Eqs

Thursday, August 12, 2021

What is a Black Hole Laser?

When you grow up as a child that does not have to worry about being fed and housed and all that, you get to think about all kinds of awesome and fantastic things: Dinosaurs! Space ships! Black holes! Lasers!

What if you could combine all of those? Would that create the ultimate awe-pocalypse? Dinosaurs flying into black holes in spaceships powered by lasers? Obviously we can't pull this off because dinosaurs are extinct, we have a hard time building spaceships (though Elon is working hard to change that) and we certainly can't fly into black holes, even though movies such as Interstellar  have teased that. But where do Lasers fit in? Well, hold on to your seats: it appears that (at least theoretically) black holes can be turned into Lasers, or maybe some already are, but we don't know it yet! 

If you are reading this post you probably already know what Lasers are, but in order to make the case for BHLs (black hole lasers), let me reintroduce you to them briefly. `Laser' is an acronym for "Light Amplification by Stimulated Emission of Radiation" (be honest, it's one of the greatest Science Acronyms of all time, because it is not tortured at all). The phenomenon relies entirely on Einstein's work concerning the probability of absorption and emission of light from atoms or molecules. Here's the title of this groundbreaking (and, as a matter of fact, easy to read) paper:

In this paper, Einstein showed how you can derive Planck's celebrated formula (the "Law of Radiation", or "Strahlungsgesetz") from first principles. Yeah, Planck had essentially guessed his law. 

Einstein made the following assumptions, given an atom or molecule that has discrete energy levels with energies \(E_1\) (the "ground state") and \(E_2\) (the "excited state"):

1. A quantum of light can be absorbed by the atom, which raises the state from ground state to excited state 
Here, \(\hbar \nu\) is the energy of the quantum that is tuned to the difference in energies: \(\hbar\nu=E_2-E_1\).

2. If an atom is in its excited state, it can spontaneously emit a quantum of energy:


3. If an atom is in the excited state and absorbs a quantum at the same time, the quantum stimulates the emission of another quantum just like it:

Using these assumptions, Einstein was able to derive Planck's law, which in turn means that these are the only processes needed to understand how radiation interacts with matter (barring quantum field-theoretic effects, of course). 

The third process, called stimulated emission, is the one that gave us Lasers. The thinking behind is (in hindsight) quite simple. Take a look at the picture above. What happened here is that the incoming quantum was doubled (copied). You don't violate the no-cloning theorem this way because the spontaneous emission process provides just about enough noise. But you already knew this. Now suppose that after you made two-out-of-one, the two quanta go on their merry way but encounter a mirror and head back:

At the same time, imagine you have a way to "pump" atoms back to their excited state from the ground state (that's what the yellow light bulb in the sketch above is supposed to do). When the two quanta encounter an atom (not necessarily the same one, there will be plenty of atoms in the gas that's enclosed in the cavity between the two mirrors), each will stimulate the emission of a clone, so from 2 make 4. Because this doubling gives rise to an exponential process, after a short while the number of quanta becomes astronomical. Now, if the quanta remain in the cavity then they would be of no use to anyone, which is why one of the mirrors is usually made semi-transparent: so that the beam of coherent quanta can leave out of that side. That's your Laser, right there:





Alright, now that you have become a Laser expert, what does this have to do with black holes? Those of you that have been following this blog from its inception maybe remember that stimulated emission is precisely the process that saves information from disappearing into a black hole (if you haven't, then start here). In a nutshell, the processes that Einstein wrote about don't just hold for light interacting with atoms, they hold for any and all particles (fermions or bosons) interacting with any matter. In particular, they hold in quantum field theory. And as a consequence, they hold for quantum field theory in curved space. They hold for stuff interacting with a black hole. So when a particle is absorbed by a black hole, it stimulates the emission of a clone outside the horizon (which is carrying a copy of the information), and so we don't have to worry about the one that disappears behind the horizon. That's it, that's the whole ballgame for solving the problem of information loss in black holes. If you take into account this process (instead of ignoring it, as Hawking did in his calculation [1]) you find that information is conserved (the capacity of the black hole to process classical information is positive). 

"Fine (I hear you saying), but this doesn't make a Laser yet"!

True. For example, where's the mirror? But, hear me out. Suppose you fell into a black hole (one that is very large, so you don't get ripped apart by the tidal forces), and suppose you brought with you some rocket engines that would allow you to hover inside the event horizon, but you don't fall towards the singularity (if there is one). Even better, imagine that there are planets inside the black hole (like in the movie Interstellar, again). Then you would "look up" to the horizon, and you would notice something strange. If you shine a light beam towards the horizon, it is reflected back. The reason for this is simple: nothing can escape the black hole, so the best that your light beam can do is "go into orbit" below the event horizon. Basically, this means that from inside of a black hole, you're looking at a white hole horizon. (I've written about this before: it is a consequence of time-reversal invariance.)

The other thing you have to keep in mind when discussing stimulated emission in quantum field theory is that you always "stimulate in pairs". The reason for this is that you need to conserve quantum numbers: for example, if you are going to stimulate an electron, you are going to have to stimulate a positron also. So for every clone that is stimulated outside of the horizon, you also stimulate an anti-clone inside of the black hole (see Fig. 1)
Fig. 1 The horizon of a perfectly absorbing black hole looks black from the outside, and white from the inside. Particle p (black) is absorbed, and stimulates the emission of the clone-anticlone pair (red).

But clearly that is not enough yet to make a Laser. So imagine then two black holes that are connected by an Einstein-Rosen bridge: a wormhole. This might look a bit like this:
Wikipedia's depiction of a Schwarzschild wormhole. 

Basically, it is two black holes connected by a "throat". We don't know if it's traversable for people, but you certainly can imagine that a particle thrown into one of the black holes might come out at the other end. 

