Monthly Archives: August 2016

Upending my equilibrium

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Few settings foster equanimity like Canada’s Banff International Research Station (BIRS). Mountains tower above the center, softened by pines. Mornings have a crispness that would turn air fresheners evergreen with envy. The sky looks designed for a laundry-detergent label.

P1040813

Doesn’t it?

One day into my visit, equilibrium shattered my equanimity.

I was participating in the conference “Beyond i.i.d. in information theory.” What “beyond i.i.d.” means is explained in these articles.  I was to present about resource theories for thermodynamics. Resource theories are simple models developed in quantum information theory. The original thermodynamic resource theory modeled systems that exchange energy and information.

Imagine a quantum computer built from tiny, cold, quantum circuits. An air particle might bounce off the circuit. The bounce can slow the particle down, transferring energy from particle to circuit. The bounce can entangle the particle with the circuit, transferring quantum information from computer to particle.

Suppose that particles bounced off the computer for ages. The computer would thermalize, or reach equilibrium: The computer’s energy would flatline. The computer would reach a state called the canonical ensemble. The canonical ensemble looks like this:  e^{ - \beta H }  / { Z }.

Joe Renes and I had extended these resource theories. Thermodynamic systems can exchange quantities other than energy and information. Imagine white-bean soup cooling on a stovetop. Gas condenses on the pot’s walls, and liquid evaporates. The soup exchanges not only heat, but also particles, with its environment. Imagine letting the soup cool for ages. It would thermalize to the grand canonical ensemble, e^{ - \beta (H - \mu N) } / { Z }. Joe and I had modeled systems that exchange diverse thermodynamic observables.*

What if, fellow beyond-i.i.d.-er Jonathan Oppenheim asked, those observables didn’t commute with each other?

Mathematical objects called operators represent observables. Let \hat{H} represent a system’s energy, and let \hat{N} represent the number of particles in the system. The operators fail to commute if multiplying them in one order differs from multiplying them in the opposite order: \hat{H}  \hat{N}  \neq  \hat{N}  \hat{H}.

Suppose that our quantum circuit has observables represented by noncommuting operators \hat{H} and \hat{N}. The circuit cannot have a well-defined energy and a well-defined particle number simultaneously. Physicists call this inability the Uncertainty Principle. Uncertainty and noncommutation infuse quantum mechanics as a Cashmere GlowTM infuses a Downy fabric softener.

Downy Cashmere Glow 2

Quantum uncertainty and noncommutation.

I glowed at Jonathan: All the coolness in Canada couldn’t have pleased me more than finding someone interested in that question.** Suppose that a quantum system exchanges observables \hat{Q}_1 and \hat{Q}_2 with the environment. Suppose that \hat{Q}_1 and \hat{Q}_2 don’t commute, like components \hat{S}_x and \hat{S}_y of quantum angular momentum. Would the system thermalize? Would the thermal state have the form e^{ \mu_1 \hat{Q}_1 + \mu_2 \hat{Q}_2 } / { Z }? Could we model the system with a resource theory?

Jonathan proposed that we chat.

The chat sucked in beyond-i.i.d.-ers Philippe Faist and Andreas Winter. We debated strategies while walking to dinner. We exchanged results on the conference building’s veranda. We huddled over a breakfast table after colleagues had pushed their chairs back. Information flowed from chalkboard to notebook; energy flowed in the form of coffee; food particles flowed onto the floor as we brushed crumbs from our notebooks.

Coffee, crumbs

Exchanges of energy and particles.

The idea morphed and split. It crystallized months later. We characterized, in three ways, the thermal state of a quantum system that exchanges noncommuting observables with its environment.

First, we generalized the microcanonical ensemble. The microcanonical ensemble is the thermal state of an isolated system. An isolated system exchanges no observables with any other system. The quantum computer and the air molecules can form an isolated system. So can the white-bean soup and its kitchen. Our quantum system and its environment form an isolated system. But they cannot necessarily occupy a microcanonical ensemble, thanks to noncommutation.

