Quantum Interference

I am not familiar with this experiment but I’ll give it a shot. Davies explanation makes a lot of sense. First, as he says, you must consider both photons along all possible paths. If the wave functions of the photons along the possible paths are 180 deg out of phase at a point in space they will cancel giving a zero probability for the photons to be at that point.Here are my assumptions:
1) The photons start in phase and travel the same distance to both detectors.
2) A photon reflected from the beam splitter will undergo a 180 deg shift in phase (yeah my E&M is probably weak, someone correct me if I am wrong).Therefore:
1)One photon on top, reflected — Cancelled by photon on bottom, not reflected2) One photon on top, not reflected — Cancelled by photon on bottom, reflected3) Both photons travel along top; one reflected, one transmitted — (Both) Cancelled by
the two photons from the bottom; one reflected one transmitted.Which leaves the only non-destructive situations as those where both photons travel along the upper or lower paths and are either both reflected or both transmitted. Both of these situations result in the photons arriving at the same detector. Just think of them as plain old sine waves. Add all the amplitudes. Where the amplitudes vanish, there is zero probability to find the photons.

I almost forgot. The two photons along the top, both reflected, cancel with the two photons from the bottom, not reflected. And the two photons from the top, not reflected, cancel with the two photons from the bottom, reflected.Therefore, I have a grand total of zero photons anywhere.bah

I had, at the least, my phase shift wrong. This derivation is taken from the website ofFrank Rioux
Department of Chemistry
Saint John's University
College of Saint BenedictOk, as Davies said you must consider the photons (labeled 1 and 2) as traveling along both the upper and lower paths. Therefore, the state of the entangled photons is give by|state) = 1/sqrt(2){|1, up)|1, down) + |1, down)|2, up)},where the wavefunction is symmetric as required by bosons.A photon traveling along the upper path will be transmitted to detector d2 and reflected to detector d1. According to Dr. Rioux the phase change upon reflection is not 180 deg but 90 deg and is required by conservation of energy. Writing the phase shift of the wavefunction as exp[ix] where x is the change in phase then gives|up) = 1/sqrt(2) {i|d1) + |d2)},where i is the imaginary number sqrt(-1).Similarly |down) = 1/sqrt(2) {|d1) + i|d2)}.Multiplying it all out,|state) = {i|1,d1)|2,d1) + i^2 |1,d1)|2,d2) + |2,d1)|1,d2) + i|1,d2)|2,d2) + i|1,d1)|2,d1)+ i|1,d1)|1,d2) + |1,d1)|2,d2) + i^2 |2,d1)|1,d2) + i|1,d2)|2,d2)}/ 2^(3/2).This reduces to|state) = 1/sqrt(2) {i|1,d1)|2,d1) + i|1,d2)|2,d2)}.Squaring the coefficient of each term and taking the square root of the magnitude gives the various probabilities:Prob of 1 at d1 and 2 at d1 = 1/2
Prob of 1 at d2 and 2 at d2 = 1/2
Prob of 1 at d1 and 2 at d2 = 0
Prob of 1 at d2 and 2 at d1 = 0.Therefore as stated, if you consider the photons as traveling along all possible paths you can show that the photons cannot be detected simultaneously at different detectors.

What happens in your version! Heh, I had to go look up what happens in the other version!

Looking at my first attempt I can see plenty of mistakes. The initial state is one where one photon travels along the lower path and one photon travels along the upper path. There is not a state where both photons travel the upper or lower path simultaneously. As such, if both photons are to be detected at the same detector, one must be reflected and one must be transmitted. The experiment I saw described actually did use 4 detectors and 2 beam splitters. Two of the set ups described by JustinCy set back to back. Is this what you are describing? Or is yours different?

Ahh, I see. I think to set up at Berkeley had a way to block the up and down paths independently.

