What we observe is not the Nature itself but the Nature exposed to our method of questioning.Even though wave mechanics was in no way the first or deepest formulation of quantum mechanics, it quickly became popular because the "wave function" looks like a classical wave and this fact makes it easier for the people to "visualize" what's going on. This "advantage" is actually a disadvantage because the visualization leads the people to the totally incorrect concept that the state vector

Natural science does not simply describe and explain Nature; it is part of the interplay between Nature and ourselves.

The first gulp from the glass of natural sciences will turn you into an atheist. But at the bottom of the glass, God is waiting for you.

Werner Heisenberg

*is*a classical wave or object of a sort, which it's not, and that it should be distinguished from (i.e. considered mutually exclusive with) nearby "similar" state vectors, which it shouldn't, and the popularity of the Schrödinger picture "helps" the people to preserve their anti-quantum misconceptions.

The oldest picture of quantum mechanics, one behind the "matrix mechanics" formulation of quantum mechanics, is the Heisenberg picture. There is no evolving wave function. Instead, it's the operators such as \(\hat L\) that evolve according to the Heisenberg equations\[

i\hbar \frac{d\hat L}{dt} = [\hat L, \hat H]

\] which are effectively "just" the classical equations with extra hats on top of all physical quantities. And in the classical limit, the Heisenberg equations (and their solutions) reduce to their classical counterparts. All the expectation values etc. are the same as in Schrödinger's picture; the two pictures differ by a time-dependent unitary transformation. The Heisenberg picture doesn't even force you to distinguish pure states and mixed states. As we will see on the double-slit example, predictions are formulated via the expectation values of \(\langle \cdots \rangle\) of the observables at the final moment, and they may be written as function(al)s of similar expectation values from the initial moment – which may be interpreted both as those coming from a pure state or a mixed state.

The double slit experiment is the ultimate toy model of quantum mechanics. If you carefully think about this experiment, you may reveal all the secrets and logic of quantum mechanics – which allows you to understand everything else, too. At least Feynman liked to say so. It is rather clear how the results of this experiment are computed (and in particular, where the interference comes from) in the Feynman path integral approach to quantum mechanics; it is also easy to understand how we solve the problem in Schrödinger's picture where a wave function is evolving in between the slits and the photographic plate.

Some time ago, I promised you to write a blog post extracting the predictions for the double slit experiment from the Heisenberg picture of quantum mechanics. Here is my first attempt to solve this task.

We consider an experiment in 2 spatial dimensions and one time; there is no \(z\) direction here. The plate is separated from the slits by the distance \(L\) in the \(y\) direction. The thin slits are separated by a small distance \(R\) in the \(x\) direction, so their coordinates are\[

x = \pm \frac{R}{2}, \quad y = 0.

\] We want to calculate the probability density that the particle lands at\[

x=S, \quad y=L

\] at the photographic plate. The particle makes it through the slits at \(t=0\). We measure the number of particles absorbed by the material around the slits at \(t=0\). We assume that no particle absorption was detected at \(t=0\) so we know that one particle is flying through the experiment.

At time \(t\gt 0\), the particle lands somewhere at the photographic plate. First of all, it is easy to calculate \(t\). The Heisenberg equations of motion tell us (i.e. can be easily solved)\[

\hat y = \frac{\hat p_y}{m} t

\] which is just like the classical equation but with the hats. Because we only assume that the slits affect the \(x\) components of the particle's position and momentum, we conclude that the momentum component \(p_y\) was already known before the particle entered the slits, at \(t\lt 0\). The particle reaches the photographic plate when \(y=L\) which implies that the time is\[

t = \frac{Lm}{p_y}.

\] There's some uncertainty about the initial momentum \(\hat p_y\); this will get translated to some uncertainty about \(t\). Unlike \(\hat p_y\), the quantity \(t\) is not an operator; even in the Heisenberg picture, \(t\) is primarily an independent variable that other things depend upon. But we may ask what is the moment of the absorption (by the photographic plate) that the apparatus will detect and record, and this

*is*an operator because it's a result of a measurement. Before a person observes the apparatus, the apparatus itself will be in a linear superposition of states with different values of the recorded \(\hat t\), and that distribution will simply mirror the distribution of \(\hat p_y\).

*Bugatti step, composed by Jaroslav Ježek in 1931 after he was impressed by a Bugatti car that won a race in Brno. The video may help you think as the fathers of quantum mechanics because someone added the Praga Piccolo Roadster 1926 car into it.*

The previous paragraph or two had a simple purpose: to eliminate the \(y\)-\(p_y\) part of the double slit problem because it has nothing to do with its essence, namely with the interference pattern we want to study. What we really want to solve is the \(x\)-\(p_x\) part of the problem, in the direction of the separation of the slits. We want to trace how the operators evolve between \({\rm time}=0\), the moment when the particle makes it through the slits, and \({\rm time}=t\) (I hope that this seemingly tautological notation isn't confusing) when it lands on the photographic plate. And we want to say lots of clarifying words about the right interpretation of all these things.

Great. You may imagine that we always work with the same and constant wave function\[

\psi(x) = \frac{1}{\sqrt{2}} \zav{ \sqrt{\delta(x+R/2)} + \sqrt{\delta(x-R/2)} }

\] The particle has a 50% probability to be at \(x=-R/2\) and 50% to be at \(x=+R/2\) at \(t=0\). I included the square root of two and the square root of the delta-function to make the state normalized to one. (The square root of the delta-function is always a very unnatural beast, however. That's why all the normalization issues below will be problematic and you're invited to fix this bug and use a more kosher initial state: some extra "width of each slit" may have to be introduced.) To make it more comprehensible, all copied \(x\) in the equation above should be replaced by \(x_{t=0}\). Note that the state vector doesn't depend on \(t\); we are working in the Heisenberg picture where the state vector is constant.

Instead, what is not constant are the operators. Between the slits and the photographic plate, we deal with a free particle. It means that\[

\hat p_x = {\rm const}, \quad \hat x = \hat x_{t=0} + \frac{\hat p_x t}{m}.

\] The momentum in the \(x\) direction is conserved when the particle is freely moving in between the slits and the plates (but not at the moment when the particle travels through the slits). The second equation was easily obtained by integrating the Heisenberg differential equation for \(d\hat x / dt\). The form of all these equations is the same as in classical physics (with extra hats). Our resulting expression for \(\hat x\) looks just like the equation for a "straight trajectory" in the spacetime. But we must remember that there is no particular slope of the trajectory because \(\hat p_x\) is an operator – it is not a well-known classical object.