We generalized the microcanonical ensemble. The generalization involves approximation, unlikely measurement outcomes, and error tolerances. The microcanonical ensemble has a simple definition—sharp and clean as Banff air. We relaxed the definition to accommodate noncommutation. If the microcanonical ensemble resembles laundry detergent, our generalization resembles fabric softener.

Detergent vs. softener

Suppose that our system and its environment occupy this approximate microcanonical ensemble. Tracing out (mathematically ignoring) the environment yields the system’s thermal state. The thermal state basically has the form we expected, \gamma = e^{ \sum_j  \mu_j \hat{Q}_j } / { Z }.

This exponential state, we argued, follows also from time evolution. The white-bean soup equilibrates upon exchanging heat and particles with the kitchen air for ages. Our quantum system can exchange observables \hat{Q}_j with its environment for ages. The system equilibrates, we argued, to the state \gamma. The argument relies on a quantum-information tool called canonical typicality.

Third, we defined a resource theory for thermodynamic exchanges of noncommuting observables. In a thermodynamic resource theory, the thermal states are the worthless states: From a thermal state, one can’t extract energy usable to lift a weight or to power a laptop. The worthless states, we showed, have the form of \gamma.

Three path lead to the form \gamma of the thermal state of a quantum system that exchanges noncommuting observables with its environment. We published the results this summer.

Not only was Team Banff spilling coffee over \gamma. So were teams at Imperial College London and the University of Bristol. Our conclusions overlap, suggesting that everyone calculated correctly. Our methodologies differ, generating openings for exploration. The mountain passes between our peaks call out for mapping.

So does the path to physical reality. Do these thermal states form in labs? Could they? Cold atoms offer promise for realizations. In addition to experiments and simulations, master equations merit study. Dynamical typicality, Team Banff argued, suggests that \gamma results from equilibration. Master equations model equilibration. Does some Davies-type master equation have \gamma as its fixed point? Email me if you have leads!

Spins

Experimentalists, can you realize the thermal state e^{ \sum_j \mu_j \hat{Q}_j } / Z whose charges \hat{Q}_j don’t commute?

 

A photo of Banff could illustrate Merriam-Webster’s entry for “equanimity.” Banff equanimity deepened our understanding of quantum equilibrium. But we wouldn’t have understood quantum equilibrium if questions hadn’t shattered our tranquility. Give me the disequilibrium of recognizing problems, I pray, and the equilibrium to solve them.

 

*By “observable,” I mean “property that you can measure.”

**Teams at Imperial College London and Bristol asked that question, too. More pleasing than three times the coolness in Canada!

Toward a Coherent US Government Strategy for QIS

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In an upbeat  recent post, Spiros reported some encouraging news about quantum information science from the US National Science and Technology Council. Today I’ll chime in with some further perspective and background.

report-cover-2The Interagency Working Group on Quantum Information Science (IWG on QIS), which began its work in late 2014, was charged “to assess Federal programs in QIS, monitor the state of the field, provide a forum for interagency coordination and collaboration, and engage in strategic planning of Federal QIS activities and investments.”  The IWG recently released a  well-crafted report, Advancing Quantum Information Science: National Challenges and Opportunities. The report recommends that “quantum information science be considered a priority for Federal coordination and investment.”

All the major US government agencies supporting QIS were represented on the IWG, which was co-chaired by officials from DOE, NSF, and NIST:

  • Steve Binkley, who heads the Advanced Scientific Computing Research (ASCR) program in the Department of Energy Office of Science,
  • Denise Caldwell, who directs the Physics Division of the National Science Foundation,
  • Carl Williams, Deputy Director of the Physical Measurement Laboratory at the National Institute for Standards and Technology.

Denise and Carl have been effective supporters of QIS over many years of government service. Steve has recently emerged as another eloquent advocate for the field’s promise and importance.

At our request, the three co-chairs fielded questions about the report, with the understanding that their responses would be broadly disseminated. Their comments reinforced the message of the report — that all cognizant agencies favor a “coherent, all-of-government approach to QIS.”

