermore, as $x \rightarrow \infty$, $E \rightarrow -\infty$ and the reverse must hold for $x \rightarrow -\infty$ with $E \rightarrow \infty.$

Furthermore, what potential $V(x)$ would give rise to such an initial wave packet? If $V(x)$ is placed back into the Schrodinger Eq., it arises within the real part of the resulting equations with Eq. 2 now given as $$-\frac{\hbar^2}{2m} (4a^2 x + 2a - k_{o}^2) + V(x) = E. \tag{4}$$ In order to keep $E$ finite, then $V(x)$ would have to approach $\infty$ for $x \rightarrow \infty$ and $V(x)$ approaching $- \infty$ for $x \rightarrow -\infty.$

So, in summary, how can this type of wave packet exist to begin with since it appears that it cannot come about in free space and comes about only for a specific potential. Yet, it is used often and under different variation as the beginning wave packet for multiple homework problems with almost any potential. Even if one where to say that the particle is localized prior to being placed into free space or into a given potential, how did it come to be localized in the first place outside potentials given by Eq. 4?

Answer 1

1. The solutions to the time-independent free Schrödinger equation $$ -\partial_x^2 \psi = E\psi$$ are the plane waves $\psi_p(x) = \mathrm{e}^{\mathrm{i}px}$. The Gaußian wavefunctions $$ \psi_{k_0,\sigma}(x) = \mathrm{e}^{\mathrm{i}(-x^2 + \mathrm{i}k_0x)/2\sigma^2}$$ are not solutions to this equation; the observation in the question that demanding that $\psi_{k_0}(x)$ solve the time-independent Schrödinger equation makes no sense and leads to contradictions is hence correct, but entirely irrelevant: The initial conditions for time evolution are not required to be solutions of the time-independent Schrödinger equation - in fact, if they are, then the time-evolution is trivial since time evolution is obtained as the solution to the time-dependent Schrödinger equation $$ -\partial_x^2\psi(x,t) = \mathrm{i}\partial_t \psi(x,t)$$ and when the initial condition $\psi(x,0)$ solves the time-independent Schrödinger equation for some $E$, then the solution of the time-dependent Schrödinger equation is just $\psi(x,t) = \mathrm{e}^{\mathrm{i}Et}\psi(x,0)$. As the question correctly observes, the time evolution of the initial condition $\psi(x,0) = \psi_{k_0,\sigma}(x)$ is instead one where the wave packet moves with $k_0$ and spreads, so it is consistent that it is not a solution to the time-independent Schrödinger equation.

2. The Gaußian wave packet is not chosen as an example because it would be produced particularly often in nature, although Gaußian beams certainly exist. It is meant to be illustrative of a number of features of quantum mechanical states, while being mathematically simple enough to not overwhelm the student with computational complexity. For example:

   • It has finite position and momentum uncertainty, and saturates the bound of Heisenberg's uncertainty principle, hence it is the "most localized" state when we look at both position and momentum space.

   • Its Fourier transform is again a Gaußian, making it simple to compare its qualities in position and momentum space with each other.

   • It can be observed to "move", and it is simple to check that the expectation values of position and momentum obey Ehrenfest's theorem.

   • For $\epsilon\to 0$, $\psi_{0,\epsilon}$ is a nascent $\delta$-function and hence narrow Gaußian wavepackets can be used to model the outcome of position (or momentum) measurements. Therefore measurements are another "source" for Gaußian wavepackets if you model the outcome of a position measurement with finite uncertainty $\sigma$ by a Gaußian of that width centered at the measured position value.

Answer 2

The simplest example of a Gaussian wave-packet is the (harmonic oscillator) coherent state $\vert\alpha\rangle$, with $\alpha\in \mathbb{C}$. Explicitly: $$ \vert\alpha\rangle = e^{-\vert\alpha\vert^2/2}\sum_{n=0}^\infty \frac{\alpha^n}{\sqrt{n!}} \vert n\rangle\, , \qquad \alpha=x_0+ip_0\, , $$ where $\vert n\rangle$ the $n$’th harmonic oscillator eigenket. In fact $\vert\alpha\rangle$ is just a harmonic oscillator ground state displaced in $x$ by $x_0$ and in $p$ by $p_0$.

It is easily seen that, since $\vert\alpha\rangle$ is a linear combinations of h.o. solutions with different energies, $\vert \alpha\rangle$ is not itself a solution to the time-independent Schrödinger equation. It turns out that $$ \vert\alpha(t)\rangle= e^{-i\omega t/2}\vert e^{-i\omega t}\alpha\rangle \, . $$ Moreover, the coherent state is stable in time, i.e. it does not spread and keeps the same shape for all $t$. Thus the harmonic oscillator potential certainly supports Gaussian states.

Modulo some technical conditions, laser light is often described as a coherent state.