Come out? But nothing can come out of a perfectly absorbing black hole, right? Well, this is both right and wrong. Physical particles cannot come out, but you can clone those particles, so copies can come out, which after all is just as good. 

Let's see how this would look like. We now take the black hole horizon from Fig. 1, and add another one like so:

Fig. 2: Two connected black holes with horizons \(H_1\) and \(H_2\). The anti-clone that travels towards \(H_2\) is reflected there, and stimulates another pair.

Because from the inside the black hole looks like a perfectly reflecting white hole, the anti-clone heads back to horizon \(H_1\), but the mathematics of stimulated emission in black holes says that the reflection creates a clone pair as well

What happens now? Well it's clear. The absorbed particle and one of the anti-clones are on a collision course leading to annihilation, but the other anti-clone will reflect on \(H_1\) and stimulate another pair. As if by magic, there are now two clones outside horizon \(H_1\), and two outside horizon \(H_2\). But the anti-clones inside the wormhole keep reflecting between the horizons just as in the optical Laser described above. Except that we don't need semi-transparent mirrors: the "Laser beams" will emanate from the horizon in a coherent manner as long as the inside of the wormhole is coherent!

Fig. 3: The wormhole Laser. Anti-clones that are reflected from the inside of the black holes stimulate emission of clones outside the respective horizons. 

So where does all this energy come from? After all, there is no pump that "charges up" this Laser, as there must be for an optical Laser. The answer is that this bill is paid for by the black hole's masses, just as it is in spontaneous emission. A detailed calculation would have to show how fast a wormhole might deplete its mass, but the calculation is already difficult for a single black hole (and it can only be made by approximating the interaction between black hole and radiation with a model, see here). 

Now to the last question: what would a BHL look like? First of all, it is clear that whatever radiation emanates from the black hole, it will look like it is coming from a disk surrounding the black hole. We also know from explicit calculations that the stimulated emission in response to absorbed material is not red-shifted (because this is "late-time" absorption). However, what happens to material that reaches the second horizon I can't say without a calculation. The important distinction for Laser light, however, is that it is coherent. If the stuff that is stimulated outside the horizon is similarly coherent, we might be able to detect this using typical Hong-Ou-Mandel interferometry of light coming from such a black hole. We've just learned how to look at light emitted from black holes using telescopes like the Event Horizon Telescope, so it might be some time before we can check if that light is really BHL light. We don't know how many black holes are actually connected to others making BHLs possible, but at least there is a chance to find out! 



[1] The reason Hawing ignored stimulated emission in his calculation of radiation coming from a black hole is that he thought that it would require energy from a black hole's rotation (the rotational energy would provide the "pump energy"). Because he treated a non-rotating black hole, he decided he could ignore the effect. It turns out that stimulated emission does not require black hole rotation. 




Monday, December 2, 2019

On quantum measurement (Part 8: Leggett-Garg Inequalities)

So Part 7 wasn't the final installment in this series after all. I know this because this is Part 8 of the quantum measurement saga. And for this one, I'm going to change the style a bit. I will begin with a fictional dialogue between two quantum physicists, called "Lenny" and "Chris". After the dialogue, I will delve more deeply into what it is that Lenny and Chris are talking about.


This is a dialogue between Lenny and Chris, two fictional physicists. Both know quantum physics. They both can calculate things. But Lenny has an advantage: he can also do experiments. He has made a discovery, and wants to show it to Chris.


Lenny: "Hey Chris, I just wrote down this equation describing a series of three consecutive measurements on the same quantum system. My equation says that when you make these three measurements 'a', 'b', and 'c', a particular sum must always be smaller or equal to 1."

Chris: "Wow Lenny, this looks great. It looks like a solid prediction. You should try and check that. You know, because of your ability to measure things. In quantum physics."

Lenny: "You know what, I'm going to do just that.

Time passes. Music is playing in the background. It is mildly annoying. A while later:

Lenny: Hey, that's actually a difficult experiment. Doing two measurements in a row is easy, but doing three is hard. Do you think I can instead do three different measurements where I just do a pair of measurements in each?"

Chris: "What do you mean?"

Lenny: "Here I'll sketch it for you on the board"



Lenny: "Here's the quantum state \(Q\), and here's the three measurements, see? Let's call them a, b, and c. And we should do each measurement at an angle \(\theta\) with respect to the previous measurement"

Chris: "Yup. Go on."

Lenny: "Well, I wanted to just do the pairs, 'ab', 'bc', and 'ac', like so:




Lenny: "See, I first do the first pair just like above. Then I do the "bc" measurement (but I don't really have to because it is really just the same as the "ab" measurement), and then I do the one at the bottom of the chalkboard, where I leave out the middle one."

Chris: "Ah but Lenny, leaving out the middle measurement changes everything. It's, like, the first lesson in quantum mechanics. Measurements change the system that is being measured. That's why we have the uncertainty relation".

Lenny: "But what if the middle measurement doesn't matter?"

Chris: "What?"

Lenny: "Hear me out. What if I did the experiment and assume that I get the same result as if I had done the measurement? As if the middle measurement didn't matter."

Chris: "You mean, like in classical physics?"

Lenny: "Yes, just as in classical physics".

Chris: "But we already know that measurements like this cannot be described by classical physics".

Lenny: "I'll just say I'm testing "classical thinking".

Chris: "Sure, you could do that. But nobody thinks that classical thinking will get the same result, so you'll never be able to publish such a result."

Lenny: "Oh but I bet I can. I just won't call it 'classical thinking'. I'll call it macrorealism. I'll say I'm testing macrorealist theories.''

Chris: "Now you're just trolling me. But go ahead knock yourself out."

More time passes. A lot of time. Finally Lenny is back. He is out of breath. The blackboard, miraculously, has not been erased.

Lenny: "Chris, you're not going to believe it. I did those three pairs of measurements and I get a result that's different from what quantum mechanics predicts for doing all three measurements!"

