Distinguishing classical physics vs. quantum physics

[houlahound]

I think you are short changing the intelligence of the average freshman. Tunnelling is the conceptually simplest thing to visualise with a simple sketch yet it is classically impossible that nobody would argue.

 

[ftr]

True enough, and nicely short. But what I was after in my OP was something a college freshman would understand.

I am an EE who is interested in physics from the early days of college and spent many many hours investigating the subject. Even with good background in classical electromagnetic being part of my field, I still struggle to have a good mental picture of what QM is as related to CM or on its own. Moreover, the conceptual difference ( not necessarily the weird things like ERP.. etc) are also interpretation dependent. But I also think with any science student who is interested in QM today with vast amount of knowledge on the internet and libraries at his fingertip should not have much problem getting the hang of it.

Ya, it is an interesting exercise, but I wouldn't worry about it.

 

[Zafa Pi]

Ya, it is an interesting exercise, but I wouldn't worry about it.

You got a pill for that? I've been dwelling about it on and off for months.

 

[secur]

Wow, I always thought that lifetime till decay was governed by an exponential distribution and was well documented by experiment. I'm not familiar with your explanation, but what is the law in that case?

Exponential decay takes over when decoherence has separated the decay modes / states, by sending off-diagonal elements of density matrix (correlations) to zero. After that happens probabilities are classical and we get exponential decay. But before they decohere, right at the beginning of the decay event ... I don't really know what it does but it's not exponential, due to interference between potential decay states.

The tail of the distribution also isn't exactly exponential. I'd be interested to hear why that is.

This non-exponentiality hasn't yet been verified experimentally AFAIK. But no doubt QM will turn out to be right, as usual.

 

[Zafa Pi]

If you really want to address someone "who has never taken physics" few of these answers will do it. If you tell them that "qp−pq=iℏqp−pq=iℏqp-pq=i\hbar" or "Alice and Bob are too far apart to communicate and neither knows what the other is doing, etc" they'll give a totally blank look - and never again ask you anything about physics!

Over the past three days I've given my story in post #1 to several high school graduates and they seem to understand it, but can not fathom a no answer. Just what I expected and wanted. Maybe you think I should want something else.

 

[Zafa Pi]

Exponential decay takes over when decoherence has separated the decay modes / states, by sending off-diagonal elements of density matrix (correlations) to zero. After that happens probabilities are classical and we get exponential decay. But before they decohere, right at the beginning of the decay event ... I don't really know what it does but it's not exponential, due to interference between potential decay states.

The tail of the distribution also isn't exactly exponential. I'd be interested to hear why that is.

This non-exponentiality hasn't yet been verified experimentally AFAIK. But no doubt QM will turn out to be right, as usual.

I like it, it sounds cool, but I don't understand, "but it's not exponential, due to interference between potential decay states."

 

[secur]

Maybe you think I should want something else.

Not at all.

I like it, it sounds cool, but I don't understand, "but it's not exponential, due to interference between potential decay states."

I don't really know what it does but it's not exponential, due to interference between potential decay states.

But the basic idea is clear. When the decay states are decohered, and obey classical probabilities, each decays at its own unique (except for degeneracy) decay rate eigenvalue. But when there's interference you get some hard-to-calculate semi-random mix of the different decay rates. It wouldn't look exponential; it might even increase at times. If you can get data at that granularity, 10^-20 seconds (typical), the graph would start off irregular. But soon it would become smooth exponential decay, because, according to MWI, this particular instance of you has become entangled with one of the decohered eigenstates. And if you believe that I've got a great deal on Nigerian gold mines you might be interested in. Please PM.

 

[vanhees71]

I think the non-exponential decay due to quantum effects has been observed in atomic transitions.

For a detailed discussion about theory on non-exponential decay, see the following paper

https://arxiv.org/abs/1110.5923

 

[houlahound]

Wooh, that is a tough paper.

 

[A. Neumaier]

I also agree that your statement in post #62, "QM predicts spectra, CM doesn't." is is accurate and elegantly short, but I doubt a freshman lit major would be familiar with the concept "spectra".

Replace ''spectra'' by ''spectral lines in a rainbow pattern obtained by shining sun light through a prism'', and they will understand. (Or they won't understand anything about classical and quantum mechanics anyway.)

 

[A. Neumaier]

Tunnelling in general is pure QM.

except that the name is completely spurious. If it were really tunneling, the rate should be independent of the height of the barrier. But is goes to zero as the barrier gets larger and larger.

 

[houlahound]

Mr Nuemaier an you comment on tunnelling of EM waves. I get confused because it was observed before QM and has a full classical explanation - is it a quantum thing?