Science funding in the US differs from elsewhere in the world. QIS is a prime example — for over 20 years, various US government agencies, each with its own mission, goals, and culture, have had a stake in QIS research. By providing more options for supporting innovative ideas, the existence of diverse autonomous funding agencies can be a blessing. But it can also be bewildering for scientists seeking support, and it poses challenges for formulating and executing effective national science policy. It’s significant that many different agencies worked together in the IWG, and were able to align with a shared vision.

“I think that everybody in the group has the same goals,” Denise told us. “The nation has a tremendous opportunity here. This is a terrifically important field for all of us involved, and we all want to see it succeed.” Carl added, “All of us believe that this is an area in which the US must be competitive, it is very important for both scientific and technological reasons … The differences [among agencies] are minor.”

Asked about the timing of the IWG and its report, Carl noted the recent trend toward “emerging niche applications” of QIS such as quantum sensors, and Denise remarked that government agencies are responding to a plea from industry for a cross-disciplinary work force broadly trained in QIS. At the same time, Denise emphasized, the IWG recognizes that “there are still many open basic science questions that are important for this field, and we need to focus investment onto these basic science questions, as well as look at investments or opportunities that lead into the first applications.”

DOE’s FY2017 budget request includes $10M to fund a new QIS research program, coordinated with NIST and NSF. Steve explained the thinking behind that request:  “There are problems in the physical science space, spanned by DOE Office of Science programs, where quantum computation would be a useful a tool. This is the time to start making investments in that area.” Asked about the longer term commitment of DOE to QIS research, Steve was cautious. “What it will grow into over time is hard to tell — we’re right at the beginning.”

What can the rest of us in the QIS community do to amplify the impact of the report? Carl advised: “All of us should continue getting the excitement of the field out there, [and point to] the potential long term payoffs,  whether they be in searches for dark matter or building better clocks or better GPS systems or better sensors. Making everybody aware of all the potential is good for our economy, for our country, and for all of us.”

Taking an even longer view, Denise reminded us that effective advocacy for QIS can get young people “excited about a field they can work in, where they can get jobs, where they can pursue science — that can be critically important.  If we all think back to our own beginning careers, at some point in time we got excited about science. And so whatever one can do to excite the next generation about science and technology, with the hope of bringing them into studying and developing careers in this field, to me this is tremendously valuable. ”

All of us in the quantum information science community owe a debt to the IWG for their hard work and eloquent report, and to the agencies they represent for their vision and support. And we are all fortunate to be participating in the early stages of a new quantum revolution. As the IWG report makes clear, the best is yet to come.

Greg Kuperberg’s calculus problem

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“How good are you at calculus?”

This was the opening sentence of Greg Kuperberg’s Facebook status on July 4th, 2016.

“I have a joint paper (on isoperimetric inequalities in differential geometry) in which we need to know that

(\sin\theta)^3 xy + ((\cos\theta)^3 -3\cos\theta +2) (x+y) - (\sin\theta)^3-6\sin\theta -6\theta + 6\pi \\ \\- 6\arctan(x) +2x/(1+x^2) -6\arctan(y) +2y/(1+y^2)

is non-negative for x and y non-negative and \theta between 0 and \pi. Also, the minimum only occurs for x=y=1/(\tan(\theta/2).”

Let’s take a moment to appreciate the complexity of the mathematical statement above. It is a non-linear inequality in three variables, mixing trigonometry with algebra and throwing in some arc-tangents for good measure. Greg, continued:

“We proved it, but only with the aid of symbolic algebra to factor an algebraic variety into irreducible components. The human part of our proof is also not really a cake walk.

A simpler proof would be way cool.”