Chris: "Actually, Lenny, I totally believe it. It is what quantum mechanics predicts after all."

Lenny: "But it's much better than that! The results are such that the equation that quantum mechanics says can never exceed one, actually does! I broke the inequality!"

Chris: "That is not possible. Show me what you did."

Lenny: "I did the measurements just like on the board."

Chris: "You did? What angles did you use?"

Lenny: "I used \(\theta\) for the angle between the first two and also between the last two, and \(2\theta\) between the first and the third, assuming the middle one does not matter."

Chris: "Yeah, you can't do that."

Lenny: "What?"

Chris: "The three experiments are supposed to mimic the experiment at the top of the blackboard, right?

Lenny: "Yes.  And as long as I prepare each of the three experiments exactly in the same manner, I can use the statistics of the three pairwise measurements to stand in for the triple one."

Chris: "That is true.  But you didn't prepare your three measurements in the same way. You used very different parameters in each. That is not allowed."

Lenny: "What do you mean? Angle \(\theta\) between the first two and the last two, and angle \(2\theta\) between the first and third. It's the same."

Chris: "It is not. Let me explain."


So what are Lenny and Chris talking about here? They are discussing making consecutive measurements on the same quantum system, something that we encountered already in Part 7, which is linked. While I describe the theory of consecutive measurements in that part, with a focus on whether or not wavefunctions collapse during measurements, that part of the theory is completely irrelevant for the following. Collapse and no-collapse pictures make the same predictions for what follows. 

Let's first set up the three measurements. For convenience, we'll take a quantum two-state system (a qubit) and all our detectors are binary detectors (they click or they don't). There are of course many ways in which we can realize such a series of measurement, and one of the simplest is via a Mach-Zehnder optical setup. You've seen versions of it several times in this series.
Fig. 1: Measuring the quantum system \(|\Psi_1\rangle\) using the classical detectors  \(A_1\), \(A_2\), and \(A_3\).
Here, the dark grey boxes with an angle written on top are polarizing beam splitters set at an angle with respect to the polarization of the initial state. You can think of these polarizing beam splitters to change the basis of the quantum state, so that for example, \(A_2\) measures the quantum state at a relative angle \(\theta_1\) compared to \(A_1\), and \(A_3\) measures at angle \(\theta_2\) with respect to \(A_2\). I'm sure you realized by now that these three measurements are just the 'a', 'b', and 'c' of the dialogue.

If you don't want to think in terms of interferometers, you can also just look at the quantum circuit in Fig. 2. 

Fig. 2: The three consecutive measurements as a quantum circuit. The initial quantum state is here written as an entangled bi-partite state. Measurements are indicated by a CNOT operation followed by a Hadamard gate that rotates the quantum state by the indicated angle.
Alright, so what is this equation that Lenny and Chris are discussing. It concerns the correlation between pairs of detectors. For example, I might want to know what the correlation is between detectors 1 and 2. I call this function \(K_{12}\). It's fairly simple to derive it, but I won't do it here. It's described in detail in the paper that I'll be linking to. Don't groan, you knew this was coming.

For example, suppose \(\theta_1=0\). This means that \(A_1\) and \(A_2\) measure in the same basis, and quantum mechanics then tells us that both detectors must agree. So \(K_{12}(0)=1\). On the contrary, if \(\theta_1=\pi/2\), then you are measuring at orthogonal angles and the correlation must vanish. Indeed quantum mechanics predicts

\(K_{12}=\cos(\theta_1)\;.\)             (1)

We can also calculate \(K_{23}\) in this manner. You get the same result, only with the angle \(\theta_2\), that is, \(K_{23}=\cos(\theta_2)\).

What about \(K_{13}\)? Well, this result will depend on whether or not we make the 'b' measurement in between. If we do it, then the result is  

                                      \(K_{13}=\cos(\theta_1)\cos(\theta_2)\;.\)             (2)

If instead we don't do it, then the third measurement will occur at the angle \(\theta_1+\theta_2\) with respect to the first, and we then should find \(K_{13}=\cos(\theta_1+\theta_2)\). 

There is, by the way, no controversy about the result of these calculations. They are easy to do. Doing or not doing a measurement matters. Everybody knows that. I assume you are nodding vigorously in assent. You did, didn't you?

Let me now write down the equation that Chris and Lenny have been discussing. 

It can be shown, using the formalism I introduced in Part 7 (or any other standard quantum measurement formalism, for that matter) that the following inequality holds:

                                     \(K_{12}+K_{23}-K_{13}\leq1\;.\)             (3)

This is only one of the so-called "Leggett-Garg inequalities", introduced by Sir Anthony Leggett and Anupam Garg [1] , but for the sake of being brief it is the only one I'm going to consider here. 

Like I wrote above, whether or not you are making two or three measurements matters. Let's find out how. First, let us check what we get for the inequality if all three measurements are performed. Using the results above, and assuming that \(\theta_1=\theta_2\equiv\theta\), that is, I'm choosing the same change in basis between the first and the second, as the second and third measurement), we get

                                 \(2\cos(\theta)-\cos^2(\theta))\leq1\;.\)             (4)

It's immediately clear that this inequality cannot be violated, because the left hand side of Eq. (4) is \(1-(1-\cos(\theta))^2\), which is less than one for all \(\theta\).  There is no way that Equation (3) can ever be violated, even if you use two different angles \(\theta_1\) or \(\theta_2\). 

But Lenny said that he did violate inequality (3), and all he had to do was not make the middle measurement. If that would be true, quantum mechanics would be broken because equation (3) should hold whether or not I'm making the middle measurement.

