TITLE: How to show existence and uniqueness of an $f: R(\omega) \to \omega$, where $R(\omega)$ is the cumulative hierarchy?
QUESTION [3 upvotes]: My questions
For part A (see below), is my reasoning correct? Is it enough to just show that $f(x) \in \omega \ \ \forall x \in R(\omega)$? I'm very new to axiomatic set theory, in particular to the cumulative hierarchy, and I don't feel I have a good grasp of it at all yet.
For part B, I don't even understand what $g(n,m)$ means. It looks like $g:\omega \to \omega$ is a function of two variables, which doesn't make sense. Could anyone elucidate?
Thanks in advance!
The problem
-Part A-
Show that there is one, and only one, function $f: R(\omega) \to \omega$ satisfying
$$f(x) = \sum_{y\in x} 2^{f(y)}$$
for each $x \in R(\omega)$. Here $R(\omega)$ is the first infinite level in the cumulative
hierarchy.
Show that this function is a bijection.
-Part B-
Find a function $g : \omega \to \omega$ such that for all $n$ and $m$ in $\omega$ and for $x$ and $y$ in $R(\omega)$ with $n = f(x)$ and $m = f(y)$ we have that $$f(\langle x, y\rangle) = g(n,m).$$
My attempt at a solution
Existence of function
By induction: We have $f(0) = 0 \in \omega$. Assume $f(x) \in \omega \ \ \forall x \in R(\alpha)$, and let $x' \in R(\alpha + 1)$. Now $x' \subset R(\alpha)$ so for any $y \in x'$ we would have $f(y) \in \omega$, and thus $f(x') = \sum_{y\in x'} 2^{f(y)} \in \omega$.
Uniqueness
Also by induction: Assume $f, g$ both satisfy the criterion. We have $f(0) = 0 = g(0)$. Assume $f(x) = g(x) \ \ \forall x \in R(\alpha)$, and let $x' \in R(\alpha + 1)$. Again we would have $x' \subset R(\alpha)$, and thus $f(x') = \sum_{y\in x'} 2^{f(y)} = \sum_{y\in x'} 2^{g(y)} = g(x')$, so $f(x) = g(x) \ \ \forall x \in R(\omega)$.
Prove $f$ is a bijection

Given an $\alpha \in \omega$, we can write it as $\alpha = 2^{m_1} + \dots + 2^{m_k}$, $m_1 < \dots < m_k$, where $m_i \in \omega \ \ \forall i$. We then have that $x = \{m_1, \dots , m_k\} \subset R(m_k + 1)$ which means that $x \in R(m_k+2)$. Thus we have found $x \in R(\omega)$ s.t. $f(x) = \alpha$, and $f$ is surjective.
Assume $f(x_1) = f(x_2) = \alpha = 2^{m_1} + \dots + 2^{m_k}$, then we must have $$f(x_1) = \sum_{y\in x_1} 2^{f(y)} = 2^{m_1} + \dots + 2^{m_k} = \sum_{y\in x_2} 2^{f(y)} = f(x_2),$$ and we must have $x_1 = \{m_1, \dots , m_k\} = x_2$, so $f$ is injective.

Find the function $g$
Not solved.
Any help appreciated!

REPLY [2 votes]: I agree with Asaf that your solution to Part A looks fine.
For Part B, I would recommend just going through the calculation1 of
$$\large
f(\langle x,y \rangle )
= f ( \{ \{ x \} , \{ x,y \} \} )
= 2^{f(\{ x \})} + 2^{f(\{x,y\})} = \cdots$$
and then make the replacements $f(x) = n$ and $f(y)=m$.

1This assumes that you are using the Kuratowski definition of ordered pairs; if you are using a different definition, make the appropriate changes.


TITLE: Suppose $ \alpha, \beta>0 $. Compute: $ \int_{0}^{\infty}\frac{\cos (\alpha x)-\cos (\beta x)}{x}dx $
QUESTION [8 upvotes]: Suppose $ \alpha, \beta>0 $. Compute:
  $$ \int_{0}^{\infty}\frac{\cos (\alpha x)-\cos (\beta x)}{x}dx $$

