TITLE: Combining a sine and cosine, with complex coefficients, into a single sine with a phase shift
QUESTION [1 upvotes]: In section 13.4 of Townsend's Modern Approach to Quantum Mechanics on the partial wave expansion for scattering, the author discusses the asymptotic expansion of the spherical Bessel functions. He writes, for the wavefunction $\psi$ at large $r$,
$$\begin{align}
\psi(\vec r)&\rightarrow \sum_l \left[A_l\frac{\sin(kr-l\pi/2)}{kr}-B_l\frac{\cos(kr-l\pi/2)}{kr}\right]P_l(\cos\theta)\\
&=\sum_l C_l \frac{\sin(kr-l\pi/2+\delta_l(k))}{kr}P_l(\cos\theta)\\
\end{align}$$
where $P_l$ are the Legendre polynomials. I am puzzled by the operation going from the first to the second line. The books says "in the last step we have combined the sine and cosine into a sine with its phase shifted by $\delta_l(k)$".
I would know how to do this is $A_l$ and $B_l$ were real, but as far as I can tell, they are complex. I do not see how to do this with complex $A_l$ and $B_l$.
In particular, $C_l$ is also complex and $\delta_l$ is real. So it appears as if somehow an expression with 4 real parameters has been turned into an expression with only 3 real parameters. Could someone explain this to me?

REPLY [0 votes]: Write
$$
A_l\sin(kr-l\pi/2)-B_l\cos(kr-l\pi/2)=A_l\frac{e^{ikr-il\pi/2}-e^{-ikr+il\pi/2}}{2i}-B_l\frac{e^{ikr-il\pi/2}+e^{-ikr+il\pi/2}}{2}.
$$
This yields
$$
-\frac{1}{2}(iA_l+B_l)e^{ikr-il\pi/2}-\frac{1}{2}(-iA_l+B_l)e^{-ikr+il\pi/2}=-\frac{1}{2}(iA_l+B_l)\left[e^{ikr-il\pi/2}+\frac{-iA_l+B_l}{iA_l+B_l}e^{-ikr+il\pi/2}\right].
$$
Now, notice that, by normalization of the wave function one should have
 $\sum_l|A_l|^2+|B_l|^2=1$ and
$$(-iA_l+B_l)(-iA_l^*+B_l^*)=|B_l|^2-|A_l|^2-2i[\operatorname{Re}(A_l)\operatorname{Re}(B_l)+\operatorname{Im}(A_l)\operatorname{Im}(B_l)].$$
Due to the normalization condition, we can choose $|B_l|^2=\cos^2\delta_l$ and $|A_l|^2=\sin^2\delta_l$. What really counts here is the relative phase between $A_l$ and $B_l$ because the wavefunction is defined aside an overall phase. We can choose $\operatorname{Im}(B_l)=0$. This finally gives
$$
e^{ikr-il\pi/2}+\frac{-iA_l+B_l}{iA_l+B_l}e^{-ikr+il\pi/2}=
e^{ikr-il\pi/2}+e^{-2i\delta_l}e^{-ikr+il\pi/2}
$$
as required. The key points of the proof are the normalization condition and an arbitrary phase available due to the properties of the wavefunction.


TITLE: Calculate $\sum_{n=1}^{\infty}\frac{(2n-1)!!}{(2n)!!\cdot 2^n}$
QUESTION [4 upvotes]: Calculate the sum $$\displaystyle \sum_{n=1}^{\infty}\frac{(2n-1)!!}{(2n)!!\cdot 2^n}$$
  where $(2n-1)!!=1\cdot 3\cdots (2n-1)$, $(2n)!!=2\cdot 4 \cdots 2n$

Using Wolframalpha, the result is $\sqrt{2}-1$. But I don't have any idea to come up with that result.
Thanks a lot.

