# Quantized energy - Photon

1. Jan 30, 2013

### jaumzaum

Everybody says energy is quantized. But for einstein-plank equation
E = h.f

If a photon could have any values of f, E would not be quantized
I know bohr orbits only accept some frequencies, but hydrogen frequencies are different from lithium or nytrogen frequencies. So what is the MINIMUM value of E? As far as I know, a photon is a quanta of energy, so what would be the quantum?

2. Jan 30, 2013

### mpv_plate

The energy in general can be arbitrary, because you can choose arbitrary frequency. But if you decide to work with one specific frequency fchosen, then h*fchosen is the quantum of that specific frequency.

3. Jan 30, 2013

### mpv_plate

Also please note one thing: the fact that only certain photons (with specific energies) can be emitted / absorbed by an atom, is not limiting the possible photon energies. The atomic energy levels are features of these atoms, but electromagnetic field does not care about that. The free e-m field can have any frequency. It is just that only certain frequencies will be absorbed by atoms, while other photons will not be affected.

4. Jan 30, 2013

### San K

is there no minimum value for the quantum?
or
is frequency quantized?
or
does quantization only make sense when in a bounded state at a specific orbital/energy level/shell in an atom?

Last edited: Jan 30, 2013
5. Jan 30, 2013

### Darwin123

I am not sure what you are talking about. You are talking about different elements. So what you wrote sounds like a contradiction.

What are the "hydrogen frequencies" of "lithium" and nitrogen?

There is a minimum value for particle number for fixed values of frequency, volume and total energy. There is also a minimum amplitude of the wave for fixed values of frequency, volume and total energy. There is no absolute value for minimum energy. If you want a smaller total energy, then you can increase the frequency and decrease the volume.

The photon is generally defined as a quanta of energy. However, that definition is a little ambiguous. When working through the mathematics, I have found it more useful to think of the photon as a quantum of amplitude. The amplitude of the wave is forced to be discrete because the number of particles is discrete. Thus, the de Broglie relations (which includes the Einstein-Planck relation) should be thought of as constraints on the amplitude of the wave.

The two definitions for "photon" often give the same final results. There is usually little experimental difference in the photon being a quantum of energy and a photon being a quantum of amplitude. However, there is a shade of difference that occasionally causes confusion especially when one counts photons.

One thing that is unclear from introductory courses in quantum mechanics is that the number of particles can be uncertain. The total energy of a wave is often precisely determined even though the frequency is not. Therefore, an uncertainty in photon energy is often manifested as an uncertainty in photon number.

"Quantization conditions" are actually constraints on the amplitude of the wave associated with the particle. They don't constrain the frequency of the wave nor do they constrain the energy of each individual photon. So there is no absolute minimum in photon energy.

The number density of particles associated with a wave increases with the square of the amplitude of the wave. The total energy is proportional to the number of particles. The number of particles is basically the total energy of the wave divided by the energy of each particle. The total energy of the wave is proportional to the amplitude squared.

The Einstein-Planck equation only determines the energy density for each type of each particle. The Einstein-Planck equation does not determine the total energy of the wave. I don't fully understand your question, but maybe I partially understand the general confusion. I conjecture that your confusion concerns issues of amplitude and photon number.

Does this sound like it could be the problem?

6. Jan 30, 2013

### jaumzaum

Yeah. Actually the question was if a single photon energy was quantized, and as you and mpv_plate said, it is not, cause a single photon can have any frequency.

But I still have a question. I mentioned the hydrogen, lithium and nitrogen frequencies as the energy levels for these atoms. As there are a finite number of elements in the periodic table, there have to be a finite number of energy levels as well as a finite number of frequencies that can be absorbed or emitted by an atom. So by that I thought frequency was quantized too. But as mpv_plate said, that works only for atoms. And the em field can have any frequency. So here is the question: How can we produce an electromagnetic wave not being by an electron jumping to another electron shell/energy level and emitting a photon? Is there any other method to do that?

7. Jan 30, 2013

### mpv_plate

For electromagnetic field, there is no minimum value of energy, but there is minimum value for amplitude. (However, if you study the quantum field theory deeper, this becomes more complicated)

Frequency is not quantized. You can have any frequency.

Bound states tend to be quantized in energy, because they may impose constraints on the possible states.

Another example: you can activate a wave in electromagnetic field inside a small metal box. You cannot have any photon (any frequency) in the box, because only specific (discrete) frequencies can fit in the constrained space.

8. Jan 30, 2013

### The_Duck

No: even a hydrogen atom has an infinite number of energy levels. For hydrogen these energy levels are -13.6 / n^2 electron volts for any integer n greater than or equal to 1.