Because of the hats, the equation that looks like a "straight line" doesn't contradict the existence of the interference pattern, as we are going to see.

*By the early 1930s, the laws of physics were known to be intrinsically probabilistic. So even Prague's famous Liberated Theater could have sung that "Life Is Just a Coincidence".*

Fine. We want to calculate the probability that at \(t=t\), the particle lands at \(x(t)=S\). Well, more precisely, the probability that it lands in the interval \((S,S+dS)\) is \(dS\) times the probability density \(\rho(S)\) that it lands around \(x(t)=S\). This density \(\rho(S)\) is the expectation value of the "not normalized projector" onto the state with \(x(t)=S\):\[

\rho(S) = \bra\psi \hat P_S \ket\psi, \quad \hat P_S = \ket{x(t)=S}\bra{x(t)=S}

\] I wrote that the projection operator is not normalized because the product of these two operators \(P_S\cdot P_{S'}\) is equal to \(P_S\) times not the Kronecker symbol but the delta-function of \(S-S'\), OK? This not normalized projection operator may also be written using a delta-function,\[

\hat P_S = \delta (\hat x - S).

\] The argument of the delta-function is an operator but it's no problem, all these functions of operators are well-defined and may be evaluated in any basis (most simply, in the basis of eigenvectors). Note that if we integrate the operator \(\hat P_S\) over some interval of values of \(S\), we get a genuine projection operator whose eigenvalues are zero and one (meaning "particle didn't land in the interval" and "particle did land in the interval", respectively). That's why the expectation value of \(\int_A^B dS\,\hat P_S\) is the probability that the particle landed in \((A,B)\).

Great. The rest is a calculation. There are many ways to proceed and I've tried several of them, to be sure that I can still calculate. The most straightforward strategy seems to employ the formula for the delta-function\[

\delta(T) = \int_{-\infty}^{+\infty} dQ\,\exp(2\pi i QT).

\] With our argument, we have\[

\hat P_S = \delta(\hat x - S) = \int_{-\infty}^{+\infty} dQ\,\exp[2\pi i Q(\hat x - S)]

\] and \(Q\) has units of the inverse distance. Don't forget that the probability density that should reveal the \(\cos(S)\)-style interference pattern is \(\langle \hat P_S\rangle\). Now, let us use our direct solution for \(\hat x\):\[

\hat P_S = \int_{-\infty}^{+\infty} dQ\,\exp[2\pi i Q(\hat x_0 + \frac{\hat p t}{m} - S)]

\] I decided to simplify the notation a bit, \(\hat p\equiv \hat p_x\) – note that this particular operator is not changing with time (free particle). It's not necessary but I chose to exploit the BCH-style formula\[

\exp(A+B) = \exp(A) \exp(B) \exp(-[A,B]/2)

\] which holds whenever \([A,B]\) is a \(c\)-number (which commutes with everything else). This is somewhat helpful because the exponentials of \(\hat x_0\) and the exponentials of \(\hat p\) which don't commute with each other are separated as different factors. With this formula (and I drop the limits on the \(Q\)-integral, to make things easier), we get\[

\eq{

\hat P_S &= \int dQ \exp(-2\pi i QS+2\pi^2 i\hbar Q^2 t/m) \times \\

&\times \exp(2\pi i Q\hat x_0) \exp(2\pi i Q \hat p t / m)

}

\] On the first line, the \(Q^2t\)-based term comes from the BCH-formula and it knows the same wisdom as the spreading of the initial delta-function in the Schrödinger picture (like in the solution to the diffusion equation, but with the extra \(i\)). Our overall normalization of the state is problematic.

Only the exponentials on the second line of this operator identity contain operators as arguments. It follows that if we calculate the probability density \(\langle \hat P_S\rangle\), we only need to surround the second line of the right hand side by the brackets.

*In 1925, as soon as Pauli found his exclusion principle and its dependence on the spin, "If You Knew SUSY" became the 3rd most popular song of the year. Fifty more years passed before physicists knew SUSY's anticommutators and at least 90 years for the LHC to actually discover it experimentally.*

Before we complete the computation of \(\langle \hat P_S\rangle\), let me emphasize a few general words. So far, the outcome – and the general intermediate result of a Heisenberg-picture calculation – was to express the observables at the final moment as function(al)s of observables at the initial moment. The same thing was true in classical physics: the quantities at the final moment are function(al)s of the quantities at the initial moment (and functions of \(t\), too). The only new aspect of quantum mechanics (in the Heisenberg picture) is that the observables are non-commuting which makes the function(al)s different (from some viewpoint, they're more complex; from another viewpoint, they are equally complicated, just different).

If the final moment observable (which we actually measure) were a function of the observables at the initial moment that we exactly knew (because we measured it at the initial moment), we could produce a "sure" prediction that this observable has a predictable value. This situation – the state vector is an eigenstate of the final moment observable (this description is OK in both pictures) – almost never occurs. We usually deal with a more general, non-eigenstate situation in which the final moment observables also depend on some unknown observables at the initial moment. At most, we may calculate the expectation values etc. The expectation value of the projection operators such as \(P_S\) are interpreted as probabilities (or, in the continuous case such as ours, probability densities).

**Finally, let's compute the interference pattern.**

The probability density as a function of the \(x\)-position \(S\) is\[

\eq{

\langle\hat P_S\rangle &= \int dQ \exp(-2\pi i QS+2\pi^2 i\hbar Q^2 t/m) \times \\

&\times \langle \exp(2\pi i Q\hat x_0) \exp(2\pi i Q \hat p t / m) \rangle

}

\] We need to compute an integral. The integrand involves an expectation value of some function of the initial state operators in the initial (or final, they're the same) state vector. In classical physics, the "state" would be fully described by the values of things like \(x,p\), and it would be "trivial" to take functions of them. Here, the functions of the noncommuting observables are "harder" and we need the expectation value of such function(al)s in the initial state.

Note that in the second line\[

\langle \exp(2\pi i Q\hat x_0) \exp(2\pi i Q \hat p t / m) \rangle

\] the second exponential is a special case the operator of the displacement \(\exp(i\hat p\cdot\Delta x / \hbar)\), so it shifts the wave function by \[

\Delta x=\frac{2\pi Q t\hbar}{m} = \frac{2\pi Q\hbar L}{p_y}

\] After the initial wave function is shifted, the phases of its two \(\sqrt{\delta}\) pieces are changed by the first exponential in our expectation values, and then the inner product of this "shifted and rephased" wave function with the original wave function is computed. Calculations of such expectation values are always the same – in the Schrödinger, Heisenberg, or Dirac picture. The Heisenberg picture we study here differs by the absence of any "evolved" wave function in the calculation.

