About the time evolution of a wave packet

[KFC]

In some texts of fundamental quantum mechanics, it introduces the wave packet by Fourier transformation of a momentum wave into a spatial version. This is easy to understand because, analogy to the optical wave, a typical beam could compose waves of more than one frequencies. The general form is something like this

$\psi(x) = \dfrac{1}{\sqrt{2\pi}} \int dk \psi(k) \exp(ikx)$

This is quite straightforward. However, I have hard time to understand why the time evolution of a general wave packet is of the following form involving $\omega(k)t$

$\psi(x, t) = \dfrac{1}{\sqrt{2\pi}} \int dk \psi(k) \exp[ikx - i\omega(k)t]$

In all texts I have in hand, above expression is given directly. From math context, it seems that it is trying to tackle the time aspect of $k$ with $\omega(k)t$ but it of above form instead of the following?

$\psi(x, t) = \dfrac{1}{\sqrt{2\pi}} \int dk \psi(k) \exp[ik(t)x]$

I am looking for more physical intuitive explanation of the term $i\omega(k)t$

 

[Twigg]

$kx-\omega (k)t$ is the general form of the phase of a plane wave traveling in the x direction. Notice it's equal to $k(x - v(k)t)$ where $v(k) = \omega (k)/ k$ is the phase velocity. Notice that if k were a function of t then your suggested wavepacket would probably accelerate and disperse, whereas a true wavepacket moves along at a constant group velocity. Also, take a look at the time-dependent schrodinger equation. Try taking $i \hbar \frac{\partial}{\partial t}$ of $exp[i(kx-\omega t)]$.

 

[KFC]

Thanks. This reminds me something learn in fundamental physics about the phase velocity and group velocity. So at the very beginning when we mention that a wave packet is composed of many plane waves of different momentum (or wavenumber), we assume the component of wave packet is plane wave so it's time evolution should involve the way including $kx - \omega(k)t$, which means each individual component should move at a certain speed (phase velocity) $\omega(k)/k$. When you say the wave packet is moving at group velocity... as I remember, group velocity is $\partial\omega/\partial k$, which is given as a constant speed in example of most texts. But in case #\omega(k)# is not linear, group velocity is not a constant and how do you understand group velocity from there? thanks.

 

[KFC]

By taking $i\hbar\partial/\partial t$, I got the $\hbar\omega\exp[i(kx -\omega t)]$, so it has unit of energy? Is that what you mean?

 

[KFC]

By taking $i\hbar\partial/\partial t$, I got the $\hbar\omega\exp[i(kx -\omega t)]$, so it has unit of energy? Is that what you mean?

By the way, I know that in quantum mechanics, we have $E=\hbar\omega$. In some text, I saw they use $E(k)$ instead of $\omega$ in the expression of time evolution. I am quite confusing. Is it always true to do that no matter what explicit form of $E(k)$ will be?

My last question is about Gaussian wave packet. If most cases dealed by the text, $\psi(k)$ is usually taken as Gaussian form. From what physical reason this should be Gaussian? One reason I can think of is that Gaussian give well defined variance of k and mean k, but beside this, any other reason why Gaussian? Is there other possible distribution of k exist in nature?

 

[vanhees71]

Well, the idea is to solve the Schrödinger equation of the free particle via Fourier transformation. I set $\hbar=1$ for simplicity:
$$\mathrm{i} \partial_t \psi(t,\vec{x})=-\frac{\Delta}{2m} \psi(t,\vec{x}).$$
Now you make the ansatz
$$\psi(t,\vec{x})=\int_{\mathbb{R}^3} \frac{\mathrm{d}^3 \vec{p}}{(2 \pi)^3} \tilde{\psi}(t,\vec{k}) \exp(+\mathrm{i} \vec{k} \cdot \vec{x}).$$
Plugging this into the Schrödinger equation and using the uniqueness of Fourier transformations leads to the simpler equation of motion for $\tilde{\psi}$:
$$\mathrm{i} \partial_t \tilde{\psi}(t,\vec{p})=\frac{\vec{p}^2}{2m} \tilde{\psi}(t,\vec{p}).$$
It has the solution, given the initial condition $\tilde{\psi}(t=0,\vec{p})=\tilde{\psi}_0(\vec{p})$
$$\tilde{\psi}(t,\vec{p})=\tilde{\psi}_0(\vec{p}) \exp(-\mathrm{i} E_{\vec{p}} t), \quad E_{\vec{p}}=\frac{\vec{p}^2}{2m}.$$
Plugging this back into the Fourier transformation gives you the answer in position space.

If you have the initial state given in position space, you just need to set
$$\tilde{\psi}_0(\vec{p})=\int_{\mathbb{R}^3} \mathrm{d}^3 \vec{x} \psi_0(\vec{x}) \exp(-\mathrm{i} \vec{p} \cdot \vec{x}).$$

 

[KFC]

Thanks for that. So does it mean the Gaussian solution is only eligible for free particle?