REPLY [6 votes]: First note that
$$(2n-1)!!=\frac{(2n)!}{2^nn!}$$
and $(2n)!!=2^nn!$, so
$$\frac{(2n-1)!!}{(2n)!!\cdot2^n}=\frac{(2n)!}{2^nn!\cdot2^nn!\cdot2^n}=\frac1{2^{3n}}\binom{2n}n\;.$$
Now use the power series
$$\frac1{\sqrt{1-4x}}=\sum_{n\ge 0}\binom{2n}nx^n\;,\tag{1}$$
the generating function for the central binomial coefficients. $\binom{2n}n\le 4^n$ for $n\ge 1$, so $(1)$ certainly converges at $x=\frac18$, and we have
$$\sqrt2=\frac1{\sqrt{1-1/2}}=\sum_{n\ge 0}\binom{2n}n\left(\frac18\right)^n=\sum_{n\ge 0}\frac1{2^{3n}}\binom{2n}n=1+\sum_{n\ge 1}\frac{(2n-1)!!}{(2n)!!\cdot2^n}\;.$$


TITLE: Is the topology of weak+Hausdorff convergence Polish?
QUESTION [5 upvotes]: Let $X$ be a compact metric space, $P_X$ the set of Borel probability measures on $X$, and $K_X$ the set of non-empty closed subsets of $X$. I will define the "topology of weak+Hausdorff convergence" to be the metrisable topology on $P_X$ whose convergence is given by
\begin{align*}
& \mu_n \overset{\text{w+H}}\to \mu \\
&\ \ \ \,  \Longleftrightarrow \\
(\mu_n \to \mu \text{ weakly, and } \, \mathrm{supp}&\,\mu_n \to \mathrm{supp}\,\mu \text{ in Hausdorff distance})\text{.}
\end{align*}
This convergence is indeed metrisable, with a metric being given by
$$ d_{\text{w+H}}(\mu,\nu) = d_\text{w}(\mu,\nu) + d_\text{H}(\mathrm{supp}\,\mu,\mathrm{supp}\,\nu) $$
where $d_\text{w}$ is any metrisation of the topology of weak convergence, and $d_\text{H}$ is Hausdorff distance.
[This is a fairly natural topology to consider when one is interested in probability measures simultaneously at the "probabilistic" level of describing the distribution of likelihoods and at the "deterministic" level of prescribing what is and is not possible.]
Although $d_\text{w}$ and $d_\text{H}$ each generate compact topologies, it is easy to see that $d_{\text{w+H}}$ is not even complete: just take $\mu_n$ to be a sequence of fully supported probability measures converging weakly to a Dirac mass; then $\mu_n$ is $d_{\text{w+H}}$-Cauchy but not (weak+Hausdorff)-convergent as $n \to \infty$. Still, it would be nice at least to be able to say that the topology of weak+Hausdorff is Polish, meaning that it is both separable and admits a complete metrisation. (So, for example, even though the Euclidean metric on $\mathbb{R} \setminus \mathbb{Q}$ is not complete, it is known that the topology on $\mathbb{R} \setminus \mathbb{Q}$ generated by the Euclidean metric is Polish.)

Is the topology of weak+Hausdorff convergence Polish?

As in the answer to Comparison of several topologies for probability measures, separability should be fairly easy: e.g. let $\mathcal{K}$ be a Hausdorff-dense countable subset of $K_X$, and for each $C \in \mathcal{K}$ let $\mathcal{P}_C$ be a weakly dense countable subset of the set of probability measures supported on $C$; then $\bigcup_{C \in \mathcal{K}} \mathcal{P}_C$ should be (weak+Hausdorff)-dense.
So the remaining question is that of complete metrisability: Is there a complete metric on $P_X$ topologically equivalent to $d_{\text{w+H}}$?

My first guess: The $\infty$-Wasserstein metric
$$ W_\infty(\mu,\nu) = \inf\{\mathbb{P}\text{-}\mathrm{ess}\sup d(\boldsymbol{\cdot},\boldsymbol{\cdot}) \, : \, \mathbb{P} \text{ is a coupling of } \mu \text{ and } \nu \} $$
generates the topology of weak-Hausdorff convergence (perhaps generalising the answer to The infinity Wasserstein distance $W_\infty$ and the weak topology) and is complete?