OK, what does it mean to *not* make a measurement? There are actually two ways in which you could do that (of course they turn out to be the same). One way to do this is to actually *do* the measurement, but you must make it at the same angle that you made the first measurement. You choose any other value, and the second measurement *is* made. But remember, in the 'ac' measurement you are not doing the second one, and since in order to simulate the triple measurement you must do all three pairwise experiments with the same parameter values, then if you do not measure at position 'b' in the third measurement (for the 'ac' correlation), you must also not make that measurement in the first of the three also (for the 'ab' correlation). And the second (for the 'bc' correlation) just the same. 

Otherwise, those three pairwise measurements do not describe the triple-measurement situation, and the inequality does not describe this set of measurements. So instead, what you should do is this:



You're not doing the 'b' measurement, so you have to choose \(\theta=0\) for the 'ab' measurement. You are doing the 'bc' measurement at angle \(\theta\). That's fine. But the angle between the first (non-measurement) and the third is then \(\theta\), not \(2\theta\). Because 0 plus  \(\theta\) is \(\theta\).

If you use these angles and plug them in to equation (3), you immediately realize that the inequality is not violated: it is exactly equal to 1. If you want to use  \(2\theta\) for the third ('ac') measurement instead, then you also have to use that angle in the 'bc' measurement, since the first measurement (at 'b') had to be done at zero angle. You can't say that you don't do a measurement (for the 'ac') experiment, yet do it in the 'ab' measurement, and believe that it describes the same experiment (the 'abc' measurement). It's like saying I use \(\theta=45\) degrees in the 'ab' measurement, but I use  \(\theta=30\) in the 'ac' measurement. You can't just change angles if you want the statistics of the pairwise measurements to reflect the full triple measurement. If you do, then the three pairwise measurements are not standing in for the triple measurement. 

I told you earlier that there was another way of *not* doing a measurement. This is making a measurement so weakly that it becomes non-existent. What's a weak measurement you ask? It's actually a simple concept, and I'll briefly explain it here.

At the risk of repeating myself, here's what happens in a measurement. You want to move the measurement device's "needle" in such a manner that it reflects the thing that you want to measure. If what you are measuring has only two states, then (for example) you want to keep the measurement device as is when the system is in one of the states, and you want to move the measurement device into a different state if the system is in the other. If you want to make sure which is which, you need to make sure that the two states of the measurement device are easily distinguishable.  In quantum physics, the best you can do (for distinguishability) is to make the two states orthogonal. Indeed that is precisely what you do in projective measurements.

But it is possible to make measurements where you don't move the measurement device's "needle" by 90 degrees (into the orthogonal direction). What if you just move it a little?

If you move the device's needle just a little, then it becomes harder to make sure what the device's state is after measurement. Say you move the needle by an angle \(\phi\). Because your device at this point is just another qubit, then in order to amplify your measurement (make it available to a classical reader) you need to measure it again. But this time using an orthogonal device. This means there is a change that the ancillary qubit will be misread. (This is not possible if the measurement was done at an angle 90 degrees, that is, a strong measurement). 

A weak measurement, therefore, is a fuzzy measurement. You might think that this is a terrible way of doing a measurement, but there are plenty of uses for this. For one, you can make this measurement more precise by repeating it many times. If you do, you can detect even very small angles \(\phi\). What you gained is that your weak measurement only weakly disturbed the quantum system. OK, so let's do some weak measurements. To prepare you for that, I'm going to magnanimously repeat strong measurements, as if you did not already read Parts 1-7. I know you did, but who can remember things from so long ago?

Here we imagine that our quantum system is prepared. This means that we took an arbitrary quantum state, and measured it with a particular device. No matter what the outcome of that "preparing" step is, we'll treat this as the first measurement (the 'a'). So we'll only have to do two more, but Lenny wants to simplify things, and do 'b' only for the first pair. Then, he'll prepare an identical quantum state with 'b' and measure 'c' (and as I said, that's really a repetition of the first measurement), and finally he'll do the 'ac' measurement. 

We do measurements by entangling using a CNOT operator 

                    \(U_1=|\theta_1\rangle \langle\theta_1|\otimes \mathbb{1} +|\bar\theta_1\rangle \langle\bar\theta_1|\otimes\sigma_x\;.\)

Here, \(|\theta_1\rangle \langle\theta_1|\) is the projector into the new basis, and \(|\bar\theta_1\rangle \langle\bar\theta_1|=1-|\theta_1\rangle \langle\theta_1|\) is the projector on the orthogonal basis state. \(\sigma_x\) is the first Pauli matrix, and just flips a qubit. 

Here's a little exercise for you: you can write the same operator as 

                               \(U_1=e^{i\pi/2 |\theta_1\rangle \langle\theta_1|\otimes \sigma_y}\;.\)

We won't do this exercise here. But if this is stunning to you, go ahead and expand the exponential and so forth, it'll be worth it. And no, it's not a typo in the exponent: it is really the y-Pauli matrix. 

The angle \(\pi/2\) in the above expression is the "flipping" part of the operator. In a weak measurement, we just use a smaller angle \(\phi\) instead. If you do that, then instead of moving your ancilla from \(|0\rangle\) to \(|1\rangle\), you instead movie it to the state

                          \(|\epsilon\rangle=\sqrt{1-\epsilon^2}|0\rangle + \epsilon|1\rangle\;.\)

Oh, and \(\cos(\phi)=\sqrt{1-\epsilon^2}\). So basically, we now repeat the entire calculation with a weak measurement 

                                  \(U_1=e^{i\phi |\theta_1\rangle \langle\theta_1|\otimes \sigma_y}\;,\)

keeping in mind that in the limit \(\phi\to\pi/2\) we return to a strong measurement, but in the limit of \(\phi\to0\) the measurement becomes so weak that it does not even take place! Just what we need!