Blocking either the up or down paths intercepts one of the photons, keeping it from interfereing with the other at the beam splitter. That way, you can check to see if the photons are reflected and transmitted with a 50% probability at the beam splitter under non-interference conditions. The advantage I see to this method is that it tests the actual setup used to to observe the effects of interference.

The website I got the derivation from pretty much spelled it out the way I have it so I don’t think it will be much more help. The lines and parentheses are my limited graphics way of trying to reproduce the typical quantum representations of things.The symbol |x) that I have [as in |state) or |up)] represents the state of the system or particle. The usual representation is actually |x> and is called a ket, but as you can see the graphics don’t really come out that good. Heh, actually they only look like crap in my word program. They actually look pretty good here. Therefore, I’m going to switch notations on you and use things like |state> and |up> instead. These kets are closely related to the wavefunctions of the system. In fact, for our purposes, we can just consider them to be the wavefunctions for the system. With that in mind|state> — This is the total wavefunction for our two photon system.|1> and |2> — These would be the wavefunctions of our two photons, photon 1 and photon 2.However, since we have to consider each photon as traveling along all possible paths each photon actually has 2 wavefunctions, one for the up path and one for the down path. I have these represented like so|1,up> — Wavefunction for photon 1 along the up path.
|1,down> — Wavefunction for photon 1 along the down path.
|2,up> — Wavefunction for photon 2 along the up path.
|2,down> — Wavefunction for photon 2 along the up path.The total wavefunction |state> is made up of a linear superposition (this means we just add them together) of all possible states of the system. This gives|state> = 1/sqrt(2) {|1,up>|2,down> + |1,down>|1,up>}The factor 1/sqrt(2) (the inverse of the square root of 2) is just a normalization factor which ensures that the total probability for the system to be found in some state is equal to one.Inside the brackets notice we do not have a states such as |1,up>|2,up> or |1,down>|2,down>. This represents the fact that one of the photons must travel along the upper path and one of the photons must travel along the lower path (in other words |1,up>|2,up> = |1,down>|2,down> = 0). This leaves the only non-zero possible states |1,up>|2,down> and |1,down>|2,up>.So what is the reality of the situation? Did photon 1 travel along the upper path and photon 2 along the lower? Or did photon 2 travel along the upper path and photon 1 along the lower?What quantum mechanics shows us is that it is not correct to consider each photon as going along a specific path. Instead we must consider each photon as going along all possible paths simultaneously. This, of course, seems absurd. However, it makes a physical difference whether the photons travel along a specific path or all possible paths and so the question of what path(s) the photons take is not a metaphysical one but an experimental one.That said, the total wavefunction for our photons must include the possibilities for each photon to travel along both the upper and lower path. This is why both the states (wavefunctions) |1,up>|2,down> and |1,down>|2,up> are included.The