Jiggle an electron back and forth: that is, run an alternating electric current through a wire. This is how we produce radio waves. The frequency of the emitted wave is equal to the frequency with which the current alternates (and we can tune this frequency to be whatever we want).

9. Jan 30, 2013

### naturale

In a vibrating string can have generic length L but the harmonics spectrum only assumes quantized wavelengths lambda_n = n lambda = n / L. An electromagnetic is like a string vibrating with arbitrary periodicity T, but the it can assume discretized energy values E_n = n E = n h /T. The energy gaps of the harmonics of this string are the photons.

10. Jan 30, 2013

### Darwin123

No.

The frequencies of a system are very often constrained to discrete values.

This is true even in classical mechanics. For instance, the resonant modes of a violin string have discrete frequencies. The boundary conditions of the string cause the wavelengths of the string to change by discrete values. This results in the frequencies changing by discrete values.

This discrete division between notes of a stringed instrument have been known since Aristotle. Probably even before Aristotle. Newton understood frequencies of vibrations on strings. However, the separate frequencies on a string were not called quanta.

The resonant frequencies of a hydrogen atom are caused by the periodic boundary conditions of the electron-wave. In that sense, they are like the waves on a violin string. However, the discrete values of frequency are not the direct result of an ad hoc hypothesis.

What is really quantized in a hydrogen atom is radius of the electron's orbit. The radius of the electrons orbit is a type of amplitude. You can think of the radius as the limit of the back and forth motion of the electron. It is this radius, which is a type of amplitude, which is quantized. The discrete values of frequency are an indirect consequence of the fact that the radius is "quantized". The frequencies aren't quantized, but the radii are quantized.

You have to be careful about the word quantized. The word is not quite synonymous with discrete. Quantization is a type of discreteness.

Maybe the word "digitized would be better. No, I take that back. There are certain qualifications to a digital system.

The "quantization" first hypothesized by Planck referred to

or
Discrete values for frequency usually make sense for bounded states. The real reason frequencies change in discrete values is because of the boundary conditions on the wave. The violin string is a good analog.

The reason that the notes of a violin string are discrete is because the violin string is fixed on both ends. Thus, the violin string has to be "bounded" in order to produce notes. Notes are bounded states! A violin string that isn't tied down does not produce separate notes.

A propose that frequencies should never be called quantized. Frequencies are merely discrete.

11. Jan 30, 2013

### naturale

http://www.amacad.org/publications/winter2002/wilczek.pdf
This is wrong. Orbitals are obtained by the Bohr-Sommerfeld quantization, which is a periodicity boundary condition for a deformed string. It is not necessary to require circular orbits. It is sufficient to ask for orbits with integer number of wavelengths, just as in a vibrating string.

see http://www.amacad.org/publications/winter2002/wilczek.pdf

12. Jan 30, 2013

### Darwin123

Sorry. I didn't mean to imply that the orbit was necessarily circular.

The Bohr-Sommerfield condition is not a periodic boundary condition for a deformed string. The Bohr-Sommerfeld condition was hypothesized before the de Broglie relationships were hypothesized. Therefore, there was no "wavelength" associated with them.

The Bohr-Sommerfeld condition (BSC)as first formulated involved orbital-angular momentum. It was a constraint on the size of the orbits. The word "radius" was perhaps wrong. However, the visual picture was of an electron in a Keplarian orbit around the nucleus. The size of that orbit had nothing to do with wavelength.

The frequency of a Keplarian orbit decreases with the size of the orbit. The size of the orbit and Kepler's Laws is what basically determines the frequency. However, the size of the orbit can vary continuously in classical physics. The reason that the frequency does not vary continuously in "old quantum theory" is that the size of the orbit can't vary continuously.

BSC was a generalization of the Planck rule for quantization. Planck assumed the atom was a harmonic oscillator. The electron in the atom was just a point charge connected to the nucleus by a type of "spring". The amplitude of this harmonic oscillator was constrained to specific and discrete values which resulted in the energy of the harmonic oscillator being constrained to discrete values. There was no electromagnetic wave and no electron wave. The frequency of the harmonic oscillator was not determined by any boundary condition. The natural frequency of the atom was the square root of the spring constant divided by the mass.

De Broglie came up with this "explanation" for the Bohr-Sommerfeld condition where the an integer number of complete cycles had to fit in on the orbit. The de Broglie explanation makes it look a bit like a string. Einstein came up with the idea that the quantization had something to do with particles. In all cases, the quantum constraints act upon the amplitude of a wave not the frequency of the wave.

The Bohr-Somerfeld quantization condition does not determine the frequency of the electromagnetic wave. BSQ applies to the electron, not to the photon. BSQ is not the same as the Einstein condition, E=hf.

