Compilation of severe errors in famous textbooks

[Demystifier]

Interesting. I've to read Ballentine's book on the Zeno effect again, but the Zeno effect isn't about collapse but it's about "stabilizing" an unstable state by some interaction. There's no need for collapse to understand it. It can be well explained within the statistical interpretation, and afaik it has been demonstrated already experimentally. I don't remember the details, but I think it was done with some metastable atomic state using a laser.

The collapse is certainly not necessary to explain the Zeno effect, but it is useful as a quick and dirty way to obtain it.

 

[martinbn]

Well, energy is conserved for any sign of $\omega^2$. Indeed, energy is conserved whenever the Hamiltonian does not have an explicit dependence on time, which is the case for any sign of $\omega^2$, as long as $\omega$ does not have an explicit dependence on time.

I still don't understand. In the case of one degree of freedom you are right. But this section is specifically for more than one degree of freedom. In that case it is not at all obvious to me. May be it is an easy calculation and I am just being silly, but I don't see. In the $s=1$ case the matrix is just a number, so the determinant is just that number and the equation for $\omega^2$ can easily be used in the expresion for the energy. In the $s>1$ case that doesn't seem so. You still have the exponentials that will decay or grow if $\omega$ isn't real. Even if it can be shown that the decay/grow isn't there, their argument is quite natural and convincing and easy. There is no error there.

 

[martinbn]

http://www.feynmanlectures.caltech.edu/III_01.html#Ch1-S8
"We choose to examine a phenomenon which is impossible, absolutely impossible, to explain in any classical way, and which has in it the heart of quantum mechanics. In reality, it contains the only mystery. We cannot make the mystery go away by “explaining” how it works. We will just tell you how it works. In telling you how it works we will have told you about the basic peculiarities of all quantum mechanics."

Feynman refers to the double slit experiment. However, most people would nowadays take the Bell tests to be the mystery of QM, not the double slit. There is interesting commentary in section 1 of https://arxiv.org/abs/1301.3274. Whitaker comments that Feynman corrected himself in his later lectures on computation https://aapt.scitation.org/doi/full/10.1119/1.4948268 "In any case, since what Feynman describes is indeed Bell's Theorem, it is extremely interesting that he adds that he often entertained himself by squeezing the difficulty of quantum mechanics into a smaller and smaller place, and he finds this place precisely in this analysis. Thus, Feynman's view is apparently clear—the content of Bell's Theorem is the crucial point that distinguishes classical and quantum physics."

"We make now a few remarks on a suggestion that has sometimes been made to try to avoid the description we have given: “Perhaps the electron has some kind of internal works—some inner variables—that we do not yet know about. Perhaps that is why we cannot predict what will happen. If we could look more closely at the electron, we could be able to tell where it would end up.” So far as we know, that is impossible. We would still be in difficulty. Suppose we were to assume that inside the electron there is some kind of machinery that determines where it is going to end up. That machine must also determine which hole it is going to go through on its way. But we must not forget that what is inside the electron should not be dependent on what we do, and in particular upon whether we open or close one of the holes. So if an electron, before it starts, has already made up its mind (a) which hole it is going to use, and (b) where it is going to land, we should find P1 for those electrons that have chosen hole 1, P2 for those that have chosen hole 2, and necessarily the sum P1+P2 for those that arrive through the two holes. There seems to be no way around this. But we have verified experimentally that that is not the case. And no one has figured a way out of this puzzle. So at the present time we must limit ourselves to computing probabilities. We say “at the present time,” but we suspect very strongly that it is something that will be with us forever—that it is impossible to beat that puzzle—that this is the way nature really is."

Feynman says something similarly erroneous in this video around 51 minutes.

Hidden variables for the double slit are possible.

To be honest I don't see your point. Can you elaborate? Nothing in what Feynman writes looks erroneous, let alone fundamentally flawed.

 

[Demystifier]

You still have the exponentials that will decay or grow if $\omega$ isn't real.

So what? The exponential grow of positive kinetic energy is accompanied with the exponential grow of negative potential energy, so that the total energy, that is the sum of positive kinetic energy and negative potential energy, is constant. If you don't believe me, solve the equations explicitly with negative $\omega^2$ and convince yourself that the total energy is indeed conserved.