REPLY [1 votes]: The set $MS=\{(\mu,K)\in P_X\times K_X:\mathrm{supp}(\mu)=K\}$ is of type $G_\delta$ in $P_X\times K_X$ and hence the weak+Hausdorff topology on $P_X$ is Polish.
Indeed, fix any countable base $\{U_n\}_{n\in\mathbb N}$ of the topology of $X$ and observe that $$MS=\bigcap_{n\in\mathbb N}\{(\mu, K)\in P_X\times K_X:\mu(U_n)>0\;\Leftrightarrow\;U_n\cap K\ne\emptyset\}.$$
So, it remains to show that for every $n\in\mathbb N$ the set
$$MS_n:=\{(\mu, K)\in P_X\times K_X:\mu(U_n)>0\;\Leftrightarrow\;U_n\cap K\ne\emptyset\}$$is of type $G_\delta$ in $P_X\times K_X$.
Observe that $(P_X\times K_X)\setminus MS_n=MS_n'\cup MS_n''$ where
$$MS_n':=\{(\mu,K)\in P_X\times K_X: \mu(U_n)>0\;\wedge\;U_n\cap K=\emptyset\}$$ and
$$MS_n'':=\{(\mu,K)\in P_X\times K_X: \mu(U_n)=0\;\wedge\;U_n\cap K\ne\emptyset\}.$$
The known properties of the topologies on the spaces $P_X$ and $K_X$ ensure that the set $\{\mu\in P_X:\mu(U_n)>0\}$ is open in $P_X$ and the set $\{K\in K_X:K\cap U_n\ne \emptyset\}$ is open in $K_X$. Then $MS_n'$ and $M_s''$ are intersections of  open and closed sets, so are of type $F_\sigma$ in $P_X\times K_X$. Then $MS_n$ is of type $G_\delta$ in $P_X\times K_X$ and so is the set $MS=\bigcap_{n\in\mathbb N}MS_n$.


TITLE: What is the value of the likelihood ratio test statistic for a given poisson distribution?
QUESTION [1 upvotes]: In past weeks, the average number of life insurance policies sold per week was $θ =
3$. However, this week he has sold 6 life insurance policies. Based on this single
observation of $x = 6$, what is the value of the likelihood-ratio test statistic for
testing the hypothesis that the salesman has suddenly seen a boost in his job, i.e.
testing $H_0 = \theta = 3$ , $H_1 = \theta > 3$.
For this, I used the likelihood ratio for poisson distribution, giving me $$\frac{\frac{3^6e^{-3}}{6!}}{\frac{6^6e^{-6}}{6!}}$$
This gives 0.3138365. Is this the right answer?

REPLY [1 votes]: You will want to reject $H_0: \theta = 3$ against $H_a: \theta > 3)$ for large values of your single observation $X.$ You can't test at exactly the 5% level because of the discreteness of the Poisson distribution.
Under $H_0,$ you have $P(X \ge 7) = .0335,$ but
$P(X \ge 6) = .0834.$ Exact Poisson computations from R
(where ppois is a Poisson CDF and qpois is
the quantile function):
qpois(.95, 3)
[1] 6
1 - ppois(6,3)
[1] 0.03350854  # 1 - P(X <= 6) = P(X >= 7)
1 - ppois(5,3)
[1] 0.08391794

So for test at level $\alpha =3.335,$ you should
reject if $X \ge 7.$ Because you observe $X = 6,$
the P-value is $P(X \ge 6\,|\,\theta = 3) = 0.834.$
In the plot below, $\alpha$ is the sum of the heights of the bars to the right of the vertical dotted line.

For such a small values of $\theta,$ depending on the desired accuracy, it may  no bet advisable to use a normal approximation. The best-fitting normal
curve is shown in blue.
However, using the normal approximation (with continuity correction), the P-value is
7.4%, so you would not reject at the 5% level.
1 - pnorm(5.5, 3, sqrt(3))
[1] 0.07445734

Addendum: Now that you have the correct likelihood ratio function $\lambda(x) = (3/x)^xe^{x-3}$ for $x \ge 3 \; (1$ for $x < 3),$ let's use R to plot it. Then we can see that $\lambda(x)$ is small (leading to rejection) for large values of $x$ as mentioned at the start.
x = 3:10;  like = (3/x)^x*exp(x - 3)
plot(x, like, pch=20, main="Likelihood Ratio")
abline(h = 0, col="green2")


TITLE: Prove that $L_p[0, 1]$ is strictly normed for $1 < p < \infty$
QUESTION [3 upvotes]: I was having trouble with the following question from Luenberg's Optimization by Vector Space Methods:

2.10 A normed space is said to be strictly normed if $\|x + y\| = \|x\| + \|y\|$ implies that $y = \theta$ or $x = \alpha y$ for some $\alpha$
a) Show that $L_p[0,1]$ is strictly normed for $1 < p < \infty$
b) Show that $X$ if $X$ is strictly normed, the solution to 2.9 (below) is unique.

