The third and final day of my Topics in QM class was harder and more advanced, and had the distinct disadvantage of not being based on a particular paper. The goal was to explain electric charge from a QM/QFT point of view; to get there, we had to discuss momentum, wave functions, etc. To wit:
How can you describe position of a particle? Need three numbers. What three numbers? Depend on my location. So particle P knows three numbers, that depend on where we're standing.
How? If I move by (column) vector [\del x, \del y, \del z], P's numbers change by [x,y,z]\mapsto [x+\del x, y+\del y, z+\del z].
This is called an action: there's an \R^3 of translations, which "act" on the \R^3 of possible states of the particle.
Now let's be mathematicians, and generalize this. But let's start wih a very simple case: let's say that P only has _one_ number, but that that number depends on where we're standing when we measure it. I.e. we have a fn \Phi: locations \to numbers.
Ok, now let's say that we move by some vector \vec{\del} = [\del x,\del y,\del z]. How does \Phi change? Well, by some function:
\Phi(location + \vec{\del}) = \Phi(loc) + f(loc,\vec{\del})
So, laws of nature seem to not depend on location. Of course, \Phi depends explicitly on location. So let's demand the next best thing: let's say that \Phi is symmetrical enough that f does not depend on location:
\Phi(location + \vec{\del}) = \Phi(loc) + f(\vec{\del})
Well, what if we move and move again?
\Phi(loc) + f(\vec\del_1 + \vec\del_2) = \Phi(loc + \vec\del_1 + \vec\del_2) = \Phi(loc + \vec\del_1) + f(\vec\del_2) = \Phi(loc) + f(\vec\del_1) + f(\vec\del_2).
So f is linear! (One student correctly asked about multiplication by constants; our functions will satisfy some smoothness condition (for \R, suffices to by cont's; for \C, suffices to be \C-differentiable), and then this \Z-linearity is enough to guarantee linearity.)
By linear algebra, a linear function f:\R^3 \to \R is exactly a row vector <f_x, f_y, f_z> s.t.
<f_x, f_y, f_z> [\del_x, \del_y, \del_z] = f(\vec\del)
(I'm writing <> for row vectors, [] for columns. This then is the usual matrix product.) Why? You can read off f_x, for instance, by stepping in one unit in the x-direction, and seeing how much \Phi changes; I'll leave it as an exercise to prove that the three numbers f_x,f_y,f_z exactly give you your function.
In QM, we call the vector <f_x, f_y, f_z> the momentum of the particle.
Let me step back and say what we've done. CM is problematic in a really important way: if a particle is over there,how can I measure it, if everything is local? So, what we've said, is that no, the "particle" is _everywhere_, and remembers a number, which we can only measure _here_.
Now, what if particle 1 turns into particle 2? Say 2 also knows a number \Phi_2, so \Phi_1(loc)\to\Phi_2(loc) as 1\to 2.
And let's say that this is as simple as possible: in particular,
\Phi_2(loc) = \Phi_1(loc) + \varphi
where \varphi does not depend on location --- i.e. is a constant.
Then
\Phi_2(loc) + f_2(\del) = \Phi_2(loc+\del) = \Phi_1(loc+\del) + \varphi = \Phi_1(loc) + f_1(\del) + \varphi
So f_1 = f_2! Conservation of momentum!
Now, we're mathematicians. Let's generalize.
Does it matter that \Phi: \to \R? What if, for instance, \Phi: locations \to S = circle? Then f(\del) is an angle change. We still demand the same linearity: f(\del_1+\del_2) = f(\del_1) + f(\del_2). By taking \del very small, we can assure that f does not go all the way around the circle, in which case it still acts like a number, and it turns out that f is still given exactly by a triple of numbers. I'll leave it as an exercise, for those who know some calculus, to make this rigorous.
By the same argument, of course, this triple, which still deserves the name momentum, is conserved.
But, other things can act on S. See, only \R (and its products, like \R^3) can act on \R, because \R is too long for something compact to act on it. But actions on S can also include S actions! So let's just imagine that, if in addition to moving in three-space, we could move in some circular direction. Then "location" includes an angle-valued component as well as the three large components: We can move by some \del(angle) = \theta, and then
\Phi \to \Phi + c(\theta)
Just as before, c should be linear:
c(\theta_1 + \theta_2) = c(\theta_1) + c(\theta_2)
But also, \theta is an angle, so \theta = \theta + 2\pi, so
c(\theta + 2\pi) = c(\theta)
And c(\theta) is also an angle, so
c(\theta) = c(\theta) + 2\pi.
So, exercise: the only linear maps satisfying this are
c(\theta) = k\theta, for k\in\Z, i.e. for k an integer!
Intuition: this is just topology! How can you wrap a circle around a circle? (Negative numbers correspond to going the wrong direction.)
This number k is called the charge of the particle! Charge is conserved exactly as it was for momentum, and this explains why charge is quantized.
Or, rather, it explains it _if_ space is (3 dimsensions) \times circle. Can it be? Well.... yes, if circle is very small, so small that atoms are bigger than size of circle.
This is called the Kaluza-Klein picture: (gravity in \R^3\times S) = (gravity in \R^3) \tensor (EM in \R^3).
BTW, in QM, physicists prefer everything to be multiplicative, not additive. So, fine: we can use \Psi(loc) = e^{\Phi(loc)}, turning all our "plus"es into "times"es. But we want \Phi to take values in a circle. In \C, we can do this via
\Psi(loc) = e^{i \Phi(loc)}
because e^{i2\pi} = 1. Then
\Psi(loc + \del + \theta) = e^{i\Phi(loc)} e^{i p.\del} e^{i k \theta}
Where p is the (row) vector of momentum, so p.\del is the dot-product, and k is the electric charge. The nice thing about using complex-valued \Psi is it lets us add them, etc., do QM, still w/ this rule. And assuming moving by (\del,\theta) is linear and that space really does have these symmetries, we get conservation of momentum, charge, etc.
