Is the entanglement of photons necessary for interference?

[jambaugh]

With respect to the phase, I have a problem. It appears that if the Hilbert space being considered is the compound Hilbert space, then the frequency should be determined by the energy of both photons, which would give a wavelength 1/2 the regular wavelength.

[/quote]...The strange thing would be that then the frequency of each photon would not agree with it's own energy. If we say that there are two types of frequency, that of the composite system and that of each individual phton, it is not very clear to me how you put both together. I guess for any simulation this should be understood before getting started.[/QUOTE]

Here is an example where I feel being a "good Copenhagenist" helps one understand what is going on. Remember the wave functions are not the physical systems but represent knowledge about the systems.
(Think strongly with the analogy of a probability distribution.)

Don't be too stuck on the idea of a physical wave-function "out there" be it for a single particle or for the entangled pair. The wave function expresses probability amplitudes for potential measurement events. In the two particle case they would be measurements of each at two independently chosen locations. Remember you can also speak of the wave-function on momentum space... or any other cross-section of classical phase-space representing a manifold of possible mutually compatible observables.

If you want to be more substantive you can work instead in the field theory where you are talking about particle number densities with the global constraint of total particle number = 2 but you must then promote your hilbert space vectors to operators. Also you are likely to need intuition built upon understanding fully the QM version before you can derive much insight from the QFT version.

But mainly I suggest you recall that the Hilbert space of the composite is a tensor product of Hilbert spaces and remember the product to sum trigonometric identities which derive from the rules for (imaginary) exponentials:

e^{i\omega_1 t} e^{i\omega_2 t} = e^{i(\omega_1 + \omega_2)t}

So the energy of the composite is simply the sum of the energies of its components (ignoring interactions) which is nicely consistent with expressing composites as products.

 

[alexepascual]

Here is an example where I feel being a "good Copenhagenist" helps one understand what is going on. Remember the wave functions are not the physical systems but represent knowledge about the systems.
(Think strongly with the analogy of a probability distribution.)
Don't be too stuck on the idea of a physical wave-function "out there" be it for a single particle or for the entangled pair. The wave function expresses probability amplitudes for potential measurement events. In the two particle case they would be measurements of each at two independently chosen locations. Remember you can also speak of the wave-function on momentum space... or any other cross-section of classical phase-space representing a manifold of possible mutually compatible observables

OK. The squared modulus of the wave function gives you the probability distribution. And we need that probability distribution in order to find out what the pattern on the screen is going to be. I am not trying to give the wave function an interpretation here. I just want to use it to calculate results. I should be able to calculate the sum of the waves that went through each slit at each point on the screen.

If you want to be more substantive you can work instead in the field theory where you are talking about particle number densities with the global constraint of total particle number = 2 but you must then promote your hilbert space vectors to operators. Also you are likely to need intuition built upon understanding fully the QM version before you can derive much insight from the QFT version.

No, I don't want to get into QFT here. If it were absolutely necessary I would, but I don't think it is.

But mainly I suggest you recall that the Hilbert space of the composite is a tensor product of Hilbert spaces and remember the product to sum trigonometric identities which derive from the rules for (imaginary) exponentials:
e^{i\omega_1 t} e^{i\omega_2 t} = e^{i(\omega_1 + \omega_2)t}

Well, that's exactly what I was saying. You just expressed it in terms of angular velocity of the phase (You are showing the time component).

So the energy of the composite is simply the sum of the energies of its components (ignoring interactions) which is nicely consistent with expressing composites as products.

I agree. Now the question is: When I write the expression for the phase as function of time in the lower arm, I need to know what energy to use. On one hand I have read that once the particles are entangled you can't consider the separate Hilbert spaces anymore. In that case I would like to use the sum of the energies of both particles, even if I am looking at only one particle there. On the other hand I couls say that because I am considering only one particle in the lower arm, I should consider only the energy for that particle. In one case the frequency is twice what it would be in the other case. So, even if I believe that the wave funtion only exists in my brain, I need numbers to plug in for p and E.
Oh! by the way.. Do you use some editor for LateX?

 

[jambaugh]

I agree. Now the question is: When I write the expression for the phase as function of time in the lower arm, I need to know what energy to use. On one hand I have read that once the particles are entangled you can't consider the separate Hilbert spaces anymore. In that case I would like to use the sum of the energies of both particles, even if I am looking at only one particle there. On the other hand I couls say that because I am considering only one particle in the lower arm, I should consider only the energy for that particle. In one case the frequency is twice what it would be in the other case. So, even if I believe that the wave funtion only exists in my brain, I need numbers to plug in for p and E.

I'm still not quite clear on what you're asking. The E and |p| are as you would expect and the same as would be defined classically. You just won't be able to write down a wave-function where E reflects the frequency.

