# What is a photon in respect to electromagnetic waves?

1. Jan 18, 2012

### khkwang

I've always thought that photons and electromagnetic waves are one in the same. And I still do, but I'm trying to get a better grasp on the idea and am finding it difficult.

1) As I understand it, they are the same. But an electromagnetic wave with definite frequency is a perfect sine wave and so goes on forever. I assumed previously that a single photon corresponded to a electromagnetic wave of definite frequency but that can't be, else the photon exists in an infinite length... so then is a photon an electromagnetic packet? If so, then a single photon must be multiple electromagnetic waves. So then what does a single electromagnetic wave of definite frequency represent?

2) I also thought that the electric field of an electromagnetic wave was related to the probability of a photon's position. If this is so, then the probability of the photon existing in a location is highest at the wave's crests and troughs and zero at the nodes. I'm positive that what I just said is wrong, but it's currently how I visualize it so please someone steer me in the correct direction.

3) Also if the electric field portion of the wave describes the photon's probability, then what does the magnetic field represent?

4) If the EM wave is just an oscillating charge, would I be able to produce an EM wave by moving an electron up and down over and over again? If so would I be producing a photon then?

What I said probably doesn't make any sense... so please if anyone could help correct me...

2. Jan 18, 2012

### Hydr0matic

They are not the same so much as different interpretations of the same phenomenon - light. Electromagnetic waves represent the classical physics view of light, and photons are a part of quantum theory's description of light.
Common mistake is to try and understand photons and EM waves as a single entity part of the same theory. They are not. There are no continuous EM waves in quantum theory and there are no photons in classical physics.
In quantum theory, the probability of finding a photon at a certain position is described by its wavefunction, which propagates through space as a wave. The probability density vector of this wave behaves the same way as an EM wave's electric field vector, but they are not the same thing.
You would, yes. This is what happens in a radio antenna for example.
According to classical physics, you will produce a continuous EM wave. According to quantum theory, you will produce lots of photons.

3. Jan 18, 2012

### khkwang

Wow okay, so I guess that means we're still at a standstill between consolidating the classical and quantum mechanical interpretations? Thank you Hydr0matic, you've helped clear up something very fundamental for me.

I do have a couple more questions though if you're willing to stick around for a moment...

1) In interference experiments such as those involving interferometers, which wave is it that is being measured? The EM wave or the wave function (as you say they are not the same thing)? If it is the EM wave, then it kind of confuses me because in a way it is determining the intensity of the light, which I'm assuming is related to the probability of a photon's arrival. So then it seems like the EM wave is related to the wave function.

2) About the uncertainty principle; If we shoot a photon through an extremely tiny hole, it's my understanding the the possible final locations smear out. But if we were to measure backwards from where the final position was to the position of the hole we can determine the velocity after leaving the hole AND the position to extreme accuracy. So we are allowed to know both the momentum and position fully at the same time, but only after the fact. Is this allowed?

4. Jan 19, 2012

### Demystifier

They are not. Photon is a state in the Hilbert space on which the electromagnetic-field operator acts. Photon is NOT an eigenstate of the electromagnetic-field operator, i.e., the value of the electromagnetic field is UNCERTAIN for a photon. Likewise, a state with a certain value of electromagnetic field contains an uncertain number of photons.

5. Jan 19, 2012

### Hydr0matic

AFAIK, there is no effort to consolidate these two. Quantum theory came about due to the failures of classical physics to explain certain phenomena, most importantly the hydrogen spectra. The hydrogen series and Rydberg formula was discovered in the 1880s, but it wasn't until 1913 that Bohr proposed his quantized atomic model. You really need to read physics from a historical perspective to understand the current state of theories.
What you measure is light. The result of the experiment can then be interpreted or explained with whatever theory you like - classical, quantum theory, etc.. or if you're ancient greek - "Zeus did it". Ofcourse, some theories are more successful than others, which is why quantum theory is used. The wave function is not a "real" physical entity like an EM wave - see this, and compare with this for example.
I would say yes.

6. Jan 21, 2012

### eaglelake

1) An interferometer experiment with waves is different than one with photons. The wave will split and travel both paths simultaneously. Likewise, both detectors will be triggered at the same time by a wave. But with photons the detectors are triggered one at a time, never simultaneously.

The electromagnetic wave is not related to the quantum probability. The electric and magnetic waves satisfy the classical wave equation, which is second order in time. They are real waves that transport energy and momentum through space-time. On the other hand, the quantum wavefunction satisfies Schrodinger’s time dependent equation, which is first order in time and which does not have real solutions. The quantum wavefunctions are necessarily complex, not real. They are defined as complex scalar products in a Hilbert space. As far as we know, they do not have energy and momentum for us to measure.

2) Once the photon hits the detection screen, we have a position measurement result and the experiment is over. Bohr called this “closure”. You can then repeat the experiment, if you like, but there is no “after the (measurement) fact”. Or, you can do a different experiment that determines the position and momentum at the tiny hole, as you suggest. That is, of course, also possible.

