DeBroglie wavelength & particle-wave duality

[Blue Scallop]

We are told to exorcise wave-particle duality because it belongs to the past, in fact prior to 1927. So it’s acceptable to believe matter is neither wave nor particle. But in exorcising wave-particle duality. Why does the deBroglie relationship of momentum being inversely proportional to wavelength still remain? If particle is both wave and particle, DeBroglie wavelength is logical. But matter is now believed as neither wave nor particle. In fact the wave is just a probability amplitude. So using the language of probability amplitude, why does smaller wavelength corresponds to higher momentum?

Physically, when you accelerate a particle to higher speed, and momentum increases, why does the wavelength decreases? Or what is the physical cause of deBroglie relationship of wavelength versus momentum?

I spent all day trying to find answers but most sources have deeply ingrained wave particle duality and we are told in PF to wipe this concept forever however deBroglie wavelength equation animate it.

And I think the reason it is not possible to exorcise wave particle duality is because the deBroglie wavelength relationship to momentum is still valid. And it’s really a wave particle duality.

 

[vanhees71]

All the ideas about "wave mechanics" remain of course valid since the math hasn't changed with the probability-amplitude interpretation of the wave function. Take the free-particle Schrödinger equation
$$\mathrm{i} \hbar \partial_t \psi=-\frac{\hbar^2 \Delta}{2m} \psi.$$
Looking at this the first idea is to try to find plain-wave solutions in order to build other solutions in terms of a Fourier transformation. So you make the ansatz
$$\psi_{\vec{k}}(t,\vec{x})=A \exp(-\mathrm{i} \omega t+\mathrm{i} \vec{k} \cdot \vec{x}).$$
Plugging this into the equation gives the dispersion relation
$$\hbar \omega=\hbar^2 \frac{\vec{k}^2}{2m}.$$

Now you need as another ingredient the idea that observables are represented as self-adjoint operators. In the position representation we use here we have
$$\hat{\vec{p}}=-\mathrm{i} \hbar \vec{\nabla}$$
and the energy (Hamiltonian)
$$\hat{H}=\frac{\hat{\vec{p}}^2}{2m}=-\frac{\hbar^2}{2m} \Delta.$$
The plain wave solutions are (generalized) eigenfunctions of these operators, because
$$\hat{p} \psi_{\vec{k}}=\hbar \vec{k} \psi, \quad \hat{H} \psi_{\vec{k}}=\frac{\vec{p}^2}{2m} \psi_{\vec{k}}.$$
If $\psi_{\vec{k}}$ was a true wave function (i.e., it's modulus squared should be integrable over $\mathbb{R}^3$, which is not the case here) it would thus describe a particle with sharp momentum and sharp energy with the energ-momenum relation given from the Schrödinger equation to be as in classical Newtonian physics. Thus you have
$$\vec{p}=\hbar \vec{k}, \quad E=\frac{\vec{p}^2}{2m}.$$
The relation to the wave length of the probability waves is now clear:
$$|\vec{p}|=\hbar |\vec{k}|=\frac{2 \pi \hbar}{\lambda}=\frac{h}{\lambda}, \quad E=\hbar \omega=2 \pi \hbar f=hf,$$
and these are the de Broglie expressions.

So the math is pretty much the same as with de Broglie's (and Schrödinger's) original approach. Only the physical interpretation has changed with the full understanding of the meaning of the equations in 1926, when Born found his probability interpretation.

 

[Blue Scallop]

All the ideas about "wave mechanics" remain of course valid since the math hasn't changed with the probability-amplitude interpretation of the wave function. Take the free-particle Schrödinger equation
$$\mathrm{i} \hbar \partial_t \psi=-\frac{\hbar^2 \Delta}{2m} \psi.$$
Looking at this the first idea is to try to find plain-wave solutions in order to build other solutions in terms of a Fourier transformation. So you make the ansatz
$$\psi_{\vec{k}}(t,\vec{x})=A \exp(-\mathrm{i} \omega t+\mathrm{i} \vec{k} \cdot \vec{x}).$$
Plugging this into the equation gives the dispersion relation
$$\hbar \omega=\hbar^2 \frac{\vec{k}^2}{2m}.$$

Now you need as another ingredient the idea that observables are represented as self-adjoint operators. In the position representation we use here we have
$$\hat{\vec{p}}=-\mathrm{i} \hbar \vec{\nabla}$$
and the energy (Hamiltonian)
$$\hat{H}=\frac{\hat{\vec{p}}^2}{2m}=-\frac{\hbar^2}{2m} \Delta.$$
The plain wave solutions are (generalized) eigenfunctions of these operators, because
$$\hat{p} \psi_{\vec{k}}=\hbar \vec{k} \psi, \quad \hat{H} \psi_{\vec{k}}=\frac{\vec{p}^2}{2m} \psi_{\vec{k}}.$$
If $\psi_{\vec{k}}$ was a true wave function (i.e., it's modulus squared should be integrable over $\mathbb{R}^3$, which is not the case here) it would thus describe a particle with sharp momentum and sharp energy with the energ-momenum relation given from the Schrödinger equation to be as in classical Newtonian physics. Thus you have
$$\vec{p}=\hbar \vec{k}, \quad E=\frac{\vec{p}^2}{2m}.$$
The relation to the wave length of the probability waves is now clear:
$$|\vec{p}|=\hbar |\vec{k}|=\frac{2 \pi \hbar}{\lambda}=\frac{h}{\lambda}, \quad E=\hbar \omega=2 \pi \hbar f=hf,$$
and these are the de Broglie expressions.