 

[blue_leaf77]

Thanks for that. So does it mean the Gaussian solution is only eligible for free particle?

Since Gaussian waveform is always square integrable, any system can have Gaussian wavefunction and not only restricted to free particle. A good example is the ground state of harmonic oscillator.

 

[KFC]

Ok. I understand vanhees71's explanation from math's point of view. Now if it is said that Gaussian waveform is not limited to free particle and it is confusing me again why Gaussian will be that typical so many physical system will have such waveform. I understand that it is square integrable but many other functional is square integrable as well. I don't understand it from physical point of view.

 

[bhobba]

I don't understand it from physical point of view.

Then you are in for a whole lot of easily avoidable confusion.

Physics, and this is especially true in QM, and very very true in QFT where refusal to accept it leads to all sorts of silly confusion that is particularly hard to dislodge, is written in the language of math. There is no such thing as a math explanation - its all just physics. Sometimes it can be explained in English, but by and large it is math.

Wikiedia explains about wave-packets quite well:
https://en.wikipedia.org/wiki/Wave_packet

Thanks
Bill

 

[blue_leaf77]

From what physical reason this should be Gaussian? One reason I can think of is that Gaussian give well defined variance of k and mean k, but beside this, any other reason why Gaussian?

To me, it's just because Gaussian function is convenient to work with. As far as I know, it's one of a few smooth, localized functions whose Fourier transform can be computed analytically. Another localized function with analytically computable FT is Lorentzian function.

 

[dextercioby]

There's a mathematically compelling argument why Gaussians are popping up all over Quantum Mechanics. They are decreasing at infinity faster than any polynomial, thus are the primal example of Schwartz's test functions. As you know, the space of Schwartz's test functions is member of a rigged Hilbert space S, L², S' which is the right one in the description of the free Galilean particles in QM. With Gaussians you build (by means of a Rodrigues formula) Hermite polynomials, a Hilbert space basis for L² (R).

 

[KFC]

I think those statements are more understandable. I am actually trying to find a reason convincing myself Gaussian has its advantage to be the choice. Thanks.

 

[bhobba]

From another thread:

But that is just one area - it has enriched many areas eg white noise analysis:
http://www.asiapacific-mathnews.com/04/0404/0010_0013.pdf

Of relation to this thread the following is quite interesting:
https://projecteuclid.org/download/pdf_1/euclid.nmj/1118796540

Thanks
Bill

 

[Twigg]

Whoops! I missed all the updates on this thread.

My last question is about Gaussian wave packet. If most cases dealed by the text, ψ(k)ψ(k)\psi(k) is usually taken as Gaussian form. From what physical reason this should be Gaussian? One reason I can think of is that Gaussian give well defined variance of k and mean k, but beside this, any other reason why Gaussian? Is there other possible distribution of k exist in nature?

I think those statements are more understandable. I am actually trying to find a reason convincing myself Gaussian has its advantage to be the choice. Thanks.

The reason for using the Gaussian wavepacket is that is a solution which minimizes the uncertainty in classical phase space: $\Delta x \Delta p = \frac{\hbar}{2}$, which is the limiting case of the Heisenberg principle. To see why this product is the "total" uncertainty, remember classical phase space for motion in one dimension has an x-axis and a p axis, so that expression is like the area of rectangle containing states that can't be distinguished in classical phase space due to quantum uncertainty. The reasoning for the gaussian being minimal uncertainty is partly due to the very interesting fact that the Fourier transform of a gaussian is another gaussian (remember, an FT and a factor of $\hbar$ is how you go from position space to momentum space). The proof of the first claim (gaussian = minimum uncertainty) is shown http://www.colorado.edu/physics/phys2170/phys2170_sp07/downloads/Gaussian.pdf. To see how the Fourier transform of a gaussian in another gaussian, I think that proof is in Griffith's, but I have my copy packed away at the moment. There's a proof here.

On an interesting side note, the other main example of a gaussian wavefunction that came up in this thread, the ground state of the quantum harmonic oscillator, is also a state of minimal uncertainty, a.k.a. a coherent state. Griffith's book definitely has a section on this I would recommend.

One caveat, as far as I know, the gaussian wavepacket is a solution for the free particle with minimal uncertainty, but I have never seen anyone prove that is the only such solution. In fact, IIRC there are other coherent states besides the ground state for the harmonic oscillator (would appreciate verification on that, anyone out there), so I wouldn't be surprised if there were alternatives for the free particle. I'm not saying there definitely are, just that it wouldn't be too surprising.

By taking iℏ∂/∂tiℏ∂/∂ti\hbar\partial/\partial t, I got the ℏωexp[i(kx−ωt)]ℏωexp⁡[i(kx−ωt)]\hbar\omega\exp[i(kx -\omega t)], so it has unit of energy? Is that what you mean?

