http://www.tcm.phy.cam.ac.uk/~gz218/Coarse Grained2012-06-12T10:30:21ZGen Zhanghttp://www.tcm.phy.cam.ac.uk/~g218/http://www.tcm.phy.cam.ac.uk/~gz218/2012/06/the-polite-fiction-of-symmetry-breaking.htmlThe polite fiction of symmetry breaking2012-06-11T20:49:25ZGen Zhanghttp://www.tcm.phy.cam.ac.uk/~gz218/<p>
The concept of spontaneous symmetry breaking (SSB) is well-studied and
used in condensed matter. Most prototypically, the first textbook
example is that of a weakly interacting Bose gas, where the global
$U(1)$ symmetry is broken to give a Bose condensed phase, with massless
Goldstone bosons as the residual evidence of that continous symmetry.
This is usually described as $\langle \phi \rangle$ being an order
parameter, and becoming non-zero, where $\phi$ is the annihilation
operator for the a particle. The intuition is clear: given a potential
like $V = |\phi|^4 - |\phi|^2$ there is a degenerate ring of non-zero
$\phi$ which has the lowest energy, and by picking a particular phase we
can minimise the energy.
</p>
<p>
The story becomes a bit messier upon closer inspection. The overall
resolution is actually not that complex, but it seems to be mostly
folklore, in that no-one has written it down.
</p>
<p>
To begin with, notice that the Hamiltonian $H$ commutes with the number
operator $N = \phi^\dagger \phi$ — we do not, after all, randomly
create or destroy rubidium atoms in a trap (losses are accounted for in
a density matrix formalism, something we will come back to). Thus, all
energy eigenstates, and particularly the ground state, will be an
eigenstate of $N$. Thus by elementary arguments, $\langle \phi
\rangle=0$, always. Fundamentally, by picking a definite phase for
$\phi$, we introduce uncertainty in $N$.
</p>
<p>
There are some conventional ways around this. One is to say that what we
really care about is off-diagonal, long-range order (ODLRO). That is,
$\langle \phi^\dagger (r) \phi(0) \rangle$ as $r \rightarrow \infty$.
This certainly is completely well-defined... except for real systems.
The first experimental systems were a few hundred to thousand atoms, and
very small. It is not clear that infinity range is applicable, and since
experimentally they did in fact condense, nor necessary. In addition,
ODLRO is usually quite hard to really compute.
</p>
<p>
The other standard method is to pass to a grand canonical ensemble, and
argue that in the thermodynamic limit this will produce the same answers
for means of macroscopic variables. This looks like it lets us have
fluctuations in $N$. This shares the same problem as the previous
method, but allows us to use easier mathematics. On the other hand, it's
just wrong: the grand canonical ensemble is defined by $H - \mu N$,
which again has eigenstates labelled by $N$ — so $\langle r \rangle$
is still zero! Indeed, what we end up with is a density matrix, with
eigenvalues defined by $\exp(H - \mu N)$ in the $N$-basis! Even there,
the expectation value of $\phi$ is still zero.
</p>
<p>
Antony Leggett, having thought about this quite hard, comes at the
problem from an experimental side — for him the important thing is to
make contact with the available experiments, which is quite reasonable.
He works without 2nd quantisation, and considers the single-particle
density matrix; he shows that in the normal phase, this has eigenvalues
of the same order, whereas in the broken symmetry phase, one eigenvalue
is near unity, with other eigenvalues of order $1/N$ or smaller. From a
theoretical point of view however, this seems to throw out so much of
the elegance in SSB as a generic mechanism of describing critical
behaviour.