beam splitter gives us two possible paths to the detectors for photons traveling along the upper and lower paths. This means that the up and down states (wavefunctions) (ie. |up> and |down>) are actually comprised of a linear superposition of two other states (wavefunctions), one leading to detector d1 and one leading to detector d2.The next question is then, how do we represent the states |up> and |down> in terms of the states leading to the two detectors |d1> and |d2>? This is where the question of phase comes in.You asked how you should try to visualize these wavefuntions along the possible paths and I said to just consider them as sine functions. A better (at least easier mathematically) way to visualize them is in terms of the exponential function such as|x> = exp[i{(kx-wt) + phase}] = e^i[(kx-wt) + phase] = exp[i(phase)] * exp[i(kx-wt)].These are known as plane waves.i is the imaginary number sqrt(-1)
k is the wavenumber and is related to the momentum of the photon.
x is the photon position.
w is the photon frequency which is related to its energy
t is time.A few things about the experimental setup are important to note.
1) The photons are identical. Therefore, their wavefunctions have identical values for k and w.2) The photons always travel the same distance to reach a given detector whether they travel along the upper or lower path. Therefore, at the detectors the wavefunctions have identical values for x and t.This implies that at the detectors the wavefunctions for the photons are identical except for the phase difference exp[i(phase)]. The difference in phase is caused by a phase shift in the wavefunction upon reflection of the photon from the beam splitter. Dr. Rioux sets the value of this phase shift at 90 deg or pi/2. For transmission through the beam splitter there is no change in phase.Since the only difference in the wavefunctions at the detectors is the difference in phase, there is no sense in writing out the factor exp[i(kx-wt)] everywhere. Therefore, it can just be symbolized |d1> and |d2>. Keep in mind though the actual wavefunctions at the detectors are exp[i(phase)]|d>.Ok, for transmission through the beam splitter there is no change in phase and we have exp[0] = 1 (transmission)For reflection we have exp[i(pi/2)] = i (reflection)Now we are in a position to construct the wavefunctions |up> and |down> in terms of the wavefunctions leading to detectors d1 and d2.For a photon traveling along the upper path it must be reflected to reach detector d1 and transmitted to reach detector d2. A superposition of the wavefunctions then gives|up> = 1/sqrt(2) {i|d1> + |d2>}.For a photon traveling along the lower path it must be reflected to reach detector d2 and transmitted to reach detector d1. A superposition of the wavefunctions then gives|down> = 1/sqrt(2) {|d1> +i |d2>}.Once again, the factor 1/sqrt(2) is just there to ensure that the probability that a photon arrives at either d1 or d2 is equal to 1.Obviously we have these up and down wavefunctions for both photon 1 and photon 2. This is where I get things like|1,up> = 1/sqrt(2) {i|1,d1> + |1,d2>} and so forth.Now just substitute these expanded wave functions back into our total wavefunction equation|state> = 1/sqrt(2) {|1,up>|2,down> + |1,down>|1,up>}and solve. Take the magnitude square of the coefficient in front of each state [for example |1,d1>|2,d1>] to get the probabilities for the various photon detector combinations.