29 July 2007
28 July 2007
Another day of QM Topics: A Finite Hidden-Variables Model
This material is from R. Spekkens In defense of the epistemic view of quantum states: a toy theory.
Heisenberg says you can't have complete knowledge of position and momentum. This is a theorem in traditional QM "wavefunction" model. We will take this as axiomatic: you can't have complete info.
Important conceptual point: At any given time, system is in definite classical state, which I will call its ontic state. _But_, this theory will also explicitly refer to our _knowledge_ of the state of the system --- I'll call this the epistemic state. In QM (and (SM) epistemic states are probability distributions. We will be more single-minded.
Let's say we have a (finite) set S of ontic states, and our epistemic state is simply a subset K of S. How much knowledge do we have? I.e. what is the "entropy" in the state K? Will turn out to relate to log_2 K.
In fact, let's define this intrinsically. We work only with yes/no questions. How many questions do I need to ask of a set S to resolve exactly the ontic state? Log_2 S. A basis of questions will be a set of questions with minimum size that's enough to resolve the ontic state exactly. Called "dim of system". You can show that the question is in a basis iff it splits S in half.
Then some questions can be answered by epistemic state, other's can't. Examples.
Knowledge of K is
max_{bases} \{ number of questions answerable \} = log_2 S - log_2 K
Lack of knowledge = log_2 K.
We demand physical law:
E.g. smallest system: \{ 1, 2, 3, 4 \}. Called an "atom". What are valid epistemic states? Cannot have epistemic state of size 3, as no question can resolve this.
What are valid physical fns? Cannot be many-to-one, by uncertainty.
What happens if you measure "Is it 12 or 34?" when it's in epistemic state 23? Say it's really 2. Then get answer 12. Do you know it's (still) in state 2? No! It we ask question again, get same answer. So def. in 1 or 2. But measuring requires interaction, so could have turned it into a 1. Called "collapsing wave fn." Now let's measure 23 v.s. 14. Could get 14. So measurements do not commute.
How about systems with two atoms? Rule: our knowledge of any subsystem should also respect axiom. Examples of valid epistemic states: tensor products, pure entangled states.
Let's do teleportation. (Act out with students as atoms.) Let's say there's an atom --- you --- that I want to transmit. Let's take two others, and collapse them into an entangled state, by measuring (in two yes/no questions) "is it I, II, III, or IV?"
atom B
|
v
4 II III IV I
3 III IV I II
2 IV I II III
1 I II III IV
1 2 3 4 <- atom A
[Question: how do we know we can measure like this? A priori, "is it I,II" need not commute with "is it I,III". But let's demand that states of maximal knowledge consistent with axioms are possible.]
Now I give one atom (say A) to my BFF, who leaves for Neptune. (We know what the entangled state it.)
Ok, now I ask unknown state C and may half B of entangled state the same question: get answer. (Remember, after measurement you're allowed to (jointly) pick and other ontic state in the epistemic state.) Important point: let's demand that physics be local, so this measurement did not affect BFF's particle.
Now I can transmit this classical info to BFF. BFF, what question did you want to know of my orig particle? Can translate now into question for your particle. Try! Compare with original.
Exercises:
Important point: QM violates Bell's inequality. As a local hidden-variables theory, our model always satisfies Bell. So different. But similar.
Heisenberg says you can't have complete knowledge of position and momentum. This is a theorem in traditional QM "wavefunction" model. We will take this as axiomatic: you can't have complete info.
Important conceptual point: At any given time, system is in definite classical state, which I will call its ontic state. _But_, this theory will also explicitly refer to our _knowledge_ of the state of the system --- I'll call this the epistemic state. In QM (and (SM) epistemic states are probability distributions. We will be more single-minded.
Let's say we have a (finite) set S of ontic states, and our epistemic state is simply a subset K of S. How much knowledge do we have? I.e. what is the "entropy" in the state K? Will turn out to relate to log_2 K.
In fact, let's define this intrinsically. We work only with yes/no questions. How many questions do I need to ask of a set S to resolve exactly the ontic state? Log_2 S. A basis of questions will be a set of questions with minimum size that's enough to resolve the ontic state exactly. Called "dim of system". You can show that the question is in a basis iff it splits S in half.
Then some questions can be answered by epistemic state, other's can't. Examples.
Knowledge of K is
max_{bases} \{ number of questions answerable \} = log_2 S - log_2 K
Lack of knowledge = log_2 K.
We demand physical law:
- systems have size 2^n (because we can only measure w/ yes/nos).
- epistemic states of maximal knowledge have knowledge = lack of knowledge.
E.g. smallest system: \{ 1, 2, 3, 4 \}. Called an "atom". What are valid epistemic states? Cannot have epistemic state of size 3, as no question can resolve this.
What are valid physical fns? Cannot be many-to-one, by uncertainty.
What happens if you measure "Is it 12 or 34?" when it's in epistemic state 23? Say it's really 2. Then get answer 12. Do you know it's (still) in state 2? No! It we ask question again, get same answer. So def. in 1 or 2. But measuring requires interaction, so could have turned it into a 1. Called "collapsing wave fn." Now let's measure 23 v.s. 14. Could get 14. So measurements do not commute.
How about systems with two atoms? Rule: our knowledge of any subsystem should also respect axiom. Examples of valid epistemic states: tensor products, pure entangled states.
Let's do teleportation. (Act out with students as atoms.) Let's say there's an atom --- you --- that I want to transmit. Let's take two others, and collapse them into an entangled state, by measuring (in two yes/no questions) "is it I, II, III, or IV?"
atom B
|
v
4 II III IV I
3 III IV I II
2 IV I II III
1 I II III IV
1 2 3 4 <- atom A
[Question: how do we know we can measure like this? A priori, "is it I,II" need not commute with "is it I,III". But let's demand that states of maximal knowledge consistent with axioms are possible.]