</p>
<p>
A truly mathematically correct justification is to add a term $-\epsilon
\phi$ to $H$, which explicitly breaks the symmetry. Then we carefully
take the limit of $N \rightarrow \infty$ first, before $\epsilon
\rightarrow 0$. Analysis of the Hilbert space shows that it breaks into
separate super-selection sectors, indexed by the phase. Indeed, for
things like ferromagnetic transitions, one can reasonably declare that
it would be unphysical for there not to be some stray fields in an
experiment which would provide a small but definite symmetry breaking
potential. Unfortunately for Bose condensation, there is no way to
introduce a non-Hermitian term such as $\phi$ into $H$ — physical loss
of atoms from the trap needs to be accounted for by a density matrix
again, and the loss term only appears in the Lindbladian, not the
Hamiltonian. Indeed, even for the ferromagnetic transition, one could
argue that this simply transfers the problem — the universe as a whole
is presumably symmetric under $SO(3)$, so the symmetry breaking term
which is classical is just an approximation to a properly invariant
quantum interaction term with a quantum environment.
</p>
<p>
So far, the analysis is standard, and may be found in Leggett's book
Quantum Liquids. What follows is the folklore.
</p>
<p>
The final analysis of the above is actually pretty close to the mark.
The missing ingredient is decoherence. A useful view point is that the
symmetry breaking system is acting as an amplifier — it turns a very
small asymmetric potential into a macroscopically observable one, and
fluctuations in the environment certainly counts. More precisely, the
ground state of the total system with its environment is a symmetric
combination of states with entanglement between the large coherent
fluctuation in the system and the much smaller fluctuation in the
enviroment. Upon taking the thermodynamic limit and tracing out the
environment, we are left with a density matrix of states equivalent to
the superselection sectors description above, justifying its use.
</p>http://www.tcm.phy.cam.ac.uk/~gz218/2010/04/topological-insulators-ii-disorder.htmlTopological insulators II: disorder2010-04-11T17:04:50ZGen Zhanghttp://www.tcm.phy.cam.ac.uk/~gz218/<p>
<a href="/~gz218/2010/04/topological-insulators.html">Last time</a> we
saw how a one particle, pure crystal theory of electrons might give rise
to quantised transverse (i.e. Hall) conductance. The theory is simple,
but rather leaves out crucial aspects of the phenomenon. Important
aspects missing are interactions and disorder. Interactions lead to <a href="http://en.wikipedia.org/wiki/Fractional_quantum_Hall_effect">fractional
quantum hall effects</a>, and we do not wish to worry about those here.
It's the usual <em>post hoc</em> justification where we see that a
single particle theory seems to work, even though precisely why is
mysterious — we ask for deliverance from <a href="http://en.wikipedia.org/wiki/Fermi_liquid">the
god of Landau</a>, etc. As far as disorder goes however, things are not
so simple. Loss of (lattice) translational invariance leads to a loss of
the Brillouin zone, and its attendant topological structures. The
classification of states by de Rham cohomology (the TKNN invariant)
doesn't work if we don't have a manifold and a differential structure.
</p>
<p>
Disorder itself is <a href="http://www.tcm.phy.cam.ac.uk/~bds10/publications/lesh.ps.gz">a
vast topic</a>. One approach is to simply write down a Hamiltonian with
a disorder potential written in, i.e. quenched, and work to compute
averages over the disorder realisations. Powerful methods include using <a href="http://www.jetpletters.ac.ru/ps/373/article_5907.shtml">supersymmetric
field theory and renormalisation</a>. However, in keeping with the idea
of simple physics, hard maths, we're going to hit the problem head on
with <a href="http://en.wikipedia.org/wiki/Noncommutative_geometry">noncommutative
geometry</a>.
</p>
<p>
Heuristically, we can build a picture via Anderson localisation.
Although we can no longer use momentum to label states, we can still
talk about the density of states for single particles. The disorder
first causes a broadening of the Landau levels. Second, we know that in
2D arbitrarily weak disorder can cause localisation, albeit with a
divergent localisation length. In a magnetic field the Landau levels are
mobility edges, and so on approaching the level, the states start
spanning the whole sample. Thus for a finite sample, around the Landau
level there are extended states. It seems reasonable to hope that as
long as the Fermi level is moving through the localised states, we still
get a quantised transverse conductance. To show this, we <em>just</em>
need to generalise the TKNN invariant to apply to a system which doesn't
have momentum operators commute with the Hamiltonian. The full details
can be found in <a href="http://arxiv.org/abs/cond-mat/9411052">arxiv:cond-mat/9411052</a>.
