In the Hanbury Brown and Twiss experiment - see picture - we write the wave-function as follows
(1) Ψ ~ |rD1A|-1|rD2B|-1 exp{2πι(|rD1A| + |rD2B|)/λ} + |rD1B|-1|rD2A|-1 exp{2πι(|rD1B| + |rD2A|)/λ},
where rD1A, rD2B, etc., are distances from sources to detectors.
Though the expression (1) doesn't seem to me correct when |Ψ|2 is used as density of probability for obtaining a joint detection in the detectors A and B. It seems to me that the states of the detectors should be taken in consideration, in which case Ψ would look differently
(2) Ψ ~ |rD1A|-1|rD2B|-1 exp{2πι(|rD1A| + |rD2B|)/λ}E1,A E2,B + |rD1B|-1|rD2A|-1 exp{2πι(|rD1B| + |rD2A|)/λ}E1,B E2,A,
where E1,A is the excited state of the detector D1 by a particle comming from source A, and E1,B is the excited state of the detector D1 by a particle comming from source B - and similarly for the detector D2. The states E1,A and E1,B differ, because in the former, the particle transmitted to the detector a linear momentum in the direction A-D1, while in the latter, the particle transmitted to the detector a linear momentum in the direction B-D1.
And though, the formula (1) proves itself correct in practcal experiment? WHY?
NOTE: I have some half-answer, i.e. it is problematic.
The formula (2) may transform into (1), if E1,A=E1,B and E2,A=E2,B , i.e. the detectors don't distinguish from which source the particle came. In this case one gets E1,A E2,B = E1,B E2,A , and the detector states factorize out of the expression.
However, Ei,A=Ei,B (i = 1, 2) seems to contradict unitarity. The detector D1 passes to the state E1,A after absorbing a linear momentum of direction A-D1, whereas to the state E1,B it passes after absorbing a linear momentum of direction B-D1 . Analogously for D2 . A unitary transformation takes two orthogonal states onto other two orthogonal states. It is not possible that for two orthogonal inputs, a detector passes to a same state.
Dear Sofia,
if I understand your question correctly, your problem comes down to the basics of the two-slit experiment. I think that your explanation with the two different states of the detectors is wrong because IF the detector states would really be different, one would be able to decide which path the photon has taken. However, if you know the path of all individual photons, then the interference pattern (thus the indistinguishability measurement) is lost since you basically collapse the quantum superposition between the photon going one path or going the other.
Once you know which path the photon has taken (which is what you assume in your 2nd equation), then you reduce your experiment to just the classical sum of the probabilities for one path and the other, which means that all 'quantum effects' are gone.
I hope my explanation can help you to better understand the experiment.
Best regards from Bayreuth,
Florian
Dear Florian,
Thanks for your answer, but it does not answer my question, which is WHY the state of each detector does not matter? The state of a detector, e.g. D1, should be different in case it absorbed a particle with linear momentum pA1, from the case it absorbed a particle with linear momentum pB1.
In writing the wave-function we take in consideration whatever object involved in the system evolution, beam-splitters, mirrors, eventual perturbing factors, etc. It is not permitted to ignore anyting. So, why the detectors, which definitely should be left in different states, are ignored?
You see, your answer describes the effect about which I ask, it does not explain the cause, WHY does it happen. There has to be a reason why the states of the detectors, which should clearly be different, are ignored.
Good question,
The answer can be simplified, it depends on the frequency of the object arriving at the double slit. That itself is dependent on the number of quintessence components (per unit time ) of the the arriving object. In the case of the electron it has 10^20 components -it's frequency.
@Andrew Worsley,
Thanks for your contribution, but the problem should not be changed.
The problem here is with the detectors. I'll tell you what I think, but I am not sure that this is the answer - otherwise I wouldn't have asked. The formula (2) in the question may transform into the formula (1), if E1,A=E1,B and E2,A=E2,B , i.e. if the detectors don't distinguish from which source the particle came. In this case one has E1,A E2,B = E1,B E2,A , and the states of the detectors factorize out of the expression.
However, how could it be that Ei,A=Ei,B (i = 1, 2)? It seems to contradict unitarity. The detector D1 passes to the state E1,A after absorbing a linear momentum of direction A-D1, whereas to the state E1,B it passes after absorbing a linear momentum of direction B-D1 . Analogously for D2 . A unitary transformation takes two orthogonal states onto other two orthogonal states. It is not possible that for two orthogonal inputs, a detector passes to a same state.
Dear Sofia,
Your worry and interest is very interesting from a pure knowledge of what is happening when you make a measurement in quantum mechanics. That is interesting in indeed but I don't see how to introduce such infomation which is not in Schrödinger equation.
Notice that if you have different values for the probability of measurement due to the device this means that you are falsifying the result or obtaining what you want. This is not an experimental result.
Perhaps with your set up you can think to distinguish two different events and therefore to associate them different probabilities. But in such a case I think that you must put this information within Schrödinger's equation and no to modify its solutions (i.e. the states and their associated eigenstates). And this must be done with the potentials or with other global (topological) information of the manifold where the potentials are defined.
Dear Daniel,
You are a wise person.
But, my question is not about probabilities, it is about the states of the detectors - did you read the note I just added? The software here displays only part of the question, one has to press on the "+" for seeing all the text. Do you see a "+" immediately above the picture? Please press it and see what I said about the detectors' states.
Now, the Hanbury-Brown and Twiss (HB&T) experiment is well-known, and the formula used there is my formula (1) in the question. As far as I can see, I falsified nothing.