*"The Beer Barrel Polka" by Mr Jaromír Vejvoda began as the purely instrumental "Modřany Suburb Polka" in the Prague radios in 1927. It took years for an improved version to appear and 7 years (1934) for the "Pity of [meaning: What a Shame that I Wasted] [My Unrequited] Love" ("Škoda lásky") Czech lyrics, "Rosamunde" German lyrics, and (even more years for) the Beer English lyrics to become popular. When it was already a "megahit" [it was arguably the world's most well-known song during WW2], the composer received... 150 Czechoslovak crowns (I mean 150, not 150 million) for his work. Even though Czechoslovakia had the strongest currency in Europe, unlike now LOL, and Prague was the most expensive city, CSK 150 wasn't much. Only the singers in the video above are West German, everyone else is Czech; it was recorded months before the fall of communism; the facades at 1:47 are scary; Vejvoda died just a year earlier.*

Fine. You may see that the inner product is only nonzero if \(\Delta x = \pm R\) or \(\Delta x=0\). This is equivalent to \(Q=\pm Rm/2\pi t\hbar = \pm Rp_y/2\pi L\hbar\) or \(Q=0\). In those cases, the inner product gives us a nonzero contribution due to a delta-function "click". It means that our integral over \(Q\) degenerates into the contributions from these three values of \(Q\). Yes, of course: the fact that there are several terms is how we get the interference. The \(Q=0\) contribution to \(\langle P_S\rangle\) simplifies to \(\langle 1 \rangle=1\) because all the arguments in the exponentials are zero.

Well, around \(Q=0\), we don't actually have \(\delta(Q)\) in the integral. We literally have just one (not infinity) at \(Q=0\). This is a symptom of our normalization problem. If the initial state were a sum of two delta-functions, and not the square roots, this problem would go away and the integral over \(Q\) would "cancel" against the delta-function. This is a more sensible way to deal with the normalization issue – the other way is to choose a finite-width slit. I chose an initial state normalized to one because it's probably easier for beginners to deal with and the overall normalization doesn't affect the interference pattern that will be obtained, anyway. At positive \(t\), such a square-root-of-delta-function state would evolve into an "infinitesimal constant" we got here times a wave function with "infinitely many maxima and minima", as we will see below, so the norm of such a later state is still \(0\times \infty =1\), if you wish.

What about the \(Q=+Rm/2\pi t\) term? The exponential of \(\hat p\) etc. manages to move the peak at \(x=-R/2\) to one at \(x=+R/2\). But the \(\hat x_0\)-based exponential adds different phases to these two peaks. It's the relative phase that matters and it gives us \(\exp(2\pi i QR)\). There is a factor of \(1/2=(1/\sqrt 2)(1/\sqrt 2)\), too. Unlike the previous paragraph, this factor of \(1/2\) is not doubled.

In this relative phase, \(\exp(2\pi i QR)\), the exponent proportional to \(QR\) is much smaller than the exponent \(QS\) from the first part of the first line, and we will neglect it. We neglect the second term on the first line for the same reason. You're invited to write down the exact results; precision may be needed to recover the normalization of the wave function (a finite number of interference peaks).

This value of \(Q\) mainly contributes \(\exp(-2\pi i QS)\) from the first factor on the first line.

Similarly, for \(Q=-Rm/2\pi t\), the relative phases get reverted and we get the multiplicative factor \(\exp(-2\pi i QR)/2\approx 1/2\). However, there's still the important phase factor of \(\exp(-2\pi i QS)\) from the first line which has the opposite value of \(Q\) now. In combination with the previous term, we obtain a cosine because \(\exp(iV)/2+\exp(-iV)/2=\cos V\). Note that in the Heisenberg picture, we directly get the real probabilities or expectation values – no complex amplitudes. That's also why we had three and not two values of \(Q\) that contributed. The argument of the cosine is\[

V = 2\pi QS = \frac{RS m}{t\hbar} = \frac{RSp_y}{L\hbar}

\] You see that the \(Q=0\) contribution \(1\) is helpful because in combination with the cosine, it's exactly needed to keep the sum non-negative. Because the probability density is \[

1+\cos(RS p_y / L\hbar)=1+\cos(2\pi RS/L\lambda)

\] where I introduced the de Broglie wavelength \(\lambda\), you may see that the separation between the minima (or maxima) is\[

\Delta S= \frac{L\lambda}{R},

\] the right result for any double-slit interference experiment that was already known to Thomas Young.

You are invited to fix all the neglected phases and normalization factors. First, I ask you to consider the non-normalizable initial state which has the delta-functions, and not their square roots. All the calculations will actually be identical as above. But you must realize that the later state will have infinitely many minima and maxima, so its norm will be infinite (just like the norm of the delta-function-based initial state).

You may also try to calculate the result for slits whose width is positive but you get much messier functions. If you do so, only a "couple" of interference minima and maxima near the center of the pattern will survive. You might also consider the problem in which \(\hat p_y\) isn't quite conserved and/or decoupled from the \(x\)-component, but as far as I know, you won't learn too much except for messier mathematical functions.

**Collapse in the Heisenberg picture**

So is there a collapse? When one observes the particle to land at \(x=S\), he simply learns that the projection operators \(\int dS' \hat P_{S'}\) over intervals including \(S\) are equal to one. This newer measurement always "overwrites" the previous measurements and it must be used as the higher-priority basis for predictions of (even more) future observations.

At any rate, there is no time-dependent wave function in this picture so it's nonsensical to talk about the "collapse" of such a wave function, too. The user of quantum mechanics is simply overwriting his knowledge about the observables – and/or their expectation values.

**Heisenberg picture as "local hidden variables" producing violations of Bell's inequality**

You may be surprised that it works so well. The observables at the final moment may be written as function(al)s of the observables at the initial moment. And the expectation values of the relevant final moment observables produce the predictions for measurable expectation values (or for probabilities, if we compute expectation values of projection operators). In some sense, the Heisenberg-picture calculation was more rudimentary than the Schrödinger-picture computation that requires you to solve "complicated" partial differential equations. We only solved the equation \(x'=p\), found that \(x=pt+x_0\), and the rest was a straightforward integration! Feynman's path integral approach is even more straightforward – all amplitudes are given by explicit (path) integrals.

Every observable may be represented by a "matrix" so you might have the following idea: Write down a Bell's theorem experiment in terms of the Heisenberg-picture field operators. They are assigned to points in the spacetime, they evolve and interact locally, but you may always express them relatively to a basis whose first (or zeroth: I mean the initial) basis vector is the initial state.

