Is this equivalent to a photon?

[hobz]

In the Feynman lectures on physics, I encountered a sum of sinusoids having almost identical frequency. The resulting sinusoid had a very small location in space, so to speak. My question is; is this resulting wave packet identical to what is called a photon?

 

Consequently, the location in space is dependent on the number of sinusoids that are summed. If the number of sinusoids is finite, the wave packet will be periodic, and there by the same "photon" (assuming that the answer to my first question is "yes") is spread, like knots on a rope.

 

What makes up all these sinusoids? (I have encountered a similar discussion in the "Beyond The Mechanical Universe"-video series, but they fail to tell us of the origin of the sinusoids. (They simply say, "If waves of different wavelengths are added... especially if the wavelengths are close together... the result can be a kind of wave concentrated in a limited region of space.")

The wavelengths that are added are [math]\frac{h}{\lambda}[/math].

Edited July 16, 2008 by hobz
Missing \ in LaTeX

[ajb]

I don't think it is a photon as photons are a feature of quantum field theory and I assume the description of the sinusoids is classical, i.e. using Maxwell's equations.

 

But then, you have not explained what you mean by "equivalent".

[swansont]

No, they're not the same thing, but I'm struggling for a good description of why. I think a wave packet is, at best, a description of a photon under some circumstances. They aren't the same thing.

[hobz]

By "equivalent" I meant, are they identical. Mea culpa for not being precise enough.

 

I am not sure that it is a classical sinusoid. I speculate that one could obtain the probability of the photon being along a direction by squaring the wave packet?

[Pete]

In the Feynman lectures on physics, I encountered a sum of sinusoids having almost identical frequency. The resulting sinusoid had a very small location in space, so to speak. My question is; is this resulting wave packet identical to what is called a photon?

No. A photon consists of a single frequency/energy. What you have described is a sum of states having different frequencies/energies.

 

Pete

[D H]

No, they're not the same thing, but I'm struggling for a good description of why. I think a wave packet is, at best, a description of a photon under some circumstances. They aren't the same thing.

"The map is not the territory."

[swansont]

"The map is not the territory."

 

Yes! Succinctly put.

 

No. A photon consists of a single frequency/energy. What you have described is a sum of states having different frequencies/energies.

 

Pete

 

A photon will also have an inherent uncertainty associated with it.

[Mr Skeptic]

Wouldn't that just be two photons?

[hobz]

A photon will also have an inherent uncertainty associated with it.

Is this uncertainty related to the wave packet? That is, the region in space "occupied" by the wave packet is related to the probability of the photon "actually" being there?

 

(I realized that the uncertainty could be interpreted as a property of the photon, and thus there is not "actual" location -- the photon is confined (by the wave packet) to a certain region of space by probability, but in a sense is in all locations probable at the same time.)

[swansont]

The wave packet is the description of the photon, it does not physically confine it.

[hobz]

The wave packet is the description of the photon, it does not physically confine it.

 

I see! So by assuming a very definite momentum for the photon, which I interpret as the ability of predicting where the photon will "go", we get a infinite sinusoid that tells us that the photon is anywhere in space!

And the inverse is also true. If the photon (in it's wave packet interpretation) is located in some region in space, it has a variety of momenta that makes it impossible to predict where it will "go"?

 

The same in other terms.

Knowing where the photon will "go" (or along which path it will travel if you will), makes it impossible to know where in space it is located.

Knowing where the photon is located, makes it impossible to know where it is "going"?

[Pete]

Knowing where the photon will "go" (or along which path it will travel if you will), makes it impossible to know where in space it is located.

Knowing where the photon is located, makes it impossible to know where it is "going"?

If you are making this statement based on the Heisenberg's Uncertainty Principle (HUP) then what you said isn't quite right. There is nothing in quantum theory that implies that you can't simultaneosly measure the position and momentum to arbitrary precision at the same time. This does not violate HUP. HUP concerns a statistical relationship between observables, i.e. it applies to a large number of measurements on an ensemble of identically prepared systems. The uncertainty is inherent in the quantum state itself and therefore as an indeterminancys which is inherent to the state itself.