(Below I am using A and B to indicate the A particle and B particle of an entangled pair)

The proper way to talk about one half of an entangled pair is with the reduced density operator. In the continuous representation (i.e. with wave-functions) this will take the form of a function of two positions:
\rho_A(\vec{x},\vec{y})
(also a function of time) (Note this is another doubling of the variables)
Think of the vectors x and y as row and column indices for a matrix.
The bifunction forms act as operators on functions via:
T[f]=g \quad \equiv \quad g(x) = \int T(x,y)f(y)dy

The full density operator for the entangled pair will then have four vector variables:
\rho_{AB}(\vec{x}_A,\vec{y}_A,\vec{x}_B,\vec{y}_B)
and the reduced density operator is obtained by "tracing over" the variables corresponding to the other half of the pair:

\rho_A(\vec{x},\vec{y}) = \iint \delta^3(\vec{x}_B-\vec{y}_B)\rho_{AB}(\vec{x}_A,\vec{y}_A,\vec{x}_B,\vec{y}_B)d^3x_B d^3 y_B

In the case of a sharp mode (for the whole entangled pair) this density operator will take the form:
\rho_{AB}( = \psi(x_A,x_B)\psi^\dag(y_A,y_B)
(henceforth assume x's and y's are vectors.)

But once you take the partial trace to get the reduced density operator you won't have such a factorable form. (It is not in a sharp mode)

Now I get to the point... At best you can write the reduced density operator as a
(classical) probability weighted average of such sharp density operators.
\rho_A(x,y) = \sum_{k}p_k\rho_k(x,y) =\sum_k p_k \psi_k(x)\psi^\dag_k(y)
Each such will be a product of wave-functions as above where each of those wave-functions has the same energy and thus frequency.
\psi_k(x) = \phi_k(x)e^{i\omega t}

You can't speak meaningfully about the "phase of the lower (A) particle" at a given point or even at a given time. When you consider only one half of an entangled pair you get something similar to a particle from a noisy source and you must work within the density operator format.

And this is my point about being "a good Copenhagenist". Get out of the habit of thinking in terms of "the phase of a particle". Its the wave-function which has a phase and without a wave-function (since the mode is not a sharp one) we can't define phase.

As the ole codger on the porch says... "Ya caint get there from here!"

Oh! by the way.. Do you use some editor for LateX?

I use both Led and WinShell (both free). Sometimes I prefer one sometimes I prefer the other. Right now Led is the main one I use.

 

[alexepascual]

I'm still not quite clear on what you're asking. The E and |p| are as you would expect and the same as would be defined classically. You just won't be able to write down a wave-function where E reflects the frequency.

(Below I am using A and B to indicate the A particle and B particle of an entangled pair)

The proper way to talk about one half of an entangled pair is with the reduced density operator. In the continuous representation (i.e. with wave-functions) this will take the form of a function of two positions:
\rho_A(\vec{x},\vec{y})
(also a function of time) (Note this is another doubling of the variables)
Think of the vectors x and y as row and column indices for a matrix.
The bifunction forms act as operators on functions via:
T[f]=g \quad \equiv \quad g(x) = \int T(x,y)f(y)dy

The full density operator for the entangled pair will then have four vector variables:
\rho_{AB}(\vec{x}_A,\vec{y}_A,\vec{x}_B,\vec{y}_B)
and the reduced density operator is obtained by "tracing over" the variables corresponding to the other half of the pair:

\rho_A(\vec{x},\vec{y}) = \iint \delta^3(\vec{x}_B-\vec{y}_B)\rho_{AB}(\vec{x}_A,\vec{y}_A,\vec{x}_B,\vec{y}_B)d^3x_B d^3 y_B

In the case of a sharp mode (for the whole entangled pair) this density operator will take the form:
\rho_{AB}( = \psi(x_A,x_B)\psi^\dag(y_A,y_B)
(henceforth assume x's and y's are vectors.)

But once you take the partial trace to get the reduced density operator you won't have such a factorable form. (It is not in a sharp mode)

Now I get to the point... At best you can write the reduced density operator as a
(classical) probability weighted average of such sharp density operators.
\rho_A(x,y) = \sum_{k}p_k\rho_k(x,y) =\sum_k p_k \psi_k(x)\psi^\dag_k(y)
Each such will be a product of wave-functions as above where each of those wave-functions has the same energy and thus frequency.
\psi_k(x) = \phi_k(x)e^{i\omega t}

You can't speak meaningfully about the "phase of the lower (A) particle" at a given point or even at a given time. When you consider only one half of an entangled pair you get something similar to a particle from a noisy source and you must work within the density operator format.