</p>
<p>
First, we should say a few words about how noncommutative geometry
works. On a usual space $X$ (a manifold say), we can consider the
functions $C(X)$ of continuous functions from $X$ to (say) the reals.
These functions from a commutative <a href="http://en.wikipedia.org/wiki/C*-algebra">$C^*$-algebra</a>.
But we can also <a href="http://en.wikipedia.org/wiki/Gelfand_representation">go
backwards</a>: start with an abstractly defined commutative algebra, ask
for the space on which these elements have representation as functions
on that space. It turns out that these two views are exactly equivalent,
and we can therefore calculate topological quantities (such as the Chern
classes) of $X$ by computations on the algebra $C(X)$.
</p>
<p>
Noncommutative geometry is what you get if you remove the restriction to
commutative algebras. It turns out that many of the geometric concepts
can be pushed through, and we get new, novel spaces in which to play. In
our case, the strategy is to take the TKNN invariant, replace anything
which mentions geometry by its algebraic equivalent, and then remove the
commutativity requirement, and hope that it remains well-defined and
integer valued. The paper above by Bellissard et al. also go through a
first principles derivation of the algebraicised Kubo conductance
formula, and show that it is related to the Chern class the same way as
the TKNN invariant and consider the corrections to the quantum hall
regime (it's an excellent paper).
</p>
<p>
To give a flavour of how it works, recall that the TKNN invariant is
basically a suitably integrated derivative of some operator by momentum.
Integration can be replaced by a suitable trace; the fact that it is
over the Brillouin zone can be considered a normalisation, i.e. it is
actually an integral divided by the area. Differentiation can be treated
algebraically by treating it as a <a href="http://en.wikipedia.org/wiki/Derivation_(abstract_algebra)">derivation</a>:
$$ \partial_{k} A \rightarrow -i[x,A], $$ relying on the commutator obeying
a Leibniz rule and $[x,k] = i$. Thus, we have again turned the
problem into a mathematical one, which means we can find clever people
to solve it for us. In this case, it can be shown that subject to states
at the Fermi level being localised, the algebraicised TKNN invariant is
still integer valued. Thus we have a very direct way to show that the
topological order is robust in the face of disorder.
</p>http://www.tcm.phy.cam.ac.uk/~gz218/2010/04/topological-insulators.htmlTopological insulators2010-04-09T14:22:50ZGen Zhanghttp://www.tcm.phy.cam.ac.uk/~gz218/<p>
Currently, <a href="http://arxiv.org/abs/1002.3895">topological
insulators</a> are very fashionable. However, the usual discussions
we've been enjoying (enduring?) have gone pretty much over the heads of
most of us little ones. I think the primary problem so far has been that
there are plenty of <em>obvious</em> things, which really should have
been repeated, but have not. Part of the problem is that people seem to
think that the integer quantum hall effect is well-understood by the
audience. Below, we hopefully introduce the necessary context.
</p>
<p>
We are dealing with pure electrons in a lattice. No phonons, no
impurities, no interactions. This means that the many-body theory
reduces in the canonical way to single particle theory — the many
body wavefunction is just a suitably antisymmetrised product of single
particle states. Further, invariance under translation by lattice
vectors imply that the single particle space separates into sectors
labelled by momentum $\mathbf{k}$, which are not mixed by the
Hamiltonian. Thus for each sector $\mathbf{k}$ we have a separate
Hamiltonian $H_\mathbf{k}$, the total Hamiltonian $H$ being the direct
sum of these, over all $\mathbf{k}$. We may assume that there are no
degenerate energy levels in any of the sub-Hamiltonians, since either
they are protected by some symmetry, or they will not be generic, i.e.
we are not in a stable phase. This is essentially just band theory.