Here is what I do:
$$\begin{align}
\int_{0}^{\infty}\frac{\cos (\alpha x)-\cos (\beta x)}{x}dx &= \int_{0}^{\infty}dx\int_{\alpha}^{\beta}\sin (yx)dy\\
&=\int_{\alpha}^{\beta}dy\int_{0}^{\infty}\sin(yx)dx\\
& \\
&   \qquad\text{let $ yx=u $}\\
& \\
&=\int_{\alpha}^{\beta}\frac{1}{y}dy\int_0^{\infty}\sin u du\\
&=\int_{\alpha}^{\beta}\frac{1}{y}dy\left( -\cos u|_{\infty}+\cos u|_0 \right)\\
&=\log\frac{\beta}{\alpha}(-\cos(\infty)+1)\\
&=\log\frac{\beta}{\alpha}-\cos(\infty)\log\frac{\beta}{\alpha}
\end{align}$$
But $ \cos(\infty) $ does not exist right? Does it mean the integral actually diverse?
Edit: The question comes from https://math.uchicago.edu/~min/GRE/files/week1.pdf
Who can point out my mistake in the above deduction?

REPLY [4 votes]: I would actually use Laplace transforms to compute this sort of an integral.
You must have used it in solving linear differential equations in the past.
It can be defined as follows:-
$$ \mathcal{L}\{f(x)\}=\int_{0}^\infty e^{-px}f(x)dx = F(p)$$
Here we are transforming the function $f$ with domain $x$ to a function $F$ with domain $p$ with a unilateral integral transform of kernel $e^{-px}$. Now consider the general Laplace transform formula given above. Differentiating both sides with respect to $p$ we get:
$$F'(p)=\int_{0}^\infty e^{-px}(-x)f(x)dx=-\mathcal{L}\{xf(x)\} \rightarrow (1)$$
Now put $G(p)$ as Laplace transform of $f(x)/x$ and obtain its differentiation using equation $(1)$:-
$$G(p)=\mathcal{L}\left\{\frac{f(x)}{x}\right\}\Rightarrow G'(p)=-\mathcal{L}\{f(x)\}=-F(p)\rightarrow (2)$$
Using the Fundamental Theorem of Calculus (relationship between derivative and integral) for $(2)$:
$$G(p)=-\int_{a}^p F(p)dp \Rightarrow \int_{0}^\infty e^{-px}\frac{f(x)}{x}dx=-\int_{a}^p F(p)dp \rightarrow (3)$$
Note that $a$ here is some constant. If $G(p) \rightarrow 0$ as $p \rightarrow \infty$ then we put $a = \infty$ and obtain the following:-
$$\int_{0}^\infty e^{-px}\frac{f(x)}{x}dx=\int_{p}^\infty F(p)dp \rightarrow (4)$$
If we let $p \rightarrow 0$ on both sides of equation $(4)$ we get the following:
$$\int_{0}^\infty \frac{f(x)}{x}dx=\int_{0}^\infty F(p)dp \rightarrow (5)$$
This is useful for us for finding the improper integral of various functions of the form $f(x)/x$ where the transform $F(p)$ is known. Now I leave the proof for the following upto you (which is elementary considering we use integration by parts):
$$\mathcal{L}\{ \cos bx \} = \int_{0}^\infty e^{-px}(\cos bx) dx = \frac{p}{p^2 + b^2} \ (p>0) \rightarrow (6)$$
For some constant $b$. Now using equation (5) and (6) we get:
$$\int_{0}^\infty \frac{\cos bx}{x}dx=\int_{0}^\infty \frac{p}{p^2 + b^2}dp \rightarrow (7)$$
Now plugging in equation $(7)$ into the integral we are required to compute:-
$$I=\int_{0}^\infty \frac{\cos \alpha x - \cos \beta x}{x}dx = \int_{0}^\infty p \left(
\frac{1}{p^2 + \alpha^2} - \frac{1}{p^2 + \beta^2} \right)dp$$
$$\Rightarrow I=\frac{\beta^2-\alpha^2}{2} \int_{0}^\infty \frac{2p}{(p^2+\alpha^2)(p^2 +\beta^2)}dp \rightarrow (8)$$
Set $v=\frac{\beta^2+\alpha^2}{2}; u=\frac{\beta^2-\alpha^2}{2}; t=p^2+u$ and using the substitutions and some further simplification:
$$I=\int_{v}^\infty \frac{u}{t^2-u^2}dt = \left[\frac{1}{2}\ln \left|\frac{t-u}{t+u}\right|\right]_{t=v}^{t=\infty}=\frac{1}{2}\ln \left|\frac{u+v}{u-v}\right|$$
Substituting the variables back and rewriting the main equation for $I$ we get:
$$\int_{0}^\infty \frac{\cos \alpha x - \cos \beta x}{x}dx = \ln \frac{\beta}{\alpha}$$


TITLE: Applications of Eckmann–Hilton argument
QUESTION [15 upvotes]: I am looking for applications of the Eckmann–Hilton argument. I found one application in Algebraic Topology where we show that the fundamental group is abelian in case of a topological group. Thank You.

