After a struggle that lasted a month or two, I understood that a very different quantum framework of physics was needed when I was 16 or 17. During that struggle, I was trying to complete a program outlined by Einstein – to describe all atomic phenomena using a model that was basically a classical field theory of a sort. I had various tricks to explain the quantization of the charge in terms of topological invariants – elementary particles were solitons (especially skyrmions, and they were wormholes of various topologies before that). But to explain the hydrogen atom, I needed to steal ideas from ordinary textbooks of quantum mechanics (I started with those for engineers) and after some moment, it became clear that I needed to steal all of quantum mechanics and the right probabilistic interpretation was then forced upon me because of a few short penetrating thoughts.
So when I was first exposed to the Feynman Lectures in Physics – a book based on Feynman's 1964 course for Caltech undergraduates – and saw his nononsense explanations of quantum mechanics etc., I had already been "converted". But it was still refreshing and insightful to see how he dealt with the problems. Incidentally, the edition of the lectures that I could get was written in Slovak (Feynman became a partly Slovak guy in my optics forever LOL) – it existed many years before the Czech translation – and the Slovak edition had 5 volumes instead of 3 volumes of the original edition.
In the previous blog post, the main physical situation that created lots of discussions was Feynman's observation of superpositions found in the benzene molecule. Caltech has made the whole Feynman lectures available in the web form.
All the "detailed" explanations of quantum mechanics appear in Volume III. Unlike most of the ordinary textbooks of quantum mechanics that start with the infinitedimensional Hilbert spaces of particles moving in space, Feynman started with finitedimensional (and especially twodimensional) Hilbert spaces, situations that are "typically" quantum mechanical.
One may say that with the help of these examples, Feynman taught things that "don't have similar enough classical counterparts" that could mislead the students. So this is arguably a better way to make the students think quantum mechanically. The examples of 2dimensional Hilbert spaces are analyzed in chapters 812 of the volume. For example, Section 10.4 describes the molecule of benzene.
Feynman never uses the word "entangled" or "entanglement" in the 1964 lectures – even though the words were coined by Schrödinger in 19351936, as less ludicrous synonyms of Einstein's 1935 term "spooky action at a distance". I believe that he just found these extra words unnecessary (like the "name of the bird", the word teaches us nothing) – and he didn't want to use words that have frequently appeared in monologues of folks like Schrödinger who said many correct things but also many incorrect things.
He also avoided the word "qubit" – in this case because the word wasn't really around in 1964 yet. But it's obvious that all his discussions of 2dimensional Hilbert spaces are possible realizations of a qubit.
Here, I want to spend some time with Section 18.3 of Volume III, i.e. Feynman's presentation of the right answers to questions posed by the EinsteinPodolskyRosen (EPR) paper about the quantum entanglement (again, Feynman has avoided this word everywhere).
The first "difference of style" you must notice is that while EPR discussed "just the general issue" of the entangled objects, Feynman picked a realworld example with all the details so that you could actually apply it to a real experiment. Feynman has always preferred to think in terms of very explicit examples. As long as one realizes that the lessons and methods generalize to many similar examples (and Feynman did realize that), I think it's better to focus on very specific examples.
To discuss entanglement, Feynman picked the decay of the positronium, a bound state of \(e^\) and \(e^+\). It looks just like the hydrogen atom except that the proton \(p\) is replaced by the equally positively charged positron \(e^+\). The relevant problem for the "relative coordinate" is isomorphic to the hydrogen problem except that the reduced mass is\[
\mu = \frac{m_e m_p}{m_e+m_p} = \frac{m_e^2}{2m_e} = \frac{m_e}{2}
\] assuming that the subscript \(p\) denotes the positron. This reduced mass is just onehalf of the reduced mass for the ordinary hydrogen, approximately \(m_e\), which is why there are many factorsoftwo inserted relatively to the hydrogen atom. For example, the ground state has energy \(6.8\eV\) instead of \(13.6\eV\) of the hydrogen atom.
As you know, the electron wants to annihilate with the positron. How quickly?
This is an issue not directly relevant for the EPR considerations – and Feynman analyzes the relative spins elsewhere – but the positronium may have the same spins of the two fermions, or opposite ones. If the spins are opposite, the resulting bound state is a singlet state with \(S=0\), the parapositronium. It almost always decays to two photons and the parity of the bound state is negative, \(P=1\). The lifetime is 124 picoseconds.
