Why is Photon Energy Quantized in Terms of Sine Wave Frequency?

[Vanadium 50]

philosophical,

We don't discuss philosophy here.

But "why are the single energy solutions sines?" is no different than "why are graphs of quadratic equations parabolas?" They are the logical consequences of the equations, and even philosophers don't doubt these consequences.

 

[PeroK]

However, my core question remains: Why is this emergent frequency specifically related to sinusoidal (harmonic) waves? Why not square waves, triangular waves, or any other waveform for that matter? I'm curious about the inherent nature of light that leads us to describe its frequency using sinusoidal waves as opposed to any other shape.

That question has been answered already and is covered in any undergraduate textbook on EM. The wavelike solutions to Maxwell's equations are (sinusoidal) oscillating electric and magnetic fields. This led to Maxwell concluding that light was EM radiation in the first place.

Once you have recovered Maxwell's equation from quantum theory- which is mathematically hard - the derivation is the same.

Also, in what Feynman book can I see what you say about how classical wavelike phenomena emerge from a probabilistic quantum theory?

The Strange Theory of Light and Matter, which you said you'd already glanced at.

 

[PeroK]

Could you point me to a demonstration or a source where it is shown that the diffraction pattern can be specifically associated with a classical sinusoidal wave of a certain frequency?

The diffraction of light depends only on the wavelength. This is covered in numerous online physics material, such as the Khan Academy.

That the light waves are oscillating sinusoidal EM fields is irrelevant to diffraction. The diffraction of light was studied long before Maxwell. All that was known was that light was a wave of a measurable speed, frequency and wavelength. Maxwell showed us what sort of wave it was.

 

[Fra]

They are the logical consequences of the equations, and even philosophers don't doubt these consequences.

Yes, so provided OP understands the answers so far (?) it seems the residual question is:

"why is energy is defined by introducing the operator "ihd/dt" on the wavefunction?"

In CM we have concepts such as momentum and energy. In QM we in a sligthly ad hoc manner redefine all generalized momenta (from hamiltons formulation of CM) to a conjugate variable.

So strictly speaking "energy" in CM is not the same thing as "energy" in QM.

But as we want a classical correspondence we keep the name.

one mystery if QM is, how can we get away with just redefine things and still make it work? And if its not he only way of doing it, what is so special about conjugate variables? This was what i tried to comment on in my other post.

Once you get past this you realise the QM terms, unlike CM terms, are sort of defined in terms of information. This is also more natural as we are in QM describing things we can not direcetly see or visualize. Therefore we have to describe instead the information about the subatomic world in terms of data we collect in the macroscopic lab. So new abstract definitions seemed required.

Some of us wonder, how come certain mathematics describes nature in simpler ways than others?.

We decide for ourselves when we think there is something more to gain by asking why.

/Fredrik

 

[vanhees71]

Quibble: Because photons do not interact directly, this interaction ( "detection"?) is central to the counting problem associated with light in a cavity "black body" at a defined temperature. So I disagree or we are drowning in the semantic sea?

@PeterDonis is right with giving the caveat that a "very dim laser light" is not a single-photon state but still a coherent state. It's with good approximation a single-mode coherent state with a frequency $\omega$, but it's not an eigenstate of the electromagnetic field energy nor a photon-number eigenstate. The photon number is Poisson distributed. If the intensity, i.e., the mean photon number is (much) smaller than one, it's "mostly a vacuum state", and you see randomly single points at a time on the screen is just due to the fluctuations of this field, but it's not a single-photon source.

Of course, nowadays, it's no problem to do the double-slit experiment with true single-photon sources. Usually you just use spontaneous parametric-down conversion to produce entangled photon pairs and use one as the photons as the "trigger", i.e., to "announce" that there's also another photon in a single-photon state which you let go through the slits. Also then you get a random point on the screen, and collecting many such single-photons on the screen the diffraction (double-slit interference) pattern builds up as long as you do the experiment in such a way that it is impossible to observe through which slit each photon went. So here you observe "wave properties" of the single-photon state. If you somehow mark the photons, e.g., with polarizers in the slits, such that you can infer which-way information (in principle, it doesn't even matter if you really measure the polarization of each photon behind the slit) the double-slit interference pattern goes away, and you get the incoherent superposition of two single-slit interference patterns. In this sense you demonstrate "particle properties" of the single photons with this variant of the experiment.

 

[hutchphd]

That said, even in the cavity case the radiation is a coherent state, not a Fock state, and so it is not in an eigenstate of energy and the E in the Planck relation still does not describe "the energy of a photon" in any sense that corresponds to an actual observation. It is an abstract "energy" that appears in the distribution function.

I fear we are in fact adrift on the semantic sea.
When the thermodynamic counting is done, the associated parameter E involves projection onto the energy eigenbasis. This is a "measurement" only on paper, so I apologize for being unclear. But that is the "energy of a photon" in my mind.