In this setup, you may interpret the matrix elements \(P_{11}(t)\) of any projection operator as "classical observables" that directly predict the probabilities associated with the projection operator \(\hat P\), right? So isn't it a violation of Bell's theorem?

Well, it's not because even though the Heisenberg-picture field operators are "local" in the physical sense – they know everything about the measurements you may do around the given spacetime point – they are not local in the Bell's theorem sense because the relevant matrices are expressed relatively to the Hilbert space that "knows" about the possible states of the whole physical system (the whole Universe, if you wish).

The clever framework of quantum mechanics guarantees that all the predicted probabilities will always be unaffected by any changes made at spacelike-separated points. If you admit – and quantum mechanics urges you to admit – that the predicted probabilities of any outcome contain "everything that has any physical sense", you may derive that the whole theory, in our case, a quantum field theory, is a perfectly local physical theory.

On the other hand, it is

*not*local by construction – I mean by the Bell's realist construction assuming that all the probabilities come from the averaging of a classical deterministic system. Nevertheless, the Heisenberg picture shows you Nature's cleverness: she can guarantee the perfect locality of all physically meaningful predictions even though the local operators are expressed relatively to states that "know about the whole Universe simultaneously" which looks non-local but within the quantum framework, it's not.

IMHO, the

ReplyDeletesolution of the double slit single particle interference is to assume all particles and we humans are entangled with an anti-material copy present in a dual system of entangled universes (material entangled to anti-material universes)

which I will call Raspberry Multiverse. Even our consciousness seems to be influenced by this system.

We are each others observers . ( see image)

In the double slit experiment I assume that we have to count with two wavefunctions, one of the material universe and the other inside the entangled anti-material universe located at long distance. See also: https://www.flickr.com/photos/93308747@N05/?details=1

Holy cow.

ReplyDeleteThree experiments using a volatile, optically resolved, skeletally rigid, deLaval nozzle-expanded, cryogenic vacuum molecular beam of identical chiral molecules. Natural camphor is cheap and easy, MW = 152.237, specific rotation +42.9 degrees. A chirality connoisseur demands

ReplyDeleteD_3-trishomocubane, pentacyclo[6.3.0.0^(2,6).0^(3,10).0^(5,9)]undecane, MW = 146.233, specific rotation +165 degrees. (The 4,7,11-triketone has MW 188.184, specific rotation +949 degrees.)1) Blow the chiral molecular beam through a single slit. We expect no racemization on the other side.

2) Blow the chiral molecular beam through a multiple slit, even or odd number of slits. Will there be racemization on the other side after wavefunction recombination with even- or odd-number of sources?

OT:

ReplyDeleteDear Luboš,"

The first gulp from the glass of natural sciences will turn you into an atheist. But at the bottom of the glass, God is waiting for you." — Werner HeisenbergI'll drink to that. :)What a superb quotation! And such great provenance! Absolutely delightful!

Nice and interesting blog Lubos! Hydrogen atom by Schrodinger method is in every textbook. I have heard that Pauli worked it out by Heisenberg method. Do you know an online or printed reference? Another thing I have heard but never pursued its detail that Hydrogen atom problem can be made into Harmonic oscillator problem by suitable transformations. Is that right? Harmonic oscillator can be more easily solved by Heisenberg method as given in textbooks.

ReplyDeleteAlso I am glad to see the last quote by Heisenberg!

Holy Raspberry Multiverse, Batman! ;-)

ReplyDeleteEven Cows should be entangled.

ReplyDeleteHi Kashyap, I will try to find Pauli's original calculation of the hydrogen atom.

ReplyDeleteBut independently of that, you may solve the hydrogen atom spectrum purely by the SO(4) symmetry, i.e. procedures involving manipulations with the operators:

http://motls.blogspot.com/2011/11/hydrogen-atom-and-so4-symmetry.html?m=1

No discussion of any wave functions here. It's very elegant and it can be solved in this way because the hydrogen atom (Kepler/Coulomb problem) is special - solvable.

And even cows may eat raspberries although they may collapse because of them later.

ReplyDeleteIf you like it, the original is

ReplyDeleteDer erste Trunk aus dem Becher der Naturwissenschaft macht atheistisch, aber auf dem Grund des Bechers wartet Gott.

I thought that he had a glass of Becher, a German Bohemian bitter herbal alcoholic beverage

https://en.wikipedia.org/wiki/Becherovka

but then I checked that Becher also means just a generic glass of chalice. ;-)

Agree, this post is all about the God quote.

ReplyDeleteIt seems to me that Schroedinger’s wave formulation was badly motivated from the beginning. He was trying to rationalize his reality view with quantum mechanics and, in doing so, fooled himself and countless others to come.

ReplyDeleteThis TRF blog is just beautiful. The absence of a time-dependent wave equation speaks volumes.

Interestingly, it was (of all people, our MW believer) David Deutsch who used Heinsenberg picture in this work.

ReplyDeleteInformation Flow in Entangled Quantum Systems

http://arxiv.org/abs/quant-ph/9906007

Abstract:

All information in quantum systems is, notwithstanding Bell's theorem, localised. Measuring or otherwise interacting with a quantum system S has no effect on distant systems from which S is dynamically isolated, even if they are entangled with S. Using the Heisenberg picture to analyse quantum information processing makes this locality explicit, and reveals that under some circumstances (in particular, in Einstein-Podolski-Rosen experiments and in quantum teleportation) quantum information is transmitted through 'classical' (i.e. decoherent) information channels.

The snark is uncalled for.

ReplyDeleteIt's a *very* big glass, and it's *free*!

Or perhaps you don't drink.

do you and uncle AI share the same psych ward? lol

ReplyDeleteA proof that for some very smart guys "shut up and calculate" is indeed the smartest thing to do.

ReplyDeletere: states that "know about the whole Universe simultaneously"

ReplyDeleteI presume that's the God at the bottom of the glass. :)

Seriously, was Heisenberg a believer in God?

Yep, he was a Lutheran:

ReplyDeletehttp://www.adherents.com/people/ph/Werner_Heisenberg.html

Thanks, Luboš. Yes, I really like that one.

ReplyDeleteBy the way, I strongly sympathise with your Mr Zeman. I like a good bellyful of beer myself. Any beer — I'm not fussy, I like 'em all, and the more the better too. But age has stolen up on me and I find I can no longer indulge myself anything like the way I used to. It's probably just as well though.

Which reminds me, tell Mr Zeman that the next time he meets an EU apparatchik he should deliver the required diplomatic message by dispatching a broken beer bottle straight into his face.