I was hooked. The cubic terms looked a little intimidating, but if I converted x and y into \tan(\theta_x) and \tan(\theta_y), respectively, as one of the comments on Facebook promptly suggested, I could at least get rid of the annoying arc-tangents and then calculus and trigonometry would take me the rest of the way. Greg replied to my initial comment outlining a quick route to the proof: “Let me just caution that we found the problem unyielding.” Hmm… Then, Greg revealed that the paper containing the original proof was over three years old (had he been thinking about this since then? that’s what true love must be like.) Titled “The Cartan-Hadamard Conjecture and The Little Prince“, the above inequality makes its appearance as Lemma 7.1 on page 45 (of 63). To quote the paper: “Although the lemma is evident from contour plots, the authors found it surprisingly tricky to prove rigorously.”

As I filled pages of calculations and memorized every trigonometric identity known to man, I realized that Greg was right: the problem was highly intractable. The quick solution that was supposed to take me two to three days turned into two weeks of hell, until I decided to drop the original approach and stick to doing calculus with the known unknowns, x and y. The next week led me to a set of three non-linear equations mixing trigonometric functions with fourth powers of x and y, at which point I thought of giving up. I knew what I needed to do to finish the proof, but it looked freaking insane. Still, like the masochist that I am, I continued calculating away until my brain was mush. And then, yesterday, during a moment of clarity, I decided to go back to one of the three equations and rewrite it in a different way. That is when I noticed the error. I had solved for \cos\theta in terms of x and y, but I had made a mistake that had cost me 10 days of intense work with no end in sight. Once I found the mistake, the whole proof came together within about an hour. At that moment, I felt a mix of happiness (duh), but also sadness, as if someone I had grown fond of no longer had a reason to spend time with me and, at the same time, I had ran out of made-up reasons to hang out with them. But, yeah, I mostly felt happiness.

Greg Kuperberg pondering about the universe of mathematics.

Greg Kuperberg pondering about the universe of mathematics.

Before I present the proof below, I want to take a moment to say a few words about Greg, whom I consider to be the John Preskill of mathematics: a lodestar of sanity in a sea of hyperbole (to paraphrase Scott Aaronson). When I started grad school at UC Davis back in 2003, quantum information theory and quantum computing were becoming “a thing” among some of the top universities around the US. So, I went to several of the mathematics faculty in the department asking if there was a course on quantum information theory I could take. The answer was to “read Nielsen and Chuang and then go talk to Professor Kuperberg”. Being a foolish young man, I skipped the first part and went straight to Greg to ask him to teach me (and four other brave souls) quantum “stuff”. Greg obliged with a course on… quantum probability and quantum groups. Not what I had in mind. This guy was hardcore. Needless to say, the five brave souls taking the class (mostly fourth year graduate students and me, the noob) quickly became three, then two gluttons for punishment (the other masochist became one of my best friends in grad school). I could not drop the class, not because I had asked Greg to do this as a favor to me, but because I knew that I was in the presence of greatness (or maybe it was Stockholm syndrome). My goal then, as an aspiring mathematician, became to one day have a conversation with Greg where, for some brief moment, I would not sound stupid. A man of incredible intelligence, Greg is that rare individual whose character matches his intellect. Much like the anti-heroes portrayed by Humphrey Bogart in Casablanca and the Maltese Falcon, Greg keeps a low-profile, seems almost cynical at times, but in the end, he works harder than everyone else to help those in need. For example, on MathOverflow, a question and answer website for professional mathematicians around the world, Greg is listed as one of the top contributors of all time.

But, back to the problem. The past four weeks thinking about it have oscillated between phases of “this is the most fun I’ve had in years!” to “this is Greg’s way of telling me I should drop math and become a go-go dancer”. Now that the ordeal is over, I can confidently say that the problem is anything but “dull” (which is how Greg felt others on MathOverflow would perceive it, so he never posted it there). In fact, if I ever have to teach Calculus, I will subject my students to the step-by-step proof of this problem. OK, here is the proof. This one is for you Greg. Thanks for being such a great role model. Sorry I didn’t get to tell you until now. And you are right not to offer a “bounty” for the solution. The journey (more like, a trip to Mordor and back) was all the money.

The proof: The first thing to note (and if I had read Greg’s paper earlier than today, I would have known as much weeks ago) is that the following equality holds (which can be verified quickly by differentiating both sides):

4 x - 6\arctan(x) +2x/(1+x^2) = 4 \int_0^x \frac{s^4}{(1+s^2)^2} ds.