There's also a triplet state, the orthopositronium, where the fermions have the same spin and the total is therefore \(S=1\) and the parity is positive, \(P=+1\). It's much more longlived, 139 nanoseconds, and decays mostly to three photons.
We will discuss the singlet state, the parapositronium, and its decay to two photons. Feynman quickly deduces that the state of these two photons (flying in opposite directions) is\[
\ket{F} = \frac{\ket{R_1 R_2}  \ket{L_1 L_2}}{\sqrt{2}}
\] He omits the factor of \(1/\sqrt{2}\) that we usually include. It's just a different convention: it's not "wrong" to work with ket vectors not normalized to unity. Note that both photons, flying to detectors \(1\) and \(2\), have the same circular polarization, \(L\) or \(R\). The total spin has to be zero but two \(R\) photons flying to the opposite directions really have the opposite angular momentum (the angular momentum has the sign of the \(L/R\) helicity times the ordinary momentum and the ordinary momenta are opposite) which add up to zero, so it's OK. (Similarly for two \(L\) photons.) And the relative sign or phase in between the two terms is \((1)\) because \[
P\ket F = \frac{\ket{L_1 L_2}  \ket{R_1 R_2}}{\sqrt{2}} = \ket F.
\] The parity of the twophoton state is negative – the parity is conserved which is why it has to be equal to the negative parity of the original positronium state. Note that the parity is a "mirror reflection" (the determinant is negative) and the righthanded screw is therefore changed to the lefthanded one. The coefficients were unchanged.
Excellent. So we may see the quantum mechanical predictions for the circular polarization. When the twophoton state\[
\ket{F} =\frac{ \ket{R_1 R_2}  \ket{L_1 L_2}}{\sqrt{2}}.
\] is measured to find the circular polarization, we see that if the photon \(1\) is \(R\), so is the photon \(2\), or both of them may be lefthanded \(L\), if the second term is realized. The correlation between the circular polarizations is "perfect" as guaranteed by the angular momentum conservation law. Even people like EPR who didn't really believe quantum mechanics would tend to agree that the angular momentum conservation law means that only \(RR\) or \(LL\) were possible if the circular polarizations are measured.
Fine. So the real controversy only arises when we decide to measure the linear polarizations of both photons – whether their electric fields oscillate in the \(x\) direction or the \(y\) direction (the electric field "names" the polarization by convention; the magnetic field oscillates along the other axis). To find the correct quantum mechanical predictions for these measurements of \(x,y\) polarizations, we simply convert \(\ket F\) from the circularly polarized basis to the linearly polarized basis.
We do it easily with the conversion formulae in between the bases:\[
\sqrt{2}\ket {R_1} = \ket{x_1} + i\ket{y_1}\\
\sqrt{2}\ket {L_1} = \ket{x_1}  i\ket{y_1}\\
\] and\[
\sqrt{2}\ket {R_2} = \ket{x_2} + i\ket{y_2}\\
\sqrt{2}\ket {L_2} = \ket{x_2}  i\ket{y_2}\\
\] The usual equations are obtained by dividing these by \(\sqrt{2}\), I just wanted to save some space. The righthanded polarized photon is a superposition of the two linearly polarized ones, with the 90degree phase shift guaranteed by the coefficient \(i\). The relative phase is opposite for the lefthanded polarization.
With the usual conventions about axes etc., we may be able to figure out which is \(R\) and which is \(L\), but even if we confused the two, it wouldn't be a big deal because \(F\) is antisymmetric with respect to \(R\) and \(L\), anyway (and produces the same probabilities for them, so if we confuse \(R\) and \(L\) everywhere, it changes nothing about predictions for \(x,y\)). You might be worried that it's very important whether we use the same relative phases for the photon \(2\) as we did for the photon \(1\). But this issue is actually also unimportant because a sign error may be included to a redefinition of the sign of \(\ket{y_2}\). Moreover, it's possible to extrapolate the relative phases so that "the convention is the same everywhere" by connecting the two photons' momenta via a 180degree rotation around the \(y\)axis.
