Double-slit experiment and watching the electrons

[forcefield]

double-slit experiment and "watching the electrons"

I'm trying to build a picture of what is happening in the double-hole experiment when you are "watching the electrons" (http://feynmanlectures.caltech.edu/III_01.html#Ch1-S6). It's mostly clear to me but I do have some questions: if you use a strong light source at the holes (so that you can see every electron at the holes), can you in that case use a long wavelength for the light source so that when you see a flash of light you can't tell which hole is closer to the flash of light ? If you can, is there interference in that case ?

(My hypothesis is that you can use such a long wavelength also in the case of strong light source and then there is no interference.)

 

[Mentz114]

If there is no interaction between your probing light and the electrons, you can get no information and the interference persists. If there is an interaction, the interference will not be present.

 

[naima]

Suppose that you have two receptors behind the slits and near them. Light interact with electrons and one of the receptors clicks.
Now close the right slit and send electrons to the left slit. The clicks will depend on the wavelength of light.
There is an amplitude "a" for a left clik and an amplitude "b" to a right click. (you have the same a and b by symmetry when you exchange the slits.
If a = 1 and b = 0 you will have a perfect which path information.
When two slits are open the probability for the electron to hit the screen at x is:
|aL(x) + b R(x)|² + |bL(x) + aR(x)|²
if a = 1 and b = 0 the probability is |L(x)|² + |R(x)|²
if a = b (with |a|² + |b|² = 1) you get
|L(x) + R(x)|²
So you see that here photons always interact but the result is not a yes/no interference
You have intermediate patterns depending on the quality of the which path information (the a and b) you get.

 

[forcefield]

If there is an interaction, the interference will not be present.

So you see that here photons always interact but the result is not a yes/no interference

So I have two answers that seem to disagree. Mentz114's answer was in line with my expectations whereas I can't follow naima's reasoning. Can we build a consensus here ?

Also I'd like to know if this experiment has been done with strong light source of long wavelength and what was the result ?

 

[naima]

I found the calculation in an old book of L Tarassov (1980)
Surprisingly i did not see it in another book nor online.

 

[naima]

We can write P(x) = |L(x)|² + |R(x)|² + k (L(x) R*(x) + R(x) L*(x)) where k (= ab* +ba*) is a real number which can be chosen between 0 and 1.

Also I'd like to know if this experiment has been done with strong light source of long wavelength and what was the result ?

Is there an experiment (in any quantic domain) where one can vary a parameter of the setup so that the intensity of interference vary?

 

[Claude Bile]

Both Mentz114 and Naima's explanations are compatible.

Mentz114 explains in "photon-by-photon" and "electron-by-electron" terms. If photon interacts with electron; coherence between the two slits is lost and interference is lost...on the other hand if the photon does not interact, the interference persists. Clearly, on average, this is a situation where there is partial coherence - which Naima outlines in more explicit terms.

Claude.

 

[naima]

I do not believe that this is the correct way to see what happens.
We have not a mixture of interacting electrons and of not interacting ones.
|a|² is not the probability of an interaction giving the which-path information:
We would have P(x) = |a|² (|L(x)|² + |R(x)|²) + |b|² |L(x) + R(x)|²
This is not the formula tarasov gives.
We have a source of photons between the slits.
Remember that when the right slit is closed electrons pass through the left slit , then |b|² is the probability that the right photon receptor clicks. Giving you a wrong which-path information.
You may have interference with always interacting electrons and photons but giving you an unreliable information.

 

[naima]

I am confused.

P(x) = |a|² (|L(x)|² + |R(x)|²) + |b|² |L(x) + R(x)|² =
(|a|²+ |b|²) (|L(x)|² + |R(x)|²) + |b|² (L(x)R*(x) + L*(x)R(x))
As |a|² + |b|² = 1 we get something like Tarasov formula.
Can we say that an unreliable information interaction is just like no interaction in this case?

 

[EskWIRED]

I think (hope?) that there exist experiments in which low energy photons have been used to detect "which-slit" information. I'd love to read the papers.

I'd even more love to see a similar experiment using massive projectiles through the slits - buckyballs perhaps - in conjunction with RF detection of the path information. I am wondering whether such methods would yield BOTH a proportion of interference patterns and "one-slit" patterns, depending upon the nature and magnitude of the individual detections. I would be fascinated if the results varied depending upon the frequency used, especially if the results shifted rapidly from one state to another as the frequency is swept.

 

[Cthugha]

I'd even more love to see a similar experiment using massive projectiles through the slits - buckyballs perhaps - in conjunction with RF detection of the path information. I am wondering whether such methods would yield BOTH a proportion of interference patterns and "one-slit" patterns, depending upon the nature and magnitude of the individual detections.

Experiments similar to that have been done as early as 1988. See Greenberger, D. M.; Yasin, A., "Simultaneous wave and particle knowledge in a neutron interferometer". Phys. Lett. A 128 (8): 391–394 (1988).

