A Visual Approach to the Delayed Choice Experiment

06 Apr 2009

8 replies [Last post]

Mon, 04/06/2009 - 12:32

doug

Offline

Joined: 03/13/2009

Posts:

The paper "Experimental realization of Wheeler's delayed-choice GadenkenExperiment" by Jacques et al. quant-ph/0610241v1 was recommended to me. I reads like a seminal paper on the topic, the one where the experimentalists "go it right". Here is the first three lines of their abstract:

The quantum "mystery which cannot go away" (in Feynman's words) of wave-particle duality is illustrated in a striking way by Wheeler's delayed-choice GedankenExperiment. In this experiment, the configuration of a two-path interferometer is chosen after a single photon pulse has entered it: either the interferometer is closed (i.e. the two paths are recombined) and the interference is observed, or the interferometer remains open and the path followed by the photon is measured. We report an almost ideal realization of that GedankenExperiment, where the light pulses are true single photons, allowing unambiguous which-way measurements, and the interferometer, which has two spatially separated paths, produces high visibility interference.

Spend some quality time with those three lines, or download the paper for more detail. The experimentalist are choosing whether the experiment is one path or two after the single photon has passed a beam splitter. The more time you spend with this paper, the more bothersome is should become! Actually, it may resign you to the first line of the abstract, that you must accept the mystery.

The first clues I spotted in working on this mystery is in the first paragraph of the text:

When the phase shift $\Phi$ between the two arms is varied interference appears as a modulation of the detection probabilities at output ports 1 and 2 respectively as $cos^2 \Phi$ and $sin^2 \Phi$ . This result is the one expected from a wave...

This is not a random source, it is so well-organized that it can be characterized by sine and cosine functions! The question then became how to generate the $cos^2 \Phi$ and $sin^2 \Phi$ using a quaternion expression? This all has to be done in the context of the rules of quantum mechanics. What you can see in an experiment is the norm of a wave function, the conjugate of one quaternion multiplied by another. This is a simple math riddle to solve:

$\psi = (cos(\omega t), sin(\omega t), 0, 0)$

$|\psi|^2 = \psi^* \psi = (cos(\omega t), -sin(\omega t), 0, 0)(cos(\omega t), sin(\omega t), 0, 0) = (cos^2(\omega t) + sin^2(\omega t), 0, 0, 0) = (1, 0, 0, 0) \quad eq ~1-2$

Two things are different from a standard approach. The angle $\Phi$ is being represented by $\omega t$ . All I really need is a dimensionless quantity that increases monotonically. Second, few people work with quaternions. Yet the complex numbers are a sub-group of quaternions. Any proof ever done with complex numbers can be done with quaternions of the form (a, b, 0, 0). Since the third and forth numbers are zeroes, they do not alter the expression. Quaternions of this form will commute with each other, a sign that the properties of complex numbers are faithfully represented. The zeros are needed by my software to create an animation in spacetime.

The wave function is in equation 1, what can be seen is in equation 2. This raises two basic questions. The first question is why can't to see what is going on in equation 1? Equation 1 can definitely be plotted like any other quaternion expression. Its "problem" is that it is free to described events that are spacelike. Spacelike events cannot be linked causally. The math machinery does what cannot be done physically.

What can be seen is in equation 2. The observer lives at zero, zero time (now), and zero in any direction no matter the choice of coordinates. The way I interprets (1, 0, 0, 0) is that it represents an event an observer can see at a later moment in time. Is there any event an observer has no chance to see? Since (0, 1, 0, 0) is happening now (t=0) at a different place (x=1), that event cannot be seen. So what if after squaring the wave function, there is a bit of t and a bit of x? That is when probability comes into play. Recall the function that got squared has events that are independent of each other. The wave function, equation 1, accurately describes the odds, which show up in the square.

What data was generated by this experiment? The data is in figure 3, reproduced here:

3a show interference over a $10 \pi$ shift in phase. Interference is completely gone in 3b. If the approach used in this web site for animating physics equations is valid, then I would have to be able to generate two animations that contained the same information content. This is a well-defined problem that I set out to solve.

The first thing is to see what equation 1 looks like, even if it is not physical.

amp.povray.100.1000.animation.scan_.gif

All motion is in the x direction. Even when we take the square, all motion will necessarily be exclusively in the x direction. That will make what is going on darn impossible to figure out! What I have done is to take the phase shift that happens in the t and x, and include it for y, so the differences can be seen. Here is what happen for the amplitude:

amp_shifted.povray.100.1000.animation.scan_.gif

So now we have a stack of similar animations, with those higher on the stack having a larger phase shift. These all look the same, but that is due to the wacky world of harmonic boundary conditions. As we shift the phase, the starting point on the tx circle at t=0 is different, but the system ends up eventually covering the same points in the tx plane, so they all look like clones even though they are not.