 

[A. Neumaier]

tunnelling of EM waves. I get confused because it was observed before QM and has a full classical explanation

Please give a reference so that I can comment.

 

[houlahound]

https://en.m.wikipedia.org/wiki/Total_internal_reflection#Frustrated_total_internal_reflection

It is leakage/tunnelling across a forbidden gap by my reckoning. Sorry about the basic level link.

 

[A. Neumaier]

https://en.m.wikipedia.org/wiki/Total_internal_reflection#Frustrated_total_internal_reflection

It is leakage/tunnelling across a forbidden gap by my reckoning. Sorry about the basic level link.

This is mathematically the same as quantum 1-photon tunneling. In both cases, the barrier height affects the tunneling efficiency, and an astromomically high barrier would produce negligible tunneling.

 

[bhobba]

In QM Born's rule is a postulate. It's unlikely that one can proof it somehow from the other postulates (see Weinberg, Lectures on Quantum Mechanics, Cambridge University Press 2012).

Just to elaborate a bit. Weinberg is of course correct. But, via Gleason, can be reduced to something that at first sight seems to have nothing to do with it, and within the formalism of QM seems almost trivial. In fact when you go through Gleason its so obvious its easy to miss - but its there. But it isn't trivial. In fact its telling us something very very deep about QM - but that requires a whole new thread.

The key point was reformulated as the Kochen-Specker theorem, but its really a simple corollary to Gleason. That it went this way has an interesting history:
https://plato.stanford.edu/entries/kochen-specker/

I believe this obscures its main import - its really the reason for the Born-Rule.

Thanks
Bill

 

[Zafa Pi]

Bell's original papers. See, for example, equation (2) here:

http://www.drchinese.com/David/Bell_Compact.pdf

Sorry for taking so long to get back to you, though you were always in my thoughts. I finally got around to perusing Bell's paper you recommended, thank you.
His proof that expectation value (e.v.) generated via hidden variables (2) cannot capture the e.v. from QM (3) (page 404) does in fact require CFD. This is in opposition to your posts #21 and #14.

For a given trial of the experiment only two measurements are made A(a,λ) and B(b,λ) = -A(b,λ) (1). However on the way to proving his nifty (15) the first equation at the top of page 406 we see three measurements A(a,λ), A(b,λ), and A(c,λ). The only way to get a third measurement in a single formula is employing CFD, we get the value of a measurement that wasn't made. That's what hidden variables allows one to do.

His proof is cute because it is short and only employs three measurements instead of the usual four as in CHSH, beside beating everyone to the punch, for which his name rings loud in the halls of QM.
I attended the - 50th anniversary of his paper - symposium in Vienna. The world came, it was was truly wonderful.

 

[PeterDonis]

The only way to get a third measurement in a single formula is employing CFD, we get the value of a measurement that wasn't made.

This is not correct. P(a, b) and P(a, c) in the formula at the top of p. 406 refer to probability distributions obtained from two sets of runs--one set with settings a, b for the two measuring devices, the other with settings a, c for the two measuring devices. Similarly, P(b, c) in equation 15 and the equation just above it refers to a third set of runs, where the settings are b, c for the two measuring devices. There are no counterfactuals at all; everything is in terms of statistics done on actual observed results. And of course when these experiments are actually done, that is exactly how they are done and how they are analyzed; nobody assumes any results for measurements that are not made.

 

[Zafa Pi]

This is not correct. P(a, b) and P(a, c) in the formula at the top of p. 406 refer to probability distributions obtained from two sets of runs--one set with settings a, b for the two measuring devices, the other with settings a, c for the two measuring devices. Similarly, P(b, c) in equation 15 and the equation just above it refers to a third set of runs, where the settings are b, c for the two measuring devices. There are no counterfactuals at all; everything is in terms of statistics done on actual observed results. And of course when these experiments are actually done, that is exactly how they are done and how they are analyzed; nobody assumes any results for measurements that are not made.

Would you say that the usual CHSH or GHZ uses CFD? If so where? If not then we have a fundamental disagreement on the nature of CFD that I can foresee may be very difficult to resolve.

BTW P(a,b) is not a probability distribution it is a number, an expectation value, allegedly = -a•b

 

[PeterDonis]

Would you say that the usual CHSH or GHZ uses CFD?

If you can give a specific reference I will take a look at it.

 

[Zafa Pi]

If you can give a specific reference I will take a look at it.

For CHSH try page 114 of Quantum Computation and Quantum Information by Nielsen and Chuang (favorite of mine and Demystifier).
I looked on the internet for GHZ and was appalled at how long and clumsy all the presentations were. I posted a presentation during a dialogue with rubi. It was simple and very short, but I can't recall the thread. Do you know of a way I can go about finding it?