 

A photon will also have an inherent uncertainty associated with it.

I disagree. Uncertainty is a statistical quantity which applies to an ensemble of identically prepared systems. A single photon cannot have an uncertainty associated with it. Uncertainty is a statistical quantity which applies only to a large number of data points.

 

Pete

Edited July 26, 2008 by Pete
multiple post merged

[swansont]

I disagree. Uncertainty is a statistical quantity which applies to an ensemble of identically prepared systems. A single photon cannot have an uncertainty associated with it. Uncertainty is a statistical quantity which applies only to a large number of data points.

 

Pete

 

You won't know where on that probability distribution your data point will lie until you actually measure it, and you can't know what the frequency was at any time other than when you measured it.

[Pete]

You won't know where on that probability distribution your data point will lie until you actually measure it, and you can't know what the frequency was at any time other than when you measured it.

I'm not sure what that has to do with what I posted. Can you clarify this for me? Thanks.

 

In the meantime consider what I posted above. You wrote A photon will also have an inherent uncertainty associated with it. That is incorrect. The problem with that assertion is that you are attempting to contribute a statistical meaning to a single measurement. Doing so makes no sense at all. Recall that uncertainty refers to a standard deviation. In fact the terms uncertainty and standard deviation are actually synonymous, i.e. they both refer to exactly the same thing.

 

In your last post you wrote

You won't know where on that probability distribution your data point will lie until you actually measure it, and you can't know what the frequency was at any time other than when you measured it.

A data point does not lie on a probability distribution. The probability distribution is determined by a collection of data. When the probability distribution is determined then and only then can the uncertainty can be determined. E.g. in the case of a gaussian probability density it will be about the same as the width of the gaussian curve. Note how the value of the uncertainty remains finite even if the errors in measurements were exactly zero. If one were to determine the momentum distribution then the width of that curve would be inversely proportional to the width of the first curve. This is meaning of the HUP.

 

Pete

[hobz]

I don't think there is experimental evidence to support your statement. As far as I can tell, a slit experiment shows us that the more we know about a photon's position (going through the slit) the less we know about where it's going. That is, we get a wider distribution on the photographic plate.

 

Saying that each individual photon has an uncertainty associated with it, is no different than saying each throw of a die has an uncertainty associated.

[Pete]

I don't think there is experimental evidence to support your statement.

On the contrary, there is overwhelming evidence of what I've just explained. If you were to actually post the definition of uncertainty then what I said would be glaringly obvious. So in your the next post I request that you, or someone else, please post the definition of uncertainty. Thank you!

 

I'm merely explaining the meaning of ther term uncertainty. Its meaning is quite often misundestood and more often than not confused with the concept of error in measurement.

As far as I can tell, a slit experiment shows us that the more we know about a photon's position (going through the slit) the less we know about where it's going. That is, we get a wider distribution on the photographic plate.

Did you notice what you did here? You actually agreed with me. The distribution on the photographic plate is a statistical oine. It is totally impossible to determine what the uncertainty is until a distribution pattern is established. Once that is done then it becomes quite apparent that the uncertainty in position cannot be defined for a single instance in the measurement of the location of a particle.

Saying that each individual photon has an uncertainty associated with it, is no different than saying each throw of a die has an uncertainty associated.

That is not what uncertainty means. In the comment you just made you were implicitly refering to probability, not uncertainty. The uncertainty of a roll of a die cannot possiblly be measured by a single roll of the die. To determine the uncertainty one keeps rolling the die and recording the number rolled. From that data one then calculates the standard deviation of the number rolled. That will give you the uncertainty in the number rolled. It sghould be quite apparent that the uncertainty (aka standard deviation) cannot be determined by a single roll, nor would it be meaningfull to even talk about unless a large number of rolls is executed and the results recorded.

 

Question: You do understand that the term uncertainty is merely a different name for the term standard deviation, right?