I just wanted you to notice that the term $exp[i(kx-\omega t)]$ represents modes which each obey the de Broglie relations. As you saw, the Hamiltonian operator gives the energy de Broglie relation. In addition, the momentum operator gives the momentum de Broglie relation.

By the way, I know that in quantum mechanics, we have E=ℏωE=ℏωE=\hbar\omega. In some text, I saw they use E(k)E(k)E(k) instead of ωω\omega in the expression of time evolution. I am quite confusing. Is it always true to do that no matter what explicit form of E(k)E(k)E(k) will be?

This gets a little confusing. If I'm not mistaken, they are treating each mode of the form $exp[i(kx-\omega t)]$ as an eigenfunction of the free particle Hamiltonian, and using E(k) to refer to the corresponding eigenvalues. It's weird because these E(k)'s do NOT correspond to the actual energy of any particle or anything like that. If taken literally, it would be taken to mean the energy of the "piece" of the particle that's in the mode with wavevector k. (Yuck!) Don't get mixed up though, the energy of the particle is the expectation value of the Hamiltonian, which is an integral over the k's: sticking to the notation of the original post, $<E> = \int_{-\infty}^{+\infty}\psi^{*}(k)E(k)\psi(k) dk$. This should always correspond to $<E> = \hbar <\omega>$, but don't mix up the modes (which in this particular case are nonphysical (but sometimes they can have physical meaning, just got to use your noggin to figure it out)) with the physical particle. When in doubt, do the integral.

 

[KFC]

Whoops! I missed all the updates on this thread.The reason for using the Gaussian wavepacket is that is a solution which minimizes the uncertainty in classical phase space: $\Delta x \Delta p = \frac{\hbar}{2}$, which is the limiting case of the Heisenberg principle. To see why this product is the "total" uncertainty, remember classical phase space for motion in one dimension has an x-axis and a p axis, so that expression is like the area of rectangle containing states that can't be distinguished in classical phase space due to quantum uncertainty. The reasoning for the gaussian being minimal uncertainty is partly due to the very interesting fact that the Fourier transform of a gaussian is another gaussian (remember, an FT and a factor of $\hbar$ is how you go from position space to momentum space). The proof of the first claim (gaussian = minimum uncertainty) is shown http://www.colorado.edu/physics/phys2170/phys2170_sp07/downloads/Gaussian.pdf. To see how the Fourier transform of a gaussian in another gaussian, I think that proof is in Griffith's, but I have my copy packed away at the moment. There's a proof here.

On an interesting side note, the other main example of a gaussian wavefunction that came up in this thread, the ground state of the quantum harmonic oscillator, is also a state of minimal uncertainty, a.k.a. a coherent state. Griffith's book definitely has a section on this I would recommend.

Hi Twigg, I will first appreciate your detail explanation on this. It clears more doubts from me. I also get some clue from where and what I should read further to get more information. It really makes sense (at least to me) to understand why Gaussian in terms of minimal uncertainty. In your context, you mention the coherent state as well, which corresponding to equal and minimal uncertainty relation in position and momentum. This has some relation to the Gaussian form as well. From this thread, I read something on the coherent and squeeze states so there are not necessary to have choose Gaussian form in some special cases like squeeze state.

One caveat, as far as I know, the gaussian wavepacket is a solution for the free particle with minimal uncertainty, but I have never seen anyone prove that is the only such solution. In fact, IIRC there are other coherent states besides the ground state for the harmonic oscillator (would appreciate verification on that, anyone out there), so I wouldn't be surprised if there were alternatives for the free particle. I'm not saying there definitely are, just that it wouldn't be too surprising.

I think this is more or less agree the possibility of having different functional form as solution. I will try to search any paper or material talking about that. Thanks.

I just wanted you to notice that the term $exp[i(kx-\omega t)]$ represents modes which each obey the de Broglie relations. As you saw, the Hamiltonian operator gives the energy de Broglie relation. In addition, the momentum operator gives the momentum de Broglie relation.
This gets a little confusing. If I'm not mistaken, they are treating each mode of the form $exp[i(kx-\omega t)]$ as an eigenfunction of the free particle Hamiltonian, and using E(k) to refer to the corresponding eigenvalues. It's weird because these E(k)'s do NOT correspond to the actual energy of any particle or anything like that. If taken literally, it would be taken to mean the energy of the "piece" of the particle that's in the mode with wavevector k. (Yuck!) Don't get mixed up though, the energy of the particle is the expectation value of the Hamiltonian, which is an integral over the k's: sticking to the notation of the original post, $<E> = \int_{-\infty}^{+\infty}\psi^{*}(k)E(k)\psi(k) dk$. This should always correspond to $<E> = \hbar <\omega>$, but don't mix up the modes (which in this particular case are nonphysical (but sometimes they can have physical meaning, just got to use your noggin to figure it out)) with the physical particle. When in doubt, do the integral.