Alright, here's the plan. We still measure the quantum state with the 'a' measurement in whatever basis. The next measurement is the middle one, so it needs to be potentially weak, so we perform it with a strength  \(\phi\), at angle \(\theta\). Just to be clear: \(\phi\) is the strength of measurement, with 0 meaning no measurement, and \(\pi/2\) meaning full-strength orthogonal measurement. The angle is still \(\theta\). The last measurement will be strong again, at angle  \(\theta\) again. The calculation for the correlation function now gives

                            \(K_{12}=(1-\epsilon^2) + \epsilon^2\cos(\theta)\) 

and you're not surprised to recover \(K_{12}=\cos(\theta)\)  when the weak measurement is strong instead. Now let's calculate 

                   \(K_{23}=(1-\epsilon^2)\cos^2(\theta) + \epsilon^2\cos(\theta)\;.\) 

You can check again that in the limit \(\epsilon\to1\) you recover the old (strong) result. Finally,
\(K_{13}\) does not depend on \(\epsilon\) because it is the 'ac' measurement and only 'b' is potentially weak. (Not that this is only true if the first measurement prepares the quantum state in an eigenstate, as opposed to a superposition, of the first detector basis.)

Let's write down the Leggett-Garg inequality for this generalized (weak) 'b' measurement:

                    \(K_{12}+K_{23}-K_{13}=1-\epsilon^2(1-\cos(\theta))^2\;.\)

In the limit of a non-existing measurement 'b' \(\epsilon\to0\), we get 

                         \(K_{12}+K_{23}-K_{13}=1\;\)

the same result we obtained when we didn't do the second measurement by choosing the first \(\theta=0\). Whichever way we choose not to do the middle measurement (because we want to simplify our work by doing only pairs of measurements), the Leggett-Garg inequalities cannot be violated. Ever. 

What does all this mean for us? Well, people have wondered over and over again why it is so easy to violate inequalities that should be observed by quantum mechanics. Some (including Leggett) have suggested that when it comes to macroscopic phenomena, quantum mechanics can't be the right theory. Of course, you might say, we use classical mechanics there. But you need to remember that classical mechanics is not a correct theory. We need to be able to describe classical objects with quantum mechanics, or at least something that both describes microscopic and macroscopic objects accurately at the same time. Others have suggested that quantum mechanics is indeed the right theory, but that we just haven't (for a variety of reasons)  been able to observe the deviations that the violation of LG inequalities implies. One of the reasons suggested by that camp is that perhaps we can never really perform perfectly strong measurements, and that this ``fuzziness" of measurements might obscure the violations [2]. Instead, we now realize that there are no paradoxes at all. Quantum mechanics correctly describes both microscopic and macroscopic physics. And we see no "weird" stuff in quantum measurement of macroscopic devices not because strong measurements can't be achieved, but because those violations simply do not exist. They are based on the faulty thinking that plagued Lenny. I do realize that there are literally hundreds of papers that follow Lenny's thinking in making pairwise experiments. All of them claim the Leggett-Garg inequalities are violated. All of them are wrong.

The preprint describing this work is [3] below. I have applied this thinking to the so-called "entropic" Leggett-Garg inequalities, which are related to the original inequalities but distinct. In [4] I show that these can't be violated either.

[1] A. J. Leggett and A. Garg, “Quantum mechanics versus macroscopic realism: Is the flux there when nobody looks?” Phys. Rev. Lett. 54, 857–860 (1985).

[2] J. Kofler and C. Brukner, “Conditions for quantum violation of macroscopic realism,” Phys. Rev. Lett. 101, 090403 (2008).

[3] C. Adami, “Leggett-Garg inequalities cannot be violated in quantum measurements,” arXiv:1908.02866 (2019)

[4] C. Adami, “Neither weak nor strong entropic Leggett-Garg inequalities can be violated,” arXiv:1908.03124 (2019).

















Saturday, July 21, 2018

How to pack your library: A guide

That's right folks, this is not a science post. After 45 posts about information, quanta, intelligence, and what not, a how-to guide? Well I only have one blog, and I didn't know where to put this stuff, that I think could be helpful to others, because I've done this several times and I learned a bunch. So now you get to read about how to move your library of precious books from one house to another.

First things first. If your library consists out of a pile of paperbacks you have largely forgotten about, then this post is not for you. In fact, I'm wondering how you even got to this page. I'm talking to people whose library looks like this:
The Upstairs Library
What you see in the pic is actually only half my library. The other half is behind you from this view, along with two shelves in the basement. It's a bit more than a thousand books. 

Thing is, it is a collection of mostly First Edition and rare books, so you would not want to just throw them into a box when you move. And I am moving. How do you make sure the books survive the trip?

I've had to move this library before (from California to Michigan), and I've learned some things from that move. One of the things you need to know about movers is that they don't really care very much about what is inside a box they are moving. They are under time pressure, and if something breaks then insurance will pay for it. So it is up to you to keep the stuff safe.

When I moved last,  I wrapped every book individually into a foam material. That protected the books well (all books survived just fine), but it was expensive and time consuming. If you have not moved like this before, I can assure you that packing material is expensive. By packing each book individually in the material that you would ordinarily wrap china, you not only buy a lot of that material (at about $15 a roll) but you also pack fewer books per box. And packing each book into foam takes time. For this move, I was looking to do better. Here's what I came up with.

First, let us think what you have to protect against. There are two main threats to rare books. One is physical injury to the box, which will impact the books inside the box from the outside. Second is rubbing of books against each other, damaging their covers or dust jackets. To protect against rubbing you do not need to wrap the books in foam, you just need to wrap them. To protect against injury to the box, you need to protect the box, not every book.  My solution was to use paper to wrap each book individually, and line the box with foam. It doesn't protect you against catastrophic box intrusion, but it should work as well as the previous method, with a substantial cost saving.

