Does the energy 6.626x10^-34J have any special meaning?

[qnt200]

We know that Planck's constant is 6.626x10-34Js. Does the energy of the same numerical value 6.626x10-34J have any special meaning. Is it perhaps the lowest possible energy or are energies less than that energy theoretically possible.

 

[PeroK]

We know that Planck's constant is 6.626x10-34Js. Does the energy of the same numerical value 6.626x10-34J have any special meaning. Is it perhaps the lowest possible energy or are energies less than that energy theoretically possible.

No, no and yes. Energy is frame dependent. Only energy differences have physical meaning.

 

[vanhees71]

There's no special meaning in the number. It's just our arbitrary choice of our metrological system, the SI. Since last year this number is simply taken as exact and defines the base units of the SI.

The "natural choice" is $\hbar=h/(2 \pi)=1$.

 

[Vanadium 50]

Units are important. Js is not J. Planck's constant is not an energy. For its numeric value to be a special energy in nature requires there be something special about the second too.

 

[vanhees71]

The Planck constant is a natural measure for action or phase-space area. The most intuitive physical picture comes from statistical mechanics, where it answers the question, what's the natural choice for a measure of the the phase-space volume.

 

[qnt200]

No, no and yes. Energy is frame dependent. Only energy differences have physical meaning.

Thanks for the interesting answer. I guess I understood well:
1. Does the energy of the same numerical value 6.626x10-34J have any special meaning. (No)
2. Is it perhaps the least possible energy. (No)
3. Are energies less than that energy theoretically possible. (Yes)

Could it be said that theoretically (and in quantum physics) the value of energy can be of any amount, from zero to more (continuous transition)?

 

[vanhees71]

Yes, as is easily seen on the example of the energy of photons, which is $E=\hbar \omega=h \nu$. Since $\nu$ can be as close to 0 as you like, there's no "smallest possible energy" in QT.

 

[qnt200]

Yes, as is easily seen on the example of the energy of photons, which is $E=\hbar \omega=h \nu$. Since $\nu$ can be as close to 0 as you like, there's no "smallest possible energy" in QT.

From this I could conclude the following:
The energy of one quantum E = hν can be of any value, because the frequency can also be of any value (> 0) but the energy is quantized. This means that it comes in small energy packets, ie in quanta of energy hν + hν + hν ...
Is this conclusion correct?

 

[weirdoguy]

but the energy is quantized.

Energy is quantized only in some systems, e.g. atoms. Energy of free photon or electron is not quantized.

 

[vanhees71]

From this I could conclude the following:
The energy of one quantum E = hν can be of any value, because the frequency can also be of any value (> 0) but the energy is quantized. This means that it comes in small energy packets, ie in quanta of energy hν + hν + hν ...
Is this conclusion correct?

The energy of a specific mode is quantized. The energy of the free em. field is not. The spectrum of the Hamiltonian is $\mathbb{R}_{\geq 0}$.

 

[qnt200]

Energy is quantized only in some systems, e.g. atoms. Energy of free photon or electron is not quantized.

Electromagnetic forces are transmitted between charged particles by exchanging photons (virtual photons). As the free electron travels, it has no quantized energy, however when it comes close to the atom of some sensor with which we want to detect it, it then exchanges virtual photons, whose energies are quantized. When interacting, we no longer consider it a free electron.
The photon when interacting with sensor atoms also has quantized energy, regardless of the fact that it did not have it as a free photon before.

All in all, it leads me to the strange conclusion, that we can never detect a continuous transition of energy, but only a quantized form of energy?

I hope this is correct?

 

[Vanadium 50]

hen exchanges virtual photons, whose energies are quantized.

That's not true. Not only are their energies not quantized, you can't even count them.

 

[weirdoguy]

And as always with virtual particles, good places to start and get rid of your misconceptions are those insights:
https://www.physicsforums.com/insights/misconceptions-virtual-particles/
https://www.physicsforums.com/insights/physics-virtual-particles/

 

[qnt200]

Thanks, that will definitely be helpful.
I would still like to ask, do we always detect quantized energy, while we are not able to detect continuous energy?

 

[PeroK]

Thanks, that will definitely be helpful.
I would still like to ask, do we always detect quantized energy, while we are not able to detect continuous energy?

Any piece of measurement apparatus has only a finite number of possible measurement readings. Take measurements of time, for example. Any measurement counts a number of cycles, but can only record an answer to a whole number of cycles. There is no piece of apparatus that could record energies on a truly continuous spectrum.

 

[vanhees71]

A continuous observable of course can only be measured with a finite resolution/accuracy. This does not imply that the measurement device measures only discrete numbers.

 

[qnt200]

A continuous observable of course can only be measured with a finite resolution/accuracy. This does not imply that the measurement device measures only discrete numbers.

Thanks. However, perhaps I should have asked, whether our measurement sensors are always based on an interaction and that interaction is always a quantized process?

 

[vanhees71]

Of course, you can only measure something by letting the measured system interact with a measurement device. I don't know what you mean by saying that this interaction is a quantized process.

 

[qnt200]

Of course, you can only measure something by letting the measured system interact with a measurement device. I don't know what you mean by saying that this interaction is a quantized process.