Problem 2.9 (attempted here, any corrections/suggestions appreciated)  is:

2.9: Let $X$ be a normed linear space and let $x_1, x_2, \ldots, x_n$ be linearly independent vectors from $X$. For fixed $y\in X$, show that there are coefficients $a_1, a_2, \ldots, a_n$ minimizing $\|y - a_1 x_1 - a_2 x_2 - \ldots - a_n x_n\|$

Question 1: could you provide feedback, if any, on my solution to 2.9?
Both 2.9 and 2.10 are introduced before we've learned Lebesgue integration or about measure or measurable functions, so there should be a solution to 2.10 that doesn't require this knowledge.
Update 1: The solution to problem 2.10a is available here. I had a question about it which I've included as a comment to the first response, if anyone is able to answer it.
Update 2: I was able to find a solution to 2.10b here, which I modified to fit the problem in the answer below.

REPLY [1 votes]: 2.10 A normed space is said to be strictly normed if $\|x + y\| = \|x\| + \|y\|$ implies that $y = \theta$ or $x = \alpha y$ for some $\alpha$.

a) Show that $L_p[0,1]$ is strictly normed for $1 < p < \infty$

The solution is available here.

b) Show that if $X$ is strictly normed, the solution to 2.9 is unique.

Here's the best solution I've found:
Let $X$ be a strictly normed space;  $y$ an element of $X$; and $\mathcal{U}\subseteq X$ the subspace generated by $x_1, x_2, \ldots, x_n \in X$. The linear approximation from $\mathcal{U}$ to $y$ using coefficients $a = (a_1, a_2, \ldots, a_n)$ is written as $u_a$. Suppose the best linear approximations of $y$ are $u_\beta$ and $u_\nu$ where $\beta \neq \nu$ and $\|y - u_\beta\| = \|y - u_\nu\| = \lambda > 0$.
If $y\in \mathcal{U}$ then $u_\beta = u_\nu = y$, and there would be exactly one linear combination of $x_1, x_2, \ldots, x_n$ equal to $y$ and $\beta = \nu$. Since $\beta \neq \nu$, the vector $y\not\in \mathcal{U}$, meaning that neither $y - u_\mu$ nor $y - u_\beta$ equal $\theta$. Additionally, since $y\not\in \mathcal{U}$, there exists no $\alpha > 0$ such that $y  = \frac{1}{1-\alpha}u_\nu - \frac{\alpha}{1-\alpha}u_\beta$ and so $y - u_\nu \neq \alpha(y - u_\beta)$ for any $\alpha$. Since $X$ is strictly normed, by the contrapositive of the definition, Minkowski's inequality is strict. So
\begin{align}
\Big\|y - \frac{1}{2}(u_\nu + u_\beta)\Big\| &= \Big\|\frac{1}{2}(y - u_\nu) + \frac{1}{2}(y - u_\beta))\Big\| \\
&< \Big\|\frac{1}{2}(y - u_\nu)\Big\| + \Big\|\frac{1}{2}(y - u_\beta)\Big\| \\
&= \frac{1}{2}\|y - u_\nu\| + \frac{1}{2}\|y - u_\beta\|  = \lambda
\end{align}
meaning that $u_{(\nu + \beta)/2}$ provides a better approximation of $y$ than $u_\beta$ or $u_\nu$, contradicting the assertion that those were the best approximations. So, $\beta = \nu$.


TITLE: Subset of Spec(A) realized as inverse image of some Spec(B)
QUESTION [5 upvotes]: Let $A$ be a (commutative) ring and $U$ be an arbitrary subset of $\mathrm{Spec}(A)$. Do there exist a ring $B$ and a ring homomorphism $\varphi\colon A\to B$ such that $\varphi^{-1}(\mathrm{Spec}(B))=U$ ? If $\varphi\colon A\to B$ and $\varphi'\colon A\to B'$ are two ring homomorpisms such that $\varphi^{-1}(\mathrm{Spec}(B))=\varphi'^{-1}(\mathrm{Spec}(B'))=U$, do there exist a ring homomorpism $f\colon B\to B'$ such that $f\varphi=\varphi'$ ?

REPLY [3 votes]: There are already two nice counterexamples. Let me just point out for question (1) a characterization of images of $\mathrm{Spec}(B)$ in $X:=\mathrm{Spec}(A)$:

A subset $U$ of $X$ is the image of some $\mathrm{Spec}(B)\to X$ if and only if $U$ is pro-constructible.

See EGA IV.1.9.5. In a noetherian space, $U$ is pro-constructible if it is an arbitrary intersection of constructible subsets (i.e. finite unions of locally closed subsets) of $X$.  A constructible subset of $\mathrm{Spec}(\mathbb Z)$ (or any irreducible noetherian scheme of dimension $1$) is either finite or contains the generic point, so a pro-constructible subset is either finite or contains the generic point. This "explains" the example of Laurent.