We can "mark" where the time for the phase is zero.
amp_shifted_marked.povray.100.1000.animation.scan_.gif
One cannot do this physically since all photons are identical, but it does help visually.

Label the two paths A and B. A is never shifted, B is shifted anywhere from $\Phi = 0 \to 10 \pi$ . The data in figure 3 of the paper involves 4 calculations:

A* B - interference (3a)
B* A - interference (3a)
A* A - no interference (3b)
B* B - no interference (3b)

Algebraically, we have:

$(cos(\omega t), -sin(\omega t), 0, 0)(cos(\omega t + \Phi), sin(\omega t + \Phi), 0, 0)$
$= (cos(\omega t) cos(\omega t + \Phi) + sin(\omega t) sin(\omega t + \Phi), cos(\omega t)sin(\omega t + \Phi) - sin(\omega t) cos(\omega t + \Phi), 0, 0)$
$(cos(\omega t + \Phi), -sin(\omega t + \Phi), 0, 0)(cos(\omega t), sin(\omega t), 0, 0)$
$= (cos(\omega t) cos(\omega t + \Phi) + sin(\omega t) sin(\omega t + \Phi), cos(\omega t + \Phi) sin(\omega t) - sin(\omega t + \Phi) cos(\omega t), 0, 0)$
$(cos(\omega t), -sin(\omega t), 0, 0)(cos(\omega t), sin(\omega t), 0, 0)$
$= (cos^2(\omega t) + sin(\omega t)^2, 0, 0, 0) = (1, 0, 0, 0)$
$(cos(\omega t + \Phi), -sin(\omega t + \Phi), 0, 0)(cos(\omega t + \Phi), sin(\omega t + \Phi), 0, 0)$
$= (cos^2(\omega t + \Phi) + sin^2(\omega t + \Phi), 0, 0, 0) = (1, 0, 0, 0) \quad eq ~3-6$

We can see the lack of ambiguity for the no interference case. The first two are going to be more complicated. For a fixed value of $\Phi$ and $\omega = 1$ , vary t and plot:

$(cos(t) cos(t + \Phi) + sin(t) sin(t + \Phi), cos(t + \Phi)sin(t) - sin(t + \Phi) cos(t), \Phi, 0)$

$(cos(t) cos(t + \Phi) + sin(t) sin(t + \Phi), cos(t)sin(t + \Phi) - sin(t) cos(t + \Phi), \Phi, 0) \quad eq ~7-8$

(one technical detail: for the way my software is designed, I had to swap the roles of x and y, as you can tell since the circle is now in the ty plane instead of the tx plane). I varied $\Phi$ from $0 \to 10 \pi$

interference_ty.povray.100.1000.animation.scan_.gif

This looks like a good copy of figure 3a in the upper right corner! That is the superposition of all states. There are 5 peaks in red, and 5 peaks in blue, just like figure 3a. Recall how all this really happens in the x direction, and only by adding the phase shift to a different spatial direction can we notice this is happening. I am uncertain of this detail, but it may be a requirement that to see interference, one must use a different spatial coordinate from the coordinate that is coherent. In other words, one is taking the nice order of the ty plane and spreading it out to see along tx.

Figure 3a and this animation are not the same obviously, since figure 3 is static. It is also concerned with counts. The times in the animation when one is certain to see an event and thus have the highest count is at the start of the circle and the end. The time when you are certain not to count any photons is at the maximal separation, when t=0.

We can appreciate why this is a tricky experiment to understand. Nothing happens in the tz plane, so that is a simple line because time marches on - albeit the line is finite because of the harmonic boundary conditions. In the ty plane we have the graph we expect for a sine and cosine working together. Yet this is a dynamical expression, so it has pairs of points rushing away from each other. There are 5 pairs in red, and 5 in blue that fly away from each other, only to find a different partner when they annihilate each other. The motion in the tx plane was for our convenience, so we could see these 10 pairs individually instead of having them all piled on top of each other.

For completeness, here is the dull no interference case:

$(cos^2(\omega t) + sin^2(\omega t), \Phi, 0, 0) = (1, \Phi, 0, 0) \quad eq 9$

no_interference.povray.100.1000.animation.scan__0.gif

No matter what the phase is, there is no interference. Good.