First, buy some boxes. What size do you need? The answer is: small, as small as you can find. You may think: "Shouldn't I get a big box where I can get as many books in as I can?" Well if you did this, it would become so heavy that your movers will either refuse to touch it, or else injure it while struggling with it, possibly for spite. So go small. In my previous move I used "book size" boxes (not all companies have them; they are one-foot cubes), but for this move, I went with the "Extra Small" size. The dimensions are 15'' long, 12'' wide, and 10'' high. In my case they were made by Home Depot, but I suspect other retailers carry that size.

I like this box because it is not high. Books are not high. In fact, almost no book is higher than 10 inches (but some are close). So you will be able to fit your books standing up in this box, and if there is some room at the top, you can put a layer of books there. If not, you fill with paper, as I'll show.

Here's a pic of some of the boxes I bought. (I ended up getting about 65 of those).

How many do you need? This is not easy to answer. I think you can get at least 15 books into such a box, up to 25 books if they are small. So, on average, think 20 books per box. I'll show you a few pics later.

OK, so let's first make the boxes. I assume you know how to tape up a standard box. I double tape the bottom (and later the top), and also tape the sides of the box. I used the tape below, after I tried many different types that all were awful. It's not too flimsy, and the cutter actually works.



Now let's line them. For that, I use dish foam, 1 foot high, separated into 1 ft squares. There are 50 squares in a roll.


One square will perfectly cover each 12 inch side of the box. For the 15'' side, I cut 3'' to the right of the perforation, leave that piece attached to the next and cut 3 inches to the left of the following perforation. It looks like this: 

Pieces 3 and 5 are 15'' and line the sides. Piece 4 is 18" and will cover the bottom, with a little overlap to the sides. Each roll will give you enough for 8 boxes, with a little bit to spare (but sometimes the beginning of the roll is unusable). 

I use the tape to glue the foam to the sides:

Do the same on all sides:

Don't worry that the bottom does not stick to the side. Once the bottom is in it'll be fine. Besides, the books will press against it.

Remember the box is 10'' high, so two inches reach up to the cover.

With the foam in, we're ready to wrap books. I ended up using paper, which is cheap and doesn't interact with the covers of the books. These rolls here have sheets that are 2 feet x 2 feet, and there's seventy in each roll. One sheet is enough for one book. You begin like this:
Fold over the bottom, then the top, then fold the left over on top of the book, then flip it over. This way, the fragile edges of the book get double coverage. Now it looks like this:
Now I just take small piece of tape to finish it off:

If you have a cat, she will probably want to assist you, by the way. I suggest wrapping a bunch of books before packing a box, because that way you can select the right sizes to optimally fill the box.

If you have fairly slim books, it does not hurt to pack two of them together, as their dust jackets will not be able to rub against each other:
Packed together, they are just like one thicker book, and you saved yourself one sheet:
You notice that I'm not very precise when wrapping, because this is altogether unnecessary. 

The roll usually has a little wave at the bottom, which is perfect to place the book on before you begin to wrap:

(Actually you can do this any which way you want. I'm just showing off this book. Yes, that's a first edition in first state dust jacket. And no you can't have it.) 

Time to put books into the box. There are two ways to do this, depending on the size of the book. Big ones go in like this:
Often there is room to put a few books on top:
Once you've done that, crumple a few sheets to make sure the box is completely "full" after closing it:

Finally, close and number:

If you are really worried about the contents, you can of course make a list for each box. I did not do this, but I did use an app called "Sortly" that allows me to take a picture of each box, name it, tag it, and a field where I can write something about the contents. For the box above, I just wrote "Nonfiction", because these books where from the nonfiction shelf. If the moving company loses a box, then I have proof that it existed, what it looked like, and I have some idea what's inside. 

I also have some oversized books (the "coffee table" kind). For those, I used bigger boxes (the "small box" type), and because these boxes are very heavy, I used the "extra strength" type.

After all is said and done, your library should look like this:


These are the 66 "extra small" boxes. There are also nine "small" boxes not in the view, for a total of 75. 

Yes, this is time-consuming. It helps a lot to develop a routine so that every step is essentially the same. That way, you can achieve something akin to an assembly line. If you have your strips of adhesive tape all ready and lined up, you can do a book in 20 seconds flat. 

In my case, all these books will go to storage first, because I won't have a library in the sabbatical apartment we will be renting. If you have to store your books too, let me give you one final piece of advice: get climatized storage. We'll put the piano into storage also, so the books and the piano will be reunited there (and of course pianos have to be stored climatized also). I'm assuming, of course, that if you have a library like mine, well then you also have a grand piano. It just comes with the territory. 







Wednesday, March 21, 2018

Black hole Lasers: The science of "Interstellar" revisited

The movie "Interstellar" has not only fascinated moviegoers: it also has created a discussion among scientists whether any and all of the science discussed in the movie is accurate. To be sure, this is not the fate that befalls your average Sci-Fi movie where the laws of physics are routinely (and often egregiously) broken over an over again. I'm sure you can find a list simply by googling appropriately. Try "physics violations in sci-fi movies", for example. 

But this movie is different. This move has famed theoretical physicist (and newly-minted Nobel laureate) Kip Thorne as an executive producer. 

Kip Thorne (source: Wikimedia)
Not only did Kip advise director Christopher Nolan about the science (including talking him out of using time travel as a plot device), but he also spent time to calculate what a black hole event horizon should look like. He is quoted as saying, for example: 

"For the depictions of the wormholes and the black hole, we discussed how to go about it, and then I worked out the equations that would enable tracing of light rays as they traveled through a wormhole or around a black hole—so what you see is based on Einstein's general relativity equations." 

This is certainly unusual for a Sci-Fi movie. Indeed, the renderings of the equations Kip provided have have been published in two papers: one aimed at the general relativity crowd, and one aimed at the SIGGRAPH audience. 