 

[Zafa Pi]

If you can give a specific reference I will take a look at it.

I rewrote it for your viewing pleasure (or not).

Physical set up for GHZ:
Alice, Bob, Carol, and Eve are all mutually one light minute apart. At noon Eve simultaneously sends a light signal to each of A, B, and C. When A receives her signal she flips a fair coin. If it comes up heads she selects (via some objective process) a value of either 1 or -1 and calls that Ah. If she flips tails she selects 1 or -1 and calls it At. This takes her less than ½ minute. Each of B and C do the same, calling their selections Bh, Bt, and Ch, Ct.

If we assume that no influence or information can go faster than the speed of light (called locality) then none of the three know what the others flipped, nor can one's selection influence another's.

GHZ Theorem: Let's assume that if only one of A, B, or C flipped a head then the product of their selections equals -1.
I.e., we assume -1 = Ah•Bt•Ct = At•Bh•Ct = At•Bt•Ch.
Then we may conclude if all three flipped heads their product would be -1. I.e., -1 = Ah•Bh•Ch.

Proof: -1 = (Ah•Bt•Ct)•(At•Bh•Ct)•(At•Bt•Ch) = At²•Bt²•Ct²•Ah•Bh•Ch which implies that -1 = Ah•Bh•Ch. QED

If Eve sent each of A, B, C one photon from the entangled triple state |ψ⟩ = √½(|000⟩ + |111⟩) and if flipping a head selects the value obtained by measuring a photon with Pauli X and flipping a tail means measuring with Pauli Y, then X⊗Y⊗Y and Y⊗X⊗Y and Y⊗Y⊗X each operating on |ψ⟩ yield -1, so the hypothesis of the Theorem is satisfied. However, X⊗X⊗X operating on |ψ⟩ gives 1, contradicting the conclusion.

Lab tests show the QM predictions are correct. So what's wrong?

Why other sites take pages of gobbledygook is beyond me. Maybe you see a reason.

 

[PeterDonis]

the entangled triple state |ψ⟩ = √½(|000⟩ + |111⟩)

Your notation is somewhat confusing; earlier you gave the eigenvalues of the measurements as 1 and -1, but in this expression it looks like they are 0 and 1. Also, which basis (X or Y, or something else) is this expression written in?

 

[Zafa Pi]

Your notation is somewhat confusing; earlier you gave the eigenvalues of the measurements as 1 and -1, but in this expression it looks like they are 0 and 1. Also, which basis (X or Y, or something else) is this expression written in?

The eigenvalues are 1 and -1. I am using the standard q-computing notation, see Nielsen and Chuang.
|ψ⟩ = √½(|000⟩ + |111⟩) = √½[1,0,0,0,0,0,0,1] in the 8D tensor product space is an eigenvector of X⊗Y⊗Y with eigenvalue -1 and is an eigenvector of X⊗X⊗X with eigenvalue 1.

 

[PeterDonis]

|ψ⟩ = √½(|000⟩ + |111⟩) = √½[1,0,0,0,0,0,0,1] in the 8D tensor product space

In which basis on that space?

 

[Zafa Pi]

In which basis on that space?

|0⟩ = [1,0], |1⟩ = [0,1] are the basic qubits, |000⟩ = |0⟩⊗|0⟩⊗|0⟩ = [1,0,0,0,0,0,0,0], |001⟩ = |0⟩⊗|0⟩⊗|1⟩ = [0,1,0,0,0,0,0,0], etc. If you don't have access to N & C any reference to q-computing should do. It is the slickest QM notation.

If the QM is a bit obscure it doesn't matter. The point is that QM satisfies the hypothesis but negates the conclusion of the Theorem. So how is that resolved?

 

[PeterDonis]

|0⟩ = [1,0], |1⟩ = [0,1] are the basic qubits

In which basis? Do you understand what that question means? There are an infinite number of possible pairs of basis qubits. Which of them are you using?

 

[zonde]

In which basis on that space?

Here is Zeilingers version in chapter 16.2 (not very long either): https://vcq.quantum.at/fileadmin/Publications/2002-12.pdf

 

[Zafa Pi]

In which basis? Do you understand what that question means? There are an infinite number of possible pairs of basis qubits. Which of them are you using?

The basis in 2D is [1,0] and [0,1], the higher dimensions are built up of tensor products. If you find this inadequate check out N&C.
But your questions are side stepping the crucial question as I said above.
As we pursue this line, I must admit I am nervous that you may use your god like powers to cut out my posts. We are a bit off topic and perhaps we should leave it with I am convinced that Bell used CFD in his argument and you are convinced he didn't.
Or we could start a separate thread on this topic, or whatever you wish.