So the math is pretty much the same as with de Broglie's (and Schrödinger's) original approach. Only the physical interpretation has changed with the full understanding of the meaning of the equations in 1926, when Born found his probability interpretation.

Thanks for such breathtaking details!
Besides the following, what are other reasons the original deBroglie wave particle duality can't work (whose Schrodinger original equation are based)?

1. Wave packets will spread with time and the idea of representing particles completely in terms of the superposition of waves is invalid.
2. Beat frequencies will not produce the spectral lines as initially implied
3. The original Schrodinger idea cannot fit rightat all into a classical framework.
Are these 3 basic listings why wave particle duality is wrong correct ? i got this from the qm intro book.

 

[jtbell]

i got this from the qm intro book.

Which QM intro book?

 

[Blue Scallop]

Which QM intro book?

the book "introducing quantum theory: a graphic guide"...

Remember when Schrodinger originally created the Schrodinger Equation and conjectured: "The image point or particle of a mechanical system can be represented by a wave group with small dimensions in every direction. Today this is called a wave packet. The image point (or particle) moves with the group velocity of the wave packet. This looks like a particle, but it's really a superposition of thousands of waves as de Broglie described."

"Schrodinger wanted to described all particles as the superposition of waves. But the grand old man of classical physics, Henrik Lorentz (1853-1928), clear minded as ever in the last few years of his life, brought him to his senses with brutal criticism of his physical interpretation."

Lorentz said to Schrodinger: "This might work in the atom, but what about the free electron? Can a wave packet really stay together and describing moving electron? and...

If any of the above are wrong. Please correct it. Or just answer in summary in english why wave particle duality doesn't work. This is a "B" thread so those itching to suggest graduate level textbooks.. please don't be tempted to do so, I just want to know the basic.. thanks!

 

[vanhees71]

Thanks for such breathtaking details!
Besides the following, what are other reasons the original deBroglie wave particle duality can't work (whose Schrodinger original equation are based)?

1. Wave packets will spread with time and the idea of representing particles completely in terms of the superposition of waves is invalid.
2. Beat frequencies will not produce the spectral lines as initially implied
3. The original Schrodinger idea cannot fit rightat all into a classical framework.
Are these 3 basic listings why wave particle duality is wrong correct ? i got this from the qm intro book.

I think the most convincing argument for the probability interpretation is that we don't see a single electron as a "smeared charge distribution" when detecting it but, e.g., as a single spot on a photoplate. This refers to 1. and 3. in your list

I'm not sure what you mean by "beat frequencies" in this context. To the contrary quantum beats are one phenomenon not correctly described by the semiclassical theory (matter desribed with QT and em. waves by classical electrodynamics) and a clear demonstration of the necessity to quantize the electromagnetic field although you come pretty far in atomic physics with the semiclassical approximation.

https://en.wikipedia.org/wiki/Quantum_beats

The most simple phenomenon indicating the necessity of em.-field quantization is spontaneous emission.

 

[Blue Scallop]

I think the most convincing argument for the probability interpretation is that we don't see a single electron as a "smeared charge distribution" when detecting it but, e.g., as a single spot on a photoplate. This refers to 1. and 3. in your list

But according to Neumaier. The electron is just that... a smeared charge distribution.. and the reason we detected it as point in the detection is an illusion. Neumaier declared in message 25 in thread https://www.physicsforums.com/threa...s-via-lie-algebras.490492/page-2#post-3280823 that:

"It doesn't. It remains smeared. But one of the electrons in the detector fires and (after magnification) gives rise to a measurable current.. This will happen at exactly one place. Thus it _seems_ that the electron has arrived there, while in fact it has arrived everywhere within its extent. If a water wave reaches a dam with a hole in it, the water will come out solely through this hole although the wave reached the dam everywhere. A detector is (in a vague way) similar to such a dam with a large number of holes, of which only one per electron can respond because of conservation of energy."

So I guess we at Physicsforums must not eliminate wave-particle duality which the rest still retain because one of our math giants, Neumaier believes in it.