As long as you believe that it's possible to write the conversion from \(\ket{R,L}\) to \(\ket{x,y}\) that has a real coefficient between \(R\) and \(x\) and between \(L\) and \(x\) and the signs are the same between \(R,L\) for a given photon, the sign conventions above are basically OK. With these formulae, it's easy to rewrite the twophoton state \(\ket F\) to the \(x,y\) bases for the photon.
It is not hard to see that in the difference \(\ket{RR}\ket{LL}\), the terms \(\ket{xx}\) cancel between \(\ket{RR}\) and \(\ket{LL}\), and so do the terms \(\ket{yy}\). On the contrary, the mixed terms \(\ket{xy}\) and \(\ket{yx}\) will double and this extra factor of two cancels against \((\sqrt{2})^2\) from the basischange. To summarize, we get\[
\ket F = \frac{ i\ket{x_1 y_2} + i\ket{x_2 y_1} }{\sqrt{2}}
\] You see that if one photon is \(x\)polarized, the other one is \(y\)polarized, and vice versa. If we had made some important sign errors in the "change of bases" or in the relative sign defining the original \(\ket F\), we could incorrectly conclude that the linear polarizations of the two photons were the same. It wouldn't be such a huge difference for the purpose of discussions about the EPR issues because what matters is that the correlation or anticorrelation would still be 100% while Einstein would predict no correlation, as we will discuss momentarily.
Feel free to do these things really carefully. With the nonzero risk that Feynman has made a mistake and the linear polarizations should be 100% correlated rather than anticorrelated, let us move on and talk about the qualitative things that really matter.
What really matters is that Einstein and collaborators believed that the probabilities have to be 25%, 25%, 25%, 25% for the results of the linear polarizations \(xx,xy,yx,yy\), respectively. That differs from our quantum mechanical prediction 0%, 50%, 50%, 0%. The experiment may be performed and it confirms that the correlated QM prediction is right while EPR were wrong: Nature really guarantees the perfect correlations or anticorrelations both in the circular polarizations and in the linear polarizations.
Feynman summarizes the wrong Einsteinlike "classical" argument to six arguments and he says that 1,2,4,6 are correct while 3,5 are wrong. The first wrong assumption 3 starts by words that sound by "locality" but the assumption is actually the whole "local realism" so 3 is wrong, indeed. One could argue that Feynman hasn't separated "locality" from "realism" as cleanly as I would. But we must realize that he was basically trying to reproduce the arguments by EPR and similar folks and these folks have surely not separated "locality" from "realism" cleanly at all. So this conflation may be said not to be Feynman's imperfection.
OK. But let me now "localize" and "quantify" the actual difference between the correct quantum mechanical calculation; and Einstein's calculation based on some "local realist" thinking. The correct quantum mechanical derivation predicts the probability of \(xx\) – both photons are \(x\)polarized – to be 0% because the coefficient in the ket vector \(F\) in front of \(\ket{x_1x_2}\) in this basis is zero, OK?
On the other hand, Einstein predicts the probability to be 25%. How does it work?
Einstein assumes that the probability of \(xx\) may be calculated from the probabilities of \(RR,RL,LR,LL\) using some "classical probabilistic reasoning". This is my contribution to the analysis "how the wrong thinking works":\[
{\rm Einstein, wrong:}\\
\eq{
P(xx)&= P(xxRR)P(RR)+P(xxRL)P(RL)+\\
&+P(xxLR)P(LR)+P(xxLL)P(LL)
}
\] Each of the four terms involves the conditional probability of \(xx\) given \(RR\), and so on.
This is how the classical reasoning works. The idea is that we may assume that one of the four states described by the circular polarization, \(RR,RL,LR,LL\), has to be objectively realized even if we don't measure the circular polarization. It seems necessary to Einstein because when the first photon is measured to be \(R\) and we know for sure that the second photon will be \(R\) as well, it seems like a proof that the second photon must be "objectively" \(R\) already before the measurement.
Because only the \(RR\) and \(LL\) states of the circular polarization are allowed, Einstein's reasoning reduces to\[
\eq{
P(xx) &= P(xxRR)P(RR)+P(xxLL)P(LL)\\
& = \frac 14 \cdot \frac 12 + \frac 14 \cdot \frac 12= \frac 14.
