Propagation of de Broglie waves

[bhobba]

IBut are there any interpretations where the Hamiltonians are not in the wave functions but in the particles? And to make it 100% on topic. What did de Broglie think about the Hamiltonian with regards to his initial de Broglie waves thoughts? Since Hamiltonian is very basic. It's possible de Broglie has thought of it.

The Hamiltonian is in Schrödinger's equation which follows from symmetry principles.

This is now way off topic and I will go no further than point you to Chapter 3 of Ballentine.

De Broglies arguments are simply working out the idea if waves can act as particles maybe the reverse is true. He cobbled together relativity and other ideas to do that. It was his PhD thesis and the examiners said they couldn't even understand it but sent it to Einstein for comment. He recognized immediately its importance and gave it his full recommendation so he got his PhD and eventually a Nobel.

You can read the exact math he used in many places.

Thanks
Bill

 

[Blue Scallop]

The Hamiltonian is in Schrödinger's equation which follows from symmetry principles.

This is now way off topic and I will go no further than point you to Chapter 3 of Ballentine.

De Broglies arguments are simply working out the idea if waves can act as particles maybe the reverse is true. He cobbled together relativity and other ideas to do that. It was his PhD thesis and the examiners said they couldn't even understand it but sent it to Einstein for comment. He recognized immediately its importance and gave it his full recommendation so he got his PhD and eventually a Nobel.

You can read the exact math he used in many places.

Thanks
Bill

Ok let's be fully on topic. About the old de Broglie waves.. he has group velocity, particle velocity and phase velocity. What is the final modern form of the different velocities as they are still mentioned in QM now..

 

[Blue Scallop]

The Hamiltonian is in Schrödinger's equation which follows from symmetry principles.

This is now way off topic and I will go no further than point you to Chapter 3 of Ballentine.

De Broglies arguments are simply working out the idea if waves can act as particles maybe the reverse is true. He cobbled together relativity and other ideas to do that. It was his PhD thesis and the examiners said they couldn't even understand it but sent it to Einstein for comment. He recognized immediately its importance and gave it his full recommendation so he got his PhD and eventually a Nobel.

You can read the exact math he used in many places.

Thanks
Bill

I always wanted to ask about this so let me ask this now as I'm quite confused about it. See:

You can see this image anywhere. What is retained in the modern form?

 

[bhobba]

Ok let's be fully on topic. About the old de Broglie waves.. he has group velocity, particle velocity and phase velocity. What is the final modern form of the different velocities as they are still mentioned in QM now..

No.

The modern view has nothing to do with what birthed it.

The wave-function, even though it has wave, in its name has nothing to do with waves - is, technobabble now follows, the expansion of the state in terms of eigenfunctions of position. Nothing like what De-Broglie or Schrödinger thought - which is why Schrödinger became so despondent - it's very mathematical. The Von-Neumann's and Dirac's of the physics world were now in charge.

Even today we still 'argue' and struggle with what it means. Great progress has been made, especially in the area of decoherence - but questions still remain and endlessly discussed.

Thanks
Bill

 

[Blue Scallop]

No.

The modern view has nothing to do with what birthed it.

The wave-function, even though it has wave, in its name has nothing to do with waves - is, technobabble now follows, the expansion of the state in terms of eigenfunctions of position. Nothing like what De-Broglie or Schrödinger thought - which is why Schrödinger became so despondent - it's very mathematical. The Von-Neumann's and Dirac's of the physics world were now in charge.

Even today we still 'argue' and struggle with what it means. Great progress has been made, especially in the area of decoherence - but questions still remain and endlessly discussed.

Thanks
Bill

Ok so de-Broglie wave is outdated.. no problem.. but here's the confusing part.. how come Valentini and others are reviving the de Broglie Pilot wave thing. Is this the same concept as the outdated de-Broglie waves? If there is difference. What's their differences so one won't mix the two concepts and get confused. Thank you Bill.

 

[bhobba]

Ok so de-Broglie wave is outdated.. no problem.. but here's the confusing part.. how come Valentini and others are reviving the de Broglie Pilot wave thing. Is this the same concept as the outdated de-Broglie waves? If there is difference. What's their differences so one won't mix the two concepts and get confused. Thank you Bill.