I'm not sure what you mean by "beat frequencies" in this context. To the contrary quantum beats are one phenomenon not correctly described by the semiclassical theory (matter desribed with QT and em. waves by classical electrodynamics) and a clear demonstration of the necessity to quantize the electromagnetic field although you come pretty far in atomic physics with the semiclassical approximation.

https://en.wikipedia.org/wiki/Quantum_beats

The most simple phenomenon indicating the necessity of em.-field quantization is spontaneous emission.

About beat frequencies. Remember Schrodinger created his Schrodinger equation after de Broglie discovery matter also has wave part. And Schrodinger conjectured that: "Frequencies of bright line spectra can now be visualized as beats between the vibrations of two other quantum states".. But Lorentz said this may not work. Do you have any reference about this as I want to understand what went out in Schrodinger mind when he thought of it.

Do they really used to think Frequencies of bright line spectra could be visualized as beats between the vibrations of two other quantum states. Please elaborate.

Just want to reflect on the historical perspective of it all. Thanks.

 

[vanhees71]

Again, if you detect a single electron with a photoplate you get one spot and not a smeared distribution. That's one of the most simple and convincing reasons why since Born (1926) we take the probabilistic interpretation of the wave function as a probability amplitude, i.e., $|\psi(\vec{x})|^2$ is the probability distribution for the position of this single electron and do not consider $\psi(t,\vec{x})$ a classical field describing the electron as a classical field, as was Schrödinger's ansatz for the interpretation of his wave function.

I don't know Lorentz's criticism about Schrödinger's interpretation in detail, but I guess one point indeed was that for a free electron an initially narrow wave packet, i.e., a situation where the electron is pretty well localized, rapidly spreads with time, which is due to the position-momentum uncertainty, i.e., a wave packet narrowly peaked in space is a superposition of a relatively broad momentum spectrum, which leads to spreading of the wave packet in position space with time. So even if you argue the electron might appear as a pointlike object while in reality it's a narrowly peaked classical wave this doesn't hold if you let this rather well localized wave packet run freely for a while. Schrödinger's interpretation would thus predict that an electron after traveling some distance should appear as a smeared trace on the photoplate which is not the case.

Concerning this "beat argument", I'm still not clear what you are talking about. Quantum beats occur in atomic physics, if you have

(a) an excited state which can decay into (at least) two lower-energy states of different energy (V-shaped spectrum). Then you have a superposition of the corresponding em. waves and an interference term.

However, for

(b) an atom with two excited states which an decay into one lower-energy state ($\Lambda$-shaped spectrum) you find that no interference occurs, when using the quantum-electrodynamical matrix elements rather than the dipole matrix elements of the semiclassical theory.

If you refer to just a decay of an excited state of the atom, already the Bohr-Sommerfeld model predicted that a photon with the energy given by the difference of the atomic energy levels is emitted. In the semi-classical theory you need a classical radiation field to make the atom undergo this emission, i.e., in semiclassical theory you have only induced emission. In the full QED, with the quantized radiation field, you also have spontaneous emission due to the coupling of the electrons with the fluctuating electromagnetic quantum field, i.e., in full QED only the ground state of the atom is really stable in the literal sense, while all other states are unstable (and thus also have a finite "natural line width").

The calculuation in the semiclassical approximation is analogous to the treatment in my Insights article

https://www.physicsforums.com/insights/sins-physics-didactics/

 

[Blue Scallop]

Again, if you detect a single electron with a photoplate you get one spot and not a smeared distribution. That's one of the most simple and convincing reasons why since Born (1926) we take the probabilistic interpretation of the wave function as a probability amplitude, i.e., $|\psi(\vec{x})|^2$ is the probability distribution for the position of this single electron and do not consider $\psi(t,\vec{x})$ a classical field describing the electron as a classical field, as was Schrödinger's ansatz for the interpretation of his wave function.

I don't know Lorentz's criticism about Schrödinger's interpretation in detail, but I guess one point indeed was that for a free electron an initially narrow wave packet, i.e., a situation where the electron is pretty well localized, rapidly spreads with time, which is due to the position-momentum uncertainty, i.e., a wave packet narrowly peaked in space is a superposition of a relatively broad momentum spectrum, which leads to spreading of the wave packet in position space with time. So even if you argue the electron might appear as a pointlike object while in reality it's a narrowly peaked classical wave this doesn't hold if you let this rather well localized wave packet run freely for a while. Schrödinger's interpretation would thus predict that an electron after traveling some distance should appear as a smeared trace on the photoplate which is not the case.

Concerning this "beat argument", I'm still not clear what you are talking about. Quantum beats occur in atomic physics, if you have

(a) an excited state which can decay into (at least) two lower-energy states of different energy (V-shaped spectrum). Then you have a superposition of the corresponding em. waves and an interference term.

However, for

(b) an atom with two excited states which an decay into one lower-energy state ($\Lambda$-shaped spectrum) you find that no interference occurs, when using the quantum-electrodynamical matrix elements rather than the dipole matrix elements of the semiclassical theory.

