I Modern View of Quantum Phenomena

[RUTA]

So in what sense this tells us where the electron was before measurement, among other properties.

We have 30+ pages in Chapter 9 of the book on this, so what I write here is very superficial. The self-consistently shared information between interacting bodily objects establishes what we know as the "classical context" and that establishes the properties of the quanta of the interactions. So, when an electron is a quantum (as opposed to a track in a particle detector, say) it is not interacting with the bodily objects in the experimental set-up and has no worldline in spacetime (or it would be a bodily object, not a quantum). That the quantum is an electron in the spacetime context in this particular experiment is established by the source from previous spacetime contexts (so you can say, "I have a source of electrons for this experiment"). All that means the electron was emitted by the source and absorbed by the detector in the spacetime context of bodily objects for the experiment. There is no "hidden reality" for the electron between source and detector otherwise.

 

[selfsimilar]

There is no "hidden reality" for the electron between source and detector otherwise.

sorry, I meant like the position of the electron in Hydrogen atom, for example.

 

[bhobba]

sorry, I meant like the position of the electron in Hydrogen atom, for example.

You miss the point of why I started this thread.

Particles are not fundamental; they emerge from fields. For fields, the question is meaningless.

While not directly related to the position question, another point was that QFT is more or less inevitable once certain general assumptions, such as Lorentz Covariance, are accepted. Moreover, our current theories are effective - not fundamental. This has implications for the reality question of the fields and, hence, even particles. It may be like basic QM. We know it is just an approximation to QFT, and is incorrect. As yet, we do not know the situation with QFT.

The interpretation question has shifted. I still adhere to the statistical interpretation of ordinary QM, but like thermodynamics, we know it is not fundamental. For QFT, my interpretation is that it is inevitable, given Wienberg's Folk Theorem.

Thanks
Bill

 

[renormalize]

Moreover, our current theories are effective - not fundamental.

1) Can you elucidate the criteria by which a fundamental theory is distinguished from an effective one?
2) Can you cite examples of known fundamental theories?
Thanks.

 

[PeterDonis]

1) Can you elucidate the criteria by which a fundamental theory is distinguished from an effective one?

An effective theory is one that is admitted to not be fundamental. Or, if you insist on a definition that doesn't use either of those two words, an effective theory is admitted to not provide a complete explanation of the phenomena it covers, but to have some aspects that need to be put into the theory to make it work but which the theory can't explain. For example, the particle masses and coupling constants in the Standard Model of particle physics. The term "effective theory" is used because of the belief that an explanation of those aspects will require some deeper theory to which the current effective theory is an approximation.

2) Can you cite examples of known fundamental theories?

I'm not aware of any. All of our best current physical theories are effective theories.

 

[bhobba]

1) Can you elucidate the criteria by which a fundamental theory is distinguished from an effective one?

Your name here, renormalize, holds the key, as I think you probably know.

In QFT, early on, there was a deep problem. We kept getting infinity. Okay - we examine the math and see that improper integrals lead to the problem, i.e., an integral to infinity gives infinity. You can get a finite answer by not using infinity, but rather a large number called a cutoff. The trouble is what large number, and what does it mean? The trick of renormalisation is that you can manipulate the equations with a cutoff, so the cutoff is not explicitly part of the equations. Dyson showed that the infinities are still there - but they have been 'swept' under the rug, so to speak. It's the kicker in the joke; if you look under a QF theorist's rug, you find all these infinities swept under it. The advantage of doing this, however, is that these terms (ie those that blow up to infinity) can actually be measured and their values used. That those values depend on the chosen cutoff is all part of the so-called renormalisation group, where measured values change depending on the cutoff. The theory is not fundamental; we need to introduce this cutoff. It is believed that this 'trick' is effective up to approximately the Planck scale. Regardless, it is a trick because it is not fundamental - what is happening above the cutoff, we currently do not know, although research is ongoing, of course. To make matters worse, even this 'trick' has its own problems - look up the Landau pole.

Of course, this fits nicely with Wienberg's Folk Theorem - we only know the 'large distance' theory - if we want the theory at smaller distances at our current level of knowledge, blank out - all we know is regardless of what it is, at distances we can access (currently), it is no surprise it is a QFT.