I may be wrong, but I guess there is a quantized energy transfer in all the interactions?

 

[vanhees71]

No, the energy transfer in a scattering event can take any value, not only discrete ones.

 

[qnt200]

No, the energy transfer in a scattering event can take any value, not only discrete ones.

Is the conclusion that a quantized energy transition can be proved by experiment, while a continuous energy transition cannot be measured and proven by experiment, correct? By measuring with more and more precise measuring devices, we will come to a limitation due to the uncertainty principle. In other words, our conclusion that there is a continuous transition of energy arose only from theoretical considerations, which are practically impossible to prove.

 

[PeroK]

Is the conclusion that a quantized energy transition can be proved by experiment, while a continuous energy transition cannot be measured and proven by experiment, correct? By measuring with more and more precise measuring devices, we will come to a limitation due to the uncertainty principle. In other words, our conclusion that there is a continuous transition of energy arose only from theoretical considerations, which are practically impossible to prove.

Systems such as atoms and molecules have discrete allowed energy levels. They do not have a continuous energy spectrum: this is the basis of spectroscopy. In such systems, we can talk about quantized transitions from one allowed energy level to another. In this case the theory can be confirmed by experiment.

A free particle, however, can in theory have any energy value. There are no discrete energy levels in the theory and no experimental evidence to contradict this. Likewise, two particles may interact (scatter off each other) and the initial and final energies are again allowed to take any value. There is no experimental evidence of quantized energy transitions in this case.

 

[vanhees71]

It's true that atoms and molecules have bound states with discrete energy levels. Nevertheless there Hamiltonian also has a continuous part of its spectrum. These states refer to scattering states.

As the most simple example take a hydrogen atom and make the simplifying assumption that the proton can be taken as infinitely heavy. Then you can describe the hydrogen atom as an electron in the static Coulomb field of the proton which is taken at rest. This leads to the well-known discrete energy levels of hydrogen and the spectral lines with frequencies given by the energy difference of the allowed transitions between these states (allowed by the usual dipole selection rules). The corresponding are $E_n=-13.6 \; \text{eV}/n^2$ ($n \in \mathbb{N}$).

Of course an electron in the Coulomb field of the proton can also be scattered, and the corresponding energy eigenstates are those with the continuous energy eigenvalues $E>0$. That describes elastic electron scattering with a proton. By definition "scattering states" are always the states in the continuous part of the Hamiltonian's spectrum, and that's why the electron can take any energy value $E>0$ not only discrete ones. Of course you can measure the energy and momenta in a scattering event as precisely as you want (or your measurement devices allow). That's what's done at the LHC and other accelerators to measure as accurately as possible the cross sections of scattering events.

 

[A. Neumaier]

Systems such as atoms and molecules have discrete allowed energy levels. They do not have a continuous energy spectrum

This is not correct.

Atoms and molecules always have a continuous spectrum. It describes the states where they break up into ions and electrons, or into two smaller molecules. In this case, the energy varies continuously since the relative kinetic energy can assume any positive value. In the spectrum, this is visible as a branch cut starting at the dissociation threshold, which is a branch point. At energies below the branch point, only a discrete spectrum is possible.

 

[qnt200]

Thank you very much for the answers.

I find it very interesting that you have separated the free particles from the complicated cases that occur with bound particles in atoms. This made the situation much more understandable to me.

Regarding free particles, the sentence: "There is no experimental evidence of quantized energy transitions in this case." it seems indicative to me.

I am aware that I lack a lot of information in order to come to the right conclusions, but logically speaking, I think the following can be said:

There is no experimental evidence for a quantized energy transition, but also vice versa, there is no experimental evidence for a continuous energy transition (unless we assume that the accuracy and resolution of our measurements are sufficient for proof)?

Obviously it all depends on the accuracy and resolution of the measurements we have achieved to date and the question of whether they are sufficient to give solid evidence. If such a quantized energy transition exists in free particles, the question is at what energy level is it visible?
Perhaps far lower than what we are able to measure.

 

[PeroK]

There is no experimental evidence for a quantized energy transition, but also vice versa, there is no experimental evidence for a continuous energy transition (unless we assume that the accuracy and resolution of our measurements are sufficient for proof)?

Obviously it all depends on the accuracy and resolution of the measurements we have achieved to date and the question of whether they are sufficient to give solid evidence. If such a quantized energy transition exists in free particles, the question is at what energy level is it visible?
Perhaps far lower than what we are able to measure.

In the end, science can only carry out a finite number of experiments with a finite number of energy values. You can always postulate that there is a quantisation below the smallest difference between any two energy values measured.

What you need is an effective theory that predicts quantisation at a certain granularity. Without that you have nothing.

It's the same with spacetime quantization. Of course it might be quantized, but unless you produce a workable theory of spacetime quantization you have nothing.

 

[qnt200]

In the end, science can only carry out a finite number of experiments with a finite number of energy values. You can always postulate that there is a quantisation below the smallest difference between any two energy values measured.

What you need is an effective theory that predicts quantisation at a certain granularity. Without that you have nothing.

It's the same with spacetime quantization. Of course it might be quantized, but unless you produce a workable theory of spacetime quantization you have nothing.

Of course, I absolutely agree with that.