You are correct, I made a typo. The correct wavefunction is as you have it|state> = 1/sqrt(2) {|1,up>|2,down> + |1,down>|2,up>}. This also seems correct. The total wavefunction describes the probability (amplitude) for the two particles to be at two particular points in space at a particular time. Suppose one of the particles has a zero probability to be at a particular point in space at a given time. Then the total wavefunction must also vanish for a state in which that particle is to be found at that particular point at the given time. This shows the wavefunctions are multiplicative not additive.*You can also think of it in terms of probabilities. What is the probability of both A and B? ( Probability of A) * (Probability of B). |d1> and |d2> give the probability (amplitudes) for the photons be to found at the distinct locations d1 and d2. Basically, they are wavefunction amplitudes evaluated at different points in space. If we labeled them the same there would be confusion as to which point in space the interference was taking place. Perhaps I should have said reduce the equation to its simplest form, or combine like terms and square the coefficients.

Sorry for the delay, I had to go out of town.Your first question is do the probabilities have to add up to one, independent of sign?.Normally, in physics, complex numbers are used to simplify the mathematics. The Schrodinger equation, however, requires that the wave function be complex. I can derive the Schrodinger equation for you if you need it, but understand the Schrodinger equation cannot be derived. It is a new theory whose validity is tested through experiment, not derived through an extension of pre-existing theories.What is the reality of a vector? It is not the vector itself, which is only a representation. A vector (in 3d space) can be defined in terms of its x, y and z components. However, upon rotation of the coordinate axis, the x, y and z components change. So what is the reality of the vector? That would be its length. The length of a vector is invariant upon rotation. The square of the vector length is given byR^2 = x^2 + y^2 + z^2 (irrespective of basis)So what is the length of a vector in complex space?Complex space is mapped like soImaginary
|
|
|
_______________RealTherefore, to determine the length of a vector in complex space one must take the dot product of the vector with its COMPLEX CONJUGATE. When one take the complex conjugate of a number one replaces i (the imaginary number) with -i.For example, for a vector of unit (1) length in complex space writtenR = 1/sqrt(2) [1 + i],its square magnitude (squared length) would be determined byR^2 = (1/2) [1+i]*[1-i].This gives a unit length of 1 regardless of rotation.Therefore, if you want to know the magnitude squared of the wave function, you must always multiply the wavefunction by its complex conjugate. This always results in a positive number. This means the probabilities will always be positive. Hope this answers your first question.2cnd questionSo why is the probability the square of the wavefunction (amplitude)?.The short answer is simple, the philosophical answerheh.AsideI am pretty much a lurker on this board and I sincerely enjoy reading the comments of those who post. I will verify (although I do not know why Buzsaw assumed in the thread on the 3 thermodynamic laws) that I am an atheist. It seems that many believers would like to incorporate science into their arguments for God, but in doing so they seem to only half-heartly embrace the findings of science. The empirical evidence for the validity of modern physics is truly outstanding. The precision with which quantum mechanics (field theory, QED, etc) can predict phenomena is truly incredible. However, what do our theories truly say? What is to be said of the western philosophy of reductionism? In the end our best theories of science point to a world of probabilities not reality. Is the sum more than its parts? Have we reached the limit where not considering the whole begins to become meaningless? Can consciousness really be understood in terms of reductionism?BahI am an experimental physicist not a philosopher.As the esteemed physicists Richard Feynman said no one understands quantum mechanics. I (personally) find it amazing that we understand anything at the quantum level at all. All our understanding stems from the understanding of how things we observe (ie. at the macroscopic level) behave. We understand how springs work, or waves on a string, and we are able to incorporate that into our understanding of atoms and subatomic particles, yet there is no a priori reason why this should be so. There is no reason why the subatomic world should act in any way we understand whatsoever.End asideAs with most of quantum mechanics, we look at things we do understand to give us guidance in the quantum mechanical world. According to Newton’s laws, the energy of an oscillating system is proportional to the square of the amplitude of the wave (oscillation). For example, consider the oscillating system described byx = A sin(wt).The velocity is described byv = Aw cos(wt).Therefore, (classically) the kinetic energy isKE = (1/2) mv^2 = (1/2) (Aw)^2 cos^2 (wt).This shows that on average the kinetic energy of the system is proportional to A^2. Since energy is conserved this also shows that the total energy must also be proportional to A^2.To use an example from electromagnetic theory, for an oscillating electric fieldE(t) = E0 sin(wt),where E0 is the maximum amplitude of the field.The energy of an electric field goes like E^2. Therefore, the average energy of the field is proportional to (E0)^2.Using the Einstein relation: Energy = nhw (The energy is quantized into a number of photons.)This means that the average number of photons at a given point is proportional to the square amplitude of the wave.So what does this mean in terms of the wavefunction for a SINGLE photon? That the square amplitude of the wavefunction must give the probability for the photon to be at that particular point in space.

I apologize, I left out mass in my equation. It should readKE = (m/2)(Aw)^2 cos^2(wt).

Also, just to tie things up, why is the wavefunction required to be complex?Without the complex nature of the wavefunction, there would be no interference term in the squared amplitude of the total wavefunction. The experiment you asked about, which started this thread, is designed to test this interference. Therefore, the complex nature of the wavefunction is required in order for the theory to correspond to reality.AsideSo what exactly is the wavefunction? It seems to dictate reality, but in and of itself, it is not real.Any ideas?

I don't really know much about string theory. Let me know if he says anything neat. Thanks

Ahh, very interesting. I think I have heard Aharonov (as in the Aharonov-Bohm effect) suggest the same thing. I think it is an interesting idea. I have never studied the Wheeler-Feynman absorber theory and always just assumed the advanced potentials in E&M violated causality. Heh, guess I have some learning to do. Thanks again. If you find out anything else please let me know.