You see, one can say that the detectors are big objects and the absorbed momenta pA-D1 and pB-D1 are shared among all the molecules in the detector' material, s.t. the difference between pA-D1 and pB-D1 is not felt. But I can show you that if we could erase the difference between two orthogonal states, we could realize superluminal communication. Though, it is known that such communication is impossible, so, it should be impossible to transform two orthogonal states of a particle, into one and the same state of some additional object - the detector in our case.
So, what shall I do with the Schrodinger equation, potentials, etc.? There are no potentials here the particles travel in the free space. What exactly you have in mind, can you be more specific?
The HB&T experiment is standard quantum mechanics, I did not invent anything. There has to be some subtility here that escapes.
With kind regards!
Dear Sofia,
I didn't explain me enough well. Perhaps it would better if go step by step
1. I had seen the small figures showing the two symmetric trajectories. That is clear, but notice that the distances are modulus an integer of 2pi. This this means that they have the same weigh if you multiply the two phases by integers. This could be interpreted as topological winding numbers (although the distances are written explicitally they don't mean a local interaction).
2. This is very different if you multiply their amplituds by coeficients as you have done. They are not within the phases and modify the whole state locally.
3. The only form to do that is coming back to initial Schrödinger equation and try to add a potential which could take into account the local configuration of the measurement device.
4. If you just add these E's coefficients you change the interaction of the state with the device and this must be done by a function able to be translated within Schrödinger equation. This is what I tried to say. You see that I don't solve your problem but I only analyse it.
Oh, Daniel, I appologize,
I complicated the things with those function f and g. I corrected that, please look again at the formulas (1) and (2) in the question.
You see, if the distances between the slits and the detectors are so big that the distance between the two slits is negligible comparing to them, then we can write
1/|rD1A| ≈ 1/|rD1B| ≈ 1/|rD1| and 1/|rD2A| ≈ 1/|rD2B| ≈ 1/|rD2|,
In this case, which is the usual case in experiments of HB&T type, in the formulas (1) and (2) the factor (|rD1| |rD2|)-1 factorizes out.
"coming back to initial Schrödinger equation and try to add a potential which could take into account the local configuration of the measurement device"
Daniel, what you say there? Can you recommend me something less difficult? The greatest minds in QM didn't succeed to do a theory of the measurement, of passing from quantum to classical. The functions E1,A, E1,B, E2,A, E2,B, are not known to us, they are very likely to be not at all quantum states, but mixtures. In that case the wave-function (2) should be replaced with a density matrix complicating a lot the formula (1) and probably erasing interference effects. The fact that the formula (1) works, tells us that, as I suggested in the NOTE, the functions Ei,A and Ei,B , (i = 1, 2), factorize out too.
"If you just add these E's coefficients you change the interaction of the state with the device and this must be done by a function able to be translated within Schrödinger equation. This is what I tried to say. You see that I don't solve your problem but I only analyse it."
I don't add the functions E, they should be there as they express the states of the detectors after interacting with the respective particle. As I said above, in very special cases they factorize out - as I assumed in the NOTE.
Still, there remains a question mark on the unitarity, as I explained in the question. If the question mark isn't obvious, I will explain.
With kind regards!
Dear Sofia,
I know that the it is difficult to find a function associated to the features of the measurement. This is difficult but, from my humble point of view, the main interesting target to get with models as the one you work.
The function that you have modified is not a wave function:
1. It is not unitary at short distances, i.e. r
Dear Daniel,
We are not interested to work at short distances, but at very big distances, because there we can approximate |riA| = |riB| (i = 1, 2).
About the units, these have to be so as ∫ |Ψ|2 dr1 dr2 = 1, where r1 and r2 are the positions of the two particles.
Dear Sofia,
The positions of the two particles r1 and r2 cannot be independent of riA and riB . Thus the integral
∫ |Ψ|2 dr1 dr2
is even more dependent of the space coordinates and distances. The result in any case is not a probability and it cannot give just the number 1. This is independent of the parametrization of the trajectories that you chose.
Dear Daniel,
What exist the slits are almost semi-spherical waves. It's not waves with definite directions of the linear momentum, p, but rather with (almost) definite value of the absolute value |p|. We can approximate the waves from the slits by 1/|r - rA| and 1/|r - rB|. But we are interested in what see the detectors.
Take in consideration that the positions of the detectors rD1 and rD2, are not fixed, only the positions of the slits, rA and rB, are fixed. Since the waves from the slits are (almost) semi-spherical, each particle may be detected at whatever position on a photographic plate. Let me remind you that the HB&T effect is not observed with fixed detectors but with wide detectors (or mobile), able to report where was detected one particle and where the other.Thus, in the formulas (1) and (2), |rDiA| and |rDiB| are variables, because |rD1| and |rD2| are variables. Therefore, in the integral
∫ |Ψ(r1, r2)|2 dr1 dr2 = 1
I used the notation r1 instead of rD1 and r2 instead of rD2, for stressing that these are variables.
"This is independent of the parametrization of the trajectories that you chose."
Since when the QM works with trajectories for particles? On a trajectory, a particle has at each time a well defined position and velocity. Since when does QT admit precise values for these two non-commuting variables?
But, let me remind you that the problem here is the states of the detectors after detection, and if factorization is possible as I made the guess in the NOTE.
With kind regards!
Dear Sofia,
I see that you have very special notation were I didn't take into account. Good luck with your research!