REPLY [38 votes]: There are tons of applications! There are also many useful reformulations. The following will be more or less in increasing level of sophistication. But first, a silly picture:

I've written a blog post which expands a lot upon this answer, if you're interested.

As you mentioned, it can be used to prove that the fundamental group of a topological group is abelian. More generally, the fundamental group of an $H$-space is commutative, for exactly the same reasons.
It also shows that the higher homotopy groups of any space are abelian. This is a "particular case" of the preceding point, because $\pi_n(X) \cong \pi_1(\Omega^{n-1}X)$ and $\Omega^{n-1}X$ (the iterated loop space) is an $H$-space (for $n-1 \ge 1 \iff n \ge 2$).
A monoid object in the category of monoids is a commutative monoid. This is literally the statement of the Eckmann–Hilton argument if you think about it, but it's a very useful reformulation, because it helps in generalizing it to other contexts. Similarly, a group object in the category of groups is an abelian group.
If $X$ is a cogroup object and $G$ is a group object in a category $\mathsf{C}$, then $\hom_\mathsf{C}(X,G)$ is an abelian group. This is a generalization of the first point: $S^1$ is a cogroup object via the pinching map, and $\pi_1(G) \cong \hom_{\mathsf{Ho}(\mathsf{Top}_*)}(S^1, G)$ for a pointed space $G$. So if $G$ is a topological group, you immediately get that $\pi_1(G)$ is commutative.
The center $Z(\mathsf{C}$) of a category $\mathsf{C}$ is defined to be the monoid of natural transformations $\operatorname{id}_\mathsf{C} \to \operatorname{id}_\mathsf{C}$. If $\mathsf{C}$ is a category with one object, AKA a monoid, it's straightforward to check that this coincides with the notion of center of a monoid: $Z(M) = \{ x \in M : xy = yx \, \forall y \}$. And just like the center of a monoid is commutative, the center of a category is commutative. This follows from the Eckmann–Hilton argument: elements of $Z(\mathsf{C})$ can either be composed horizontally or vertically, and the two are compatible.
Another notion of "center of a category" exists: when $(\mathsf{C}, \otimes, 1)$ is a monoidal category, you can consider the monoid $Z(\mathsf{C}, \otimes) = \operatorname{End}(1)$ of endomorphisms of the unit object $1$. This has two multiplications: the usual composition, and the tensor product (if $f,g : 1 \to 1$, then $f \otimes g : 1 \cong 1 \otimes 1 \xrightarrow{f \otimes g} 1 \otimes 1 \cong 1$). With a bit of work you can show that this second law is well-defined, and compatible with composition, thus by the Eckmann–Hilton argument $\operatorname{End}(1)$ is commutative.
A generalization of the two notions of the center of a category I talked about is the following. Let $\mathsf{C}$ be a $2$-category, then the endomorphisms of the identity $\operatorname{End}(\operatorname{id}_x)$ of any object $x \in \mathsf{C}$ has two multiplications, vertical and horizontal composition. Thus the two laws are the same and it's a commutative monoid.

To recover the center $Z(\mathsf{C})$, consider the $2$-category $\mathsf{Cat}$ of categories and let $\mathsf{C} \in \mathsf{Cat}$, then $Z(\mathsf{C}) = \operatorname{End}(\operatorname{id}_\mathsf{C})$.
To recover the center $Z(\mathsf{C}, \otimes)$, consider the monoidal category $(\mathsf{C}, \otimes)$ as a $2$-category $\Sigma \mathsf{C}$ with a single object $*$ (this is the "suspension"), then $Z(\mathsf{C}, \otimes) = \operatorname{End}(\operatorname{id}_*)$.

(The following is more advanced, and purposefully vague to not get bogged down in technical details.)

A category is a category. (I didn't manage to get Markdown to start the list at 0, but consider this to be the zeroth case.)
A monoidal category is (more or less) a monoid object in the category of categories.
A $2$-fold monoid object in the category of categories (ie. a category $\mathsf{C}$ with two bifunctors $\otimes_1, \otimes_2 : \mathsf{C}^2 \to \mathsf{C}$ that are "compatible" in the Eckmann–Hilton sense) is a braided monoidal category, not a symmetric monoidal category, as one might expect!
But a $3$-fold monoidal category is a symmetric monoidal category, and it stabilizes: a $k$-fold monoidal category is symmetric for $k \ge 3$.