</p>
<p>
The lattice induces the topology of a torus in reciprocal space. The
sub-Hamiltonians $H_\mathbf{k}$ can be seen as a mapping from the torus
$T^d$ to self-adjoint operators on a suitable Hilbert space. For
physically reasonable suitations, the sub-Hamiltonians $H_\mathbf{k}$
should be continuous as functions of $\mathbf{k}$. We can now invoke
some algebraic topology, and ask how many equivalence classes exist of
such mappings, where equivalence is defined to be any continuous
deformation without causing any of the $H_\mathbf{k}$ to become
degenerate by having a level-crossing, i.e. how many phases exist. This
is a well-defined mathematical question, and has a well-defined answer
in terms of the <a href="http://en.wikipedia.org/wiki/Chern_class">Chern
classes</a>, which physically are called TKNN invariants. It unimportant
for us what they are, but just that they exist and may be computed for
real situations.
</p>
<p>
Note that in a magnetic field the above is not strictly true. A uniform
magnetic field causes orthogonal translations to become non-commutative.
We are therefore unable to simultaneously label the states with $k_x$
and $k_y$ (say in 2D). However, at particular values of $B$-field the
commutator vanishes — exactly those at which the quantum hall
plateaus exist. The TKNN invariant also turns out to be proportional to
the transverse conductance. The different integer filling factors are
thus proper phases, separated by some complex quantum phase transition.
In practise, impurities actually produce most of the observed
phenonmenon (i.e. the broad plateaus), but these can be thought of as
simply fighting against the magnetic non-commutativity and maintaining
the phase even for not exactly correct field strengths.
</p>
<p>
Now, since we are concerned with insulators, we may assume that our
Fermi level lies in a gap; this is also true of the integer quantum hall
states. The vacuum outside our sample is also such a state. It has, by
assumption, a different value for the TKNN invariant. Thus somewhere on
the edge, there will be a level crossing, and the gap will close forming
a conducting edge state. Thus we see that topological phases will
necessarily be accompanied by edge states.
</p>
<p>
This understanding is still deficient in a very crucial way. The construction
above relies strongly on single particle behaviour and momentum (or at least pseudomomentum) being a
good quantum number. As mentioned, magnetic fields which are not exactly
integer quanta per unit cell cause problems --- in fact we get fractional
quantum hall effects, where the interaction leads to intrinsically
many-body effects. Furthermore, it is not clear that these effects are impervious
to impurities — to say that it is pretected <em>because</em> it is
topological is getting things mixed up; we can only declare it to be
topological because it is insensitive to impurities. Indeed, in the case of the
integer quantum hall effect impurities actually extend the "radius of convergence"
of the phase.
</p>http://www.tcm.phy.cam.ac.uk/~gz218/2010/01/higher-derivative-theories.htmlHigher derivative theories2010-01-20T11:54:00ZGen Zhanghttp://www.tcm.phy.cam.ac.uk/~gz218/<p>
We tend to not use higher derivative theories. It turns out that there
is a very good reason for this, but that reason is rarely discussed in
textbooks. We will take, for concreteness, $L\left(q,\dot q, \ddot
q\right)$, a Lagrangian which depends on the 2nd derivative in an
essential manner. Inessential dependences are terms such as $q\ddot q$
which may be partially integrated to give ${\dot q}^2$. Mathematically,
this is expressed through the necessity of being able to invert the
expression $$P_2 = \frac{\partial L\left(q,\dot q, \ddot
q\right)}{\partial \ddot q},$$ and get a closed form for $\ddot q
\left(q, \dot q, P_2 \right)$. Note that usually we also require a
similar statement for $\dot q \left(q, p\right)$, and failure in this
respect is a sign of having a constrained system, possibly with gauge
degrees of freedom.