It is possible to construct a Gaussian wavepacket for a large collection of potentials. Let $\vert\psi\rangle$ be the ket for the initial wavepacket, so that $$ \psi(x;a,k_0)=\langle x\vert \psi\rangle = {\cal N}e^{-ax^2}e^{-ik_0x}\, . $$ and simply expand $$ \psi(x;a,k_0)=\sum_n c_n\phi_n(x) $$ where $c_n=\int dx \phi^*_n(x)\psi(x;a,k_0)$. By construction the system at $t=0$ is a Gaussian state but at later $t$ is has evolved to $$ \Psi(x;a,k_0,t)=\sum_n c_n e^{-iE_nt/\hbar}\phi_n(x) $$ which is no longer guaranteed to be Gaussian.

Answer 3

The main reason that Gaussian packet is discussed in quantum texts is because it is easy to analyze mathematically and illustrate some basic concepts. I think the other answers have characterized the packet from this point of view quite well.

The real wave packets are not necessarily Gaussian. However, for a packet centered about a specific value of momentum, i.e., being close to plane wave states but with some momentum uncertainty, Gaussian might be a good approximation. That is, if we assume that the momentum of a particle is distributed according to the Gaussian Law: $$ w(k)=\frac{1}{\sqrt{2\pi\sigma_k^2}}e^{-\frac{(k-k_0)^2}{2\sigma_k^2}}, $$ so that the wave function is $$ \psi(x)=\int dk w(k)e^{ikx}=e^{ik_0x}e^{-\frac{x^2\sigma_k^2}{2}}. $$

Note that this ansatz has nothing to do with how the normal law usually arises in statistics - from adding many contributions, deviating from average by small random values. E.g., light emitted by many atoms may have Gaussian distribution of frequencies, which is referred to as inhomogeneous broadening. However, this broadening is due to the randomness unrelated to quantum mechanics, and such light cannot be described by a wave function (but rather by a density matrix.) A homogeneously broadened light originating from a single atom has Lorentzian distribution.

Answer 4

There are several concepts that are combined here (both in the OP and some of the answers) that can lead to confusion. The OP's example in quantum mechanics is very basic and does not specify the details in the physics.

In general, the solutions of the Schrödinger equation are mutually orthogonal and can be used as a basis to expand any arbitrary initial state. However, such an initial state will then evolve in a nontrivial way as a function of time so that the shape of the initial state is not in general maintained. So, one can use a Gaussian initial function even if the eigenstates are Gaussian shaped functions. One would simply make an expansion in terms of the eigenstates that would sum up to a Gaussian function. When this state is then allowed to evolve one would typically see a broadening of the Gaussian function.

Another aspect that may introduce confusion is the fact that physical systems have different degrees of freedom. Not all these degrees of freedom behave like quantum degrees of freedom associated with the Schrödinger equation.

An example that is often used (which is presented by @ZeroTheHero) is the physical harmonic oscillator. If such a system is treated as a quantum system, then the position and momentum are quantum degrees of freedom. However, in quantum optics, for example, the quantum degrees of freedom are the photon-number degrees of freedom and have nothing to do with position and momentum. So the optical beams in quantum optics can have Gaussian shapes while the particle number degrees of freedom of the quantum optical state could be described by completely different state. It is therefore important to take note of the physical quantum quantities that are involved in the system.

Answer 5

Welcome to the discovery that "math is not equal to physics". In math we can do all kinds of stuff that is hard to do in experimental physics. The trouble with QM 101 is that it introduces you to a way of calculating quantum mechanical systems that is NOT representative of what happens on the bench. When we are measuring the properties of a hydrogen atom, for instance, we are not actually stopping the electrons in them cold to see the shape of the orbitals. Instead we are performing a scattering experiment with light to measure the optical spectrum of hydrogen. The proper theoretical framework for that would be the S-matrix and not the wave function. Physicists realized this very early on, and the scattering approach to QM became "standard" just a few years after the Copenhagen interpretation and the Schroedinger equation were introduced. The trouble with this "upgrade" is that it suffers from a number of non-trivial problems that are not easy to deal with, like IR-divergences even for the most trivial Coulomb scattering problem (that's an effect that exists already in the classical version, so it's not something that one can just ignore). So instead of introducing the beginner to the "real" theory, we are starting off with what I would consider a toy theory. That is, educationally, the right thing to do because most physicists can live with the simplifications of the wave function approach and von Neumann's solution theory for their entire lives. What is not completely right is that they usually don't tell you that either in the lectures or the textbooks. You won't be taught how to work on the "hard" problems until you take a graduate course in quantum field theory... which few physicists do, for all I can tell.

One can, of course, create approximations for Gaussian wave packets in experiments by spatial/spectral shaping. It's usually just not important. How a quantum is localized in space and time does not have any effect on the physical interactions that we are trying to investigate. The experimental physicist will usually optimize experiments for the least errors in one parameter, e.g. time resolution or frequency resolution (location/momentum etc.), i.e. delta functions or plane waves. The Gaussian solution is the mathematically optimal compromise in both variables simultaneously. There are experiments where this is required, but they are less common, at least in my personal experience.