Unfortunately I am not aware of any free copy. It turns out that you get a duality relation between which-way information and interference pattern visibility called Englert-Greenberger duality relation. See the Wikipedia entry for the fundamental journal references: http://en.wikipedia.org/wiki/Englert–Greenberger_duality_relation.

 

[1832vin]

try frensel lens

 

[naima]

I found another easy experiment.
photons pass through two slits. a polarizer stands behind each slit.
When the polarizers are parallel (sin² = 0) there is no which-path information and fringes are well seen.
we can rotate one of the polarizer. fringes progressively disappear.
Here photons allways pass and interact through the polarizers. So fringes are not a question of yes/no interaction.
read this

 

[forcefield]

So fringes are not a question of yes/no interaction.

I'm not convinced in the case of this thread where we are sending electrons through the slits.

 

[Cthugha]

I'm not convinced in the case of this thread where we are sending electrons through the slits.

By using a double slit and a reasonably narrow photon or electron source you prepare a scenario with reasonably well defined relative phases. The question of whether you see fringes at the detection screen or not depends on whether this phase gets messed up along the way. The extreme cases of no phase distortion and complete phase randomization correspond to perfect and no fringes, but there are always ways to distort the phase just a bit or inside a narrow range of values which just gives you reduced fringe visibility (and of course little which-way information).

 

[naima]

I'm not convinced in the case of this thread where we are sending electrons through the slits.

Suppose that the screen is shuttered when photons are not detected by the detectors near the screen. we will have a pattern due to the only electrons that are observed. they all interacted with photons.
Now let the photons wavelength greater than the distance between the slits. we cannot see details smaller than that distance. so we have no which path information.
In this case all electrons interacted but we have interference.

 

[forcefield]

By using a double slit and a reasonably narrow photon or electron source you prepare a scenario with reasonably well defined relative phases. The question of whether you see fringes at the detection screen or not depends on whether this phase gets messed up along the way. The extreme cases of no phase distortion and complete phase randomization correspond to perfect and no fringes, but there are always ways to distort the phase just a bit or inside a narrow range of values which just gives you reduced fringe visibility (and of course little which-way information).

Does that mean that if you prepare the electrons with complete phase randomization, you get no interference ?

Suppose that the screen is shuttered when photons are not detected by the detectors near the screen. we will have a pattern due to the only electrons that are observed. they all interacted with photons.
Now let the photons wavelength greater than the distance between the slits. we cannot see details smaller than that distance. so we have no which path information.
In this case all electrons interacted but we have interference.

Is that because your experiment is different from the one explained by Feynman ?

 

[naima]

Did feynman say something about watching the electrons?

 

[Cthugha]

Does that mean that if you prepare the electrons with complete phase randomization, you get no interference ?

No. What matters is the phase difference you get from the two paths via the two slits. Even if you prepare the electrons with random initial phase the phase difference between the paths via the two slits is independent of the initial phase.

What you can do is randomize this phase difference. With light the easiest way to achieve that is using a large light source and place it directly in front of rather broad slits. If the possible paths from all the positions on the surface of the light source to all the possible positions at one slit differ by way more than one wavelength of the light you will not get an interference pattern as the relative phases get randomized. That works for electrons, too, of course.

 

[forcefield]

Did feynman say something about watching the electrons?

Check the link I posted in post #1.

 

[naima]

Sorry I did not see it.

Feynman says:
"when we make the wavelength longer than the distance between our holes, we see a big fuzzy flash when the light is scattered by the electrons. We can no longer tell which hole the electron went through! We just know it went somewhere! And it is just with light of this color that we find that the jolts given to the electron are small enough so that P′12 begins to look like P12—that we begin to get some interference effect."

That is what i said. You have the answer to your question.

When wavelength increases interaction are still there but you cannot see through which slit the electron passes.
You begin to see interference .
the "a" and "b" are functions of wavelength. it would be interesting to know the function.

 

[forcefield]

So whether we have interference is not just about whether we interact with the electrons. Also it is not about whether we know the path taken by the electron. Can you say that it is about whether we change the wave into a particle before final detection ?

 

[naima]

It is about how light interact with photons and about what we can learn about paths
Feynman tells you to calculate. I think that it is a good advice. Words are not enough to understand QM
One needs hilbert spaces operators and so on.

 

[naima]

i delete this post. formulas were wrong

 

[naima]

When there is a goog lightning, electrons are always detected so we need two 2-dimensional Hilbert spaces.
H1 has a |L> and |B> basis. One vector for each slit the electron passes through.
Suppose that there is a photon detector near each slit i call these detectors 1 and 2.
I need another hilbert space H2 with basis |1> and |2>
|1> correspond to a photon seen near the left slit and |2> near the other.