Using the above equality (and the equivalent one for y), we get:

F(\theta,x,y) = (\sin\theta)^3 xy + ((\cos\theta)^3 -3\cos\theta -2) (x+y) - (\sin\theta)^3-6\sin\theta -6\theta + 6\pi \\ \\4 \int_0^x \frac{s^4}{(1+s^2)^2} ds+4 \int_0^y \frac{s^4}{(1+s^2)^2} ds.

Now comes the fun part. We differentiate with respect to \theta, x and y, and set to zero to find all the maxima and minima of F(\theta,x,y) (though we are only interested in the global minimum, which is supposed to be at x=y=\tan^{-1}(\theta/2)). Some high-school level calculus yields:

\partial_\theta F(\theta,x,y) = 0 \implies \sin^2(\theta) (\cos(\theta) xy + \sin(\theta)(x+y)) = \\ \\ 2 (1+\cos(\theta))+\sin^2(\theta)\cos(\theta).

At this point, the most well-known trigonometric identity of all time, \sin^2(\theta)+\cos^2(\theta)=1, can be used to show that the right-hand-side can be re-written as:

2(1+\cos(\theta))+\sin^2(\theta)\cos(\theta) = \sin^2(\theta) (\cos\theta \tan^{-2}(\theta/2) + 2\sin\theta \tan^{-1}(\theta/2)),

where I used (my now favorite) trigonometric identity: \tan^{-1}(\theta/2) = (1+\cos\theta)/\sin(\theta) (note to the reader: \tan^{-1}(\theta) = \cot(\theta)). Putting it all together, we now have the very suggestive condition:

\sin^2(\theta) (\cos(\theta) (xy-\tan^{-2}(\theta/2)) + \sin(\theta)(x+y-2\tan^{-1}(\theta/2))) = 0,

noting that, despite appearances, \theta = 0 is not a solution (as can be checked from the original form of this equality, unless x and y are infinite, in which case the expression is clearly non-negative, as we show towards the end of this post). This leaves us with \theta = \pi and

\cos(\theta) (\tan^{-2}(\theta/2)-xy) = \sin(\theta)(x+y-2\tan^{-1}(\theta/2)),

as candidates for where the minimum may be. A quick check shows that:

F(\pi,x,y) = 4 \int_0^x \frac{s^4}{(1+s^2)^2} ds+4 \int_0^y \frac{s^4}{(1+s^2)^2} ds \ge 0,

since x and y are non-negative. The following obvious substitution becomes our greatest ally for the rest of the proof:

x= \alpha \tan^{-1}(\theta/2), \, y = \beta \tan^{-1}(\theta/2).

Substituting the above in the remaining condition for \partial_\theta F(\theta,x,y) = 0, and using again that \tan^{-1}(\theta/2) = (1+\cos\theta)/\sin\theta, we get:

\cos\theta (1-\alpha\beta) = (1-\cos\theta) ((\alpha-1) + (\beta-1)),

which can be further simplified to (if you are paying attention to minus signs and don’t waste a week on a wild-goose chase like I did):

\cos\theta = \frac{1}{1-\beta}+\frac{1}{1-\alpha}.

As Greg loves to say, we are finally cooking with gas. Note that the expression is symmetric in \alpha and \beta, which should be obvious from the symmetry of F(\theta,x,y) in x and y. That observation will come in handy when we take derivatives with respect to x and y now. Factoring (\cos\theta)^3 -3\cos\theta -2 = - (1+\cos\theta)^2(2-\cos\theta), we get:

\partial_x F(\theta,x,y) = 0 \implies \sin^3(\theta) y + 4\frac{x^4}{(1+x^2)^2} = (1+\cos\theta)^2 + \sin^2\theta (1+\cos\theta).