Now I give one atom (say A) to my BFF, who leaves for Neptune. (We know what the entangled state it.)
Ok, now I ask unknown state C and may half B of entangled state the same question: get answer. (Remember, after measurement you're allowed to (jointly) pick and other ontic state in the epistemic state.) Important point: let's demand that physics be local, so this measurement did not affect BFF's particle.
Now I can transmit this classical info to BFF. BFF, what question did you want to know of my orig particle? Can translate now into question for your particle. Try! Compare with original.
Exercises:
- cloning violates uncertainty.
- study triplets of atoms --- prove the impossibility of triply-entangled state.
- "dense coding". A priori, although an atom secretly knows two bits of information, it seems impossible to actually saturate the channel: even if I can give an atom to my BFF, I can't use it to transmit more than 1 bit, right? Because I have to tell him some particular direction along which to measure the
spinepistemic state. But if we start with an entangled pair, he can keep one half, and then without touching his atom, I can manipulate my atom and hand it to him in such a way that I can transmit two bits. Work out the details of this.
Important point: QM violates Bell's inequality. As a local hidden-variables theory, our model always satisfies Bell. So different. But similar.
27 July 2007
Quantum Mechanics in Pictures, or How I spent my summer vacation
After a two-week camping trip, I visited Canada/USA Mathcamp for one week. While there, I studiously avoided any Harry Potter spoilers (upon returning home, I dutifully took a one-day break from the rest of life to read the book), offered an evening activity focused on mind-body-movement-partnering exercises called "Learn to Fly", and hung out with some of the best people on the planet. Oh, and I taught a three-day course on topics in quantum mechanics, with a fun structure: I set myself the challenge of preparing three one-day talks, which, while all part of the same large story, could be understood and appreciated entirely separately. I more-or-less succeeded, although certainly I would edit the talks before giving them again. Partly to preserve them for posterity, and partly so that I could share them with you all, I thought it might be nice to post my lecture notes here; the first is in this post, and the later two will be in posts of their own.
Day One was on "Quantum Mechanics in Pictures", and is based largely on B. Coecke's paper Kindergarten Quantum Mechanics.
We begin by setting up a graphical notation with which to describe science. We have systems = particles, which we draw by vertical lines with upward-pointing arrows, and which we label by the kind of system=particle that the line represents. States, costates, operations, and numbers are, respectively, downward-pointing triangles with single lines coming out the top, upwards-pointing triangles with single lines coming out the bottom, squares with a line coming out each end, and diamonds with no lines. By "line", I always mean labeled directed line i.e. system.
"Numbers" live in some system --- for CM, numbers are 0 or 1; for SM, probabilities; for QM, numbers are complex ---, and we can multiply them by putting them next to each other. We can arrange our system in some state; a "costate" is really a measurement, asking "how much is this system having this property?". Operations are functions: we plug in a state of system A, and get out a state of system B.
Given two systems, we can combine them ("tensor product") by drawing their lines next to each other, and interpreting it as a single line --- like putting two pipes together into one big tube. This corresponds to having a joint system of A and B. In general, we can have "operations" from any number of systems to any other. We can also hook these objects together to form composite objects. Lines are allowed to cross each other --- we don't care about whether on top or on bottom --- and the usual laws hold. In general, we can tell what kind of object we have (A-state, function from A to B, etc.) just by inspecting the incoming and outgoing lines. E.g. a costate is a function from A to "nothing", i.e. to numbers. If I were speaking in abstract language, I'd say that we are working in a symmetric braided monoidal category.
This will be Axiom 0: that we have these types of objects and can combine them in these natural ways. Axiom 0 is required of any type of "physics": any scientific theory should accommodate systems, states, joint systems, manipulations, measurements, etc.
Quantum Mechanics admits two kinds of "turning upside down". The first kind gives us Axiom 1: Given any X: A -> B, we can form its adjoint X† : B -> A. In general, I will be sloppy and drop the dagger when I'm taking the adjoint a state (to get a costate). The adjoint of a number is called its "complex conjugate". Given a state S, its adjoint is the measurement "how S-like is it? We can define the dot product by drawing two systems meeting at a dot, meaning "turn the second one into its adjoint and compose". Taking the adjoint is some kind of "time reversal".
Axiom 2: for any physical evolution --- i.e. time running forward --- i.e. anything we can actually do _in a lab_ --- X:A->B, we have X^{-1} = X†. This is called "unitarity". Pictures: this says that time evolves deterministically, in such a way that relative similarity does not change. Note: This is not true in the physics of _math_. We could make a theory of "functions"; these can be many-to-one, so definitely not unitary.
There's another kind of turning upside down. Adjoints reversed operators, etc., but not spces. Axiom 3 pt. 1: to every system A, a "dual" system A*, which we still label A but draw the arrow pointing down, s.t. A** = A. Axiom 3 pt. 2: Existence of Bell (co)states; pictures; satifying rule.
Then we have a "teleportation protocol": I have a state. I want to transmit all info in state to you. But there's no reason to believe that I can _measure_ all info (c.f. Heisenberg). One protocol: just carry particle to you. Bad: maybe you're in China, and we're just talking by phone. Better: before you leave, we make entangled pair (BFFs sharing dollare bill). Then I _measure_ "in what way is my state like LH part of Bell state" and transmit that classical info. Advertisement: we will explaion this in more detail tomorrow.
_But_: no cloning theorem. (Star Trek). A "cloning protocol" would be a physicial thing we could do X(s,e) = s,s for a fixed e and any s. But then by unitarity and gluing we get a contradiction, unless only numbers are 0 and 1. E.g. CM. If more numebrs are available, no.