The set electron + photon is in the tensor product H = [itex]H1 \otimes H2.
A basis of this 4 dimensional space is |L>|1>, |R>|1>, |L>|2>, |R>|2>
A generic vector in H is a|L>|1>+b |R>|1>+c|L>|2>+d |R>|2>
If the vector is normalized, this corresponds to an amplitude «a*» that the electron passes through left slit and seen near it, an amplitude «*b*» that it passes through the right slit but is seen near the other one and so on.
As the setup is symmetric we have a = d and b = c.
So the state is (a|L>+b |R>)|1> + (a|L>+b |R>)|2>

When we measure where the photon is seen (|1> or |2>) the state vector collapses on
(a|L>+b |R>)|1> or on (a|L>+b |R>)|2> with equal probabilities.
As these vectors are orthonormal, the probalility that the electron hits the screen at a given point is
|aL(x)+b R(x)|² + |aL(x)+b R(x)|²
So the visibility of fringes depends on the values of a and b in the statistical mixture.

If the light is low we have to extend H2 with two vectors |L>[3> and |R>|3> where |3> corresponds to
a no-click on the detectors. This looks like a POVM.
The values of a and b depend on the wavelength of the photon.
I do not know what are a(wl) and b(wl)
We have a(0) = 1, b(0) = 0 (no interference).
If di is the distance between the slits we have no information when wl = di,
so i suppose that a(di) = b(di)
What about them when wl > di*? do they remain equal*?

I found that the disappearance of the fringes is due to the recoil of the electron but i found no formula with the limit wl = di.

 

[forcefield]

i found no formula with the limit wl = di.

Did you notice Cthugha's post ? According to that, it's not about the distance between the slits, it's about the distance differences from the light source to the slits. Makes sense to me.

 

[naima]

I used the ² "little 2 " (under escape) for sqarring in my posts, they appear now as unrecognized character.
Could anyone repair this bug?

As i was looking for how "b" depends on the wl wavelength, i found that $\frac {|a| - |b|}{|a| + |b|}$ has a name
It is Visibility. look here
when |b| = 0 V = 1 so fringes are well seen. when V = 0 you cannot see the fringes.

The V(wl) function is not so simple that i thought.
you can find it by googling "loss of coherence tan" or "Fringe Visibility and Which-Way Information: An Inequality"

I thought that the Visibility was null when wl = d. Feynman wrote
"So now, when we make the wavelength longer than the distance between our holes, we see a big fuzzy flash when the light is scattered by the electrons. We can no longer tell which hole the electron went through!"
Formulas seem to say something else.
Do you understand this?

 

[naima]

Merry Christmas,

I found three links with (he same formula of fringes Visibility:
jump to p 12
jump to p 5
look at fig 7

It gives the curve of the electron fringes visibility as a function of the light wavelength.+
It is surprising that fringes disappear then reappear and so on.

 

[naima]

There is a problem with the third link (the one with fig 7). When i click it the pdf is loading but i cannot read it.
if you have the same problem: save it on your desk quit the forum and read it from your computer.
Please tell me if you succeed. It is worthwhile
It is the best paper i found about watching the double slit. it was written by Daniel Walls who worked with Glauber (Nobel prize).
the Visibility(u) is $\frac{3}{2}(\frac{sin u}{u}+\frac{cos u}{u^2}-\frac{sin u}{u^3})$
you can verify that lim u-> 0 is = 1

You can find the complete derivation of this formula by googling "decoherence from spontaneous emission ole steuermagel and harry paul"
if the same problem occurs save it also!

 

[DrChinese]

Here is an alternative link to the Tan & Walls paper (the 3rd link in post #28):

http://www.physics.arizona.edu/~cronin/Research/Lab/some%20decoherence%20refs/LossOfCoherenceInInterf.pdf

 

[naima]

Do you know if S M TAN is Shina Tan? He looks very young.

 

[Cthugha]

Do you know if S M TAN is Shina Tan?

No. S M Tan is Sze Meng Tan.

 

[naima]

thank you.
There is a 3/2 factor in the visibility formula. From which identity does it come?

 

[naima]

We begin with an integral from zero to pi of a function of the angle $\theta$.
Using x = cos ($\theta$.) visibility of the fringes becomes
$A\int_{-1}^1 (1+x^2) e^{i u x} dx$.
Integrating it twice by parts we get
$4A [\frac{sin u}{u}+\frac{cos u}{u^2}-\frac{sin u}{u^3}]$.
"u" being the distance of slits divided by the wavelength of the photon when it decreases to zero the limit of visibility must be 1. We need so a normalization factor and the good formula is
$3/2 [\frac{sin u}{u}+\frac{cos u}{u^2}-\frac{sin u}{u^3}]$.

We can compute the visibility when u = 1. It is when we watch the electrons with light having for wavelength the distance
between the slits. We have V(1) = 3/2 cos(1) = 0.8
So the visibility is still very good!