If you refer to just a decay of an excited state of the atom, already the Bohr-Sommerfeld model predicted that a photon with the energy given by the difference of the atomic energy levels is emitted. In the semi-classical theory you need a classical radiation field to make the atom undergo this emission, i.e., in semiclassical theory you have only induced emission. In the full QED, with the quantized radiation field, you also have spontaneous emission due to the coupling of the electrons with the fluctuating electromagnetic quantum field, i.e., in full QED only the ground state of the atom is really stable in the literal sense, while all other states are unstable (and thus also have a finite "natural line width").

The calculuation in the semiclassical approximation is analogous to the treatment in my Insights article

https://www.physicsforums.com/insights/sins-physics-didactics/

Thanks..

Some other time for the beat frequency as I'm still analyzing Schrodinger thinking a century ago.

I'd like to ask about atomic orbitals now.
I know Bohr model is already obsolete. Here the first orbital corresponds to 1 (h/2pi), 2nd orbital is 2*(h/2pi), 3rd is 3*(h/2pi) and so on where larger orbitals have larger angular momentum. deBroglie discovery of matter wave gives the 2pi the wavelength justification.

Now 90 years of Born probabilistic interpretations later. I think we no longer have to think higher orbitals have longer probability waves. But still Bohr postulate that higher energy electron occupies higher orbital is still valid, see for example http://www.astronomynotes.com/light/s8.htm

In our modern QED.. why does higher energy corresponds to higher orbital.. like when the photons have more energy, the wave functions become fatter? and it doesn't mean the energy of the photons is really located in higher orbitals.. only the wave function is fatter overall?

 

[vanhees71]

I think you should first get a good understanding of non-relativistic QT. Perhaps I shouldn't have started to also include QED in my arguments, although it's the most complete theory relevant for the atomic physics you seem to discuss.

The discrete energy levels of the hydrogen atom in the non-relativistic approximation have energy eigenvalues
$$E_n=-\frac{13.6 \; \text{eV}}{n},$$
where $n \in \mathbb{N}=\{1,2,3,\ldots \}$. Each level is $n^2$ fold degenerate due to the high symmetry of the Kepler problem. The observed emission and absorption spectrum is given by differences of these energies, as was already found empirically by Balmer in the 19th century.

I don't know what you mean by "the wave functions become fatter". It's also not clear to which quantity you refer to when you talk about the "wave function of a photon". Despite some attempts to describe even photons with a wave function, that's a more or less ill defined concept since photons do not even admit a position observable in the usual sense. This somewhat shaky concept is completely absent from the standard formulation of QED.

 

[Blue Scallop]

I think you should first get a good understanding of non-relativistic QT. Perhaps I shouldn't have started to also include QED in my arguments, although it's the most complete theory relevant for the atomic physics you seem to discuss.

The discrete energy levels of the hydrogen atom in the non-relativistic approximation have energy eigenvalues
$$E_n=-\frac{13.6 \; \text{eV}}{n},$$
where $n \in \mathbb{N}=\{1,2,3,\ldots \}$. Each level is $n^2$ fold degenerate due to the high symmetry of the Kepler problem. The observed emission and absorption spectrum is given by differences of these energies, as was already found empirically by Balmer in the 19th century.

A century ago. They thought of these orbital things because they either imagined like Bohr initially that the electrons were like planetary orbits. Or in the case of de Broglie.. they thought of it as actual waves around the atoms. So when you make atoms absorb photons.. the energy would still occupy next orbitals. But with our now probability interpretations. I don't think one can say the electron is stable in an orbital due to the probability waves.

Or when you make the atoms absorb energy. Why is there an electronic transition to higher orbital. There are now no orbits or de Broglie waves to make it transfer to higher orbital. And probability waves are not really there.. so what happens in electronic transition in our modern understanding?

I don't know what you mean by "the wave functions become fatter". It's also not clear to which quantity you refer to when you talk about the "wave function of a photon". Despite some attempts to describe even photons with a wave function, that's a more or less ill defined concept since photons do not even admit a position observable in the usual sense. This somewhat shaky concept is completely absent from the standard formulation of QED.

 

[vanhees71]

We calculate transition probabilities. E.g., for absorption you can use the semiclassical theory, i.e., you consider an electromagnetic wave interacting with the atom and calculate with time-dependent perturbation theory the probability for transitions from one atom-energy eigenstate to another higher one. See my Insights article for the case of the photoelectric effect.

https://www.physicsforums.com/insights/sins-physics-didactics/

 

[Blue Scallop]

We calculate transition probabilities. E.g., for absorption you can use the semiclassical theory, i.e., you consider an electromagnetic wave interacting with the atom and calculate with time-dependent perturbation theory the probability for transitions from one atom-energy eigenstate to another higher one. See my Insights article for the case of the photoelectric effect.

https://www.physicsforums.com/insights/sins-physics-didactics/

I read it. So these atom-energy eigenstate doesn't really correspond to lower and higher orbital of the electrons? in plain English, what cause the orbital to be higher or lower? or are these remnants of the ancient particle-wave duality thing by de Broglie, and the atoms just absorb and emit energy without the electrons actually in higher or lower orbital?