The people who wrote this paper knew all the equations presented here. This is how they drew the solid lines in figure 3. What I have done is to use a spacetime representation that can consistently keep all the work in spacetime instead of venturing into some form of wave function abstraction. When time plays directly with space, fun things can happen. The solution to the mystery of quantum mechanics is to let time and space play together.

Tue, 04/28/2009 - 20:34

dougsweetser

Offline

Joined: 03/13/2009

Posts:

Complex proofs with quaternions

Hello Lowell:

I am not skilled at doing rigorous proofs. A slight change I would make in your description is that three complex numbers are subgroups of quaternions. A subset is part of set theory. Being a subgroup requires a few more constraints, namely that subgroup has the properties of a group: (a, b, 0, 0) times (c, d, 0, 0) is always part of the group, there is an identity (1, 0, 0, 0), and an inverse (a, -b, 0, 0)/((a^2 + b^2) so long as zero is omitted.

A good question forces me to be even more precise. The complex numbers are a subfield. That means that the numbers are a group under the addition operator, and a group under multiplication modulo zero. For addition, (a, b, 0, 0) + (c, d, 0, 0) remains in the group, there is the identity, (0, 0, 0, 0), the inverse (-a, -b, 0, 0).

My statement can be generalized. I choose the easy example, where two of three imaginary numbers are zero. So long as the quaternions commute, they will behave like complex numbers. The more general statement would be that quaternions that point in precisely the same direction will have the same properties as complex numbers.

Sun, 05/03/2009 - 16:55

Lowell

Offline

Joined: 04/25/2009

Posts:

Re: Complex proofs with quaternions

'Linnaeus-ian' mathematical zoo of algerbraic things, ha, ha! But good point about Sets and Groups!

My statement can be generalized. I choose the easy example, where two of three imaginary numbers are zero. So long as the quaternions commute, they will behave like complex numbers. The more general statement would be that quaternions that point in precisely the same direction will have the same properties as complex numbers.

Yes, Simplicity is Golden!
So things like (a,b,0,0) in general alway commute?
This is way cool...there is the identity, (0, 0, 0, 0), the inverse (-a, -b, 0, 0).

__________________

Lowell

__________________

Lowell

Sun, 05/03/2009 - 19:04

doug

Offline

Joined: 03/13/2009

Posts:

Re: Complex proofs with quaternions

Hello Lowell:

Yes, (a, b, 0, 0) will always commute with (c, d, 0, 0) because they both point in the (i, 0, 0) direction.

The complex numbers are a field, meaning they are a group under addition - with the inverse and identity you cited correctly - and a group under multiplication, where the identity is (1, 0, 0, 0) and the inverse is (a, -b, 0, 0)/(a^2 + b^2) so long as the additive inverse (zero) is excluded.

What appears to be happening in physics is that for one part of a problem, people characterize the story using groups under addition. For another aspect, they use groups under multiplication. Nature does both in stereo. Her books are open, honest, and hard to read.

Sun, 05/03/2009 - 20:51

Lowell

Offline

Joined: 04/25/2009

Posts:

Re: Complex proofs with quaternions

Doug,

When you say,

What appears to be happening in physics is that for one part of a problem, people characterize the story using groups under addition. For another aspect, they use groups under multiplication. Nature does both in stereo. Her books are open, honest, and hard to read.

Are you referring to the additions and multiplications found in the field theory text books? Ex. 3 x 3 x 3 = 16 + 8 + 2 + 1, where the x's and +'s are in cirlces. It's been some time since I've studied how tensors and Young's Tableau relate to this group stuff.

So regardless of whether the physics is quantum (internal) or classical, are saying that the rules for using groups are biased depending on what is being described? as you state, "what part of a problem is analyzed."

Is this your whole line of reasoning for using H's in attempting unification?

__________________

Lowell

__________________

Lowell

Sun, 05/03/2009 - 21:44

doug

Offline

Joined: 03/13/2009

Posts:

Re: Complex proofs with quaternions

Hello Lowell:

What I was thinking about specifically, and not writing it out, was the covariant derivative of the standard model:

$\mathcal{D}_{\mu} = \partial_{\mu} - i g_1 \frac{Y}{2} B_{\mu} - i g_2 \frac{\tau^i}{2} W^i_{\mu} - i g_3 \frac{\lambda^a}{2} G^a_{\mu} \quad eq ~1$

That was taken from page 77 of "Modern Elementary Particle Physics" by Gordon Kane. The g's are coupling constants. The Y, $\tau$ and $\lambda$ are the generators of group symmetries U(1), SU(2), and SU(3) respectively. The B, W, and G are the 4-vector potentials. Note that the last two have "internal symmetries" that march nicely instep with the generators (the 'i: 1-3' and 'a:1-8' indicies). So Y and B should be put together as the quaternion $\frac{A}{|A|}$ . For the weak term, mashup tau and W with exp(A-A^*) . For the strong which needs 8 elements for its lie algebra, go with $(\frac{A}{|A|} exp(A-A^*))^* \frac{B}{|B|} exp(B-B^*)$ . This is my "smaller-than-the-standard-model" proposal. You can see that if I drop that potential into an action, then the field equations that result would have the symmetries for light, the weak, and the strong force.