 

[martinbn]

So what? The exponential grow of positive kinetic energy is accompanied with the exponential grow of negative potential energy, so that the total energy, that is the sum of positive kinetic energy and negative potential energy, is constant.

No, the exponetials are the same and can be factored out. So you have an exponetial times a bounded term. So the whole expression will grow or decay. It looks like this $e^{\lambda t}\times\text{bounded stuff}$. The $\lambda$ is zero only of the $\omega$ is real, otherwise it is negatve or possitive, so the whole energy will change with time.

If you don't believe me, solve the equations explicitly with negative $\omega^2$ and convince yourself that the total energy is indeed conserved.

That's what I am doing, but I can see how it works only in the case of one degree of freedom. The case at hand doesn't seem that way.

 

[Demystifier]

It looks like this $e^{\lambda t}\times\text{bounded stuff}$.

Almost, but not quite. It looks like
$$E_0 +e^{\lambda t}(B_{\rm kinetic}+B_{\rm potential})$$
where the $B$-terms are bounded. But $B_{\rm potential}<0$ (because $\omega^2<0)$ and in fact $B_{\rm kinetic}+B_{\rm potential}=0$, so the full energy is $E_0$, which is a constant.

That's what I am doing, but I can see how it works only in the case of one degree of freedom.

Can you at least see that for one degree of freedom you get my form above?

 

[martinbn]

Almost, but not quite. It looks like
$$E_0 +e^{\lambda t}(B_{\rm kinetic}+B_{\rm potential})$$
where the $B$-terms are bounded. But $B_{\rm potential}<0$ (because $\omega^2<0)$ and in fact $B_{\rm kinetic}+B_{\rm potential}=0$, so the full energy is $E_0$, which is a constant.Can you at least see that for one degree of freedom you get my form above?

Yes, my point exactly. For one degree it is simple. For many, I don't see it.

 

[Demystifier]

Yes, my point exactly. For one degree it is simple. For many, I don't see it.

OK, so we agree that for one degree the energy is conserved, right? To solve the equations explicitly for many degrees, you have to diagonalize the Hamiltonian. For inspiration in the case of 2 degrees, see e.g. my http://de.arxiv.org/abs/1702.03291 Sec. 3 (and ignore the $y$-dependence).

 

[martinbn]

OK, so we agree that for one degree the energy is conserved, right? To solve the equations explicitly for many degrees, you have to diagonalize the Hamiltonian. For inspiration in the case of 2 degrees, see e.g. my http://de.arxiv.org/abs/1702.03291 Sec. 3 (and ignore the $y$-dependence).

Suppose it is diagonal to begin with. Then the matrices $k_{ij}$ and $m_{ij}$ are diagonal and have diagonal elements $k^i$ and $m^i$. Then the possible values for $\omega^2$ are $\frac{k^i}{m^i}$. But it is only one of them, because they (L&L) are looking for solutions of the form $x_l=A_le^{i\omega t}$. In this decoupled case the energy is the sum of the energies of the individual systems and only one of them will be zero. The others will not, so you still have the exponential decay or increase if $\omega$ is not real. My guess is that your mistake is that you assume that $x_l=A_le^{i\omega_l t}$ with different omegas so that all terms will cancel.

 

[Demystifier]

My guess is ...

Why don't you just solve the equations of motion explicitly and completely, instead of guessing? When you do that, and when you insert the solution into the expression for the full Hamiltonian, you will see that all the time-dependent terms cancel.

 

[martinbn]

Why don't you just solve the equations of motion explicitly and completely, instead of guessing? When you do that, and when you insert the solution into the expression for the full Hamiltonian, you will see that all the time-dependent terms cancel.

Erm, I did. And they don't cancel. Did you try it? Take two degrees of freedom with different k's and m's, then it is obvious that one term will cancel and the other will be left.

 

[martinbn]

Just add that in your paper sec3. you are looking at the case where the k and the m are the same for both subsystems. Then it all works out. In general though it will not.