 

[vanhees71]

You can translate "orbital" with "state". If the atom is in an energy eigenstate, its energy has a determined value, implying that an eigenstate with a higher eigenvalue describes an atom at a higher energy.

All your troubles are the wrong idea to learn quantum theory by looking at the history of its development. That's the worst didactics to learn the subject, because you try to comprehend an ununderstood issue rather than to learn the modern status of our knowledge which is much more consistent than the "old quantum mechanics" of Planck, Einstein, Bohr, and de Broglie.

 

[Blue Scallop]

You can translate "orbital" with "state". If the atom is in an energy eigenstate, its energy has a determined value, implying that an eigenstate with a higher eigenvalue describes an atom at a higher energy.

This is what I wanted to know. So higher orbital doesn't mean the electrons are really in higher orbital? So these are all wrong?

http://www.astronomynotes.com/light/s8.htm

http://montessorimuddle.org/2012/02/01/emission-spectra-how-atoms-emit-and-absorb-light/

Most of our chemistry teach this. So please confirm (by yes or no) if higher orbitals don't mean the electrons are really in higher orbital because without a "yes' or "no" answer, I can't read in between the line of your reply. Thanks.

All your troubles are the wrong idea to learn quantum theory by looking at the history of its development. That's the worst didactics to learn the subject, because you try to comprehend an ununderstood issue rather than to learn the modern status of our knowledge which is much more consistent than the "old quantum mechanics" of Planck, Einstein, Bohr, and de Broglie.

 

[Blue Scallop]

I think orbitals still exist.. or there would be no pauli exclusion principles.. no nuclear bombs because uraniums won't split or all elements can split.. the orbitals esp the outer orbital is what created chemical bonding.. and these subshells like 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s, 4f, 5d, 6p, 7s, 5f, 6d, 7p still exist..

Maybe the question is not whether orbitals exist.. but what is the equivalent of orbitals in QED? Whatever, there is still outer and inner orbitals in QED, right? or is it like in Matrix Mechanics where we must cease to visualize the atoms and that means all those subshells things are really illusion?

 

[vanhees71]

Orbitals exist in the sense that you can plot the position probabilities $|\psi(\vec{x})|^2$. As with the notion of "wave-particle duality", you have to give the word "orbital" just the modern meaning in terms of probability distributions for position given the state of your considered particle (in this case bound-state energy-eigensolutions for the Coulomb problem), and everything is fine. It's not an "orbit of an electron". That's also easily refuted by observational evidence: A hydrogen atom in the ground state doesn't occur as a little disk as in the Bohr-Sommerfeld model but rather a little sphere (with blurred boundary), as predicted by the Schrödinger equation.

The Pauli principle tells you that in the approximation, where you neglect the interaction among the many electrons in the atom you can put only one electron per energy-angular-momentum state (i.e., 2 electrons for s states, 6 for p states, etc.), taking into account also the spin 1/2 of the electron. This has not so much to do with position, since position is not too well defined for the energy eigenstates.

 

[Demystifier]

We are told to exorcise wave-particle duality because it belongs to the past, in fact prior to 1927. So it’s acceptable to believe matter is neither wave nor particle. But in exorcising wave-particle duality. Why does the deBroglie relationship of momentum being inversely proportional to wavelength still remain? If particle is both wave and particle, DeBroglie wavelength is logical. But matter is now believed as neither wave nor particle. In fact the wave is just a probability amplitude. So using the language of probability amplitude, why does smaller wavelength corresponds to higher momentum?

Physically, when you accelerate a particle to higher speed, and momentum increases, why does the wavelength decreases? Or what is the physical cause of deBroglie relationship of wavelength versus momentum?

I spent all day trying to find answers but most sources have deeply ingrained wave particle duality and we are told in PF to wipe this concept forever however deBroglie wavelength equation animate it.

And I think the reason it is not possible to exorcise wave particle duality is because the deBroglie wavelength relationship to momentum is still valid. And it’s really a wave particle duality.

The problem with "wave-particle duality" is the duality part. It is, of course, true that we need both the concept of wave and concept of particle to understand modern quantum physics. But it is not true that we need duality. Wave and particle are not dual to each other.

 

[Blue Scallop]

Orbitals exist in the sense that you can plot the position probabilities $|\psi(\vec{x})|^2$. As with the notion of "wave-particle duality", you have to give the word "orbital" just the modern meaning in terms of probability distributions for position given the state of your considered particle (in this case bound-state energy-eigensolutions for the Coulomb problem), and everything is fine. It's not an "orbit of an electron". That's also easily refuted by observational evidence: A hydrogen atom in the ground state doesn't occur as a little disk as in the Bohr-Sommerfeld model but rather a little sphere (with blurred boundary), as predicted by the Schrödinger equation.