This is a higher version of the Eckmann–Hilton argument. In Higher category theory you have similar generalizations, by considering $k$-fold monoidal $n$-categories. You should look at the table 21 in Baez & Dolan's article cited below.
An $E_n$-algebra (ie. an algebra over the little disks operad) is, in some sense, a (strongly) homotopy-associative algebra, which is homotopy commutative "up to level $n$". One can make an alternative definition in the spirit of Eckmann–Hilton: consider a symmetric monoidal $(\infty, 1)$-category $\mathsf{C}$. Then an $E_n$-algebra in $\mathsf{C}$ can be defined as an object $A$ equipped with $n$ compatible multiplications (cf. Lurie below).
The link with braided monoidal categories can also be made here. The nerve of an $n$-fold monoidal category is an $E_n$-algebra (cf. Balteanu–Fiedorowicz–Schwänzl–Vogt below).

Useful references, and also interesting articles related to what I said at the end:

Eckmann–Hilton argument at Wikipedia.
These two blog posts by John Baez: number 1, number 2. The second one has a very nice graphical representation of the argument.
John C. Baez and James Dolan. “Higher-dimensional algebra and topological
quantum field theory”. In: J. Math. Phys. 36.11 (1995), pp. 6073–6105. ɪssɴ: 0022-2488. ᴅᴏɪ: 10.1063/1.531236. arXiv: q-alg/9503002 [math.QA]. MR1355899.
C. Balteanu, Z. Fiedorowicz, R. Schwänzl, and R. Vogt. “Iterated monoidal categories”. In: Adv. Math. 176.2 (2003),
pp. 277–349. ɪssɴ: 0001-8708. ᴅᴏɪ: 10.1016/S0001-8708(03)00065-3. arXiv: math/9808082 [math.AT]. MR1982884.
Jacob Lurie. “On the classification of topological field theories”. In: Current developments in mathematics, 2008. Int. Press, Somerville, MA, 2009, pp. 129–280. arXiv: 0905.0465 [math.CT]. MR2555928.


TITLE: Cleaning Already Mounted Microspheres on an AFM
QUESTION [12 upvotes]: I am running an experiment that uses a microsphere (Diameter ~ 50um) mounted onto the end of an AFM cantilever.  I run several experiments in a row, but somewhere during a sequence, the microsphere acquires some non-organic dirt that effects the outcome of the experiment.
Does anyone know - or can suggest - a way to clean the ball while it is still mounted onto the AFM?  Or a way to clean the ball at all?  We have a UV plasma cleaner, but it is not doing the job.
Thanks

REPLY [1 votes]: You can use general metal cleaning processes used in microelectronics - which is basically washing in acids and then in DI water. You obviously should choose chemicals which does not dissolve your sphere or cantilever.
If sphere is SiO2 then sulfuric, nitric or HCl acids should do the job. Not sure which of these would be safe for cantilever.


TITLE: Partial derivatives with transformation matrix
QUESTION [1 upvotes]: I have the following function:
$ f(x,y) = 2xy + \frac{x}{y}$
I need do determine if it is differentiable at $ a = (1,1) $.
Calculating the partial derivatives and plugging the values, I get that the linear transformation corresponds with multiplying by :
$ [ 2y + \frac{1}{y}, 2x - \frac{x}{y^2}] = [ 3, 1]  $
so as $[h,k] \to [0,0]$, I have:
$f(1 + h, 1 +k) - f(1,1) - \lambda([h,k]) = \frac{k^2+hk+2hk^2}{1+k} $
thus I need to show:
$\lim \limits_{[h,k] \to [0,0]} \frac{\frac{k^2+hk+2hk^2}{1+k}}{||[h,k]||} = 0 $
when $||[h,k]||$ is the euclidean norm.
This is probably needing some clever algebraic trick which I cannot seem to find..