13. Jan 31, 2013

### naturale

The amplitude of a quantum wave is fixed by the Born rule, i.e. unitary modulo square. That is, we have string vibrating always with amplitude 1. The only variables are the fundamental frequency of the vibration (length of the string) and how the harmonics are populated.

To obtain the frequency spectrum of a string you can use BSQ without the Planck constant. If the fundamental period is T and the string is homogeneous
$\int f_n dt = f_n T = n$, that is $f_n = n / T$
the quantization of the mode with period T, say, in a Black Body radiation is
$\int E_n dt = E_n T = h n$, that is $E_n = n h / T$

BSQ can be used to quantize photon, in this case the difference is the bose statistic in the population of the harmonics, and the fact that the photon is a particle with zero mass lambda = c T. The effect of the Coulomb potential of an hydrogen atom is a distortion of the spectrum. In our analogy this correspond to a non homogeneous string

14. Jan 31, 2013

### marksesl

This gets to an issue I was asking elsewhere. Is there an electromagnetic frequency independent from the frequency contained within the photons? It seems odd that the photon's probability wave is one in the same as the wave that we think of that focuses in cameras, or makes colors in our brain. Somebody told me that Maxwell’s equations deal with electromagnetic radiation, which has nothing to do with the probability waves of the photons, but nobody can show me any distinction.

15. Jan 31, 2013

### marksesl

This is also my exact question, but I'm still confused after reading the answers. All kinds of equations have constants; that doesn’t mean all those things are quantized. Pi is a constant relating the circumference of a circle to its diameter. Does that mean circles are quantized? Of course not. There has to be some relationship between frequency and energy, so Planck’s constant just turned out to be that number. I just don't understand how Planck's constant is identified with quantum amounts. Please explain and give clear examples.

16. Jan 31, 2013

### f95toli

In many cases it does. While we dont' talk about quantized for purely mathematical relationships, there are number of quantized phenomena in physics. In electromagnetics you have the abovementioned photon but also charge (1e=1.6022e-19 C or 2e depending on whether or not ou are working with normal metals or superconductor) and the magnetic flux quanta (2e^-15 Wb). Sometimes we also talk about resistance quanta (from the Hall effect)
All of of these can be directly measured because there are periodic phenomena where the period is set by the the quanta or a fraction thereof.
Note that the fact that they are periodic does not stop you from measureing e.g. a charge of 0.2e (which can happen because of screening).

In the case of e you have e.g. Coloumb blockade, for resistance the quantized Hall effect and for the magntic flux quanta Aharononv-Bohm rings or (more common) superconducing SQUIDs.

17. Jan 31, 2013

### marksesl

Ok, I got it figured out. It's explained here:

In E=hv that number can only be multiplied by a whole number.
So, the equation actually becomes E=nhv. For blue light, for example, the value for hv is about 3. So, only 3, 6, 9, 12 electron volts are allowed for blue light.

Last edited by a moderator: Sep 25, 2014
18. Jan 31, 2013

### f95toli

This is not correct. Blue light has an enery of around 3 eV, 6eV would be outside the visible par of the spectrum. Again, "quantized" does not mean that the light can only take on values that are integer multiplies of a specific number.

Your equation is the equation for e.g. multi-photon excitation of a transition in an atom.

Last edited by a moderator: Sep 25, 2014
19. Jan 31, 2013

### marksesl

Ok, so a blue photon is just always 3eV. Correct?

"The classical frequency of your light determines the quantization of the photons (as packets of h*nu energy). You can vary the classical frequency of your light continuously, and for every value it takes, you get a different quantized energy for your photons."

So what is the classical frequency of light as opposed to the frequency contained in the photons?

20. Jan 31, 2013

### Staff: Mentor

The frequency associated with the photon equals the classical frequency of the light that it corresponds to.

The n=3 to n=2 transition in hydrogen produces photons with energy 1.9 eV, frequency f = E/h = 4.59 x 10^14 Hz, and wavelength λ = c/f = 6.53 x 10^-7 m = 653 nm.

If we have a few bazillion of these photons (give or take), we have a classical electromagnetic wave with that frequency and wavelength.

How much is a bazillion? Consider sunlight at the Earth's surface. Hold up a 1 m^2 screen facing directly towards the sun on a clear day, and in one second it will receive about 1500 joules of electromagnetic energy. The light contains all visible wavelengths, of course, but let's pretend it's monochromatic with wavelength 653 nm. Then each photon carries 1.9 eV = 3.04 x 10^-19 J of energy, so one second's worth of light on the screen contains about 1500 / (3.04 x 10^-19) = 4.93 x 10^21 photons.

Last edited: Jan 31, 2013
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