The uncertainty principle does not prohibit knowing both momentum and position at the same time. But it does prohibit repeating the same experiment a great many times and always getting the same value for momentum and always getting the same value for position. Then the uncertainty would be zero for both momentum and position in the same experiment. And that is impossible.

7. Jan 22, 2012

### Hydr0matic

This would only be the case if you had a perfect source, perfect identical beamsplitters and perfect identical detectors. In reality, you do not. if you performed this experiment and both detectors were triggered simultaneously, that would not imply that light is a wave. One could simply argue that they were hit by two seperate photons. Likewise, if only one detector is triggered, this does not imply that light is quantized. The two split waves will likely never have identical intensity, and if their intensity borders on what is detectable, only one of them will be detected.
I've never heard this interpretation. From wiki:
This says the principle is applicable on a single particle at any given time.

8. Jan 23, 2012

### ytuab

Sorry. I want to know how the next experiment (Fake of Bell inequality violation) has been thought since it was published.

Fake violation of Bell tests reinforce inportance of closing loopholes.

In the current study, the scientists showed that Eve can send strong, classical light pulses, with a polarization of her choice, into both Alice and Bob’s photon detectors at the same time. This classical light produces photocurrents that are interpreted as photons. Therefore, Alice and Bob are unknowingly measuring classical light pulses, which means that some of the coincidences that they count are not due to quantum entanglement but to Eve’s manipulation.

---

This very bright pulse causes sufficient photocurrent to cross the detection threshold only in one of the four polarizer settings. In the other three polarizer settings, the polarizer partially blocks light, the photocurrent stays below the threshold and the detector remains blind. Thus, Eve classically controls exactly which of the four polarizations Alice and Bob register.

So this experiment showed that the classical light can maipulate photon detection in photodetector (above or below detection threshold ) ?

9. Jan 26, 2012

### eaglelake

When discussing the fundamental principles of quantum mechanics we always assume ideal experiments to purposely exclude effects caused by any limitations in the devices used in the experiments. It is understood, then, that any non-classical behavior is due to the quantum nature of the experiment and not due to any human error or experimental imperfections.

We are currently able to do real experiments with only one photon in the experimental apparatus at any time. Only one detector is ever triggered. This is an experimental fact. We never see the two detectors in a Mach-Zehnder interferometer, for example, triggered simultaneously when only one photon is present. This seems obvious if there is only one photon available to do anything. A wave, being associated with a continuum, should trigger both detectors at least some of the time, even with imperfect devices.

The effect is more dramatic with photons hitting a detection screen, which is a continuum of detectors. In a one-photon interference experiment, the one photon produces one dot on the screen. We do not observe the total distribution of dots all appearing simultaneously. (As an aside, how would we get a zillion photons from the original one??) The interference pattern is built up one dot at a time, not continuously as a wave would do. The result obtained in a single measurement is always a single dot on the detection screen.

The point is this – Some experiments with light exhibit particle properties. This has been demonstrated in many experiments done over many years. That is what is being discussed here. That is why quantum mechanics was invented. The wave nature of light cannot explain the results of all these experiments. Nor can we explain all such experiments as being due to imperfections in our instruments or due to human ignorance, as you suggest.

Quantum mechanics is indeterminate. There are many different possible results of a quantum measurement and, generally, quantum mechanics does not predict which result will happen. It only predicts the probability of getting each possible result. Each result is an eigenvalue of the observable being measured, For example, if we perform a position measurement, there are very many locations where the particle can be found. If we repeat the same experiment many times we generate a statistical distribution of all the dots that contains the entire eigenvalue spectrum of the position operator. The position does not have a unique value as it does in classical physics where the same experiment always yields the same result. This is what we mean when we say that the position is uncertain. (Bohr, among others, preferred to say the position is “indeterminate”, which I believe is more descriptive and less confusing.) The uncertainty principle is a consequence of the purely statistical nature of quantum events.

Theoretically, the uncertainty in position is defined to be the root-mean-square deviation from the mean value of all the measurement results, called the standard deviation in ordinary statistics: $$\Delta x = \sqrt {\left\langle {\left. {\psi \left| {\left( {\hat x - \left\langle {\left. {\hat x} \right\rangle } \right.} \right)^2 } \right|\psi } \right\rangle } \right.}$$ Likewise the uncertainty in momentum is:
$$\Delta p_x = \sqrt {\left\langle {\left. {\psi \left| {\left( {\hat p_x - \left\langle {\left. {\hat p_x } \right\rangle } \right.} \right)^2 } \right|\psi } \right\rangle } \right.}$$ ]. Notice that the uncertainties depend on the wavefunction $$\psi (x).$$ Every wavefunction gives uncertainties that satisfy $$\Delta x\Delta p \ge \hbar /2$$. Notice, also, that the uncertainties depend on the operators $$\hat x$$ and $$\hat p$$.

This is how we calculate uncertainties. It is these definitions, along with the commutation relation $$\left[ {\hat x,\hat p_x } \right] = i\hbar$$, that gives the uncertainty relation [tex}\Delta x\Delta p_x \ge \hbar /2[/tex]
I apologize for having to resort to the mathematical formalism, but, in the hope of minimizing misconceptions, the exact definitions should be used in any discussion of the uncertainty principle.