REPLY [1 votes]: Since the partial derivatives are continuous at $(1,1)$, $f$ is differentiable there.Here's a proof of the fact  that $f'(1,1)(x,y)=3x+y$ from the definition. Let $\varphi(x,y)=3x+y$. Note that\begin{align}f(x,y)-f(1,1)-\varphi\bigl((x,y)-(1,1)\bigr)&=2xy+\frac xy-3x-y+1\\&=\left(2-\frac1y\right)(x-1)(y-1)+\frac{(y-1)^2}y.\end{align}Now, take $\varepsilon>0$. If $\bigl\lVert(x,y)\bigr\rVert<\frac14$, then $\lvert y-1\rvert<\frac14$ and so $\lvert y\rvert>\frac34$ and $\left\lvert2-\frac1y\right\rvert<\frac23$. So\begin{align}\left\lvert f(x,y)-f(1,1)-\varphi\bigl((x,y)-(1,1)\bigr)\right\rvert&=\left\lvert\left(2-\frac1y\right)(x-1)(y-1)+\frac{(y-1)^2}y\right\rvert\\&\leqslant\frac23\lvert x-1\rvert\lvert y-1\rvert+\frac43\lvert y-1\rvert^2.\end{align}So, take $\delta>0$ such that $\delta<\frac12$ and that $\delta<\frac\varepsilon2$. Then, if $\lVert(x,y)\rVert=r<\delta$, you have $\lvert x-1\rvert,\lvert y-1\rvert\leqslant r$ and\begin{align}\frac{\left\lvert f(x,y)-f(1,1)-\varphi\bigl((x,y)-(1,1)\bigr)\right\rvert}{\lVert(x,y)\rVert}&\leqslant\frac23r+\frac43r\\&=2r\\&<2\delta\\&<\varepsilon.\end{align}


TITLE: How does temperature affect a semiconductor band gap?
QUESTION [2 upvotes]: From this link, I've read that

An increased interatomic spacing decreases the potential seen by the electrons in the material, which in turn reduces the size of the energy bandgap.

Can this statement be explained more clearly?

REPLY [8 votes]: Let us take it one step at a time, when the temperature increases the vibration energy of atoms increases causing the distance between them to increase. I hope that is clear.
Now we know from solid state physics that electrons exist in bands rather than discrete levels for single atoms. The electrons in the valance band are the outermost electrons, which are responsible for chemical bonding and conduction of heat/current. All the characteristics of electronic bands can be obtained by solving Schrödinger equation for the electrons in that band. That requires extensive mathematical treatment.
The electrons in avalanche band “feel” or “see” electric potential coming from the atoms/ions they are associated with and from repulsing each other. Let us focus at the potential coming from atoms/ions. Electrons when they are moving pass by atoms/ions periodically. So they experience what is called a periodic potential, have a look at the following figure.  Now when the interatomic distance increases, the periodic potential becomes weaker on average (the parts in which the potential is zero become larger and wider). So the potential energy of electrons decreases. Keep in mind that the potential energy is the product of electron charge which is minus by the potential which is always less than or equal zero, so the minimum potential energy of electrons is when the potential is zero.

The potential is one of many factors determining the band characteristics. It is not easy to see the direct correlation between the potential and the band gap size.  The best way to understand the impact of reducing potential on band gap can is by solving Schrodinger equation. It is difficult to solve it but luckily people have solved it for us. The effect of increasing the interatomic spacing is shown in the next figure.

As you can see when the interatomic distance increases we move toward the right in the above figure.  The width of the band gap decreases when going to the right until it vanishes. If the intermolecular distance becomes very large, that corresponds to dissolving solid into separate atoms where atomic levels are restored.
To put everything in one picture, have a look at this report. It might not be easy for you to understand the mathematical expressions if that is not your field of knowledge, but it will show you what forces are taken into account, what assumptions are made, and how band structure changes as function of interatomic forces. It will also show you how tricky it is to solve Schrodinger equations for electronic bands.
I hope that was useful


TITLE: given two sets how many ways can we choose 2 subsets of same length?
QUESTION [0 upvotes]: Given two sets (not neccesarily of same size), how many ways can we choose two subsets of same size, each from a different set? Order here doesn't matter and empty subsets count too, also I don't need the answer which includes binomial coefficients. I tried thinking about the Cartesian product of both sets' partitions and how we can eliminate pairs with different lengths but couldn't figure it out.
Example:
given two sets {0,1} and {2,3,5} there are ${2\choose 0}*{3\choose 0}+{2\choose 1}*{3\choose 1}+{2\choose 2}*{3\choose 2}$ possible combinations

REPLY [1 votes]: Suppose $|A| = n, |B| = m$ (and that $n \leq m$). We want to pick a two subsets of size $k$, one from $A$ and one from $B$. Note that we can do this in $\binom{n}{k} \binom{m}{k}$ ways; summing over all $k$ gives $\sum\limits_{k=0}^n \binom{n}{k} \binom{m}{k}$.
But it turns out that there is a closed-form expression for this. Note that $\binom{n}{k} = \binom{n}{n-k}$. So our sum is equal to $\sum\limits_{k=0}^n \binom{n}{n-k} \binom{m}{k}$. In other words, how many ways can we choose $k$ elements of $B$ and $n-k$ elements of $A$ as $k$ ranges over all possible values? But this merely asks how many ways are there to pick $n$ elements from $A \cup B$ (assuming the sets are disjoint), which is $\binom{m+n}{n}$. Intuitively, this counts the number of ways to choose $n-k$ elements of $A$ to exclude and $k$ elements of $B$ to include.