The Pauli principle tells you that in the approximation, where you neglect the interaction among the many electrons in the atom you can put only one electron per energy-angular-momentum state (i.e., 2 electrons for s states, 6 for p states, etc.), taking into account also the spin 1/2 of the electron. This has not so much to do with position, since position is not too well defined for the energy eigenstates.

Thanks for the clarification. At least my quantum chemistry books are still accurate for non-relativistic processes. I just realized that in higher energy states, the shapes become lobes and rings, due to the interaction of the quantum effects between the different atomic particles. I thought it was just the simple de Broglie wavelength ring shapes in different layers but only probabilistic Bornwise. I was asking what maintained the shapes since there was no longer standing waves.. and realized now what maintain the lobes and rings or dumbells are the quantum interactions. Good.

 

[vanhees71]

But by definition the energy eigenstates are "standing waves", i.e., static configurations, because indeed the Schrödinger equation tells you that the time evolution of the wave function is given by
$$\psi(t,\vec{x})=\exp(-\mathrm{i} \hat{H} t) \psi(t=0,\vec{x}).$$
If now $\psi(t=0,\vec{x})=u_E(\vec{x})$,
$$\psi(t,\vec{x})=\exp(-\mathrm{i} E t) u_E(\vec{x}),$$
i.e., a standing wave, and the position probability distribution is
$$P(t,\vec{x})=|\psi(t,\vec{x})|^2=|u_E(\vec{x})|^2=P(\vec{x}),$$
i.e., time independent, i.e., an energy state stays the same state under time evolution, i.e., it describes static situations.

 

[Blue Scallop]

The problem with "wave-particle duality" is the duality part. It is, of course, true that we need both the concept of wave and concept of particle to understand modern quantum physics. But it is not true that we need duality. Wave and particle are not dual to each other.

Thanks for the tips. I guess I will never mention about Bohr orbits or even de Broglie standing waves anymore when sharing to newbies. I'll start right away with quantum chemistry sort of thing.. that is probability clouds.

(but then Vandees71 mentioned the probability clouds have standing waves too.. hmm)

also instead of the words "wave-particle duality".. pls suggest a new word to replace "duality".. so its "wave-particle <blank>"

 

[weirdoguy]

quantum mechanics :P It's good to just stick with that.

 

[Mentz114]

Thanks for the tips. I guess I will never mention about Bohr orbits or even de Broglie standing waves anymore when sharing to newbies. I'll start right away with quantum chemistry sort of thing.. that is probability clouds.

(but then Vandees71 mentioned the probability clouds have standing waves too.. hmm)

also instead of the words "wave-particle duality".. pls suggest a new word to replace "duality".. so its "wave-particle <blank>"

Synthesis ?

 

[Blue Scallop]

But by definition the energy eigenstates are "standing waves", i.e., static configurations, because indeed the Schrödinger equation tells you that the time evolution of the wave function is given by
$$\psi(t,\vec{x})=\exp(-\mathrm{i} \hat{H} t) \psi(t=0,\vec{x}).$$
If now $\psi(t=0,\vec{x})=u_E(\vec{x})$,
$$\psi(t,\vec{x})=\exp(-\mathrm{i} E t) u_E(\vec{x}),$$
i.e., a standing wave, and the position probability distribution is
$$P(t,\vec{x})=|\psi(t,\vec{x})|^2=|u_E(\vec{x})|^2=P(\vec{x}),$$
i.e., time independent, i.e., an energy state stays the same state under time evolution, i.e., it describes static situations.

Before 1927. The following is the idea of atoms.

After the Born probability interpretation. We have:

In all pictures in the internet. There are no longer standing waves present. But is the truth there is standing waves inside the probability cloud? like:

?? So although de Broglie standing waves were replaced with probability clouds.. the standing waves are still there, only you can know the location of the particle only by probability as per Born interpretation whereas de Broglie thought the entire standing wave is the particle (?)

(Is the above description accurate if I'd state it to laymen??)

Also someone please show me good drawing of a probability cloud (orbitals) with standing waves present inside. Thanks!

 

[Demystifier]

also instead of the words "wave-particle duality".. pls suggest a new word to replace "duality".. so its "wave-particle <blank>"

One could talk about wave and particle aspects of quantum objects. But I would not use a hyphen between wave and particle because the hyphen suggests that two aspects are somehow relevant on an equal footing, which they aren't.

 

[vanhees71]

Thanks for the tips. I guess I will never mention about Bohr orbits or even de Broglie standing waves anymore when sharing to newbies. I'll start right away with quantum chemistry sort of thing.. that is probability clouds.