To boot, Kip Thorne not only is an advisor: he co-wrote the initial script of the movie (that originally had Steven Spielberg attached as director). But according to Kip, the final story in "Interstellar" only bears a fleeting resemblance to his initial script.   

So if "Interstellar" is so infused with Kip's science, why would there be a need to go "revisit the science of 'Interstellar', as this blog post titillatingly promises.

Black holes, that is why! 

"Interstellar" has black holes front and center, of course. And having Kip Thorne as an advisor is probably as good as you can get, as he is co-author of the magnum opus "Gravitation", and also wrote "Black Holes and Time Warps" for the less mathematically inclined. I have both volumes, I should disclose. I believe my copy of "Black Holes" is signed by him.

Having said all this, and granted my admiration with his science and his guts (he defied federal funding agencies by writing proposals on closed time-like loops) I have a bone to pick with the science depicted in the movie.

Lord knows, I'm not the first one (but perhaps a tad late to the party). So just give this long moribund blog post a chance, will you?

This is no time to worry about spoilers, as the movie came out a while ago. The story is complex, but a crucial element of the story requires traveling into, and then somewhat miraculously out of, a black hole.

When you get past the event horizon, as our hero Joseph Cooper (played by Matthew McConaughey) does, could there be any way back (as depicted in the movie)? At all? Without breaking the laws of physics and therefore credulity at the same time?

This blog post will tell you: "Yes actually, there is", but it is not an answer that Kip Thorne, or anyone else involved with the Interstellar franchise, might cozy up to. It is possible, but it involves making a people-Laser. Read on if that's not immediately obvious to you.

I am going to ask and answer the question: "If something falls into a black hole, and that black hole is connected to another black hole for example by a wormhole, can you come out on "the other side"?

But I have to issue a quantum caveat: I'm going to assume that the two black holes are connected via quantum entanglement as well. Black holes that are connected this way have been considered in the literature before (Google "ER=EPR" if you want to learn more about this).
Two black holes connected by a wormhole (Source: Wikimedia).

The main point for us here is that the two "mouths" of the "Einstein-Rosen bridge" (that's what the two black holes that are connecting two different regions of spacetime are called) are quantum coherent. And of course, you figured out that "ER" stands for "Einstein-Rosen", and "EPR" abbreviates "Einstein-Podolski-Rosen", the three authors that investigated quantum entanglement and its relation to quantum reality in the famous 1935 paper. Now, previously people had argued that the wormhole connecting the two black holes would not be stable (it would collapse), and that anyway it could not be traversed. But later on it was shown (and Kip Thorne was involved in this work) that wormholes could be stabilized (maybe using some exotic type of matter), and possibly could also be traversable. So I'm not going to debate this point: I'll stipulate that the wormhole is indeed stable, and traversable. What I'm concerned with is the escape. Because remember: "What goes on inside of a black hole stays inside the black hole"?

"So what's this about a people-Laser?"

Hold your horses. Let me get some preliminaries off my chest first. You've been to my blog pages before, right? You've read about what happens to stuff that falls into black holes, no? If any of your answers is "Umm, no?", then let me gently point your browser to this post where I explain to you the fate of quantum information in black holes. In the following, I will shamelessly assume you have mastered the physics in that post (but I will gently issue reminders to those whose memory is as foggy as mine own).

So what you the reader of my pages know (that, alas, most of the people working in the field have yet to discover) is that when you throw something in a black hole, several things happen. The thing (and we usually think of a particle, of course) is either absorbed or reflected (depending on the particle's angular momentum). Let's say it is absorbed. But Einstein taught us in 1917 that something else happens: the vacuum makes a copy of what you threw in, along with an anti-copy.

I pause here for those readers that just struggled with an onset of what feels like an aneurysm.

To those readers: please read the post on the "Cloning Wars", which explains that, yes, this happens, and no, this does not violate the no-cloning theorem.

A copy and an anti-copy? Well, yes, if you're gonna create a copy and you don't feel like violating conservation laws (like, pretty much all of them) then you have to create a copy and an anti-copy. If you throw in an electron, an electron-positron pair will be created (Einstein says: "stimulated"), where the positron is the anti-copy. If you throw in a proton, Einstein's stimulated emission process near the black hole will create a proton-anti-proton pair. The copy stays outside of the black hole, and the anti-copy now finds itself inside of the black hole, alongside the original that the black hole just swallowed. 

So let's just keep a count, shall we? Inside the black hole we have the original and the anti-copy, outside we have the copy. You can use the outside copy to make sure the laws of information conservation aren't broken, as I have argued before, but today our focus is on the stuff inside, because we imagine we threw Cooper into the wormhole. The black hole dutifully responds by stimulating a Cooper clone on the outside of the black hole, while the original Cooper, alongside an anti-Cooper, is traveling towards the singularity, which in this case connects this black hole to another one, far far away. Here's a handy diagram to keep track of all the Coopers.
A Cooper is absorbed at horizon \(H_1\) (black), stimulating the emission of a Cooper pair (red)

At this point I feel I should have a paragraph arguing that the vacuum could really produce something as complex as a Cooper-pair (see what I did there?) via stimulated emission. This is not a terrible question, and I'm afraid we may not really be able to answer that. It sure works for elementary particles. It should also work on arbitrary pure quantum states, and even mixed ones. We don't have an apparatus nearby to test this, so for the sake of this blog post I will simply assume that if you can somehow achieve coherence, then yes the vacuum will copy even macroscopic objects. Just take my word for it. I know quantum.

But this interlude has detracted us from Cooper and his anti-twin traveling through the suitably stabilized wormhole, towards the event horizon on the other black hole that the entry portal is connected with, in both an ER and an EPR way. They are now inside of a black hole, yearning to be free, and let's imagine they have the time to ponder their existence. (They are not holding hands, by the way. That would be utter annihilation).