TITLE: Finding velocity when mass is $8.0$ cm from equilibrium
QUESTION [1 upvotes]: I have the question "A mass is tethered between two springs. It is displaced by $10$ cm then released. The mass oscillates with $SHM$ with a frequency of $0.55$ Hz. Calculate the velocity when the mass is $8.0$ cm from the equilibrium position."
I used the formula
Velocity = total displacement / total time.
$$V = 0.08 / 1.82$$
And the answer I get is $V = 0.04$ m/s.
Is this correct ?

REPLY [0 votes]: This is simple harmonic motion.
Hence, the displacement is given by the equation:
$$s=A \cos{(\omega t)} \tag{1}$$
Where $A$ is the amplitude, and $\omega$ is the angular frequency.
If we differentiate both sides with respect to $t$, we obtain:
$$v=-A \omega \sin{(\omega t)} \tag{2}$$
Where $v$ is the velocity.
Reminder:
$$\omega = 2\pi f=1.1\pi$$
Where $f$ is the frequency.
We evaluate a time $t$ where the displacement $s=8 \text{ cm}=0.08 \text{ m}$ using equation $(1)$.
$$0.08=0.1\cos{(\omega t)}$$
$$\frac{4}{5}=\cos{(1.1 \pi \cdot t)}$$
$$t=\frac{\arccos\left(\frac{4}{5}\right)}{1.1\pi}$$
One solution is:
$$t \approx 0.186212 \text{ s}$$
We do not need to know all solutions, since for each of these solutions, the velocity is the same (Simple Harmonic Motion).
We now can substitute this value of $t$ into our expression for the velocity $v$ on equation $(2)$.
$$v=-0.1\cdot 1.1\pi \cdot \sin(1.1\pi \cdot t) \approx -0.207345 \text{ ms}^{-1}$$
Hence, the velocity when displaced by $8.0 \text{ cm}$ from the equilibrium position is:
$$|v| \approx 0.21 \text{ ms}^{-1}$$


TITLE: Do matter and antimatter undergo relativistic effects identically?
QUESTION [2 upvotes]: As far as I know, the Dirac equation describes quantum particles moving at relativistic speeds. Since the dirac equation predicts antimatter as well, we should expect matter and antimatter to undergo relativistic effects identically. But has this been experimentally proven?

REPLY [1 votes]: There's no evidence that matter and antimatter experience different relativistic effects. As Wiki puts it, "No Lorentz violations have been measured thus far, and exceptions in which positive results were reported have been refuted or lack further confirmations."
But people are searching. See Wiki for Antimatter tests of Lorentz violation.


TITLE: 3-regular claw free graphs
QUESTION [1 upvotes]: A graph is said claw-free if it contains no induced subgraph isomorphic to $K_{1,3}$. Prove that every 3-regular claw-free graph has a 2-factor such that this 2-factor only has cycles of length 3 or 4.
I think that maybe if we consider the 2-factor with the maximum number of cycles, this could be the answer but I do not know how to prove this.

REPLY [1 votes]: Consider a vertex $a$ in the $3$-regular graph and its three neighbours $b,c,d$. By claw-freeness, there must be an edge between at least one pair out of $b,c,d$. Say there is an edge $bc$, then $b$ and $c$ have exactly one other neighbour each, $e$ and $f$ respectively.
If all of $d,e,f$ are distinct, we include the cycle $abc$ in the $2$-factor; we call the subgraph induced by $abc$ a type A fragment. If two of $d,e,f$ are the same, say $e=f$, the vertex $e=f$ is adjacent to exactly one other vertex $g$, and we include the cycle $abec$ in the $2$-factor; the subgraph induced by $abec$ is a type B fragment.
Since the choice of $a$ was arbitrary, every vertex in the $3$-regular claw-free graph belongs to a fragment, so the graph may be partitioned into fragments that can each be covered by one $3$- or $4$-cycle. This proves the existence of a $2$-factor with only cycles of length $3$ and $4$.