 

[vanhees71]

Just to clarify the terminology:

In a cavity (which is easier because of the somewhat delicate "infinite-volume limit" in QFT) you have
$$\hat{H}=\sum_{\lambda} \sum_{\vec{k}} |\vec{k}| \hat{N}(\vec{k},\lambda).$$
A complete set of energy eigenstates of the free em. field thus is given by the corresponding Fock basis,
$$|\{N(\vec{k},\lambda) \}_{\vec{k,\lambda}} \rangle=\prod_{\vec{k},\lambda} \frac{1}{\sqrt{N(\vec{k},\lambda)!}} \hat{a}^{\dagger N(\vec{k},\lambda)} |\Omega \rangle.$$
The energy eigenvalues are
$$E[\{N \}]=\sum_{\lambda,\vec{p}} |\vec{k}| N(\vec{k},\lambda),$$
i.e., each occupied single-photon state contributes $E_{\vec{k}}=|\vec{k}|=\omega$.

The thermal state is given by the statistical operator,
$$\hat{\rho}=\frac{1}{Z} \exp(-\hat{H}/T)$$
with the partition sum
$$Z=\prod_{\vec{k}} \left [\frac{1}{1-\exp(-|\vec{k}|/T)} \right]^2,$$
where each factor comes from summing over $N(\vec{k},\lambda) \in \{0,1,2,\ldots\}$. With the two cases $\lambda=\pm 1$ giving the same factor (that's why each factor is squared in the product).

A single-mode coherent state is given by
$$|\alpha,\vec{k},\lambda \rangle=\sum_{n=0}^{\infty} \exp(-|\alpha|^2/2) \frac{\alpha^n}{n!} \hat{a}^{n}(\vec{k},\lambda) |\Omega \rangle.$$
The photon numbers are Poisson distributed,
$$P[N(\vec{k},\lambda)=n]=\exp(-|\alpha|^2) \frac{|\alpha|^{2n}}{n!}$$
with
$$\langle n \rangle =\langle n^2 \rangle-\langle n \rangle^2=|\alpha|^2$$
The mean energy is
$$\langle E \rangle=|\alpha|^2 \omega, \Delta E=|\alpha| \omega.$$

What is now the precise question, we try to answer in this thread, concerning either of these states, loosely mentions in this thread?

 

[PeterDonis]

When the thermodynamic counting is done, the associated parameter E involves projection onto the energy eigenbasis.

But if you do that with a state that is not an energy eigenstate, such as a coherent state, you get either a superposition of multiple energy eigenstates with different amplitudes, or an expectation value (depending on what you mean by "projection"). Neither of these are "the energy of a photon".

But that is the "energy of a photon" in my mind.

I disagree. See above.

 

[vanhees71]

The energy of a single photon is of course well defined, because "a photon" is described by the single-photon Fock state, where $N(\vec{k},\lambda)=1$ and $N(\vec{k}',\lambda')=0$ for all $\vec{k}' \neq \vec{k}$, $\lambda \neq \lambda'$. This is an eigenstate of the em.-field energy with the said eigenvalue, $\omega=|\vec{k}|$. In the case of the now discussed "cavity QED" this is a proper state (in contradistinction to the infinite-volume case, where a plane wave never can represent a proper state).

 

[PeterDonis]

The energy of a single photon is of course well defined

For a Fock state, yes, as you say. But that does not mean the term "the energy of a photon" can be used in a meaningful sense for states that are not Fock states, which is what other posters in this thread are trying to do.

 

[hutchphd]

Neither of these are "the energy of a photon".

But they are the energies to which Planck was referring in the quantum relation Planck compounded for a cavity black box which was the original question. Semantics, see.

 

[PeterDonis]

they are the energy Planck was referring to

"They" doesn't make sense. Planck wasn't referring to a "they". He was referring to a single thing, not multiple things. Unfortunately there is no single thing that has the properties Planck originally assumed. The fact that Planck proposed his relation with some idea in mind does not mean that idea is still considered to be valid.

the relation Planck compounded, which was the original question

I do not read the OP's question as being about Planck's original proposal, but about our best current understanding of what the Planck relation means--or doesn't mean, which is more to the point.

 

[hutchphd]

They: more than one individual energy.

 

[PeterDonis]

They: more than one individual energy.

More than one individual energy of what? And don't answer "a photon", because the whole point is that there is no such thing as "a photon" in the states that are relevant for black-body radiation (i.e., coherent states), which is the scenario that Planck's proposal was intended to apply to.

 

[hutchphd]

Please see #69 . Semantics I fear.

 

[PeterDonis]

Perhaps an alternate response to the OP question will help: the actual physical law that Planck proposed was not $E = h \nu$, it was his black-body radiation law:

$$
B_\nu(\nu, T) = \frac{2 h \nu^3}{c^2} \frac{1}{\exp \left( \frac{h \nu}{k T} \right) - 1}
$$

(Note that we now understand this law as the law for bosons; for fermions the last term in the denominator on the RHS is $+1$ instead of $-1$.)

Planck proposed the formula $E = h \nu$ as a heuristic justification for the above law, but the actual physics is the above law, and even though we now do not think Planck's original heuristic justification was valid, the above law is still valid (we just derive it now using different reasoning from the reasoning Planck originally used).

 

[hutchphd]

Yes but the fundamental Ansatz was regarding the counting.

 

[PeterDonis]

Thread closed for moderation.

 

[PeterDonis]

The OP question has been addressed. Thread will remain closed. Thanks to all who participated!