But back to the good Herr Doktor Heisenberg.

I didn't notice it at first but I wonder now if the quotation might have lost something in the translation to English (which I also found elsewhere and seems to be common). I'm thinking maybe it's too literal?

From what I have read about him, Heisenberg was a highly cultured man. Given the weight of his intent here, the word "gulp" seems somewhat out of place — with its hint of the possible, and highly incongruous, vulgarity of the drinker. But my German isn't good enough to appreciate the nuance of the original so I can't tell how it was intended to come across. I take your point about 'Becher' though, which only adds to my concern here.

Do you think "taste" or "sip" might better replace "gulp"? Or maybe it's just a matter of irregular conjugation: I taste, you slurp, he gulps? :)

Far be it from me to suggest an alternative translation, but do you—or any other German speaker here—think that maybe something like the following might capture the flavour of the original?

"

The first [heady] sip from the flute of natural sciences makes one an atheist. But at the bottom of the glass the deep imbiber finds God awaiting him."OK then, I think I'll just stick with the original German and let others risk the translation.

Hello.

ReplyDeleteIf the first two sentences of the abstract were untrue it would imply acausality, would it not?

ReplyDeleteThanks Lubos. Now chemists have to learn Heisenberg picture!

ReplyDeleteIn my understanding Faster Than Light (thus certainly instantaneous) propagation of information always creates causality paradoxes.

ReplyDeleteBut apparently there are other nuances - even with the light-speed transport is there a physical carrier of information?

Deutsch addresses teleportation saying "The very term ‘teleportation’ was chosen by the discoverers of the phenomenon

(Bennett et al. (1993)) because it was deemed to be a spectacular example of

information from one location A appearing at another location B without being

carried there in any physical

object travelling from A to B – i.e.

without information flow."

The molecule looks very fragile, but excluding that, I'm not sure why racemization would be expected.

ReplyDeleteSeparately, what is the 3rd experiment?

Trunk would be translated as draught, it does imply a serious pull, not merely a sip.

ReplyDeleteBecher is likewise a more substantial container than a glass or a flute, perhaps a mug or a big beaker.

"The absence of a time-dependent wave equation speaks volumes." This sentence penetrated my philosophical skull very nicely, so than you Gene! :-)

ReplyDeleteI always suspected you of being a Hindu! ;-)

ReplyDeleteI don't agree that it's badly motivated. It's very useful picture and its simplicity to grasp is a good thing. The bad thing is that regretfully even a lot of physicists not give enough attention to different ways to consider the quantum theory that make its different features more manifest. And hence they don't understand them properly.

ReplyDeleteAnd that's not only a problem with Schroedinger/Heisenberg pictures. Worse - what about not knowing anything about open quantum systems? I've discovered that it's surprisingly widespread problem. By the way you can instantly detect this sort of ignorance if the "quantum intepreter" talks only about pure state vectors.

Frankly the paragraph about the collapse is confusing. Obviously except of some simple cases all previous measurements influence the probability distribution for the last measurement (you have to take into account the whole history operator - temporary ordered product of all projectors) so I don't get what you've meant by "overwriting".

ReplyDeleteFor me all the "collapse problem" ends with realization that its just a transition to the conditional probabilities.

Amazingly, searching on relevant terms on Google, this is now the top and the only reference to the actual calculation in Heisenberg formalism. I stopped after 5th or 6th page because all of the links were irrelevant generic babble and getting more irrelevant with every page.

ReplyDeleteWhat about the calculation with the initial state at laser, where individual photons depart toward slits?

In this computer age one would expect a nice calculation/simulation, taking care of all the deltas (i.e. with Gaussian packets) to be available somewhere on YouTube, no?

Thanks for the feedback. It *is* misleading because it's oversimplified. At some moment, I wanted to write a full discussion what happens if one does additional measurements at the intermediate times, and takes their results into account. Then I decided it would really double the article and this added topic is independent from the topic announced in the title. So the bulk of this (almost) calculation of the double slit experiment does use the initial wave function, but not the later one, and one is expected to use the wave function from the latest measurement, after the "last collapse".

ReplyDeleteBut one doesn't really have to use any wave function from intermediate times at all. If one calculates the probabilities with the extra knowledge of some intermediate measurements, one computes a conditional probability, and the conditions may be guaranteed by sandwiching all the operators in between two copies of the projection operators on what was seen (the condition), in the chronological order (most future conditions are close to the sandwiched operators), and with an extra normalization (we divide it by the same sandwich with 1, instead of the operator of predictive interest, in the middle). I should write those things sometime later.

My Google search gives me

ReplyDeletehttp://physics.stackexchange.com/questions/165036/double-slit-experiment-in-the-heisenberg-picture

where someone tried to answer the question by a link, and it was closed! ;-)

He was brought up as a Lutheran and stayed one but it's always debatable whether this scientists' belief is the same thing as the non-scientists' beliefs in the same church. I just read an interview with Richard Dawkins - who is on a tour here in Czechia - where he discusses similar things in a way I would probably agree with.

ReplyDeleteYour clarification of Gene's point isn't that bad, I think. The bad implications of the Schrödinger's picture for people's interpretations *could* perhaps be blamed on the narrow-mindedness of their perspectives and knowledge only, so the picture itself doesn't hurt.

ReplyDeleteOpen systems are ignored by "fundamental physicists" because the apparently "sufficient" fundamental laws seem to describe closed systems, and the open systems equations are just an implication of these fundamental laws in a less clean context.

Of course, I agree that knowing something about the less clean context is often important to actually understand the physics of phenomena one should understand, to connect the fundamental laws with the observations properly.

Maybe. Maybe it wouldn't helpful for them. Chemists aren't those who want to get the most accurate or proper understanding of the underlying laws. They are the people who work to poison other human beings, which is a slightly different motivation. ;-)

ReplyDeleteBefore I opened the full comment, I thought that the picture was the caricature of a pig we used to draw:

ReplyDeletehttp://petisee.ex-bloguje.cz/img/prase.jpg

I agree about collapsing related to raspberries if I translate this to the cyclic raspberry multiverse hypothesis:

ReplyDeletehttp://vixra.org/pdf/1503.0186v2.pdf

Dear Lubos,

ReplyDeletemany thanks for the great article. I think paedogically this is really helpful and it would make sense that at least the little bit advanced text books would contain such calculations. However I think that the school books for first contacts with quantum mechanics will not be rewritten and the machinery to understand this calculation is quite considerable compared to the Schroedinger one, which can be understood by a kid. So I would support the statement of OON that the simplicity of the Schroedinger picture is a good thing and the lack of familiarity of people with different formulations and aspects including decoherence is the problem.