Axiom 4: For any two states a and b, can get from one to another "continuously", in the sense that it's cont's when you dot with some (co)state c. This is true in SM. There exist "mixed states" where you don't have complete knowledge. SM allows the following cloning: measure state exactly, then clone it. But this _does not_ clone the "mixed state" unless I perversely then _forget_ the new info. In QM, axiom is amended: Axiom 4': Can cont'sly transmorm from any _pure_ state to any other _pure_ state going through only _pure_ staets.
"Pure" state can be well-defined; intuition is "state of maximum knowledge". We will give example of this tomorrow. It is a theorem that SM (= CM + mixed states = Axioms 0 through 3 plus technicalities) _plus_ Axiom 4' is equivalent to QM (in particular, forces complex numbers), proved in L. Hardy Quantum Theory From Five Reasonable Axioms.
Day One was on "Quantum Mechanics in Pictures", and is based largely on B. Coecke's paper Kindergarten Quantum Mechanics.
We begin by setting up a graphical notation with which to describe science. We have systems = particles, which we draw by vertical lines with upward-pointing arrows, and which we label by the kind of system=particle that the line represents. States, costates, operations, and numbers are, respectively, downward-pointing triangles with single lines coming out the top, upwards-pointing triangles with single lines coming out the bottom, squares with a line coming out each end, and diamonds with no lines. By "line", I always mean labeled directed line i.e. system.
"Numbers" live in some system --- for CM, numbers are 0 or 1; for SM, probabilities; for QM, numbers are complex ---, and we can multiply them by putting them next to each other. We can arrange our system in some state; a "costate" is really a measurement, asking "how much is this system having this property?". Operations are functions: we plug in a state of system A, and get out a state of system B.
Given two systems, we can combine them ("tensor product") by drawing their lines next to each other, and interpreting it as a single line --- like putting two pipes together into one big tube. This corresponds to having a joint system of A and B. In general, we can have "operations" from any number of systems to any other. We can also hook these objects together to form composite objects. Lines are allowed to cross each other --- we don't care about whether on top or on bottom --- and the usual laws hold. In general, we can tell what kind of object we have (A-state, function from A to B, etc.) just by inspecting the incoming and outgoing lines. E.g. a costate is a function from A to "nothing", i.e. to numbers. If I were speaking in abstract language, I'd say that we are working in a symmetric braided monoidal category.
This will be Axiom 0: that we have these types of objects and can combine them in these natural ways. Axiom 0 is required of any type of "physics": any scientific theory should accommodate systems, states, joint systems, manipulations, measurements, etc.
Quantum Mechanics admits two kinds of "turning upside down". The first kind gives us Axiom 1: Given any X: A -> B, we can form its adjoint X† : B -> A. In general, I will be sloppy and drop the dagger when I'm taking the adjoint a state (to get a costate). The adjoint of a number is called its "complex conjugate". Given a state S, its adjoint is the measurement "how S-like is it? We can define the dot product by drawing two systems meeting at a dot, meaning "turn the second one into its adjoint and compose". Taking the adjoint is some kind of "time reversal".
Axiom 2: for any physical evolution --- i.e. time running forward --- i.e. anything we can actually do _in a lab_ --- X:A->B, we have X^{-1} = X†. This is called "unitarity". Pictures: this says that time evolves deterministically, in such a way that relative similarity does not change. Note: This is not true in the physics of _math_. We could make a theory of "functions"; these can be many-to-one, so definitely not unitary.
There's another kind of turning upside down. Adjoints reversed operators, etc., but not spces. Axiom 3 pt. 1: to every system A, a "dual" system A*, which we still label A but draw the arrow pointing down, s.t. A** = A. Axiom 3 pt. 2: Existence of Bell (co)states; pictures; satifying rule.
Then we have a "teleportation protocol": I have a state. I want to transmit all info in state to you. But there's no reason to believe that I can _measure_ all info (c.f. Heisenberg). One protocol: just carry particle to you. Bad: maybe you're in China, and we're just talking by phone. Better: before you leave, we make entangled pair (BFFs sharing dollare bill). Then I _measure_ "in what way is my state like LH part of Bell state" and transmit that classical info. Advertisement: we will explaion this in more detail tomorrow.
_But_: no cloning theorem. (Star Trek). A "cloning protocol" would be a physicial thing we could do X(s,e) = s,s for a fixed e and any s. But then by unitarity and gluing we get a contradiction, unless only numbers are 0 and 1. E.g. CM. If more numebrs are available, no.
Axiom 4: For any two states a and b, can get from one to another "continuously", in the sense that it's cont's when you dot with some (co)state c. This is true in SM. There exist "mixed states" where you don't have complete knowledge. SM allows the following cloning: measure state exactly, then clone it. But this _does not_ clone the "mixed state" unless I perversely then _forget_ the new info. In QM, axiom is amended: Axiom 4': Can cont'sly transmorm from any _pure_ state to any other _pure_ state going through only _pure_ staets.
"Pure" state can be well-defined; intuition is "state of maximum knowledge". We will give example of this tomorrow. It is a theorem that SM (= CM + mixed states = Axioms 0 through 3 plus technicalities) _plus_ Axiom 4' is equivalent to QM (in particular, forces complex numbers), proved in L. Hardy Quantum Theory From Five Reasonable Axioms.
18 July 2007
The problem of time
One of my campers asked me about the problem of Time: why is it different from space, why can't there be hidden time dimensions.
Here's my response; please comment with corrections/additions:
Here's my response; please comment with corrections/additions:
The problem of Time is a very difficult one, and you've asked some deep questions, many of which do not yet have good answers. I'll answer bits of them, but my answers will be necessarily incomplete.
First of all, why aren't there hidden time dimensions? In fact, lots of physicists have asked about this, and a priori there could be hidden time dimensions. In special/general relativity, the only difference between space and time dimensions is a minus sign somewhere; you could make a manifold with two time dimensions and d spatial dimensions if you want. Then you can still do general (and special) relativity just fine --- the laws are perfectly consistent with any (nonnegative) numbers of time and space dimensions.