The reason Nature uses this: it is a complete decomposition of the symmetries of (1, 0, 0, 0) in spacetime. I will presume you have taken a look at the animations of these symmetries on this site, possibly after smoking some quality weed, which visually demonstrates these groups show life on a unit sphere.

Mon, 05/04/2009 - 03:32

Lowell

Offline

Joined: 04/25/2009

Posts:

Re: Complex proofs with quaternions

Yes, I've watched the animation video on this...and my mind is limited enough without the weed. the Y and G quaternion expressions look suspiciously alike, is this related to the fact the the first looks like the "perfect circle" and the second looks like the "perfect sphere?" (also the fact that the one describing W is a sphere that slightly disintegrates?

The expression (1,0,0,0) looks isotropic spatially and that it preserves the speed of light "ct." Is this like the Poincare group?

With respect to W, the exp(A-A*),does this particular form for the quaternion have anything to do with the parity violation of the weak force?

__________________

Lowell

__________________

Lowell

Wed, 05/06/2009 - 21:31

doug

Offline

Joined: 03/13/2009

Posts:

NFC (No Frigging Clue)

Hello Lowell:

At this time, I have No Frigging Clue (NFC) how this work impacts partity violations.

That is all I should say about being in an NFC state, but thought I should talk about that situation, so I can reference this post later when someone askes a different, well-justified question.

A unified field theory is obligated to answer all outstanding questions in physics, as well as be consistent with all current experimentally confirmed results. If the proposal does not do so, then it is incomplete.

As I sat through a few talks by real particle physicists showing data on top-top quark decay measurements at Tevetron, the only place that makes them fresh, it is easy to feel overwhelmed by how little math machinery I have constructed. I can accept that someone trained in these obscure arts would dismiss the few widgets in my collection.

When deep-woods lost on a question, I look for someone who has framed the question well, read that, then let the problem sit filed away in my head for years. An example involves SU(3). I read that the unit quaternions can be represented by the exponential, exp(A-A^*) . It helps my confidence to read anwswers. I also read that the group U(1) is a normalized complex number, so that was trivial to apply to the house of three complex numbers that share a real, the quaternions. Put the to together, and one has electroweak symmetry with $\frac{A}{|A|} exp(A - A^*)$ .

I was on a roll. All I had to do was bring in SU(3). I had two of three players of the standard model. It might sound abzurd, but it was about two years before I tripped into the idea of taking the conjugate of one normalized quaternion and multiplying it by another. The snail pace is due to this being a sliver-of-my-time project and having no collaborators, a biproduct of the time I can put into the work. Due to my skimpy training, reading the technical literature is heavy sledding. I also feel inscure in that I don't know how to prove the isomorphism, to show to all the world that the group multiplication table for SU(2) is no different from A* B.

I got one good question out of the April APS meeting. The moderator wondered if I had done any calculations with W's and Z's. That is a fair question that at this time have no idea how to start writing anything related to the question on a sheet of paper (I do have both paper and writing tools, but a random approach is unlikely to work within my lifetime, or the Universe's life). I may drop him an email asking for advice on self-study.

The parity problem is filed away. I should search out a few more good reference that characterize the issue. Quaternions certainly have an innate handedness. How the SU(2) animation relates to real beta decay which varies from atomic isotope to isotope is undoable today. Instead, I stay practical. My next animations will be polynomials because I have done them before, and I recall some theorem saying once you have the polynomials, you can recreate almost any function one can construct. That will probably take me a month or so, but the tools are there in the latest release.

Doug

Tue, 04/28/2009 - 19:01

Lowell

Offline

Joined: 04/25/2009

Posts:

Re: A Visual Approach to the Delayed Choice Experiment

Any proof ever done with complex numbers can be done with quaternions of the form (a, b, 0, 0).

This is an interesting statement. It should be obvious concerning proofs of the complex numbers, as you say the C's are a just a subset of H's. I know this is not the topic of this post, but since you are an expert H's, could you post demonstrating some of these proofs.

__________________

Lowell

__________________

Lowell