 

[Zafa Pi]

Here is Zeilingers version in chapter 16.2 (not very long either): https://vcq.quantum.at/fileadmin/Publications/2002-12.pdf

Zeilinger's version and mine are the same. Where he has |H⟩₁ |H⟩₂ |H⟩₃ I have |0⟩|0⟩|0⟩ = |0⟩⊗|0⟩⊗|0⟩ = |000⟩ as per Nielsen & Chaung.

 

[PeterDonis]

The basis in 2D is [1,0] and [0,1]

Still doesn't answer the question. I need actual directions in space that describe the orientation of the basis vectors.

Where he has |H⟩₁ |H⟩₂ |H⟩₃ I have |0⟩|0⟩|0⟩ = |0⟩⊗|0⟩⊗|0⟩ = |000⟩ as per Nielsen & Chaung.

Which means that your "0" means "horizontal" and your "1" means "vertical", as in polarization directions for photons. That answers the question. (Actually, you also used "X" and "Y" to describe measurements, so I assume that "X" also means horizontal and "Y" also means vertical.)

 

[zonde]

Actually, you also used "X" and "Y" to describe measurements, so I assume that "X" also means horizontal and "Y" also means vertical.

X is diagonal basis H'/V' (H'=-1, V'=+1)
$|H'\rangle=\frac{1}{\sqrt 2}(|H\rangle+|V\rangle)$ and $|V'\rangle=\frac{1}{\sqrt 2}(|H\rangle-|V\rangle)$
and Y is circular polarization basis L/R
$|R\rangle=\frac{1}{\sqrt 2}(|H\rangle+i|V\rangle)$ and $|L\rangle=\frac{1}{\sqrt 2}(|H\rangle-i|V\rangle)$
it's in the link I gave in post #98

Here is Zeilingers version in chapter 16.2 (not very long either): https://vcq.quantum.at/fileadmin/Publications/2002-12.pdf

There is a heuristic principle how to make sense of this GHZ state. All phases between H and V modes have to sum up to $0$.
H' state gives phase $0$, V' gives phase $\pi$, R state gives phase $\pi/2$ and L state gives phase $3\pi/2$.
With this we get allowed combinations:
H'LR gives $0+3\pi/2+\pi/2=2\pi$
H'RL gives $0+\pi/2+3\pi/2=2\pi$
V'LL gives $\pi+3\pi/2+3\pi/2=4\pi$
V'RR gives $\pi+\pi/2+\pi/2=2\pi$

 

[Zafa Pi]

X is diagonal basis H'/V' (H'=-1, V'=+1)

X = Pauli X = σ_x = $\begin{pmatrix}0&1\\1&0 \end{pmatrix}$

Y is circular polarization basis L/R

Y = Pauli Y = σ_y = $\begin{pmatrix}0&-i\\i&0\end{pmatrix}$

And for completeness Z = σ_z = $\begin{pmatrix}1&0\\0&-1\end{pmatrix}$ has eigenvector |0⟩ = [1,0] = $\begin{pmatrix}1\\0\end{pmatrix}$ with eigenvalue 1, and eigenvector |1⟩ = {0,1] with eigenvalue -1.
These are the most common 2D measurement operators

 

[Zafa Pi]

Still doesn't answer the question. I need actual directions in space that describe the orientation of the basis vectors.

The vector [1,0] in the xy plane appears to be pointing horizontally off to the right, while [0,1] appears to be pointing vertically upward. That's why Zeilinger used the notation |H⟩ for {1,0}, where as I used the standard q-computing notation of |0⟩

Which means that your "0" means "horizontal" and your "1" means "vertical", as in polarization directions for photons. That answers the question. (Actually, you also used "X" and "Y" to describe measurements, so I assume that "X" also means horizontal and "Y" also means vertical.)

No, that isn't the case. The 0 in |0⟩ is an atavistic homage to the classic bit 0, while the 1 in |1⟩ refers to the bit 1. The X and Y I defined in post #103
If you are really interested in the quantum computing notation you should take a look at Nielsen & Chuang, it's quite nice and far easier to read than say Ballentine, for example.

I suspect you have lost interest in our original CFD question.

 

[PeterDonis]

I suspect you have lost interest in our original CFD question.

No, I just wasn't familiar with your notation or the references you gave. I am looking at the Pan & Zeilinger paper you linked to (which gives a much better brief and explicit description of its notation) and will probably have further comments once I have digested it.