</p>
<p>
In any case, the non-degeneracy leads to the Euler-Lagrange equations in
the usual manner: $$\frac{\partial L}{\partial q} -
\frac{d}{dt}\frac{\partial L}{\partial \dot q} +
\frac{d^2}{dt^2}\frac{\partial L}{\partial \ddot q} = 0.$$ This is then
fourth order in $t$, and so require four initial conditions, such as
$q$, $\dot q$, $\ddot q$, $q^{(3)}$. This is twice as many as usual, and
so we can get a new pair of conjugate variables when we move into a
Hamiltonian formalism. We follow the steps of Ostrogradski, and choose
our canonical variables as $Q_1 = q$, $Q_2 = \dot q$, which leads to
\begin{align} P_1 &= \frac{\partial L}{\partial \dot q} -
\frac{d}{dt}\frac{\partial L}{\partial \ddot q}, \\ P_2 &=
\frac{\partial L}{\partial \ddot q}. \end{align} Note that the
non-degeneracy allows $\ddot q$ to be expressed in terms of $Q_1$, $Q_2$
and $P_2$ through the second equation, and the first one is only
necessary to define $q^{(3)}$.
</p>
<p>
We can then proceed in the usual fashion, and find the Hamiltonian
through a Legendre transform: \begin{align} H &= \sum_i P_i \dot{Q}_i -
L \\ &= P_1 Q_2 + P_2 \ddot{q}\left(Q_1, Q_2, P_2\right) - L\left(Q_1,
Q_2,\ddot{q}\right). \end{align} Again, as usual, we can take time
derivative of the Hamiltonian to find that it is time independent if the
Lagrangian does not depend on time explicitly, and thus can be
identified as the energy of the system.
</p>
<p>
However, we now have a problem: $H$ has only a linear dependence on
$P_1$, and so can be arbitrarily negative. In an interacting system this
means that we can excite positive energy modes by transferring energy
from the negative energy modes, and in doing so we would increase the
entropy — there would simply be more particles, and so a need to
put them somewhere. Thus such a system could never reach equilibrium,
exploding instantly in an orgy of particle creation. This problem is in
fact completely general, and applies to even higher derivatives in a
similar fashion.
</p>http://www.tcm.phy.cam.ac.uk/~gz218/2010/01/central-limit-theorem.htmlCentral Limit Theorem2010-01-14T11:52:41ZGen Zhanghttp://www.tcm.phy.cam.ac.uk/~gz218/<p>
There's a proof of the <a href="http://en.wikipedia.org/wiki/Central_limit_theorem">Central
Limit Theorem</a> which I am very fond of, which is not often seen in
textbooks. It is a sort of renormalisation argument. In a way, it's not
even a rigorous proof — often the way with renormalisation —
but in conjunction with the more classical proofs it lends a powerful
insight.
</p>
<p>
Without loss of generality, let's consider a whole bunch of identical,
univariate variables $X_n$, each with zero mean. Thus we have quite
trivially that a sum of $N$ of them will still have zero expectation,
and a variance of $N$.
</p>
<p>
Now, rather than summing all of them at the same time, we do it in
steps, and renormalise along the way to see what happens. So concretely,
let $X$ have a distribution given by $f$, which is assumed to be
sufficiently well-behaved for whatever we need to do. Then $X+X \sim f
\star f$, where the multiplication is a convolution. This is our
coarse-graining step, so we still need to re-scale, so that we get back
a univariate distribution: $$f^\prime(x) = \sqrt 2 (f \star f)(\sqrt 2
x).$$ Convolutions are easiest to take in Fourier space:
$$\widetilde{f^\prime}(k) = \left[ \tilde{f}(k/\sqrt{2}) \right]^2.$$ It
is then fairly trivial to check that the univariate Gaussian
$\widetilde{f^*}(k) = e^{-k^2/\sqrt{2}}$ is a fixed point.