 

[formodular]

If you are you referencing the argument after equation 23.8, there is no error - they prove the claim directly there, but the physical argument is also correct as the Lagrangian will then explicitly depend on time and so Noether does not let you even state conservation of energy, but the Lagrangian was assumed to be time-independent from the beginning, so that Noether then gives energy conservation.

 

[bhobba]

Simon's paper is almost as "ancient" as von Neumann and Wigner's result was when Simon wrote that.

Speaking of Von-Neumann I am surprised nobody mentioned his famous no-go theorem on Hidden Variables in his Mathematical Foundations of QM (I wish I could say I picked it up when I read it, but didn't). It was only universally picked up years later, due to Von-Neumann's well deserved reputation and not carefully checking its assumptions. Greta Hermann did, but she was dismissed. I hope it was because of the respect Von-Neumann had - not because she was a woman. Either way not one of sciences finest hours. Then their was his scathing rebuke of the Dirac Delta function. Rather than say we need further developments in math to make sense of it, which has now been done (admittedly requiring mathematicians like Grothendieck whose mathematical reputation is the equal of Von-Neumann himself) he simply dismissed it as a fiction. Von-Neumann is one of my heroes, being more that just a great mathematician, but that even rarer beast, a polymath, however like all human beings perfect he was not.

Thanks
Bill

 

[Keith_McClary]

It was only universally picked up years later

Simon (as a "predoctoral fellow") found and corrected an error of von Neumann and Wigner (see Examples and Remarks B). He also has:

Note added in proof: There is a minor technical flaw in the proof of Theorem 2 ... We have really only proven that w = 0 outside a sufficiently large sphere.

BTW, I messed up the link to his paper, this should work:
On Positive Eigenvalues of One-Body Schrodinger Operators

Nobody's perfect.

 

[Demystifier]

Erm, I did. And they don't cancel. Did you try it? Take two degrees of freedom with different k's and m's, then it is obvious that one term will cancel and the other will be left.

I find it hard to believe, because there is a general theorem telling that energy is conserved whenever the Hamiltonian does not have an explicit time dependence. In our case the Hamiltonian is essentially
$$H=\frac{p_1^2}{2}+\frac{p_2^2}{2}-\alpha_1^2\frac{x_1^2}{2}-\alpha_2^2\frac{x_2^2}{2}$$
where $\alpha_i^2=-\omega_i^2$ are positive constants. Hence the Hamiltonian does not have an explicit time dependence so the theorem should apply. Indeed, since the Hamiltonian is diagonalized, i.e. there is no coupling between $x_1$ and $x_2$, the Hamiltonian is simply
$$H=H_1+H_2$$
so each $H_i$ is conserved separately, as in the case of only one degree of freedom. Since this contradicts your result, can you present the essential steps of your calculation?

Anyway, here is mine calculation. The equations of motion (in the Newton form) are
$$\ddot{x}_i(t)=\alpha_i^2x_i(t)$$
so the general solution is
$$x_i(t)=c_{i+}e^{\alpha_i t}+c_{i-}e^{-\alpha_i t}$$
where $c_{i\pm}$ are arbitrary constants. Hence
$$x_i^2(t)=c_{i+}^2e^{2\alpha_i t}+c_{i-}^2e^{-2\alpha_i t}+c_{i+}c_{i-}$$
$$\dot{x}_i(t)=\alpha_i(c_{i+}e^{\alpha_i t}-c_{i-}e^{-\alpha_i t})$$
$$\dot{x}_i^2(t)=\alpha_i^2(c_{i+}^2e^{2\alpha_i t}+c_{i-}^2e^{-2\alpha_i t}-c_{i+}c_{i-})$$
Therefore
$$\dot{x}_i^2(t)-\alpha_i^2x_i^2(t)=-2\alpha_i^2c_{i+}c_{i-}$$
is time independent, so finally we have the energy
$$H=\sum_{i=1,2} \frac{1}{2}\left( \dot{x}_i^2(t)-\alpha_i^2x_i^2(t) \right) =-\sum_{i=1,2} \alpha_i^2c_{i+}c_{i-}$$
which is time independent, Q.E.D.