But problems arise in quantum field theory. Many QFTs have what are called "anomalies": while the original classical laws are symmetric in a certain way, it becomes impossible to quantize and keep the symmetry. So, for instance, without a Higgs field, it's impossible to quantize a chiral gauge theory with massive fermions (most of the universe is a non-chiral gauge theory, which can be quantized with massive fermions, but the Weak Force is chiral, meaning that if you look at it in a mirror, you get different effects; this is why physicists believe that there are Higgs bosons). There's a very important example in string theory: the _only_ anomaly-free bosonic string theories exist in 26 dimensions (25 space, one time), and the _only_ anomaly-free superstring theories are in 10 (9+1) dimensions.
In fact, for an anomaly-free QFT --- i.e. a QFT that respects special relativity --- you cannot have both at least two space and at least two time dimensions. You could give up special relativity, but then you'd be giving up the entire ballgame: the whole idea of QFT is to respect a symmetry that definitely appears in nature.
It seems that abstractifying (not a word) time is different from abstractifying space. Geometry takes properties of real space, and idealizes them into abstract spaces; of course, the whole point of special/general relativity is to take properties of spacetime and idealize them into abstract spacetimes.
What about phase space? Don't we normally abstractify space into phase space, and import time in the most stupid "\times\R" way? This is where the meat of the matter is. In some sense, I'd argue that time is inherently _more_ abstract than space. "Time" is a measurement of how things _change_ --- a function should be thought of as a one-dimensions object (with a "front" or "in" end, and "back" or "out" end; functions can be glued together so long as you respect this orientation) pointing in the "time" direction. What's going on with things like "phase space" is that phase space exactly describes the _state_ of a particle, and you need a time dimension for the state to change. Thus Hamiltonian formalism in classical or quantum mechanics.
Incidentally, Lagrangian formalism is manifestly special-relativistic invariant --- in Lagrange (equals "least action principle"), the law is that particles evolve in such a way as to minimize a certain total path length. This is perfectly easily expressible in a spacetime, and that expression is probably more natural than in just space, because the path the particle takes is through spacetime. Lagrangian formalism is what most modern QFT folks use.
One other point is worth making: Roger Penrose, one of the most brilliant (and crackpottish) physicists alive today, invented a physics called "Twistor Theory". For twistor theory, he has to replace SO(3,1), the symmetry group of (our) three space dimensions and one time dimension, with SO(4,2), which most naturally describes a universe with two time dimensions. Twistor theory, which takes light rays as basic rather than spacetime points, has not been widely adopted; it's not clear to me what the philosophical corollaries of this change are.
Sorry I don't know any references. Hope this gives you some more ideas where to look.
30 June 2007
Marie Antoinette was right!
Or, rather, Marie-Therese was, when she encouraged the commoners to eat rich, expensive desserts. I recently procured a copy of Rose Levy Beranbaum's The Cake Bible, a book I highly recommend. Ultimately, I hope to try every cake she suggests; over the last two weeks, I began that project with the first five of her recipes, in order, and also a few of her cakes later on. We've had almost one cake a night — no mean feat for a family of five.
It was with great pleasure, then, that I came upon an easy vegan gem. Beranbaum loves her butter, and her ingredient-sorted categories include All Egg Yolks, All Egg Whites, and Safe For Passover, but not, for instance, Low Fat (low saturated fat, yes, e.g. meringue). So when I saw a recipe from the '50s — she says it's an original, although I will paraphrase — that was safe for my godmother's birthday (my godmother tries to avoid dairy), I knew I had to try it. It turned out fantastic.
Chocolate Cake
Preheat oven 350°F.
In a medium bowl, whisk together
In bowl of standing mixer with padle (or by hand in a large bowl, I guess), combine dry ingredients:
Prepare two 8-inch cake pans, by (spray) greasing, and then lining the bottoms with circles of baker's parchment, and greasing again. Scrape batter (very liquidy) into prepared pans; it should only fill about 1/3 full (cake expands significantly). Bake 20 to 25 minutes.
Having cherries on hand, and having made a cake with brandied cherries earlier last week, I combined some cherry preserves and the brandy-cherry-syrup in a small frypan over low heat to dissolve. After turning out the first cake, I poured a thin layer of glaze over it, so that I could glue the top layer on. I then poured more glaze on the top of the cake, so that little bits rolled down the sides, and lined the outer edge of the cake top with cherry halves.
* Note: I used only 1/3 cup mayonnaise, since I misread the recipe. It worked fine. More will only make it richer. And I'm sure that vegan mayonnaise substitutes, such as Nayonnaise would work just as well. I would even be interested in trying this cake with Wildwood Garlic Aioli — I've had a mayonnaise cake made with garlic aioli before (a plain yellow cake; the cook also substituted rice flour for cake flour to make it gluten free, and it still turned out good), and Khymos suggests that chocolate and garlic might go well together with the addition of coffee. This recipe would definitely be good with some added instant espresso (I'd put in between a tsp and a Tbsp in the first step with the cocoa powder).
It was with great pleasure, then, that I came upon an easy vegan gem. Beranbaum loves her butter, and her ingredient-sorted categories include All Egg Yolks, All Egg Whites, and Safe For Passover, but not, for instance, Low Fat (low saturated fat, yes, e.g. meringue). So when I saw a recipe from the '50s — she says it's an original, although I will paraphrase — that was safe for my godmother's birthday (my godmother tries to avoid dairy), I knew I had to try it. It turned out fantastic.
Chocolate Cake
Preheat oven 350°F.
In a medium bowl, whisk together
- 1/3 cup unsweetened cocoa powder
- 1 cup boiling water
- 1 tsp vanilla
- 3/4 cup mayonnaise*.