 

To summerize - Individual measurements of observable quantities does not have an uncertainty associated with them. Only quantum states themselves have an intrinsic uncertainty associated with them. Different quantum states -> Different values of the uncertainty of a particular observable.

 

Pete

Edited July 27, 2008 by Pete

[swansont]

I'm not sure what that has to do with what I posted. Can you clarify this for me? Thanks.

 

In the meantime consider what I posted above. You wrote A photon will also have an inherent uncertainty associated with it. That is incorrect. The problem with that assertion is that you are attempting to contribute a statistical meaning to a single measurement. Doing so makes no sense at all. Recall that uncertainty refers to a standard deviation. In fact the terms uncertainty and standard deviation are actually synonymous, i.e. they both refer to exactly the same thing.

 

I'm not talking about measurement statistics.

 

If I get a photon from a transition with a particular lifetime, there will be an energy width associated with that transition, and the photon I get will likewise have an uncertainty in its energy. You can't say the photon has a particular frequency until you measure it.

[Pete]

I'm not talking about measurement statistics.

Huh?????

 

Please please post the definition of of the term uncertainty as it pertains to quantum mechanics.

 

Pete

[hobz]

That is not what uncertainty means. In the comment you just made you were implicitly refering to probability, not uncertainty.

You are of course correct. I meant probability, not uncertainty. But the wave packet is an uncertainty (to my understanding). The probability of the photon being "present" is spread out in space, and is not measurable to arbitrary precision at the same time as the momentum?

This makes it hard to understand your statement

There is nothing in quantum theory that implies that you can't simultaneosly measure the position and momentum to arbitrary precision at the same time.

because I thought that HUP prevented this, so to speak.

[swansont]

Huh?????

 

Please please post the definition of of the term uncertainty as it pertains to quantum mechanics.

 

Pete

 

 

What I'm talking about is the lack of precision in the variable in question, because it's described by wave mechanics.

[Pete]

What I'm talking about is the lack of precision in the variable in question, because it's described by wave mechanics.

Uncertainty is not a lack of precision, The uncertainty of an observable has a precise mathematical expression which is a function of the state vector. Do you know what that expression is and how its value is determined from measurements made in the lab?

 

And it need not be related to a wave function. E.g. its possible that a state can exist for spin in which the uncertainty in the spin is zero, even though the value of a particular measurement of spin can only give one of two values.

 

You are of course correct. I meant probability' date=' not uncertainty. But the wave packet is an uncertainty (to my understanding).

[/quote']

The wave packet is not an uncertainty. Consider the example I gave above, i.e. a Gaussian wave function. I have a diagram in one of my web pages which shows this curve. See Figure 1 in

http://www.geocities.com/physics_world/qm/gauss_state.htm

 

The width of the curve equals the uncertainty in position. If you have a different wave function then you will have a different probability density and, in general, the width of the curve and thus a different uncertainty in position. Notice that I have not said anything about measurement yet. The question now arises as how does one measure the probability distribution? To do this one simply considers measurements from a ensenmble of N identical experiments (where N is "large"). One then determines how many times the position of the particle was measured at a given position (actually one marks out a large number of finite "bins" and then counts how many times the position falls within each bin. The probability of the particle being measured to be found at X (within an interval of width dx) divided by N. This gives an approximation of the probability density. Once that is found one can then calculate the uncertainty in position. The mathematical quantity [math]\Delta X[/math] is given in Eq. (6) at

http://www.geocities.com/physics_world/qm/probability.htm

The probability of the photon being "present" is spread out in space' date=' and is not measurable to arbitrary precision at the same time as the momentum?

[/quote']

Yes. It is the probability that is spread out, not the photon.