 

[martinbn]

What you've written is correct, but irrelevant. You have two different $\omega_j$, as I guessed, and you wagged your finger at me for doing so. L&L do something else. They look for a solution of the form $x_j=A_je^{i\omega t}$. And if you put that into the energy the time dependence will remain if your $\alpha_j$ are not the same.

 

[Demystifier]

What you've written is correct, but irrelevant. You have two different $\omega_j$, as I guessed, and you wagged your finger at me for doing so. L&L do something else. They look for a solution of the form $x_j=A_je^{i\omega t}$. And if you put that into the energy the time dependence will remain if your $\alpha_j$ are not the same.

If they look for a solution in which both modes have the same frequency, then such a solution simply doesn't exist. In fact, the modes which are not solutions don't conserve energy even with real $\omega$. But modes which are not solutions are nonphysical simply because they are not solutions, so it's irrelevant whether they conserve energy or not.

 

[vis_insita]

But you don't make explicitly wrong statements about collapse, because you accept "collapse" at least in the sense of information update and you do not deny the quantum Zeno effect. In that sense you are not like Ballentine.

Ballentine does not deny the Quantum Zeno effect either. I think you misunderstood him. In

Ballentine, L.E. Found Phys (1990) 20: 1329. https://doi.org/10.1007/BF01883489.

he has a short discussion of the Quantum Zeno effect. His conclusion is "Thus the Quantum Zeno effect actually occurs for this system. [This refers to what is reported in https://doi.org/10.1103/PhysRevA.41.2295 ] But it is misleading to explain it as being due to a "collapse of the wave function" caused by measurement. No "collapse" actually occurs; rather the excitation of the atom is impeded by the string perturbation of the optical pulses and the coupling to the radiation field. Moreover [...], the effect occurs regardless of whether or not any measurement of the emitted photons is actually made."

Do you think there is anything wrong with these statements? If not, what other explicitly wrong statements do you have in mind?

Also, I believe there is no reason to accept the collapse even in the sense of "information update". The update of information is properly handled without a collapse by means of conditioning upon past measurement results. (As is argued in his textbook and also in the paper above.)

The collapse is certainly not necessary to explain the Zeno effect, but it is useful as a quick and dirty way to obtain it.

This somewhat contradicts your earlier statement that there are experiments that "show that the collapse exists." There would be nothing wrong with criticising a "quick and dirty" derivation of the Quantum Zeno effect if there is a cleaner one available.

 

[Demystifier]

Ballentine does not deny the Quantum Zeno effect either. I think you misunderstood him. In

Ballentine, L.E. Found Phys (1990) 20: 1329. https://doi.org/10.1007/BF01883489.

he has a short discussion of the Quantum Zeno effect. His conclusion is "Thus the Quantum Zeno effect actually occurs for this system. [This refers to what is reported in https://doi.org/10.1103/PhysRevA.41.2295 ] But it is misleading to explain it as being due to a "collapse of the wave function" caused by measurement. No "collapse" actually occurs; rather the excitation of the atom is impeded by the string perturbation of the optical pulses and the coupling to the radiation field. Moreover [...], the effect occurs regardless of whether or not any measurement of the emitted photons is actually made."

Do you think there is anything wrong with these statements? If not, what other explicitly wrong statements do you have in mind?

That's perfectly fine, but that's not how he explained it in the book.

Also, I believe there is no reason to accept the collapse even in the sense of "information update". The update of information is properly handled without a collapse by means of conditioning upon past measurement results. (As is argued in his textbook and also in the paper above.)

It's not necessary to use collapse in the sense of information update, but I think it's also not wrong to use it that way.

This somewhat contradicts your earlier statement that there are experiments that "show that the collapse exists." There would be nothing wrong with criticising a "quick and dirty" derivation of the Quantum Zeno effect if there is a cleaner one available.

Perhaps it was not obvious from the context, but I distinguish a "true" collapse from a FAPP collapse (essentially, an update of information). The latter makes sense even when the former doesn't exist.

 

[vis_insita]

That's perfectly fine, but that's not how he explained it in the book.