Substituting x and y with \alpha \tan^{-1}(\theta/2), \beta \tan^{-1}(\theta/2), respectively and using the identities \tan^{-1}(\theta/2) = (1+\cos\theta)/\sin\theta and \tan^{-2}(\theta/2) = (1+\cos\theta)/(1-\cos\theta), the above expression simplifies significantly to the following expression:

4\alpha^4 =\left((\alpha^2-1)\cos\theta+\alpha^2+1\right)^2 \left(1 + (1-\beta)(1-\cos\theta)\right).

Using \cos\theta = \frac{1}{1-\beta}+\frac{1}{1-\alpha}, which we derived earlier by looking at the extrema of F(\theta,x,y) with respect to \theta, and noting that the global minimum would have to be an extremum with respect to all three variables, we get:

4\alpha^4 (1-\beta) = \alpha (\alpha-1) (1+\alpha + \alpha(1-\beta))^2,

where we used 1 + (1-\beta)(1-\cos\theta) = \alpha (1-\beta) (\alpha-1)^{-1} and

(\alpha^2-1)\cos\theta+\alpha^2+1 = (\alpha+1)((\alpha-1)\cos\theta+1)+\alpha(\alpha-1) = \\ (\alpha-1)(1-\beta)^{-1} (2\alpha + 1-\alpha\beta).

We may assume, without loss of generality, that x \ge y. If \alpha = 0, then \alpha = \beta = 0, which leads to the contradiction \cos\theta = 2, unless the other condition, \theta = \pi, holds, which leads to F(\pi,0,0) = 0. Dividing through by \alpha and re-writing 4\alpha^3(1-\beta) = 4\alpha(1+\alpha)(\alpha-1)(1-\beta) + 4\alpha(1-\beta), yields:

4\alpha (1-\beta) = (\alpha-1) (1+\alpha - \alpha(1-\beta))^2 = (\alpha-1)(1+\alpha\beta)^2,

which can be further modified to:

4\alpha +(1-\alpha\beta)^2 = \alpha (1+\alpha\beta)^2,

and, similarly for \beta (due to symmetry):

4\beta +(1-\alpha\beta)^2 = \beta (1+\alpha\beta)^2.

Subtracting the two equations from each other, we get:

4(\alpha-\beta) = (\alpha-\beta)(1+\alpha\beta)^2,

which implies that \alpha = \beta and/or \alpha\beta =1. The first leads to 4\alpha (1-\alpha) = (\alpha-1)(1+\alpha^2)^2, which immediately implies \alpha = 1 = \beta (since the left and right side of the equality have opposite signs otherwise). The second one implies that either \alpha+\beta =2, or \cos\theta =1, which follows from the earlier equation \cos\theta (1-\alpha\beta) = (1-\cos\theta) ((\alpha-1) + (\beta-1)). If \alpha+\beta =2 and 1 = \alpha\beta, it is easy to see that \alpha=\beta=1 is the only solution by expanding (\sqrt{\alpha}-\sqrt{\beta})^2=0. If, on the other hand, \cos\theta = 1, then looking at the original form of F(\theta,x,y), we see that F(0,x,y) = 6\pi - 6\arctan(x) +2x/(1+x^2) -6\arctan(y) +2y/(1+y^2) \ge 0, since x,y \ge 0 \implies \arctan(x)+\arctan(y) \le \pi.

And that concludes the proof, since the only cases for which all three conditions are met lead to \alpha = \beta = 1 and, hence, x=y=\tan^{-1}(\theta/2). The minimum of F(\theta, x,y) at these values is always zero. That’s right, all this work to end up with “nothing”. But, at least, the last four weeks have been anything but dull.

Update: Greg offered Lemma 7.4 from the same paper as another challenge (the sines, cosines and tangents are now transformed into hyperbolic trigonometric functions, with a few other changes, mostly in signs, thrown in there). This is a more hardcore-looking inequality, but the proof turns out to follow the steps of Lemma 7.1 almost identically. In particular, all the conditions for extrema are exactly the same, with the only difference being that cosine becomes hyperbolic cosine. It is an awesome exercise in calculus to check this for yourself. Do it. Unless you have something better to do.