 

For purposes of illustration let's consider a very simple system such as a spin-1/2 system. Let the base kets be the eigenkets of the [math]S_z[/math] operator. There are only two possible eigenvalues of this operator, +hbar/2 and -hbar/2. For simplicity denote these base kets as |+> which corresponds to the spin-up state eigenstate and |-> which corresponds to the spin-down eigenstate. Now consider the spin-up egienstate of the operator [math]S_x[/math]. Denote this eigenstate as [math]|S_x: +>[/math]. Let the system be in the eigenstate of [math]S_x[/math]. This can be expressed in terms of |+> and |-> as

 

[math]|S_x;+> = \frac{1}{\sqrt{2}}|+> + \frac{1}{\sqrt{2}}|->[/math]

 

Now think about what one obtains when one measures [math]S_z[/math] when the system is in this state. Keep in mind that there are only two possibilities of spin that can result from measurement of the spin. What do you think the uncertainty in [math]S_z[/math] is? I.e. what would you expect the value of [math]\Delta S_z[/math] to be? Do you think it would be zero or non-zero?If its non-zero then how would you reconcile this with the fact that there are only two possible eigenvalues of [math]S_z[/math] which could result be measured in the lab?

 

What if the system was initially in the state |+> instead. Would [math]\Delta S_z[/math] be zero or non-zero?

 

Note: These are basically rhetorical questions since the general reader of this post might not know the math/theory of quantum mechanics well enough to calculate the actual values of the uncertainty.

 

Back to the original question...

In the Feynman lectures on physics, I encountered a sum of sinusoids having almost identical frequency. The resulting sinusoid had a very small location in space, so to speak.

Can you tell me where in the Feynman lectures that is located?

My question is; is this resulting wave packet identical to what is called a photon?

No. The particle property of a photon refers to the fact that a photon has no spatial dimensions. The wave packet defines a probability distribution. The smaller the wave packet the more locallized the photon is in space. The wave function is a superposition of various quantum states, each state corresponding to a photon of a particular value of energy having a specific phase.

What makes up all these sinusoids?

Each sinusoid corresponds to a different state where a state is defined by a freqency and phase. Only when a measurement is done will the state fall into an eigenstate corresponding to the observable measured. The wave packet will then cease to exist. Ideally the wave function becomes a Dirac delta function.

 

Pete

Edited July 29, 2008 by Pete
multiple post merged

[swansont]

It is the probability that is spread out, not the photon.

 

How do you distinguish the two?

[Pete]

How do you distinguish the two?

Measurement. A single photon can be localized within a region arbitraryl smaller than the width of the wave as defined by the uncertainty [math]\Delta x[/math].

I should mention that what I've said above applies to non-relativistic quantum mechanics. If the subject is about photons then quantum electrodynamics is the correct theory to apply and that's not something I've learned ... yet.

Measurement. A single photon can be localized within a region arbitraryl smaller than the width of the wave as defined by the uncertainty [math]\Delta x[/math].

I should mention that what I've said above applies to non-relativistic quantum mechanics. If the subject is about photons then quantum electrodynamics is the correct theory to apply and that's not something I've learned ... yet.

 

Pete

 

To graphically illustrate this point I created a diagram which I'm sure plenty of you have already seen if you've studied quantum mechanics. Please see

http://www.geocities.com/physics_world/qm/image_gif/qm15-im-01.gif

 

This is a diagram which shows a screen where photons hit one at a time and their positions are recorded. Those positions are shown by the dots. Let [math]\delta x[/math] denote the error that exists in measuring the position of the photon. The physical meaning of the wave function is that the magnitude of the sqaure of it is proportional to the probability of measuring the photons position at a location x within an interval of dx.

 

The quantity [math]\Delta x[/math] gives an idea of how the photons are not predetermined to be located at a particular place. Most of the photons are near the vertical axis and the probability of them being found further from the axis decreases with distance. The "spread" of the area where "most" of them are contained is denoted by [math]\Delta x[/math]. It is [math]\Delta x[/math] that is the uncertainty and not [math]\delta x[/math] as one might think.