And this is my point about being "a good Copenhagenist". Get out of the habit of thinking in terms of "the phase of a particle". Its the wave-function which has a phase and without a wave-function (since the mode is not a sharp one) we can't define phase.

As the ole codger on the porch says... "Ya caint get there from here!"


I use both Led and WinShell (both free). Sometimes I prefer one sometimes I prefer the other. Right now Led is the main one I use.

I realize that I don't have a full understanding of the situation. However, I am not convinced about your approach. The Copenhagen interpretation does no forbid using the wave function to describe the evolution of a system before it interacts with a macroscopic apparatus. Now, perhaps I don't understand very well the wave function of a pair of entangled particles, but I looks to me that using the density matrix would be a little like sweeping the problem under the rug. After the photons have been created, their evolution in time should be described by the wave function. After one of the photons has been detected, then maybe you should consider tracing over the density matrix. I will definitely have to think about this, but I think it may be a little premature right now. I could instead focus on what happens before detection.
I am trying to see how I can make sense of your argument without including the density matrix part. Let me see if the following agrees with the way you are thinking:
The wave function of the pair of photons is a function of the space coordinates for both photons and time coordinate (in a non-relativistic approach). So, for a particular combination of positions for both particles we have a single complex number which evolves in time with a frequency determined by the sum of the energies of both photons. If we look at the photon that goes towards the slits, and we want to assign to each point in space for that photon a wave function value which we can use to determine the probability of finding the particle at a particular position, we run into the problem that we don't know exactly where the other particle is.
Now, doing a trace on the density matrix would give you a mixture, but here the composite state is still a pure state evolving unitarily. If you had the degrees of freedom of the other photon getting entangled with the environment, according to the decoherence theory, you would have to do a trace. But in this case we don't have that situation yet.
Going back to the state vector, I would reconsider what I said about not knowing where the other particle is. There is some uncertainty along the path of the particle (perpendicular to the waves) and this needs to be considered. But if we ignore that for a first aproximation, then we have to consider any lateral uncertainty in the other particle. But it happens that due to the momentum entanglement, the lateral position of the second particle should be correlated to the position of the first. With respect to the position along the path, we could just use the speed of light x time (ignoring uncertainty).
In an x-y graph (configuration space) where x and y are the 1-D positions of both particles (I think this may be closer to your original idea for a simulation) If we ignore uncertainties and consider propagation along rays (This could be objected but I don't think it matters here), then we would have a parametric curve where each point corresponds to the configuration at a particular time. And there should be value for the wave function at each point. Does this make sense?
Thanks for the tip about the LateX editor.

 

[jambaugh]

I can only repeat what I've said before and I will try to with better examples but think about this some more before we go on.

 

[alexepascual]

I can only repeat what I've said before and I will try to with better examples but think about this some more before we go on.

OK. I'll give it some thought. Thanks again.

-Alex

 

[alexepascual]

I can only repeat what I've said before and I will try to with better examples but think about this some more before we go on.

I have been studying a little about the details of entangled photons but I haven't made much progress. Actually I haven't spent enough time on it. But I'll continue to look into it and eventually I'll understand it.
On the other hand there is something I thought about yesterday. If an atom emits a photon, then the photon will be entangled with some degrees of freedom of the atom. So, applying your argument to this situation, being that it is entangled to something else, a description of this photon in some basis (position for example) should be given by a reduced density matrix and it should not produce interference if it goes through a double slit.
But it happens that light is frequently produced by emision from atoms and it does produce interference after going through double slits. I know this example is a little complicated and I would not want at this point to get into the details, as this would not help me in understanding the more basic facts. But I was wondering if in this example you would consider the photon as producing interference or not (of course after a few from the same transition have gone through the slits).

 

[jambaugh]

I have been studying a little about the details of entangled photons but I haven't made much progress. Actually I haven't spent enough time on it. But I'll continue to look into it and eventually I'll understand it.
On the other hand there is something I thought about yesterday. If an atom emits a photon, then the photon will be entangled with some degrees of freedom of the atom. So, applying your argument to this situation, being that it is entangled to something else, a description of this photon in some basis (position for example) should be given by a reduced density matrix and it should not produce interference if it goes through a double slit.
But it happens that light is frequently produced by emision from atoms and it does produce interference after going through double slits. I know this example is a little complicated and I would not want at this point to get into the details, as this would not help me in understanding the more basic facts. But I was wondering if in this example you would consider the photon as producing interference or not (of course after a few from the same transition have gone through the slits).