TITLE: How can one write the electron wavefunction in a periodic lattice as $\psi(x)=\sum_GC_{k-G}e^{i(k-G)x}?$
QUESTION [1 upvotes]: The following is taken from Kittel's solid state physics.
An electron moving in a one-dimensional lattice of lattice constant $a$, suffers a periodic potential $U(x)$ where $U(x+a)=U(x)$. The periodicity of $U(x)$ implies $$U(x)=\sum_G U_Ge^{iGx}.$$ where $G=G_n=2\pi m/a$, the reciprocal lattice vectors. Due to periodic boundary condition, the wavefunction satisfies the periodicity condition $\psi(x+L)=\psi(x)$ so that $$\psi(x)=\sum_k C_ke^{ikx}$$ where $k=k_n=2\pi n/L$. Substituting these expansions into time-independent Schrodinger equation for the electron in the lattice, we get $$\Big(\frac{\hbar^2 k^2}{2m}-E_k\Big)C_k+\sum_GU_GC_{k-G}=0.$$
How does the equation above mean that $\psi(x)$ given above can be rewritten as $$\psi(x)=\sum_GC_{k-G}e^{i(k-G)x}?$$

REPLY [3 votes]: Let $\sim$ be an equivalence relation on the set of possible $k$'s, where $k\sim k' \iff k-k' = G$ for some reciprocal lattice vector $G$.  That is, $\{k_0,k_{\pm N},k_{\pm 2N},\ldots\}$ is one equivalence class, $\{k_1,k_{1\pm N},k_{1\pm 2N},\ldots\}$ is another, and $\{k_{-1},k_{-1\pm N},k_{-1\pm 2N},\ldots\}$ is a third.  We can label each equivalence class by the smallest $k$ (in absolute value) contained within it; then $k_0=0$ is the representative of the first class, $k_1 = \frac{2\pi}{L}$ is the representative of the second, and $k_{-1} = -\frac{2\pi}{L}$ is the representative of the third.
By convention, we choose the set of $N$ distinct representatives to be
$$FBZ := \left\{-\frac{\pi N}{L}, -\frac{\pi (N-2)}{L},\ldots,\frac{\pi (N-2)}{L}\right\}$$
Noting that $L=Na$, this can also be written
$$FBZ:= \left\{-\frac{\pi}{a},-\frac{\pi}{a}+\frac{2\pi}{L}, \ldots ,-\frac{\pi}{a} + (N-1)\frac{2\pi}{L}\right\}$$
which spans the range $\big[-\frac{\pi}{a},\frac{\pi}{a}\big)$.  $FBZ$ is called the first Brillouin zone.  In particular, observe that every possible value of $k$ is related to exactly one $k\in FBZ$ via translation by some reciprocal lattice vector.

This being the case, a generic wavefunction $\Psi$ can be decomposed as
$$\Psi(x) = \sum_{\text{all }k}C_k e^{ikx} =  \sum_{k\in FBZ}\left( \sum_{G} C_{k-G} e^{i(k-G)x}\right) = \sum_{k\in FBZ} \psi_k(x)$$
where
$$\psi_k(x) \equiv \sum_{G} C_{k-G} e^{i(k-G)x} = e^{ikx}\sum_G C_{k-G}e^{-iGx} \equiv e^{ikx} u_k(x)$$
with $u_k(x)\equiv \sum_G C_{k-G}e^{-iGx}$ a manifestly periodic function with period $a$. Furthermore, if we look carefully at the equation
$$\left(\frac{\hbar^2 k^2}{2m}-E\right)C_k = -\sum_G U_G C_{k-G}$$
we see that it only couples $C_k$'s from within the same equivalence class. That is, the Hamltonian operator is "block diagonal" and does not couple these $\psi_k$'s together:
$$H \Psi(x) = \sum_{k\in FBZ} H_k \psi_k$$
As a result, we can solve the Schrödinger equation for one $k\in FBZ$ at a time.  The functions $\psi_k$ are called Bloch waves.