 

I was discussing this subject a while back with some people and this same subject came up. I created the following web page to explain myself on this topic. The page is at

http://www.geocities.com/physics_world/qm/probability.htm

 

It is interesting to review the example I of a single fair die. If someone asked you "What is the uncertainty in the measurement in the number shown on the die?" One might answer "Zero since when the die is rolled and I see the face I am 100% sure of what number I read." The actual answser is that there is a finite undertainty in the number of dots on the face. The example demonstrates the meaning of all this and why one has to be sure of what one is asking. The two questions

 

Q1) What is the uncertainty in the measurement in the number shown on the die?

 

Q2) What is the error in measurement of the number which appears on the die's face when it is rolled.

 

The answer to Q1 is 31.4. The answer to Q2 is zero.

 

Pete

Edited July 31, 2008 by Pete
multiple post merged

[swansont]

Measurement. A single photon can be localized within a region arbitraryl smaller than the width of the wave as defined by the uncertainty [math]\Delta x[/math].

 

But before you do the measurement, you don't know where the photon is. There is an uncertainty (as in, a lack of certainty) of the photon's path. If we happen to have sent it through a barrier with some slits in it, the photon will end up at some location that we can't predict except with a probability. (When we do this many times, of course, the interference pattern will emerge.) But we do not know the path of the photon (or electron, or whatever we used) was — we can't say it went through one slit or the other.

 

So uncertainty has a particular meaning in the context of a measurement. In other contexts, however, it has meaning as well. As I said: a lack of certainty.

 

Now, to the situation of measuring conjugate variables — yes, the Heisenberg relation will emerge when you measure an ensemble. But what of a single measurement, were to attempt to measure both variables simultaneously. You use instruments that have insanely precise measurement capabilities, and you get some answer with no instrument error at all. Does that mean you know those two variables exactly? My understanding is no, you don't. You can't say that the electron was in that exact place, with that exact momentum, just because your instrument gave you that reading — the electron could have been in some region given by [math]\Delta x[/math] and given you that position reading, and could have had a range of momenta given by [math]\Delta p[/math] and given you that momentum reading, because that's the best you can think about how localizes an electron is in position and momentum space. The question "what is the position of the electron" loses meaning at that level.

[Pete]

But before you do the measurement, you don't know where the photon is. There is an uncertainty (as in, a lack of certainty) of the photon's path.

When you use the term uncertainty in that way you are using it to mean something different than it means in uncertainty in x etc. You are using it to mean something synonymous to probability. Perhaps some authors do use the term uncertainty in some instances to mean the same thing that you do, I just haven't see anyone do that. I do know that [math]\Delta x[/math] is not defined that way though. Using it to mean two different things in the same context would lead to all sorts of problems. For this reason the term probability is used for this instead.

 

I also recall that I originally mentioned this due to someone implying that one can't precisely measure an observable quantity.

Does that mean you know those two variables exactly? My understanding is no, you don't.

People rarely, if ever, use the term "know" in quantum mechanics because it raises all sorts of philosophical problems. All that one can speak of is what is measured, its probability, is expectation and its standard deviation (i.e. uncertainty).

You use instruments that have insanely precise measurement capabilities, and you get some answer with no instrument error at all. Does that mean you know those two variables exactly? My understanding is no, you don't. You can't say that the electron was in that exact place, with that exact momentum, just because your instrument gave you that reading ..

This is equivalent to saying that when you roll a fair die and a six comes up that you really don't know its a six.

...the electron could have been in some region given by [math]\Delta x[/math] and given you that position reading, and could have had a range of momenta given by [math]\Delta p[/math] and given you that momentum reading, because that's the best you can think about how localizes an electron is in position and momentum space. The question "what is the position of the electron" loses meaning at that level.

That is not what quantum mechanics has to say about this. Quantum mechanics would say that the position is not determined until you measure it where at that point it becomes determined (if it wasn't already in an eigenstate of position). This phenomena is known as the collapsing of the wave function.

 

And all this talk of uncertainty only applies to quantum states which are not eigenstates of the observable. If a state is in an eigenstate of, say, energy E, then there is no uncertainty in energy in that the value that will be measured has a 100% probability of having that value.

 

Pete

Edited August 1, 2008 by Pete
multiple post merged