(but then Vandees71 mentioned the probability clouds have standing waves too.. hmm)

also instead of the words "wave-particle duality".. pls suggest a new word to replace "duality".. so its "wave-particle <blank>"

I suggest to use the word "particle" or "quantum particle", if it is necessary to emphasize the quantum aspect.

 

[vanhees71]

In all pictures in the internet. There are no longer standing waves present. But is the truth there is standing waves inside the probability cloud? like:

These are standing waves, as I've shown above.

 

[Blue Scallop]

These are standing waves, as I've shown above.

Are they standing waves caused by the invisible Born probability waves or are they standing waves caused by the charges in the electron? Or standing waves caused by fields? What about particles without charges. Let's just focus on non relativistic none qft concepts meantime..thanks

 

[vanhees71]

I don't know, what's your problem and why you call probability waves "invisible". They are measurable and the probabilistic content is measured all the time.

The basic principles of QT are as follows:

(a) There is a Hilbert space. Pure states are represented by rays in this Hilbert space or equivalently by a projection operator $\hat{P}_{\psi}=|\psi \rangle \langle \psi|$ with a normalized vector $|\psi \rangle$ ($|\psi \rangle$ is determined up to an unimportant phase factor, thus the states are represented by rays not the vectors themselves)
(b) Observables are represented by essentially self-adjoint operators defined on a dense subspace of the Hilbert space.
(c) When an observable $A$ is measured precisely, the value is an eigenvalue of the self-adjoint operator $\hat{A}$ that represents it
(d) If $|a,\beta \rangle$ are a complete orthonormal set of eigenvectors of $\hat{A}$ to the eigenvalue $a$, then the probability to find the value $a$ when measuring $A$ on a system prepared in the state $\hat{P}_{\psi}$ is given by
$$P(a|\psi)=\sum_{\beta} \langle a,\beta|\hat{P}_{\psi}|a,\beta \rangle=\sum_{\beta} |\psi(a,\beta)|^2.$$

If you have observables with a continuous spectrum the sums become integrals and the eigenvectors are generalized distribution valued vectors.

That's it. According to the standard minimal interpretation there's no other meaning to the formal building plots of QT than that.

 

[jtbell]

Are they standing waves caused by the invisible Born probability waves or are they standing waves caused by the charges in the electron? Or standing waves caused by fields?

Have you ever compared the mathematical description of a standing wave on e.g. a vibrating string with the mathematical description of a hydrogen orbital?

A standing wave (as opposed to a traveling wave) on a vibrating string can be described by an equation of the form $y(x,t) = f(x)g(t)$ e.g. $y(x,t) = \cos(kx)\cos(\omega t)$.

$f(x)$ gives the overall "envelope" or "shape" of the wave, which does not change with time. $g(t)$ gives an oscillating time-dependence which is the same for every point on the wave. The amplitude of the oscillation at each point x is $f(x)$. The oscillations at all points are in step ("in phase") with each other.

[If you're not already acquainted with the above, I suggest you study standing waves in classical mechanics.]

The wave function for a single hydrogen orbital with quantum numbers n,l,m is $\Psi_{nlm}(r,\theta,\phi,t) = \psi_{nlm}(r,\theta,\phi) e^{-i(E_n/\hbar)t}$.

$\psi_{nlm}(r,\theta,\phi)$ gives the overall "envelope" or "shape" of the orbital, which does not change with time, and is analogous to $f(x)$ in the standing wave above. $e^{-i(E_n/\hbar)t} = \cos[(E_n/\hbar)t] + i \sin[(E_n/\hbar)t]$ gives an oscillating time-dependence which is the same for every point of the orbital. The amplitude of the oscillation at each point $r,\theta,\phi$ is $\psi_{nlm}(r,\theta,\phi)$. The oscillations at all points are in step ("in phase") with each other.

This description is completely analogous with a standing wave on a string, except for the number of spatial dimensions and the use of complex numbers in the time-dependent part.

You can find tables of $\psi_{nlm}$ for different sets of quantum numbers, in many textbooks and on many web pages, e.g. here:

http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/hydwf.html

So e.g. for the ground state (n,l,m = 1,0,0) we have $$\Psi_{100}(r,\theta,\phi) = \frac{1}{\sqrt{\pi} a_0^{3/2}} e^{-r/a_0} e^{-i(E_0/\hbar)t}$$ where $a_0$ is the Bohr radius and $E_0$ is the ground-state energy of hydrogen.

The pictures that you see in textbooks (like the ones you showed above) are an attempt to show the "shapes" of the probability distributions $\left| \Psi_{nlm}(r,\theta,\phi,t) \right|^2 = \left| \psi_{nlm}(r,\theta,\phi) \right|^2$. They are analogous to pictures of the "envelope" of a standing wave on a string. To be complete, they should include some kind of "shading" or "fuzziness" to indicate the variations in probabilty density with position.