Thanks
Bill

 

[selfsimilar]

You miss the point of why I started this thread.

The post was really intended for RUTA saying that his system does not seem to add too much to understand the conceptual problems of QM.

 

[RUTA]

The post was really intended for RUTA saying that his system does not seem to add too much to understand the conceptual problems of QM.

Again, I'm presenting our particular completion of the quantum reconstruction program (QRP), so this is not an idiosyncratic approach to resolving the conceptual problems of QM. There is a vast literature on the QRP and we have published several papers, blogs, and a book on our particular completion thereof, which deflates the 'big' and 'small' measurement problems and shows how the "big three" characteristics of QM (randomness, superposition, and entanglement per Brian Greene) follow from the observer-independence of h, as required by the relativity principle and Planck's radiation law, in exact analogy with SR per Rovelli's challenge. Accordingly, QM is complete as possible and there is no need for "spooky actions at a distance" or retrocausality or superdeterminism or the violation of intersubjective agreement or infinitely many realities. All this strikes me as significant progress in resolving the conceptual problems of QM, but I'm biased of course :-)

 

[gentzen]

I guess I don't understand what you're talking about then. The rotation is always in the plane perpendicular to the beam but the beam direction is arbitrary. The Pauli matrices allow you to specify that rotation plane in any direction relative to your Cartesian coordinates. This is all in accord with homogeneous spacetime per the Lorentz group.

If I have two entangled particles in "two beams" with known fixed direction, then I want to measure their spin, not the spin of two other particles corresponding to different "two beams" with different known fixed directions.

I could try to bent a given beam with some combination of electric and magnetic fields. That would probably also affect the spin in some "in principle predictable" way. And this part would exhibit a complex relation to special relativity. How sensitive the spin would react to minute details of the electric and magnetic fields and the particle momentum is also an interesting question (which probably also has a known answer).

 

[RUTA]

If I have two entangled particles in "two beams" with known fixed direction, then I want to measure their spin, not the spin of two other particles corresponding to different "two beams" with different known fixed directions.

I could try to bent a given beam with some combination of electric and magnetic fields. That would probably also affect the spin in some "in principle predictable" way. And this part would exhibit a complex relation to special relativity. How sensitive the spin would react to minute details of the electric and magnetic fields and the particle momentum is also an interesting question (which probably also has a known answer).

But you're just changing the direction of the beam, you're still going to make your SG spin measurement in the plane perpendicular to that beam. I don't see what this accomplishes physically.

 

[gentzen]

But you're just changing the direction of the beam, you're still going to make your SG spin measurement in the plane perpendicular to that beam. I don't see what this accomplishes physically.

It measures the spin of the particle in some specific direction, as it was initially before the bending of the beam.

(I am still a bit confused about the number of parameters. An arbitrary direction in 3D space would still only be two continuous parameters. Maybe I should try to look up the posts where that topic popped up last time.)

 

[RUTA]

It measures the spin of the particle in some specific direction, as it was initially before the bending of the beam.

(I am still a bit confused about the number of parameters. An arbitrary direction in 3D space would still only be two continuous parameters. Maybe I should try to look up the posts where that topic popped up last time.)

Suppose you start with the state $|\psi\rangle=|z+\rangle$, then you make a SG measurement along $\hat{b}$ making an angle of $\theta$ with respect to $\hat{z}$. It doesn't matter what coordinates you use for the $\hat{b}\hat{z}$ plane, your distribution of $\pm 1$ outcomes (in units of $\frac{\hbar}{2}$) is given by P(+1) = $\cos^2{\frac{\theta}{2}}$ and P(-1) = $\sin^2{\frac{\theta}{2}}$ so that the average is $\cos{\theta}$, i.e., what you would expect per simple Newtonian mechanics if the spin angular momentum vector for the quantum state $|\psi\rangle=|z+\rangle$ is $+1\hat{z}$. And since that direction is arbitrary, I'm not sure how you expect these results would change if you somehow changed the direction of the beam.

 

[gentzen]

Suppose you start with the state $|\psi\rangle=|z+\rangle$, then you make a SG measurement along $\hat{b}$ making an angle of $\theta$ with respect to $\hat{z}$.