Of course, it is $$\left| \psi \right|^2$$ that predicts the statistical distribution of all the measurement results. When we see the statistical distribution of many repeated position measurements, we know there is an uncertainty in position. If we always get the same position in repeated measurements, then, when we calculate the position uncertainty, we get $$\Delta x = 0$$, and we say the position is absolutely certain, as in classical physics. If the statistical spread is clustered around only one location, then it is or less uncertain. If the dots are scattered over a wide range of positions, then the position is more uncertain. In any case, you must repeat the experiment many times to determine whether the position is certain $$\Delta x = 0$$ or whether it is uncertain $$\Delta x \ne 0$$.

We intentionally use “certain” in place of words like “accurate” and “precise”, which refer to a single measurement and which can be misleading in this discussion. A single measurement tells us nothing about uncertainties. Uncertainties are not measured values. When a particle hits a detection screen we see a dot on the screen that gives the particle’s position at the instant it hit. We have a value for the position. Yet, there is no way to obtain the uncertainty in position from that number.

We repeat, for emphasis, THE UNCERTAINTY PRINCIPLE IS ABOUT UNCERTAINTIES. (I apologize for yelling, but too many discussions ignore this essential fact.) It is not about the actual values of position and momentum obtained in single measurements. The uncertainty principle tells us that there is no experiment, and no wavefunction, for which both position and momentum are certain. The uncertainty principle also states that an uncertainty in position is accompanied by an uncertainty in momentum, but in no case will the product of the uncertainties be less than $$\hbar /2$$.

The part of the Wiki article you site is

In quantum mechanics, the Heisenberg uncertainty principle states a fundamental limit on the accuracy with which certain pairs of physical properties of a particle, such as position and momentum, can be simultaneously known. In layman's terms, the more precisely one property is measured, the less precisely the other can be controlled, determined, or known.

IMHO, a more meaningful statement is

In quantum mechanics, the Heisenberg uncertainty principle states a fundamental limit on the product of certain pairs of uncertainties, such as the position uncertainty and the momentum uncertainty. There is no experiment in which the product of those uncertainties can be less than $$\hbar /2$$. The more certain we are of one property, the more uncertain is the other.

Best wishes

10. Jan 26, 2012

### Hydr0matic

When discussing classical physics and waves, the single detections are caused by the limitations in the devices, so we can't assume ideal experiments.
See section 2.1 in Generation of single photons and correlated photon pairs using InAs quantum dots.
If what you're saying is in fact "an experimental fact", a concept called "suppression of the two-photon probability" would not exist.
Also see this article (published three days ago) about a new innovative single photon light source. If creating single photons is as easy as you believe, why are physicists still working on improving these sources?
A "continuum of detectors" cannot exist. There will always be a limit on resolution, as well as a limit on minimum energy required to excite cells.
You seem to be defining a photon by what a single photon source emits and a single cell in a detector detects. You are aware that light can hit a detector and not set it off? See Quantum efficiency. Just because light waves diffracted through a slit will spread over an area of the detector doesn't mean multiple cells has to go off simultaneously. If a "single photon" source is used, the initial intensity borders on what can be detected even without the diffraction. So if you add diffraction, it's only the peak in the distribution that's intense enough to trigger a detector cell.
Where did I suggest that? The OP question was "What is a photon in respect to electromagnetic waves?", not "is light a particle or a wave?". As you can see in a previous post, I already explained that "Quantum theory came about due to the failures of classical physics to explain certain phenomena". But in order to understand the difference between photons and electromagnetic waves you have to explain the concepts and predictions of both theories.
Why not? According to any "individual system" interpretation (e.g. Copenhagen), the uncertainty in position at that instant would be confined to the extent of the detector cell that was hit.
That's one interpretation, one of many. The interpretation you're describing seems to be the Ensemble interpretation. Which, ironically, is the one closest to classical physics where particles have real properties and no wavefunction collapse is necessary. But this is not the most common view, so you shouldn't state it as if it was fact.

11. Jan 27, 2012

### Demystifier

While it is true that there are many interpretations, the ensemble interpretation of the uncertainty principle (UP) is the only interpretation of UP which can be directly tested experimentally. So from an experimental point of view, the ensemble view of UP is more than a mere interpretation; it is an experimental fact!

12. Jan 27, 2012

### f95toli

One reason is that most good single photon sources operate at unusual wavelengths, i.e. wavelengths where you have to use expensive and difficult to use equipment (because e.g. it requires cooling to cryogenic temperatures). What is needed for commercial applications are good sources and detectors that work at standard telecom wavelengths so that normal fibre networks can be used. The equipment also needs to be cheap and reliable.
There is also a need for sources and detectors that work at other frequencies (lets say MW frequencies).
Remember that the field of single photonics is mainly driven my cryptography, and that is very much reflected in the work that is done.The foundations of QM is very much a non-issue for most people.

Also, yes you will always have dark counts etc. ; but it is down to the skill and experience of the people performing the experiments to take that into account. There is no such as a perfect experiment, and in any real-world situation there are effects that needs to be taken into consideration and corrected for.