Whether the photon entangled with the atom will or will not form an interference pattern depends on which variables in particular are being entangled. Also remember that when we do use atomic sources of photons there is a selection process going on. For example the photons coming from a laser are produced by stimulated emission and only those with very confined momenta are produced in great numbers. There is a selection process here which is distinct from say trying to see an interference pattern from photons emitted by arbitrary excited atoms (even if those atoms are all initially produced in the same excited sharp quantum mode).

Ultimately you should ask in specific cases "what do we know about the photons?" That knowledge if it exists comes from a physical selection process. Photons coming out of a laser are photons we know quite a bit about because one specific mode of emission has been amplified dramatically through the stimulated emission process.

 

[alexepascual]

Whether the photon entangled with the atom will or will not form an interference pattern depends on which variables in particular are being entangled. Also remember that when we do use atomic sources of photons there is a selection process going on. For example the photons coming from a laser are produced by stimulated emission and only those with very confined momenta are produced in great numbers. There is a selection process here which is distinct from say trying to see an interference pattern from photons emitted by arbitrary excited atoms (even if those atoms are all initially produced in the same excited sharp quantum mode).
Ultimately you should ask in specific cases "what do we know about the photons?" That knowledge if it exists comes from a physical selection process. Photons coming out of a laser are photons we know quite a bit about because one specific mode of emission has been amplified dramatically through the stimulated emission process.

I was not thinking about a laser. I was rather thinking about an atom whose conduction electron has been excited and then it spontaneously transitions to a lower level. You may consider any source of energy that is sufficient to elevate that electron to a level that will allow the desired transition. I understand sodium for example has two famous lines that are very close together (589.0nm and 589.6nm) . Of course sodium has other lines. But they are weaker and you could post-select those two using a monochromator or a filter. With respect to what we "know" about the photons, I guess we only know that they have a narrow wavelength spectrum centered around some known value (589.3nm in the case of sodium) that corresponds to that line and that's all. I had also in mind considering the emission from a single atom and one photon at a time.

 

[jambaugh]

I was not thinking about a laser. I was rather thinking about an atom whose conduction electron has been excited and then it spontaneously transitions to a lower level. You may consider any source of energy that is sufficient to elevate that electron to a level that will allow the desired transition. I understand sodium for example has two famous lines that are very close together (589.0nm and 589.6nm) . Of course sodium has other lines. But they are weaker and you could post-select those two using a monochromator or a filter. With respect to what we "know" about the photons, I guess we only know that they have a narrow wavelength spectrum centered around some known value (589.3nm in the case of sodium) that corresponds to that line and that's all. I had also in mind considering the emission from a single atom and one photon at a time.

Yes but don't confuse the spectral lines with interference patterns. Take the light from one spectral line of a sodium lamp and put it through a double slit. You won't get an interference pattern because the light is not coherent.

More precisely to the question at hand, there is no way knowing when or in what direction the emitted photon has gone nor when or in what direction the emitting atom has recoiled. They are entangled so these positions are correlated and the magnitude of the momentum is known so you know the relationship between when the emission occur and the distance the two ejecta are from the original position but until you observe one you have no information about the other.

You might get an interference pattern by putting a double slit far enough away but you are selecting out only those cases where the photon got to the double slit instead of Timbuktu. You are thereby making a measurement = "preparing" the photon which breaks the entanglement.

I say "might" as I am not sure if even then you'll get one. It depends on how one is setting up the repeated trials to get enough data to see the pattern.

 

[alexepascual]

Yes but don't confuse the spectral lines with interference patterns. Take the light from one spectral line of a sodium lamp and put it through a double slit. You won't get an interference pattern because the light is not coherent..

My understanding was that you don't need coherent light to get interference. Even though the separate photons are incoherent because they come from different atoms, each phton will interfere with itself. So you can have light from any source that produces a fairly narrow line and it will produce interference even if it is totally incoherent. I understand that you first have to make the light pass through a narrow slit so that all the waves arrive at the slits at the same angle. Do you agree with this?

More precisely to the question at hand, there is no way knowing when or in what direction the emitted photon has gone nor when or in what direction the emitting atom has recoiled. They are entangled so these positions are correlated and the magnitude of the momentum is known so you know the relationship between when the emission occur and the distance the two ejecta are from the original position but until you observe one you have no information about the other.

I think I agree on this.

You might get an interference pattern by putting a double slit far enough away but you are selecting out only those cases where the photon got to the double slit instead of Timbuktu. You are thereby making a measurement = "preparing" the photon which breaks the entanglement.

Does a measurement break the entanglement? In a EPR experiment, measuring the photon on one side does not seem to break the entanglement because you can always expect the other photon to have the correlated polarization (when you measure on the same axis).

I say "might" as I am not sure if even then you'll get one. It depends on how one is setting up the repeated trials to get enough data to see the pattern.

I agree with this.