In bowl of standing mixer with padle (or by hand in a large bowl, I guess), combine dry ingredients:
- 2 cups (sifted) cake flour
- 1 cup sugar
- 2 tsp baking soda
- 1/2 tsp salt.
Prepare two 8-inch cake pans, by (spray) greasing, and then lining the bottoms with circles of baker's parchment, and greasing again. Scrape batter (very liquidy) into prepared pans; it should only fill about 1/3 full (cake expands significantly). Bake 20 to 25 minutes.
Having cherries on hand, and having made a cake with brandied cherries earlier last week, I combined some cherry preserves and the brandy-cherry-syrup in a small frypan over low heat to dissolve. After turning out the first cake, I poured a thin layer of glaze over it, so that I could glue the top layer on. I then poured more glaze on the top of the cake, so that little bits rolled down the sides, and lined the outer edge of the cake top with cherry halves.
* Note: I used only 1/3 cup mayonnaise, since I misread the recipe. It worked fine. More will only make it richer. And I'm sure that vegan mayonnaise substitutes, such as Nayonnaise would work just as well. I would even be interested in trying this cake with Wildwood Garlic Aioli — I've had a mayonnaise cake made with garlic aioli before (a plain yellow cake; the cook also substituted rice flour for cake flour to make it gluten free, and it still turned out good), and Khymos suggests that chocolate and garlic might go well together with the addition of coffee. This recipe would definitely be good with some added instant espresso (I'd put in between a tsp and a Tbsp in the first step with the cocoa powder).
25 June 2007
A note on ultrafilters and voting.
Terence Tao has a very nice discussion of ultrafilters in his most recent post. In his comment thread, I posed two problems (and provided an embarrassingly incorrect solution to one of them), which I will repeat here; then I'll write about another aspect of Tao's post.
Problem (1): Construct (with Axiom of Choice, of course) a bounded sequence and an ultrafilter so that the sequence has no large convergent subsequences.
Problem (2): Open-ended. Continuum Hypothesis, I've heard it claimed, is (at least sufficient to guarantee that) ultrapowers by different (nonprinciple) ultrafilters are equivalent. I first heard (a version of) this claim from Conway, and I believe he suggested that this was actually equivalent to CH, although that could be a faulty memory. A commenter says that it's not the case that two nonprinciple ultrafilters are themselves isomorphic (in the sense of there existing a bijection on \N that takes one ultrafilter to the other), but it's not clear to me why not. So, why not? And, more generally, is there a reasonably easy way to see the first claim in this paragraph?
Lastly, I would like to point out another extremely cool fact that Tao breezes through. To explain ultrafilters, Tao uses an analogy with voting theory: an ultrafilter is a way of deciding which side wins when infinitely many people vote.
This, then, suggests a proof of Arrow's Theorem.
Let me explain.
Given a set A of alternatives, and a set P of people, let's define a social welfare function (SWF) as a function that takes as input each person's (linear) rankings of the elements of A and gives as output a global ranking of the elements of A (the output may have ties). There are many conditions a "fair" SWF might satisfy; Arrow singles out three:
By the way, although originally we allowed SWF to produce ties in the output, properties P and IIA are sufficient to guarantee (if A has at least three alternatives) that ties never occur. Indeed, say a certain input yields a tie between x and y. Let X be the set of people who prefer x to y, and Y the set of people who prefer y to x. Taking another alternative z, we can, using IIA, insert it so that all of X prefer x>z>y and all of Y prefer z>y>x; then IIA says x and y are still tied, and P says z>y and hence also x; inserting so that, for X, x>y>z, and for Y, y>z>x, then P says z<y and hence also x; but the relative rankings of z and x are the same in each insertion, so IIA gives us the desired contradiction. Thus, from now on, I will assume that our SWF does not result in ties.
Arrow's theorem says that, if A has size at least three, and if P is finite, then the only SWF satisfying both P and IIA is a single-person dictatorship. Since M is also natural, and since the book* I'm glancing at for my definitions does so too, I will show that SWFs satisfying all three conditions are single-person dictatorships.
Well, an ultrafilter \F can give us a SWF: if you want to know the relative ranking of x versus y, just see whether the set of people who put x above y is \F-large (i.e. in \F) or not. This binary relation is then definitely an ordering, because of ultrafilter properties: for instance, if x > y and y > z in the output, then there are \F-large sets X and Y in P, but then X \cap Y is \F-large, and so since the individual rankings are linear orderings, we must have x > z in the final ranking. Such a SWF obviously satisfies properties IIA and M (and hence P) above.
On the other hand, a SWF \F satisfying IIA and M gives us an ultrafilter. How? Let's pick two particular elements a and b in A. Then for any X \subseteq P, consider feeding \F a collection of rankings so that everyone in X put a > b everyone else puts b > a. If the SWF declares a > b, then declare X to be an a,b-large subset of P. IIA makes this well-defined, and M guarantees that supersets of large sets are large. But this definition does not depend on the choices of a and b: given another c, inserting so that X ranks a>b>c and everyone else ranks b>c>a, then by P, \F yields a>b>c, and so X is a,c-large; similarly, it's c,b-large by a different insertion, and by continuing this type of argument you can get that an a,b-large X is x,y-large for any x,y\in A.
Is the finite intersection of two large sets also large? It suffices to show that the division of a large set into two sets yields a large set and a small set. Say X = Y \cup Z is large, with Y and Z disjoint; then assign Y as b>a>c and Y gets a>c>b and everyone else gets c>b>a. As X is large, certainly \F assigns a>c. Of the three possible (relative) places for \F to put b, either b>c, in which case Y is large, or a>b, in which case Z is large.
Thus \F generates an ultrafilter on A. Writing A! for the set of permutations on A, it's clear how a SWF=ultrafilter \F acts: take our P-sequence of points in A!, and take the (ultra)limit. In the case when A is finite, then the ultraproduct of A! is just A! — i.e. the limit is just another permutation.