</p>
<p>
We can view this step as a transform on the space of distributions, and
so it makes sense to linearise about this fixed point and look at what
happens to small perturbations: \begin{align*} \widetilde{f^*}(k) +
\widetilde{g_n}(k) &\rightarrow \widetilde{f^*}(k) + 2
\widetilde{g_n}(k/\sqrt{2}) \widetilde{f^*}(k/\sqrt{2}) +
\mathrm{h.o.t.} \\ &= \widetilde{f^*}(k) + \lambda_n \widetilde{g_n}(k)
\end{align*} Which has solutions as $\widetilde{g_n}(k) = (ik)^n
\widetilde{f^*}(k)$ with eigenvalues $\left|\lambda_n\right| =
2^{1-n/2}$; these correspond to mathematically meaningful perturbations
if and only $n$ is an integer greater than 0, for reasons of convergence
and normalisation. That still leaves $n=1$ or $n=2$ as being relevant
and marginal; the former correspond to shifting the mean, the latter to
changing the variance — and since those are not allowed by
assumption, we find that the Gaussian is a stable fixed point.
</p>
<p>
Notice that this does not say anything about the size of basin of
attraction, so if another fixed point existed it could cause finite
perturbations to actually flow away. Indeed, this makes it not quite a
proper proof. On the other hand, this procedure gives the actual rate of
convergence to a Gaussian, something that the classical proofs do not
give.
</p>http://www.tcm.phy.cam.ac.uk/~gz218/2010/01/product-representation.htmlProduct representation2010-01-10T11:47:06ZGen Zhanghttp://www.tcm.phy.cam.ac.uk/~gz218/<p>
In dealing with complex analytic functions, it is often quite handy to
represent them as their Taylor expansions, i.e. a summation
representation. We can generalise a little bit and expand even around
non-essential, isolated singularities with a Laurent series. However,
quite often we really care about the zeros of a function (e.g. <a href="http://en.wikipedia.org/wiki/Leeâ€“Yang_theorem">Yang-Lee
circle theorem</a> on zeros of the grand partition function for
Ising-esque models), and extracting those out of summations is unwieldy.
Therefore, it would be much nicer to have a product representation
instead.
</p>
<p>
For polynomials, such a representation is obvious, and unique —
the fundamental theorem of algebra guarantees the factorisation into
linear factors: $$p(z) = p(0) \prod_{\mathrm{finite~}n} \left(1 -
\frac{z}{z_n}\right). $$ The various $z_n$ are then the location of the
zeros. We would like to extend this to more general functions.
</p>
<p>
However, in general this is difficult, and non-unique (see <a href="http://en.wikipedia.org/wiki/Weierstrass_factorization_theorem">Weierstrass
factorization theorem</a> for an existence statement). For one
particular subset however, we can create an effective procedure for
manufacturing these: functions with only simple isolated zeros and no
poles.
</p>
<p>
So let $f(z)$ be such a function. If $f$ has a zero of order $m$ at
$z=0$, then we can divide out by $z^m$ and get something without a zero
at the origin, and so without loss of generality that's what we'll
assume. Let $z_n$ be the location of remaining (infinite number of)
zeros. Then $g(z) = [\ln f(z)]^\prime = f^\prime(z)/f(z)$ has only
simple isolated poles with unit residues at $z_n$; thus if we find a
summation representation of $g$ we could integrate and exponentiate to
find a product representation for $f$.