Readers interested in the Heisenberg formulation might like to know that Dirac's "Lectures on Quantum Field Theory" published by the Belfer Graduate School of Science, Yeshiva University, New York, 1966 shows how to calculate the anomalous magnetic moment of the electron and the Lamb shift in QED entirely in the Heisenberg picture.

ReplyDeletePauli's paper can be found in the van der Waerden compilation:

ReplyDeletehttps://archive.org/details/SourcesOfQuantumMechanics

By the way, Harvard has a colloqium with Wojciech Zurek

ReplyDeleteonline with the title

"Decoherence and Quantum Theory of the Classical".

http://media.physics.harvard.edu/video/id=COLLOQ_ZUREK_32315

It is quite nice although the tendency of Zurek to invent new words such as Einselection and Quantum Darwinism can also be distracting.

Sure, the religious beliefs work differently for each person and also for each intellectual group (rationalists, mystics, cultivated, unlearned, clever, brainless...), the thing is that theist/atheism is distributed evenly between groups, I guess that Dawkins does not like that cultured scientists as Heissember or Planck were not only religious but also were tolerants with other people beliefs.

ReplyDeleteIn Heisenberg picture as you clearly pointed out the

ReplyDeleteobservables are the most important and they are given by the operators (matrices).

Nevertheless, for the entanglement we have at least 2 observables. The

entanglement means that the matrices have some correlation, namely since we can

look different final observables, the corresponding coefficients are

correlated. As somebody mentioned, I would really like to see how the operators

(matrices) in the entanglement case

(Heisenberg picture)!

Now, you may say ok this is QM and that is how it is.

However, as humans we have like to ask deeper why ? What is so special about

initial observable that matrices become highly correlated in the final

observables?

And isn’t it strange that on one hand we have in QM uncertainty

relation between different operators (non-commutation). But on the other hand have

the possibility to prepare the observables such that the final ones are always

highly correlated in the final measurement (complementar certainty !)

Did ever somebody discussed possible relation between the

two namely uncertainty principle and entanglement? For example, is one

necessary for the other, or they are unrelated as such.

Thank you.

ReplyDeleteThis is what Werner Heisenberg wrote in a letter to Wolfgang Pauli in 1926:

ReplyDelete"The more I think of the physical part of the Schrödinger theory, the more detestable I find it. What Schrödinger writes about visualization makes scarcely any sense, in other words I think it is shit. The greatest result of his theory is the calculation of matrix elements."

Pigs are proof that God exists. Every part pf a pig is delicious - its ears, snout, chuckle, bung, and even its trotters.

ReplyDeleteHeisenberg was also an accomplished pianist, as was Planck.

ReplyDeleteThe experiment testifies as to the symmetry of wavefunction recombination. Everybody talks big, Nobody has ever looked. Theory is a whore - it will prove anything for grant funding. Look up arXiv for Gran Sasso superluminal muon neutrinos. It was proven to be true at least ten dfiferent ways.

ReplyDeleteD_3-trishomocubane is a stabilomer, a deep thermodynamic hole. You dump the precursor into strong acid and cook the crap out of it. Cp and BZQ starting materials to final product in 39% yield in 30 gram preps.The third experiment. is identical to the first two, but using the triketone. The hydrocarbon is exceptionally geometrically chiral (Petitjean QCM). The triketone is exceptionally electronically chiral (huge specific rotation). Use the C_60 diffraction apparatus. collect product, read specific rotations before and after.

I did not find it confusing at all.

ReplyDeleteA gravity wave would also be a physical carrier; would it not?

ReplyDeleteHi, if you really want it, I will happily write a purely Heisenberg-picture description of the EPR experiments or Bell's theorem, when I have time. Most of the usual treatment is about the operators, anyway, and the time-dependence of states doesn't play any important role in EPR or Bell's theorem - it's virtually non-existent - so things are effectively in the Heisenberg picture already.

ReplyDeleteIt is hard for me to understand your "why" question, especially because the same correlations are analogously created by the evolution both in classical and quantum mechanics. The question is really a mathematical one, and the solution to the mathematical problem is easy, isn't it?

Even in classical physics, one may find observables that are perfectly correlated in the final states in analogous experiments. Quantum mechanics is "richer" because it has a bigger space of observables - operators, matrices, and there are many ways to choose the pairs so that there is a perfect correlation, and so on.

All these things easily and directly follow from the most rudimentary laws of QM so I don't really think it's right for you to say that those particular features of the simple example experiments are "deep".

Sure. In the context of the paper it all depends on what you choose to physically represent the qubit and operations on its state.

ReplyDeleteBTW, I cited it mostly because I was surprised to find the calculation in Heisenberg picture dealing with the 'modern stuff', such as quantum information. The manifest locality of information in this picture mystified me a bit.

Other curiosities:

Deutsch had to defend his article as not some new formulation of QM but as quite ordinary QM in Heisenberg picture.

Nowhere in the article does he mention MW. In fact, I believe that Lubos would agree with the most statements in this paper. Yet Wiki page on teleportation says that this calculation was done 'in the context of MWI'.

Hi, I didn't like that talk. I don't have time to watch it completely but it seems that he considers states as something real. There shouldn't be any complex explanation of wave function collapse, because it is not a real physical process, as wave function is not something like a phase space point. Also you can't derive Born's rule, as it is the only thing that connects formalism to reality, if you claim that you can derive born rule from something else, then you at some point you must give additional meaning to wave function. Correct me if I am wrong.

ReplyDeleteHi Lubos,

ReplyDeleteonce more I admire your deep insight into QM AND your patience!

Erwin

Hi John,

ReplyDeleteZurek is one of the people which are known for their good understanding of the quantum classical transition and I don't think there are any obvious logical flaws in his talk. Whether you like it is still subjective. If you look carefully he still has an axiom which makes contact with reality so his derivation of Born's rule does not come from nothing. I am not that excited with the derivation of Born's rule myself because I think it is more a mathematical game to find mininal axioms and not something new and deep about quantum mechanics. But it is still interesting.

Danke, Herr Schrödinger. :-)

ReplyDeleteDear Mikael,

ReplyDeleteThanks for your answer. Of course whether I like it is subjective, therefore I asked you. The reason why I asked you is that, in the beginning of the talk he talks about deriving born rule which he shouldn't be able to. I see that you agree with me. Of course you can define other axioms and then "derive" born rule.