In any case, Arrow's theorem is thus equivalent to the statement that there are no nonprinciple ultrafilters on a finite set.
Some discussion is warranted. In justifying that a SWF satisfying the desired criteria is an ultrafilter, I all but proved Arrow's theorem. Given that the statement that "there are no nonprinciple ultrafilters on a finite set" is rather trivial, this is not too surprising. Also, I had to use the fact that A had at least three elements throughout the argument, because Arrow's theorem fails when there are only two options (for instance, Majority Rules works). So it's not immediate to translate Arrow's criteria into the definition of an ultrafilter. Possibly, though, my argument could be cleaned a bit to make it more immediate. Also, it seems that monotonicity is required in guaranteeing that our SWF gives an ultrafilter, but Arrow's theorem holds even if you drop that condition.
On the other hand, this analysis does allow us to understand Arrow-style voting systems with infinitely many people. Or with continuously many people. Or other topologies. And, you know the US is very large — if we could use Axiom of Choice in practice, then perhaps we can pretend that the population of the US is infinite and pick a nonprinciple ultrafilter to be our voting system.
* For a very fine text, suitable for someone who's never seen any mathematics or someone who's seen a lot, on Arrow's Theorem and related topics, I highly recommend Mathematics and Politics: Strategy, Voting, Power, and Proof by Alan D. Taylor (Springer, 1995).
Problem (1): Construct (with Axiom of Choice, of course) a bounded sequence and an ultrafilter so that the sequence has no large convergent subsequences.
Problem (2): Open-ended. Continuum Hypothesis, I've heard it claimed, is (at least sufficient to guarantee that) ultrapowers by different (nonprinciple) ultrafilters are equivalent. I first heard (a version of) this claim from Conway, and I believe he suggested that this was actually equivalent to CH, although that could be a faulty memory. A commenter says that it's not the case that two nonprinciple ultrafilters are themselves isomorphic (in the sense of there existing a bijection on \N that takes one ultrafilter to the other), but it's not clear to me why not. So, why not? And, more generally, is there a reasonably easy way to see the first claim in this paragraph?
Lastly, I would like to point out another extremely cool fact that Tao breezes through. To explain ultrafilters, Tao uses an analogy with voting theory: an ultrafilter is a way of deciding which side wins when infinitely many people vote.
This, then, suggests a proof of Arrow's Theorem.
Let me explain.
Given a set A of alternatives, and a set P of people, let's define a social welfare function (SWF) as a function that takes as input each person's (linear) rankings of the elements of A and gives as output a global ranking of the elements of A (the output may have ties). There are many conditions a "fair" SWF might satisfy; Arrow singles out three:
- Pareto (P): if the input consists of identical lists, then this list is the produced output,<\li>
- Independence of Irrelevant Alternatives (IIA): given two separate inputs such that, person-by-person, the relative rankings of two particular alternatives x and y are the same, then the outputs have the same relative rankings of alternatives
- Monotonicity (M): Say the output has x higher-ranked than y. Then if some people who originally ranked y above x switch their positions, then the output will still have x above y. Except for the possibility of ties, this property is strictly stronger than property P.
By the way, although originally we allowed SWF to produce ties in the output, properties P and IIA are sufficient to guarantee (if A has at least three alternatives) that ties never occur. Indeed, say a certain input yields a tie between x and y. Let X be the set of people who prefer x to y, and Y the set of people who prefer y to x. Taking another alternative z, we can, using IIA, insert it so that all of X prefer x>z>y and all of Y prefer z>y>x; then IIA says x and y are still tied, and P says z>y and hence also x; inserting so that, for X, x>y>z, and for Y, y>z>x, then P says z<y and hence also x; but the relative rankings of z and x are the same in each insertion, so IIA gives us the desired contradiction. Thus, from now on, I will assume that our SWF does not result in ties.
Arrow's theorem says that, if A has size at least three, and if P is finite, then the only SWF satisfying both P and IIA is a single-person dictatorship. Since M is also natural, and since the book* I'm glancing at for my definitions does so too, I will show that SWFs satisfying all three conditions are single-person dictatorships.
Well, an ultrafilter \F can give us a SWF: if you want to know the relative ranking of x versus y, just see whether the set of people who put x above y is \F-large (i.e. in \F) or not. This binary relation is then definitely an ordering, because of ultrafilter properties: for instance, if x > y and y > z in the output, then there are \F-large sets X and Y in P, but then X \cap Y is \F-large, and so since the individual rankings are linear orderings, we must have x > z in the final ranking. Such a SWF obviously satisfies properties IIA and M (and hence P) above.
On the other hand, a SWF \F satisfying IIA and M gives us an ultrafilter. How? Let's pick two particular elements a and b in A. Then for any X \subseteq P, consider feeding \F a collection of rankings so that everyone in X put a > b everyone else puts b > a. If the SWF declares a > b, then declare X to be an a,b-large subset of P. IIA makes this well-defined, and M guarantees that supersets of large sets are large. But this definition does not depend on the choices of a and b: given another c, inserting so that X ranks a>b>c and everyone else ranks b>c>a, then by P, \F yields a>b>c, and so X is a,c-large; similarly, it's c,b-large by a different insertion, and by continuing this type of argument you can get that an a,b-large X is x,y-large for any x,y\in A.
Is the finite intersection of two large sets also large? It suffices to show that the division of a large set into two sets yields a large set and a small set. Say X = Y \cup Z is large, with Y and Z disjoint; then assign Y as b>a>c and Y gets a>c>b and everyone else gets c>b>a. As X is large, certainly \F assigns a>c. Of the three possible (relative) places for \F to put b, either b>c, in which case Y is large, or a>b, in which case Z is large.
Thus \F generates an ultrafilter on A. Writing A! for the set of permutations on A, it's clear how a SWF=ultrafilter \F acts: take our P-sequence of points in A!, and take the (ultra)limit. In the case when A is finite, then the ultraproduct of A! is just A! — i.e. the limit is just another permutation.