Its never gone away - we have DBB.

But Dymytifyer is the person to ask about that.

BTW in DBB its not a wave - its a particle guided by a wave.

Thanks
Bill

 

[Blue Scallop]

Its never gone away - we have DBB.

But Dymytifyer is the person to ask about that.

BTW in DBB its not a wave - its a particle guided by a wave.

Thanks
Bill

Ah. So the difference with the original de Broglie wave is the original de Broglie wave considered particle as wave only and no particle.. yet DBB has both particle and wave.. but then de Broglie invented the pilot wave too.. did he never mean it's both particle and wave?

 

[vanhees71]

One has to be careful with the various "velocities" related to waves. One must be aware, what they mean and under which circumstances one can use these notions. Take e.g., "group velocity". A general solution for the time evolution of a wave equation often can be written as
$$\psi(t,\vec{x})=\int_{\mathbb{R}^3} \mathrm{d}^3 \vec{k} \frac{1}{(2 \pi)^{3/2}} \tilde{\psi}_0(\vec{k}) \exp[-\mathrm{i} \omega(\vec{k}) t + \mathrm{i} \vec{k} \cdot \vec{x}].$$
The reason is that often the Laplace operator occurs in the wave equation, and the plane waves are a complete set of generalized orthonormal functions.

E.g., for the free Schröinger equation (with $\hbar=1$),
$$\partial_t \psi(t,\vec{x})=-\frac{1}{2m} \Delta \psi(t,\vec{x}),$$
you get the dispersion relation
$$\omega(\vec{k})=\frac{1}{2m} \vec{k}^2.$$
To make sense of the group velocity you need the assumption that $\tilde{\psi}_0(\vec{k})$ is narrowly peaked around $\vec{k}_0$. Then one can expand the exponent in powers of $\vec{k}-\vec{k}_0$:
$$\mathrm{i} [\vec{k} \cdot \vec{x}-\omega(\vec{k})t] \simeq \mathrm{i} [\vec{k} \cdot \vec{x}-\omega_0 t - t (\vec{k}-\vec{k}_0) \cdot \vec{\nabla}_k \omega(\vec{k}_0)].$$
Then you get with $\vec{v}_g=\vec{\nabla}_k \omega(\vec{k}_0)$
$$\psi(t,\vec{x}) \simeq \exp[-\mathrm{i} (\omega_0 -\vec{k}_0 \cdot \vec{v}_g) t+\mathrm{i} \vec{k}_0 \cdot \vec{x}] \int_{\mathbb{R}^3} \mathrm{d}^3 \vec{k} \tilde{\psi}_0(\vec{k}) \exp[\mathrm{i} \vec{k} \cdot (\vec{x}-t \vec{v}_{\text{g}})],$$
i.e., finally
$$\psi(t,\vec{x}) \simeq \exp[-\mathrm{i} (\omega_0 -\vec{k}_0 \cdot \vec{v}_g) t+\mathrm{i} \vec{k}_0 \cdot \vec{x}] \psi_0(\vec{x}-\vec{v}_{\text{g}} t).$$
This means, in this approximation up to a phase the initial wave packet moves undeformed with group velocity $\vec{v}_{\text{g}}$. The position-probability density is
$$|\psi(t,\vec{x})| \simeq |\psi_0(\vec{x}-\vec{v}_{\text{g}} t)|.$$
It is important to note that this only holds for sufficiently narrow-peaked $\tilde{\psi}(\vec{k})$, i.e., for a particle with a quite well prepared momentum!

With the dispersion relation for the free-particle Schrödinger wave you get the expected result
$$\vec{v}_{\text{g}}=\frac{1}{m} \vec{k}_0,$$
i.e., the wave packet moves with this velocity, which is pretty intuitive from the classical particle picture.

One must, however be aware that this interpreation relies on the assumptions made, i.e., that you have a wave packet that is narrowly peaked in momentum space, such that dispersion can be neglected.