}
\] The probability simply has to be 25% if relativity is true, Einstein thought. However, the calculation above assumes not only locality but also "realism". As Feynman also points out, Einstein incorrectly assumes that the question "whether the photons are in \(RR,RL,LR\) or \(LL\)" has a sharp answer even if we don't measure the circular polarization. But this is simply not true. Questions about properties of objects generally only have sharp answers if the answers are measured by an actual measurement. It matters whether an observer actually observes something or not.
So the calculation using the conditional probabilities above is incorrect. Is there a way to "fix it by a localized fix"? Yes, there is. The correct quantum mechanical calculation that determines the probability of \(xx\) is actually very similar to Einstein's wrong calculation using the conditional probabilities above. But it differs from the "formula with four terms" in one respect: we have to calculate the probability amplitudes, not the probabilities themselves, and only calculate the probabilities at the very end out of the probability amplitudes.
Let me repeat the wrong Einstein's fourterm calculation of the probability using the conditional probabilities:\[
\eq{
P(xx)&= P(xxRR)P(RR)+P(xxRL)P(RL)+\\
&+P(xxLR)P(LR)+P(xxLL)P(LL)
}
\] The correct quantum mechanical formula for the corresponding probability amplitude is very similar:\[
\eq{
\langle xx \ket F &= \langle xx \ket {RR} \bra {RR} F \rangle + \langle xx \ket {RL} \bra {RL} F \rangle\\
&+\langle xx \ket {LR} \bra {LR} F \rangle + \langle xx \ket {LL} \bra {LL} F \rangle
}
\] Where does it come from? I have simply "sandwiched" the completeness relation\[
1 = \ket {RR} \bra {RR} + \ket {RL} \bra {RL} +\ket {LR} \bra {LR} + \ket {LL} \bra {LL}
\] in between \(\bra {xx}\) and \(\ket F\). This "sandwiching" had the effect of converting the known probability amplitudes stored in the state vector \(\ket F\) from the circularly polarized bases to the linearly polarized bases.
But my correct quantum mechanical formula is very analogous to Einstein's because the correct Born formula for the quantum mechanical probability is\[
P_{QM}(xx) = \abs{ \langle xx \ket F }^2
\] and similarly\[
P_{QM}(RR) = \abs{ \langle RR \ket F }^2
\] and so on. The probability is just the squared absolute value of the probability amplitude. So what Einstein wanted to be calculated is just the squared absolute value of \(\langle xx \ket F\). However, when we square the absolute value of this correct amplitude which I copyandpaste here again,\[
\eq{
\langle xx \ket F &= \langle xx \ket {RR} \bra {RR} F \rangle + \langle xx \ket {RL} \bra {RL} F \rangle\\
&+\langle xx \ket {LR} \bra {LR} F \rangle + \langle xx \ket {LL} \bra {LL} F \rangle,
}
\] we have to use things like \((A+B)^2 = A^2 +2AB+B^2\) and aside from the (four) squared terms that look like four Einstein's terms, we also get all the (twelve) mixed terms such as \(2AB\)! These are terms that Einstein's classical derivation – one that produced \(P(xx)=1/4\) – has overlooked.
We may summarize the situation by the statement that the "classical reasoning" assuming that objects have objective properties even if these properties are not measured is exactly equivalent to omitting all the mixed terms i.e. exactly equivalent to neglecting all the quantum interference!
Once you realize that the correct predictions e.g. for the double slit experiment require the interference and mixed terms, then you should be able to understand that this is equivalent to admitting that one can't assume that one of the "possible outcomes of a measurement" are objectively realized even in the absence of a measurement.
According to quantum mechanics, this is simply not the case. When a physical object is measured, one obtains a sharp result – one of the eigenvalues of the corresponding operator – but just a picosecond earlier, this outcome just wasn't and couldn't have been decided yet. If the observable isn't measured, the object finds itself in a superposition determined by complex probability amplitudes and the right way to translate these probability amplitudes to predictions differs from the classical logic, one that assumes that "some properties objectively exist even if they're not measured".
Dark matter rises to its biggest challenge… and succeeds! (Synopsis)

“Faced with the choice between changing one’s mind and proving that there
is no need to do so, almost everybody gets busy on the proof.” J. K.
Galbraith L...
13 hours ago
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