By the way there is a very nice talk by Sidney Coleman in the case you are interested:

http://media.physics.harvard.edu/video/?id=SidneyColeman_QMIYF

Dear John,

ReplyDeletethe very word "derive" means that there are other axioms you derive it from. If you think that somebody like Zurek does not know the proper meaning of the word "derive" then no, I don't agree wth you.

Ok, that link is the famous quantum mechanics in your face leture of Sidney Coleman which I know and I think which was even commented by Lubos in an own blog post. Indeed, it is a very good lecture and recommended to anybody who has not seen it.

Dear Mikael,

ReplyDeleteWhat I mean is following, when people talk about derivation of born rule they usually mean that there is some deterministic mechanism behind it. I say that this is impossible, that is quantum mechanics intrinsically probabilistic. All derivations you can conceive are circulatory. So this is my point.

You have also said there is no deep meaning behind that derivation. Now I know Zurek has done very important work on decoherence (but haven't read), which is why I asked you about the video.

Dear John,

ReplyDeleteno, he isn't deriving the Born rule from some deterministic mechanics. People who try that are usually crackpots, although even the great Gerard 't Hoot is among them. The axioms Zurek is using appear in minute 53:36. I think he is making the implicit assumption that quantum mechanics is probabilistic and then interpreting the coefficients as probability amplitudes becomes the only way to make sense of the formalism.

Great free, dowloadable as PDF (ahppens to be important for me) collection! Fun reading already from the first pages. Thanks.

ReplyDelete"she can guarantee the perfect locality of all physically meaningful predictions even though the local operators are expressed relatively to states that "know about the whole Universe simultaneously" which looks non-local but within the quantum framework, it's not."

ReplyDeleteI have to admit that this is one of the most profound statements on QM I have ever seen. Congratulations Lubos!

Me, OTOH, I must admit that, I'm sorry, I don't really understand this part. I understand that Hilbert space is huge but that's about it.

ReplyDeleteApologize for not understanding, but I still do not see why there would be any racemization here.

ReplyDeleteWhy would the presence of one or more slits have any effect on the structure of the molecules involved?

As was Lubos. ;-)

ReplyDeleteIf you admit – and quantum mechanics urges you to admit – that the predicted probabilities of any outcome contain "everything that has any physical sense", you may derive that the whole theory, in our case, a quantum field theory, is a perfectly local physical theory.

ReplyDeleteSo it Is possible, after the initial wave function is shifted, it launches the hammer mechanism and kills Wigner's friend instead of the cat? :0)

Yes it's a slightly odd way of stating things by Deutch et. al.

ReplyDeleteAs I understand it, in Quantum Encryption/Teleportation there's always a physical carrier for the classical transmission of part of the state information from site A to site B.

Maybe he's referring to the fact that the shared-secret/full-state can't be recovered by interception of the physical/classical carrier alone?

Still, the entanglement between the qbits at sites A and B also had to be prepared at some point in the past as a prior resource, so it's more about intrinsic/un-fakeable correlation than any kind of "unusual transportation", no?

So a draught from a flagon, rather than a sip from a flute. :-)

ReplyDeleteWell, he was German rather than French, right?

"Disznótor", (Pig slaughter party) is a big deal in Hungary...

ReplyDeleteAnd indeed that poor fellow is truly delicious from head to toe! :D

Dear Liam, what you write sounds nice except for a key bug. By causality (included in classical GR), nothing in the interior can ever affect things outside the black hole - such as the thermality of its radiation. Well, in some sense, it does, and there are some nonlocalities at least in some descriptions, but you surely can't view a black hole as just another object with a hot interior.

ReplyDeleteThanks, Kashyap!

ReplyDeleteTony, the point of the sentence was to discuss the idea that the operators' matrix elements in the Heisenberg picture are interpreted as "classical observables".

They evolve according to local equations of motion, so it would look like a "local realist" picture that nevertheless violates Bell's inequality and similar things - because it's quantum mechanics.

But to make the picture explicit, one has to choose a basis, and express all the (field) operators with respect to this basis. The basis also contains states with different properties (values of other operators) at other, spacelike-separated points.

But in QFT, the field operators for spacelike-separated points still evolve "independently" - when it comes to all (probabilistic) physical predictions. However, this picture isn't a counterexample to Bell's theorem because the assumptions of Bell's theorem don't allow the matrix elements to depend on the choice of basis describing properties from other points.

None of these dependences on the basis actually affect any physical predictions for a given region but this statement is "nontrivial" while Bell's theorem assumes that the locality is guaranteed in a "trivial", classical way, which it's not in quantum mechanics, not even in the Heisenberg picture.

Hi Lubos, thanks for your reply.

ReplyDeleteIt really does not matter, but my intention where never to spread irrational fog. Quiet the opposite, I am great fun of Heisenberg picture and all the things that follow form it.

Just you also have to understand as well, that people will try to improve their understanding of the phenomena such as QM and not to mention some useful thinks may come out of it. Is that possible time will tell. after all, there is no guarantee that QM is the final story. But is the best we have so far.

Can anyone point me to possible source on Bohr opinion about the Schrodinger work ? Did quick search on google but found nothing interesting.

ReplyDeleteBohr had better things to do than to write detailed reviews of a Schrödinger but I think that e.g. the second paragraph on the page below is an example of the thing you were looking for.

ReplyDeletehttps://books.google.cz/books?id=3OjdRDagCPsC&pg=PA145&lpg=PA145&dq=bohr+%22about+schr%C3%B6dinger%22&source=bl&ots=Ep1RxWmN_e&sig=plqhhcXwIQUO_A-xW5UBOHXgK5I&hl=cs&sa=X&ei=Dsw0VbfXFcOBywPF94CADQ&ved=0CDAQ6AEwAg#v=onepage&q=bohr%20%22about%20schr%C3%B6dinger%22&f=false

LOL

ReplyDeleteBB: David Cassidy "Beyond Uncertainty" 2009 provides what you are looking for. it is complicated by the evolution in time of the theory and of the personal relationships and egos.

ReplyDeleteI don't have the ability to understand this article ATM. So the only constructive comment I can provide is just a minor grammatical correction: In the first Heisenberg quotation, you keep the definite article 'the' before nature (i.e., 'the Nature'), which is unidiomatic. But in the second quotation, you correctly remove the definite article 'the' before 'Nature' (i.e., 'describe and explain Nature'). I'm curious why you (perhaps consciously) thought you needed to make this distinction, as I've seen you unnecessarily include the definite article 'the' in many of your blog posts. Best wishes.

ReplyDeleteDear Mike, I copied all the quotes including the articles or their non-existence, and while I have limited knowledge of English, I have already learned about the issues of the articles in front of Nature.