</p>
<p>
Now consider: $$\frac{1}{2\pi i} \oint_{\Gamma_n}
\frac{g(z^\prime)}{z^\prime-z} dz^\prime = g(z) + \sum^n_j \frac{1}{z_j
- z}$$ where $\Gamma_n$ is a contour which encloses the closest $n$
poles to the origin. Then $$g(z) - g(0) = \frac{z}{2\pi i}
\oint_{\Gamma_n} \frac{g(z^\prime)}{z^\prime(z-z^\prime)} dz^\prime +
\sum_j^n \left(\frac{1}{z-z_j} + \frac{1}{z_j}\right).$$ Thus if we can
find a sequence of contours $\Gamma_n$ such that $g$ remains bounded on
them, the integral will converge to zero as $n \rightarrow 0$. In that
case, we find $$g(z) = g(0) + \sum_n \left(\frac{1}{z-z_n} +
\frac{1}{z_n}\right).$$
</p>
<p>
So now we can return to factorising $f$. Integrating $g$ gives $$\ln
f(z) - \ln f(0) = cz + \sum_n \left[ \ln\left(1-\frac{z}{z_n}\right) +
\frac{z}{z_n} \right]$$ where $c = g(0)$; re-exponentiating gives $$f(z)
= f(0) e^{cz} \prod_n \left(1 - \frac{z}{z_n}\right) e^{z/z_n}.$$
</p>
<p>
As an example, consider $f(z) = \sin(z)/z$. Then $g(z) = -1/z +
\cot(z)$; we can pick contours $\Gamma_n$ which sit between the poles in
$\cot z$, and our procedure will converge. A bit of limit work shows
that $g(0) = 0$ and $f(0) = 1$. The zeros sit at $z_n = n \pi$, with $n$
being any non-zero integer. Thus we find \begin{align*}
\frac{\sin(z)}{z} &= \prod_{n \neq 0} \left(1 - \frac{z}{\pi n}\right)
e^{z/n\pi} \\ &= \prod_{n=1}^{\infty} \left(1 - \frac{z^2}{\pi^2
n^2}\right). \end{align*}
</p>http://www.tcm.phy.cam.ac.uk/~gz218/2009/12/mathbbc-succ-mathbbr.html$\mathbb{C} \succ \mathbb{R}$2009-12-28T11:46:19ZGen Zhanghttp://www.tcm.phy.cam.ac.uk/~gz218/<p>
Suppose I have two functions $f$ and $g$, on the real numbers (or some
compact interval if it makes you feel warmer inside). Is it reasonable
to expect that if their derivatives match at some point, i.e.
$f(x)=g(x)$, $f^\prime(x)=g^\prime(x)$, and so on, then they are equal?
Furthermore, suppose this is true for all values $y \le x$?
</p>
<p>
As a counter-example, consider the function: $$f(x) = \begin{cases}0 & x
\le 0 \\ \exp(-1/x) & x \gt 0 \end{cases}$$ This function is continuous
at $x=0$, and its derivatives there are all zero. In other words, at
$x=0$ it "exactly looks like" the constant function $g(x)=0$.
</p>
<img src="http://upload.wikimedia.org/wikipedia/commons/b/b4/Non-analytic_smooth_function.png" alt="f(x)">
<p>
In a way, it seems perverse that we can't "sense" the impeding rise as
we move through the origin. Another way to say it is that the
derivatives of $f$ are well defined, and so can be used to form a Taylor
series; however, the function does not equal its Taylor expansion, even
though the latter exists.
</p>
<p>
This all makes a bit more sense when discussed in the context of complex
analysis. The function $f(z)$, regarded as a function over $\mathbb{C}$
has an essential singularity at $z=0$. This is an example of the fact
that although smooth functions over the reals seems nicer, from an
elementary point of view, smooth functions over the complex numbers
enjoy more "globally" nice properties, e.g. over $\mathbb{C}$ the
existence of a Taylor expansion is the same as being smooth and defined
almost everywhere (see <a href="http://en.wikipedia.org/wiki/Holomorphic_functions_are_analytic">Wikipedia's
entry on this theorem</a>).
</p>http://www.tcm.phy.cam.ac.uk/~gz218/2009/12/lets-get-serious.htmlLet's get serious2009-12-23T11:45:35ZGen Zhanghttp://www.tcm.phy.cam.ac.uk/~gz218/<p>
The intention of this blog is to have a record of interesting papers,
ideas and discussions. Regular features will probably be David's Fairy
Tales, TCM seminars, correlated systems lunches and maybe even some
biology.
</p>
<p>
A central part of this blog will be the existence of maths, so here some
rough tests. Inline: $\sqrt{1-\xi^2}$. Display: $$\zeta(2) =
\sum_{n=1}^\infty \frac{1}{n^2} = \frac{\pi^2}{6}.$$
</p>