What is it like inside of a black hole, anyway? It's a question I have been asked many times, be it on a Reddit AMA or when giving presentations called "Our Universe" to elementary school children. (Black holes always come up, because, as I like to say, they are like dinosaurs.)

If you can ignore the crushing gravity (which you can if the black hole you inhabit is big enough, and you are far enough away from the center) then the black hole doesn't look so different from the universe you are used to. But there is a very peculiar thing that happens to you. Now, if you happened to inhabit a decent-sized planet like Earth (not a black hole), then if you shoot a small rocket up into the sky, it falls back down somewhere far away. If you can make a much more powerful rocket, it may go up but, when coming back down, will miss the surface of the planet. And keep missing it: it is actually in orbit. But if your rocket is even more powerful, it could leave your planet's gravitational field altogether.

But when you live inside of a black hole, gravity is so strong that no rocket is powerful enough to leave orbit. So you decide to fire a ray-gun instead because light surely goes faster than any rocket, but you then realize that gravity also bends light rays. And because you're in a black hole, the best you can do is that your light ray (after going up and up) basically goes in orbit around the black hole. Just about where Schwarzschild said the event horizon would be.

So you see, when you are inside a black hole, nothing can go out. Everything you shoot out comes back at you (so to speak). There is actually a word for something like that: an area of space-time that you cannot penetrate: it is called a white hole.

So basically, if you are inside a black hole the rest of the universe looks to you like a white hole.

I've mentioned this before in the last paragraph of a somewhat more technical post, so if you want to revisit that, please go ahead, because it has some interesting connections to time-reversal invariance.

But now we understand the Cooper-pair's predicament. No Cooper can escape the black hole, because the horizon they are looking at is a white-hole horizon. Everything is lost, right?

Well, not so fast. In the move Interstellar, some Deus-Ex-Machina extracts Cooper from the black hole, but how could this work in a world where the laws of physics aren't just mere suggestions?

The answer is obvious, isn't it? You hurl anti-Cooper at the White Hole Horizon.

Stimulated emission was able to breach the horizon in the first place, by creating a Cooper-pair. Can anti-Cooper do the same thing from the inside? We now treat the horizon that separates the black hole interior from the outside as a white-hole horizon. And if you do the math right (and I did in the cloning paper) the anti-clone that is hurled into the white-hole horizon will create an anti-anti-clone on the other side. That anti-anti-clone is, of course, a clone: it is Cooper himself, resurrected on the outside. Somewhat stunned, we assume. Here's what this would look like:

The anti-Cooper that was produced at horizon \(H_1\) stimulates the production of a Cooper-anti-Cooper pair as it reflects off of the White Hole Horizon \(H_2\)

This is great: the wormhole resident anti-Cooper has stimulated a Cooper outside horizon \(H_2\), which is how Cooper can see the light of day once more. The anti-Cooper that is stimulated at the same time annihilates the Cooper that was absorbed initially.  But hold on.

We now have two Coopers. The one stimulated at horizon \(H_1\), and the one stimulated at horizon \(H_2\). Doesn't this violate the no-cloning theorem? I'll tell you in a minute that the answer is no, but not before I have to let you in on a secret: it's getting much worse for the Coopers.

You see the anti-Cooper that is reflected at the White Hole horizon \(H_2\)? It is on its way back to Horizon \(H_1\). And guess what happens if anti-Cooper is reflected there? You're right, it stimulates another Cooper-pair!
Reflection at horizon \(H_1\) creates another Cooper-pair

Now there are two Coopers at horizon \(H_1\) and one at horizon \(H_2\)! And, there are now two anti-Coopers on their way to \(H_2\). You can guess what happens from here on out.

Internal reflection of anti-Coopers stimulates Coopers at both horizons, giving rise to a Cooper-Laser

As long as anti-Coopers can reflect without loss at the horizons, the wormhole is going to produce beams of Coopers at both horizons. And if these Coopers are coherent (as we have assumed here), then these are Cooper-Lasers!

Now, to answer all the questions that are racing through your head:

1. No, we are not violating the no-cloning theorem, for the same exact reason why the no-cloning theorem is not violated by standard Lasers. Of course you know that Laser stands for "Light-Amplification via Stimulated Emission of Radiation". So the process that saves information conservation in black holes is the same that is responsible for Lasers, and if we were to connect two black holes via a wormhole, then you have created a black hole laser. The reason why Lasers do not violate the no-cloning theorem is that spontaneous radiation is also formed at the interface (I've omitted those in the depictions here for clarity, but the math has them, as do the diagrams in the original post on information-conservation in black holes). This spontaneous emission is, of course, just Hawking radiation. A tiny amount of this radiation is random Cooper pairs. 

2. The energy to produce all these Coopers is donated by the black holes that form the wormhole in the first place. All this stimulation of Coopers just makes it evaporate much faster than via the ordinary spontaneous emission of pairs (Cooper and non-Cooper alike). For standard Lasers, you actually have to provide this energy by creating an "inversion" that puts the atoms in the gas within the reflecting cavity into excited states. That's not needed for black hole Lasers: they come with their own energy source! 

3. Is there a paper that describes black hole Lasers? The answer is, for once, no. While it is obvious that wormholes that are connected coherently would make a black hole Laser, I also don't think referees would buy it. Such is the state of publishing in this area. After all, the community is still ignoring the fact that stimulated emission gives rise to a non-vanishing information-transmission capacity, seven years after the paper that showed this appeared (and seventeen years after this appeared as preprint). 

4. Could we observe black hole lasers? I don't know the answer to this. Black holes sure do spew out all kinds of stuff, but we mostly assume that these are jets that are fed from an accretion disk. To test whether this is instead stuff that is stimulated from inside a black hole, we would have to test whether there is some coherence in these beams. I suppose this could be done in principle, by taking advantage of the Hanbury-Brown-Twiss effect. I doubt anyone has tried it, or will.