ReplyDeleteWhat I meant by 'I've seen you unnecessarily include the definite article "the" in many of your blog posts' are sentences such as 'I am obviously no expert in the Indian law and traditions' - the definite article 'the' here is unidiomatic. A native speaker would formulate the sentence about Indian law as 'I am obviously no expert in Indian law and traditions.' I bring all this up not to be a pedant who gleefully points out the (insignificant) grammar errors of others, but simply to be helpful. I like your English prose style; and I don't believe you have 'limited knowledge of English'; you likely know more about formal English grammar than I do.

ReplyDeleteIs that the Christian God that Heisenberg refers to in the quotation?

ReplyDeleteThanks Andre.

ReplyDeleteI think, you would find Lubos’ reply to your question satisfactory. As one of the readers requested, I would also read with lot of interest if and when Lubos writes about entanglement in Heisenberg picture, although

ReplyDeleteI do not expect any paradox to come out of that! The fact that no useful information (one that is not already known to the two observers) can travel faster than light between

the entangled points is a completely satisfactory argument for me for locality.As for the other issue "know about the whole Universe simultaneously" I understand, if you have infinite accuracy, then you will see everything in the universe entangled! I am not sure if Lubos agrees with this. This is little bit of

stretch on his statements.

I think he was suggesting serious engagement, rather than casual flirtation.

ReplyDeleteI would like to see a post about quantization. When different quantization schemes are equivalent, when there is more than

ReplyDelete“The first gulp from the glass of natural sciences will turn you into an atheist, but at the bottom of the glass God is waiting for you.”

ReplyDelete― Werner Heisenberg

“The reality we can put into words is never reality itself.”

― Werner Heisenberg

It strikes me that it is possible these two statements are equivalent in intent, and refer the the existence of a reality that is beyond the realm of scientific inquiry and human understanding (i.e., that is ineffable), and, if science is the study of nature, you might have to call it supernatural. You could call that reality God or you can call it Dave, but human understanding of it promises to continue to approach nil.

It would be interesting to know if, when speaking of an interplay with nature, whether he was referring to the Nature itself or to the Nature exposed to our method of questioning. If the former it would seem to leave us with the possibility of an interplay with the Nature itself that is itself ineffable and not susceptible to description (since describing it would be describing some attribute of the Nature itself) and hence might be conceived of as being mystical in nature.

Or perhaps .. Don't drink enough.

ReplyDeleteIt would be nice to know what kind of spectrum, if any, you get from gravitational collapse where BH doesn't form, just a much smaller and denser neutron star. Of course, neglecting all the other 'ordinary radiation' from such a star.

ReplyDeleteAccounting for backreaction and looking at evaporation stage would also be nice, but he said it is extremely difficult.

I understand that Dejan's observer never sees event horizon in his foliation but at one point would-be-horizon must start decreasing because BH has formed, albeit in another foliation, and is evaporating.

Wait and see, I guess.

Thanks for your reply. I was a bit mystified by what Unruh said, even if jokingly.

Dear Lubos,

ReplyDeleteI can tell you what Werner Heisenberg would say to you in this case. He would point out, that he would say or write this things in German, and "die Natur" is correct, while "Natur" without definite article simply doesn't work. So we are probably looking at a literal translation. Czech must demand the definite article as well in front of nature I guess if you separately had to learn this rule. Not that anything like this is important.

I believe Drink is best translation for trunk, The first drink is therefore what Heisenberg wrote. However I will defer to the German hypothetical particle, Dilaton, if he has another interpretation. :-)

ReplyDeleteCzech language has no definite articles at all according o

ReplyDeleteSeconded.

ReplyDeleteWhat an interesting publication, Lubos. It seems that all of the various objections to Bohr’s view were based on nothing more than discomfort. Still are.

ReplyDeleteObviously, you should not take the time to write a “Heisenberg” description of anything for BB unless he (she?) is willing to do a lot more work first. You have already done quite enough, Lubos.

ReplyDeleteDear Mikael, as Rehbock correctly says, Slavic languages don't use any definite articles at all.

ReplyDeleteYour mentioning Czech sounds dumb, anyway, given the fact that Heisenberg had nothing to do with Czechia or Czech things; and I wrote that I copied-and-pasted the quotes from the web and you may easily verify that the quotes appear thousands of times in the very form I wrote.

Interesting quotations from Heisenberg. Who'd say he was such a religious douchebag? It speaks volumes about his basically worthless discoveries that (mis)led science in the wrong direction for a century or so. Just because we don't understand something it doesn't mean a deity made it so. Religion ought to be outlawed for scientists. Wow. I'm shocked, actually. What an idiot. Physicists actually consider him a founding father of modern mechanics? Unbelievable. An imbecile, really. Yuck.

ReplyDeleteHeisenberg was undoubtedly one of the five most important and deepest 20th century scientists.

ReplyDeleteOn the other hand, evidence seems overwhelming that by giving the birth, your mother couldn't have positively contributed to the mankind, so I will just politely thank you for your last comment on my blog.

Czechs may have indefinite articles after a drought from a mug, no?

ReplyDeleteLOL, I think that we don't have indefinite articles even in that special condition, either.

ReplyDeleteNo articles. At most, we can count nouns - like one dog, two dogs (rather rare, not mandatory for a/the dog) - and we may point to a particular object (this dog, that dog).

But otherwise, it is just [...] dog! It's stupid to add extra words in front of him or her or it.

When a dog is barking, why wasn't it "the" dog? It may very well have been. A/the difference between these two cases is that in the case of "a dog", one doesn't know which particular one the dog was. But whether one knows "which dog" one dog is seems to be so subjective, arbitrarily, and irrelevant for the claim that [...] dog is barking.

Moreover, I think that the articles are being inserted in non-Slavic languages because the defective construction of these languages would lead to omnipresent confusions between nouns and verbs (or other types of words).

For example, "United States" could very well mean that the United Airlines are stating - saying - something. These confusions appear virtually everywhere, and that's why "the" in front of the United States reduces the frequency of such misunderstandings.

In recent years, I wouldn't have used "the" in front of "Nature" in the first Heisenberg quote. Maybe it was used by Heisenberg when he translated his quote to Heisenberg English. maybe these definite articles were added by someone else. I don't really think that the answer to this question is important.

This is a fine illustration of how atheists and anti-theists can let their ideology make them just as moronic and biased against reality as the most asinine religious. There is nothing inherently more rational or scientific about atheism - it is merely the inverse prejudice of theism.

ReplyDelete