What exactly are you referring to? I don't remember any discussion of the Quantum Zeno effect in the book. He criticises the collapse on the grounds that continuous observation cannot completely halt the time evolution of the system (called "the watched-pot-paradox" by him), which seems to be implied by the collapse postulate. But this is not what happens with the Quantum Zeno effect.

It's not necessary to use collapse in the sense of information update, but I think it's also not wrong to use it that way.

The paper I cited discusses some instances in which it is wrong (or ambiguous). For sequential measurements the conditional probability for obtaining the result $a$, given some earlier result $b$ only agrees with the collapse postulate if -- paradoxically -- the first measurement didn't change the state of the system.

Perhaps it was not obvious from the context, but I distinguish a "true" collapse from a FAPP collapse (essentially, an update of information). The latter makes sense even when the former doesn't exist.

True, but if Ballentine's arguments are correct, then the collapse can be used only in some idealized situations, but certainly not FAPP.

 

[martinbn]

If they look for a solution in which both modes have the same frequency, then such a solution simply doesn't exist. In fact, the modes which are not solutions don't conserve energy even with real $\omega$. But modes which are not solutions are nonphysical simply because they are not solutions, so it's irrelevant whether they conserve energy or not.

No, you are still not reading it carefully. The amplitudes are chosen so that you have a solution.

 

[Keith_McClary]

This may still be controversial:

The Casimir Effect and the Quantum Vacuum
R. L. Jaffe (2005)

I have presented an argument that the experimental confirmation of the Casimir effect does not establish the reality of zero point fluctuations. Casimir forces can be calculated without reference to the vacuum and, like any other dynamical effect in QED, vanish as α → 0. The vacuum-to-vacuum graphs (See Fig. 1) that define the zero point energy do not enter the calculation of the Casimir force, which instead only involves graphs with external lines. So the concept of zero point fluctuations is a heuristic and calculational aid in the description of the Casimir effect, but not a necessity

 

[weirdoguy]

This may still be controversial:

Why?

 

[Demystifier]

This may still be controversial:

The Casimir Effect and the Quantum Vacuum
R. L. Jaffe (2005)

It's not a book but since you mentioned it, I think I have resolved the controversy completely. First I proved a no-go theorem showing what the Casimir effect is not
http://de.arxiv.org/abs/1605.04143and then I explained what it really is
http://de.arxiv.org/abs/1702.03291

 

[Demystifier]

No, you are still not reading it carefully. The amplitudes are chosen so that you have a solution.

I have presented my calculation in detail and you said that it's correct but irrelevant. Please present your calculation in detail, because otherwise I will never understand what's your point.

 

[Demystifier]

What exactly are you referring to? I don't remember any discussion of the Quantum Zeno effect in the book. He criticises the collapse on the grounds that continuous observation cannot completely halt the time evolution of the system (called "the watched-pot-paradox" by him), which seems to be implied by the collapse postulate. But this is not what happens with the Quantum Zeno effect.

Watched pot and quantum Zeno are two names for the same effect.

 

[martinbn]

Then their was his scathing rebuke of the Dirac Delta function. Rather than say we need further developments in math to make sense of it, which has now been done (admittedly requiring mathematicians like Grothendieck whose mathematical reputation is the equal of Von-Neumann himself) he simply dismissed it as a fiction. Von-Neumann is one of my heroes, being more that just a great mathematician, but that even rarer beast, a polymath, however like all human beings perfect he was not.

Thanks
Bill

That sounds interesting. Where can we read more about it? By the way, I am not sure about the history, but I think that Grothendick was not involved here. You need to credit Schwartz.

 

[martinbn]

I have presented my calculation in detail and you said that it's correct but irrelevant. Please present your calculation in detail, because otherwise I will never understand what's your point.

For this there is no need for any calculation other than what is written in the book. They look for a solution of the form $x_k(t)=A_ke^{i\omega t}$. They plug it in the Euler-Lagrange equations, the exponentials cancel because they are the same. This give linear equations for the $A_k$'s, which will have a non-zero solution if $\omega$ is chosen so that the corresponding determinant is zero. It is all written in the book!

 

[vis_insita]

Watched pot and quantum Zeno are two names for the same effect.