It's a good exercise to solve the initial-value problem for a Gaussian wave packet as the initial state exactly. Most conveniently one gives it in terms of the momentum representation,
$$\tilde{\psi}(\vec{k})=N \exp \left (-\frac{(\vec{k}-\vec{k}_0)^2}{4 \sigma_k^2} \right ),$$
where $N$ is an unimportant normalization factor and $\sigma_k$ the standard deviation of the momentum components. Here the time-evolution integral can be easily solved exactly since it's a Gaussian integral, and you can also easily compare it to the approximation involving the group velocity $\vec{v}_{\text{g}}$.

 

[Blue Scallop]

No.

The modern view has nothing to do with what birthed it.

The wave-function, even though it has wave, in its name has nothing to do with waves - is, technobabble now follows, the expansion of the state in terms of eigenfunctions of position. Nothing like what De-Broglie or Schrödinger thought - which is why Schrödinger became so despondent - it's very mathematical. The Von-Neumann's and Dirac's of the physics world were now in charge.

Even today we still 'argue' and struggle with what it means. Great progress has been made, especially in the area of decoherence - but questions still remain and endlessly discussed.

Thanks
Bill

Do you know what is the best book and biography about de Broglie with focus on his thinking process from the matter waves thesis days to proposing the pilot waves after Born suggested the probability interpretation in 1927? Something like... de Broglie, Life and Thoughts? anyone?

 

[OurBladesAreSharp]

The 'interference patterns' associated with the famous 'double-slit' experiment, are really patterns left on screens. The question is, what is causing those patterns? The answer is, sub-atomic particles. But if the particles are fired one-at-a-time, then how can they form an 'interference pattern'? How is a single particle interfering with itself? That seems to be the issue.

Think of the particle as a wave function which has non zero value at both slits and interferes with itself later to produce interference patterns

 

[vanhees71]

Well, but this is not the right picture either since it's not what is observed. If you send a single particle with some quite well-defined momentum (such that in the wave picture you get nice interference patterns) through a double slit, you won't see the interference pattern at the slit as suggested if you interprete the (position) space function as a classical field entity, i.e., rather than finding some smeared distribution you find a single spot on the screen, which suggests that the electron is rather something more particle like. This contradiction was dubbed "wave-particle duality", but it's an obsolete concept. Very quickly after de Broglie's PhD thesis, where he suggested the introduction of "matter waves" and which was so unusual at the time that the PhD committee sent it to Einstein for evaluation. Since Einstein enthusiastically liked it, de Broglie rightfully got his PhD and later the Nobel prize for his ideas.

Nevertheless, the classical wave picture is obviously wrong from the observation just described. On the other hand, due to the very fact that there are indeed interference patterns occurring when sending many electrons through the double slits (all carefully prepared in the same way with a pretty well-defined momentum) it's as obviously wrong to interpret electrons in terms of classical particles.

That's why Schrödingers wave function had to be intepreted differently to make sense, and this was the breakthrough for quantum theory as a consistent picture of matter and is due to Born, who interpreted the wave function as "probability amplitude", i.e., after solving the Schrödinger equation, given a Hamiltonian (including the interaction of the electron with the material forming the slits, usually simplified by infinitely high potential wells, which is reformulated simply to boundary conditions as in optical diffraction theory) you just know the probability distribution for a particle to be at any position at a given time $t$:
$$P(t,\vec{x}) \mathrm{d}^3 \vec{x}=|\psi(t,\vec{x})|^2 \mathrm{d}^3 \vec{x}$$
is the probability to find at time $t$ the electron in a little volume $\mathrm{d}^3 \vec{x}$ around position $\vec{x}$.

The surprising finding is that although there's only one electron running through the double slit at any time you find the interference pattern when running enough electrons through. This shows, it's not an interference phenomenon of two or more electrons but it's an interference effect of probability waves for a single electron. One has to get used to it. There is no simpler way to somehow explain this fact about electrons but it's, according to present knowledge, the only consistent description of this observed fact. In other words the, admittedly counterintuitive, basic rules of QT are fundamental laws of Nature that can not derived from "simpler" or "more intuitive" laws.