In any case, Arrow's theorem is thus equivalent to the statement that there are no nonprinciple ultrafilters on a finite set.
Some discussion is warranted. In justifying that a SWF satisfying the desired criteria is an ultrafilter, I all but proved Arrow's theorem. Given that the statement that "there are no nonprinciple ultrafilters on a finite set" is rather trivial, this is not too surprising. Also, I had to use the fact that A had at least three elements throughout the argument, because Arrow's theorem fails when there are only two options (for instance, Majority Rules works). So it's not immediate to translate Arrow's criteria into the definition of an ultrafilter. Possibly, though, my argument could be cleaned a bit to make it more immediate. Also, it seems that monotonicity is required in guaranteeing that our SWF gives an ultrafilter, but Arrow's theorem holds even if you drop that condition.
On the other hand, this analysis does allow us to understand Arrow-style voting systems with infinitely many people. Or with continuously many people. Or other topologies. And, you know the US is very large — if we could use Axiom of Choice in practice, then perhaps we can pretend that the population of the US is infinite and pick a nonprinciple ultrafilter to be our voting system.
* For a very fine text, suitable for someone who's never seen any mathematics or someone who's seen a lot, on Arrow's Theorem and related topics, I highly recommend Mathematics and Politics: Strategy, Voting, Power, and Proof by Alan D. Taylor (Springer, 1995).
19 June 2007
Money
It takes a lot of money to run a Presidential campaign. In 2000, Bush raised and spent a little under two hundred million dollars; Gore only got 150 million (Center for Responsive Politics). In 2004, each of the major candidates raised and spent between three and four hundred million dollars. For the 2008 race, so far Clinton has raised the most money (36 million, ibid.) and Obama is next (Romney is a close third). Mr. Clinton personally raises one hundred thousand dollars a night for his wife's campaign (New York Times). Each of the major candidates will likely (need to) raise at least half a million dollars by the general election.
Of course, all presidential candidates are reasonably affluent. Romney is the richest, with between 190 and 250 million dollars to his name, and Clinton is next, with 10 to 50 million dollars, mostly from her husband's lecture circuit (CBS News). Campaign finance laws restrict the amount of money you or I may donate to any given candidate (for 2008, individuals may not donate more than $2,300 per candidate per election), but politicians are allowed to spend as much of their own money as they want.
Which leads, of course, to Michael Bloomberg, mayor of New York City. Today he announced, while campaigning in California, that he has changed his party affiliation from Republican to Independent (he was a Democrat before running for mayor). Almost certainly he will run for President; he certainly enjoys stoking that fire. But without a major party to provide funding and machinery, does he stand a chance? To find a third-party candidate who did better than Ross Perot (close to 20% in 1992, splitting the right and assuring Clinton a win), you have to go all the way back to Roosevelt in 1912; Nader never cracked 5% (New York Times). Except that Bloomberg is Forbes' 44th richest American, with 5.5 billion dollars (Wikipedia). He will likely run on the Independence ticket --- the same one that sponsored Ross Perot, and on whose ticket he ran (along with the GOP endorsement) for mayor. He supports abortion rights and gay marriage, and is staunchly pro-business. And, it's rumored, he has already set aside $1 billion for a presidential campaign (ibid.). This is easily enough to be competitive.
Bloomberg is not, of course, my favorite of the presidential candidates, but I certainly wouldn't mind him in the White House. I'm amused by what his friend George Stephanopoulos said about his potential run (ibid.; Stephanopolous in italics):
1. 70% of the nation would need to feel as though the country is moving in the wrong direction. Check.
2. Both nominees would need to have disapproval ratings in the 40% range. Clinton and Romney.
3. 40% of the country would need to be open to a third party candidate. Likely?
4. 20-25% of the country would need to be "open to Mike Bloomberg." Time will tell.
Of course, all presidential candidates are reasonably affluent. Romney is the richest, with between 190 and 250 million dollars to his name, and Clinton is next, with 10 to 50 million dollars, mostly from her husband's lecture circuit (CBS News). Campaign finance laws restrict the amount of money you or I may donate to any given candidate (for 2008, individuals may not donate more than $2,300 per candidate per election), but politicians are allowed to spend as much of their own money as they want.
Which leads, of course, to Michael Bloomberg, mayor of New York City. Today he announced, while campaigning in California, that he has changed his party affiliation from Republican to Independent (he was a Democrat before running for mayor). Almost certainly he will run for President; he certainly enjoys stoking that fire. But without a major party to provide funding and machinery, does he stand a chance? To find a third-party candidate who did better than Ross Perot (close to 20% in 1992, splitting the right and assuring Clinton a win), you have to go all the way back to Roosevelt in 1912; Nader never cracked 5% (New York Times). Except that Bloomberg is Forbes' 44th richest American, with 5.5 billion dollars (Wikipedia). He will likely run on the Independence ticket --- the same one that sponsored Ross Perot, and on whose ticket he ran (along with the GOP endorsement) for mayor. He supports abortion rights and gay marriage, and is staunchly pro-business. And, it's rumored, he has already set aside $1 billion for a presidential campaign (ibid.). This is easily enough to be competitive.
Bloomberg is not, of course, my favorite of the presidential candidates, but I certainly wouldn't mind him in the White House. I'm amused by what his friend George Stephanopoulos said about his potential run (ibid.; Stephanopolous in italics):
1. 70% of the nation would need to feel as though the country is moving in the wrong direction. Check.
2. Both nominees would need to have disapproval ratings in the 40% range. Clinton and Romney.
3. 40% of the country would need to be open to a third party candidate. Likely?
4. 20-25% of the country would need to be "open to Mike Bloomberg." Time will tell.
Subscribe to:
Posts (Atom)