The watched-pot-paradox discussed by Ballentine is not the same as the Quantum Zeno effect. The paradox is an absurd consequence of the "quick-and-dirty"-explanation of the Quantum Zeno effect as due to a "collapse" caused by measurement. Abandoning the collapse avoids the absurd consequence and resolves the paradox, without denying the real Quantum Zeno effect.

 

[Demystifier]

For this there is no need for any calculation other than what is written in the book. They look for a solution of the form $x_k(t)=A_ke^{i\omega t}$. They plug it in the Euler-Lagrange equations, the exponentials cancel because they are the same. This give linear equations for the $A_k$'s, which will have a non-zero solution if $\omega$ is chosen so that the corresponding determinant is zero. It is all written in the book!

What you just said has absolutely nothing to do with violation of energy conservation. If you cannot (or do not want to) present an explicit calculation showing violation of energy conservation, then I see no point in further discussion.

 

[Demystifier]

The watched-pot-paradox discussed by Ballentine is not the same as the Quantum Zeno effect. The paradox is an absurd consequence of the "quick-and-dirty"-explanation of the Quantum Zeno effect as due to a "collapse" caused by measurement. Abandoning the collapse avoids the absurd consequence and resolves the paradox, without denying the real Quantum Zeno effect.

If we agree that (according to quantum Zeno) frequent measurements slow down the decay, then I see nothing absurd in the idea that quantum Zeno in the continuous measurement limit (that is, the watched pot) can, in principle, stop the decay completely.

 

[martinbn]

What you just said has absolutely nothing to do with violation of energy conservation. If you cannot (or do not want to) present an explicit calculation showing violation of energy conservation, then I see no point in further discussion.

No, this was just a response to your claim that you don't get a solution. With this solution you plug in the energy and you can see that if $\omega$ is not real you will have an overall exponential factor that makes the energy to either decay or increase, and because you have only one $\omega$ you cannot get cancellations for more than one degree of freedom. I thought we cleared that, and your only objection was that it is not a solution, hence my comment.

 

[vanhees71]

I'm a bit puzzled about this long debate. It's enough to consider only the one-dimensional case. You have a Hamiltonian system (I'll add the mass to get the dimensions correct) with a Hamiltonian that is not expclicitly time dependent. Thanks to Noether the total energy thus must be conserved. Indeed we have
$$H=\frac{p^2}{2m}-\frac{\alpha^2}{2} q^2.$$
Then the EoM. reads
$$\dot{p}=-\partial_q H=\alpha^2 q, \quad \dot{q}=\partial_{p} H=\frac{p}{m}.$$
This gives
$$m \ddot{q}=\alpha^2 q$$
with the general solution
$$q(t)=A \exp(\lambda t) + B \exp(-\lambda t), \quad \lambda=\frac{\alpha}{\sqrt{m}}.$$
Then
$$p=m \dot{q}=m A \lambda \exp(\lambda t) - m B \lambda \exp(-\lambda t).$$
Finally
$$H=\frac{p^2}{2m}-\frac{\alpha^2}{2} q^2=-2 A B \alpha^2=\frac{p_0^2}{2m}-\frac{\alpha^2}{2} q_0^2 = \text{const}.$$

 

[martinbn]

@vanhees71 The discussion is about the argument of L&L. What they do is to look for a solution of the form $x_k=A_ke^{i\omega t}$. The $\omega$ has to be a root of some polynomial. Can it be non-real? The argument is that it has to be real, otherwise there the energy will have an overall factor of the form $e^{\lambda t}$ with $\lambda$ real, which contradicts conservation of energy. @Demystifier said that this argument is erroneous because the time dependence will cancel out. To support his claim he showed and example. The problem was that in his example the solution was of the form $x_k=A_ke^{i\omega_k t}$, where the omegas are different and you get a cancellation. That is incorrect, more accurately irrelevant, in L&L the solution has the same $\omega$. Then he said that with the same $\omega$ you cannot get a solution, which is also wrong. Of course you cannot get a solution with non-real $\omega$, but that is their claim as well. So I am still unconvinced that they have done anything wrong.