There are no smaller waves "hidden" inside these diagrams, as you attempted to draw in one of your diagrams. $\Psi$ at all points in an orbital oscillates together, in step with each other, according to the position-independent time dependence $e^{-i(E_0/\hbar)t}$.

 

[jtbell]

$\Psi$ at all points in an orbital oscillates together, in step with each other, according to the position-independent time dependence $e^{−i(E_0/ℏ)t}$

I'd better modify this. I'm feeling nit-picky this morning. $\psi_{nlm}(r,\theta,\phi)$ can be either positive or negative. In regions where it is negative, $\Psi_{nlm}(r,\theta,\phi,t)$ oscillates in the opposite sense because of that minus sign, even though the time-dependent factor is the same. Some orbital diagrams indicate this by using e.g. different colors for the "positive" and "negative" regions.

This is analgous to a standing wave on a string. In my example above, alternating sections of the string oscillate in opposite directions because $f(x) = \cos(kx)$ can be either positive or negative, even though the same $g(t) = \cos(\omega t)$ applies everywhere.

 

[vanhees71]

I'm not sure, but aren't orbital diagrams defined with $|\psi|^2$ rather than $\psi$, which is complex anyway, so you'd need to draw two orbital diagrams, one for the real and one for the imaginary part, but what does this tell me? The wave function itself is not observable but only the probability distribution $|\psi|^2$.

 

[Blue Scallop]

I don't know, what's your problem and why you call probability waves "invisible". They are measurable and the probabilistic content is measured all the time.

But aren't probability waves the wave function itself? And I said the wave function was invisible because it didn't really exist in our world.. remember that 2 properties of a particle has already 6 dimensional wave function.. which occurs in configuration space.. and we have been taught repeatedly that wave function was not objective.. so we never imagine the wave function can form standing wave in the orbital even when not measured or observed.. or if it is continuously measured... what serves as the observer... potential in the nucleus or the neighboring electrons? i'll reflect on yours and jtbell excellent arguments below and contemplate on it for a day. Thanks a lot.

The basic principles of QT are as follows:

(a) There is a Hilbert space. Pure states are represented by rays in this Hilbert space or equivalently by a projection operator $\hat{P}_{\psi}=|\psi \rangle \langle \psi|$ with a normalized vector $|\psi \rangle$ ($|\psi \rangle$ is determined up to an unimportant phase factor, thus the states are represented by rays not the vectors themselves)
(b) Observables are represented by essentially self-adjoint operators defined on a dense subspace of the Hilbert space.
(c) When an observable $A$ is measured precisely, the value is an eigenvalue of the self-adjoint operator $\hat{A}$ that represents it
(d) If $|a,\beta \rangle$ are a complete orthonormal set of eigenvectors of $\hat{A}$ to the eigenvalue $a$, then the probability to find the value $a$ when measuring $A$ on a system prepared in the state $\hat{P}_{\psi}$ is given by
$$P(a|\psi)=\sum_{\beta} \langle a,\beta|\hat{P}_{\psi}|a,\beta \rangle=\sum_{\beta} |\psi(a,\beta)|^2.$$

If you have observables with a continuous spectrum the sums become integrals and the eigenvectors are generalized distribution valued vectors.

That's it. According to the standard minimal interpretation there's no other meaning to the formal building plots of QT than that.

 

[vanhees71]

This is all philosophical gibberish (SCNR). The wave function has a probabilistic meaning and is thus measurable on ensembles of equally prepared systems. For me anything reproducibly measurable "exists" objectively (and nothing else!).

 

[Blue Scallop]

This is all philosophical gibberish (SCNR). The wave function has a probabilistic meaning and is thus measurable on ensembles of equally prepared systems. For me anything reproducibly measurable "exists" objectively (and nothing else!).

That's a great idea. But I wonder if Bohr believed what you believe. I'll ponder on it. It is mentioned in the book "Quantum: Einstein, Bohr and the Great Debate about the Nature of Reality":

"Once the momentum of particle A is measured, it is possible to predict accurately the result of a similar measurement of the momentum of particle B as outlined by EPR. However, Bohr argued that that does not mean that momentum is an independent element of B's reality. Only when an 'actual' momentum measurement is carried out on B can it be said to possesses momentum. A particle's momentum becomes 'real' only when it interacts with a device designed to measure its momentum. A particle does exist in some unknown but 'real' state prior to an act of measurement. In the absence of such a measurement to determine either the position or momentum of a particle, Bohr argued that it was meaningless to assert that it actually possessed either."

I'm thinking if there are ensembles of equally prepared system where no measurements were made.. then would the properties even exist. For a lone hydrogen atom with a nucleus and electron.. the electron exists because the nucleus observed it.. without the nucleus, maybe the electron would cease to exist as per Bohr belief? Note this is the heart of orthodox QM.