Suppose you start with the state $|\psi_{\alpha,\beta}\rangle=\alpha |z+\rangle+\beta|z-\rangle$. Then there is a measurement that will always give +1, and never -1. I want to realize that measurement. When you measure the state $|\psi\rangle=|z+\rangle$ with that measurement, you will get P(+1)=$|\alpha|^2$ and P(-1)=$|\beta|^2$.

 

[RUTA]

Suppose you start with the state $|\psi_{\alpha,\beta}\rangle=\alpha |z+\rangle+\beta|z-\rangle$. Then there is a measurement that will always give +1, and never -1. I want to realize that measurement. When you measure the state $|\psi\rangle=|z+\rangle$ with that measurement, you will get P(+1)=$|\alpha|^2$ and P(-1)=$|\beta|^2$.

I would have to know what the phases $\alpha$ and $\beta$ mean physically. For example, in the double-slit qubit where you are illuminating the slits equally and in phase your state is $|\psi\rangle = \frac{1}{\sqrt{2}} \left( |S1\rangle + |S2\rangle \right)$ (S1 = slit 1, S2 = slit 2). So, when you do a "which-slit" or "position" measurement (you put your detector screen right up against the slits, say) half the time you get an outcome behind S1 and half the time you get an outcome behind S2. When you do an "interference" or "momentum" measurement (you put your detector screen far from the slits relative to the slit separation) every outcome lands in a constructive interference band. If you add a $\pi$ phase plate behind S2, the state is now $|\psi\rangle = \frac{1}{\sqrt{2}} \left( |S1\rangle - |S2\rangle \right)$ and the "position" measurement outcomes are the same, but the previous "momentum" measurement outcomes change where the bright bands are now dark and the dark bands are now bright.

 

[RUTA]

Suppose you start with the state $|\psi_{\alpha,\beta}\rangle=\alpha |z+\rangle+\beta|z-\rangle$. Then there is a measurement that will always give +1, and never -1. I want to realize that measurement. When you measure the state $|\psi\rangle=|z+\rangle$ with that measurement, you will get P(+1)=$|\alpha|^2$ and P(-1)=$|\beta|^2$.

You got me wondering how to show that no matter what orientation I choose for $\hat{b}$ in the xy plane, when $\hat{b}$ makes an angle $\theta$ with $\hat{z}$ the distribution of outcomes will be P(+1) = $\cos^2{\frac{\theta}{2}}$ and P(-1) = $\sin^2{\frac{\theta}{2}}$ averaging to $\cos{\theta}$. Let me start with $|\psi\rangle = \cos{\frac{\theta}{2}}|z+\rangle + \sin{\frac{\theta}{2}}|z-\rangle$ for $\hat{b}$ in the xz plane, which gives me P(+1) = 1 when I measure in the $\hat{b}$ direction, i.e., $|\psi\rangle = |b+\rangle$ (use Eq 17 in the Answering Mermin's Challenge paper to get $\sigma_b$ then show $\sigma_b |\psi\rangle = |\psi\rangle$). Now I rotate this state about the z axis by $\alpha$ to get $|\psi\rangle = e^{i \alpha/2} \cos{\frac{\theta}{2}}|z+\rangle + e^{-i \alpha/2}\sin{\frac{\theta}{2}}|z-\rangle$ (see Eq 24 in the Answering Mermin's Challenge paper), which does give those probabilities for any $\alpha$. To get a bit of a feel for this, let $\theta = 90^{\circ}$ (again, we're in the xz plane), then $|\psi\rangle = \frac{1}{\sqrt{2}}|z+\rangle + \frac{1}{\sqrt{2}}|z-\rangle = |x+\rangle$, which you can see from $\sigma_x |\psi\rangle = |\psi\rangle$. Now let $\alpha = 270^{\circ}$ and you get $|\psi\rangle = (-1 + i)\frac{1}{2}|z+\rangle + (-1 - i)\frac{1}{2}|z-\rangle = |y+\rangle$, which you can see from $\sigma_y |\psi\rangle = |\psi\rangle$.