TITLE: Conjugate prior of a normal distribution with unknown mean
QUESTION [1 upvotes]: I'm following these notes to compute the conjugate prior of a normal distribution with unknown mean and known variance. At some point they claim:
$p(D|\mu) \propto exp(-\frac{n}{2\sigma²}(\overline{x}-\mu)^2) \propto N(\overline{x},\theta,\frac{1}{n})$
but then they claim that the natural conjugate prior has the form:
$p(\mu) \propto exp(-\frac{n}{2\sigma_0^2}(\mu-\mu_0)^2) \propto N(\mu|\mu_0,\sigma_0^2)$
Is this mathematics?
Contrast this with these other notes that write the likelihood:
$L(\mu|x_1,\ldots,x_n) = c exp(-\frac{n(\mu-\overline{x}}{2\sigma^2}-\frac{\sum (x_i-\overline{x})^2}{2\sigma^2})$
then they claim that one should focus on the term $exp(-\frac{n(\mu-\overline{x}}{2\sigma^2})$ which is said to be clearly propotional to a normal distribution of $\mu$!!!
What is going on here? How can I compute the conjugate prior of a normal distribution with unknown mean?
References:
A good account of this case can be found in De Groot's "Probability and Statistics"

REPLY [1 votes]: Assume you model the likelihood through the sufficient statistic $\bar{X}$:
\begin{equation}
P(D|\mu) \propto \textrm{Exp}(-\frac{n}{2\sigma²}(\overline{x}-\mu)^2) \propto N(\overline{x},\theta,\frac{1}{n})
\end{equation}
And use the following as a prior for $\mu$:
\begin{equation}
P(\mu) \propto \textrm{Exp}(-\frac{n}{2\sigma_0^2}(\mu-\mu_0)^2) \propto N(\mu|\mu_0,\sigma_0^2)
\end{equation}
We need to show that the product of these two distributions has a kernel which is the kernel (as a function of $\mu$) of a normal distribution; so we can ignore all of the normalizing constants and any terms not containing $\mu$.
\begin{equation}
\begin{split}
P(D|\mu)P(\mu) & \propto \textrm{Exp}(-\frac{n}{2\sigma²}(\overline{x}-\mu)^2) \textrm{Exp}(-\frac{n}{2\sigma_0^2}(\mu-\mu_0)^2) \\
 & = \textrm{Exp}(-\frac{n}{2}[\frac{1}{\sigma^2}(\bar{X}^2 - 2\bar{X}\mu + \mu^2)+\frac{1}{\sigma_0^2}(\mu^2 - 2\mu\mu_0 + \mu_0^2)])
\end{split}
\end{equation}
Throwing away terms which do not contain $\mu$, we get:
\begin{equation}
\begin{split}
\textrm{Exp}(-\frac{n}{2}[\frac{1}{\sigma^2}(2\bar{X}\mu + \mu^2)+\frac{1}{\sigma_0^2}(\mu^2 - 2\mu\mu_0)]) & \propto \textrm{Exp}(-\frac{n}{2}[\mu^2(\frac{1}{\sigma^2} + \frac{1}{\sigma_0^2}) +\mu(\frac{2\bar{X}}{\sigma^2}-\frac{2\mu_0}{\sigma_0^2})])
\end{split}
\end{equation}
Now to get a quadratic in $\mu$ (and thus an appropriately specified normal distribution) we complete the square:
\begin{equation}
\begin{split}
 \textrm{Exp}(-\frac{n}{2}[\mu^2(\frac{1}{\sigma^2} + \frac{1}{\sigma_0^2}) +\mu(\frac{2\bar{X}}{\sigma^2}-\frac{2\mu_0}{\sigma_0^2})]) & =  \textrm{Exp}(-\frac{n}{2}[a(\mu-h)^2 + k])
\end{split}
\end{equation}
For some $a$, $h$, and a term $k$ which does not involve $\mu$ and can be removed.
Specifically, ignoring $k$, we have that $a = (\frac{1}{\sigma^2} + \frac{1}{\sigma_0^2})$, and that $h = -\frac{(\frac{2\bar{X}}{\sigma^2}-\frac{2\mu_0}{\sigma_0^2})}{2(\frac{1}{\sigma^2} + \frac{1}{\sigma_0^2})}$.
This gives us our mean ($h$), and the reciprocal of our variance (up to rescaling by the $n$ factor present):
\begin{equation}
\begin{split}
 \textrm{Exp}(-\frac{n}{2}[\mu^2(\frac{1}{\sigma^2} + \frac{1}{\sigma_0^2}) +\mu(\frac{2\bar{X}}{\sigma^2}-\frac{2\mu_0}{\sigma_0^2})]) & \propto  \textrm{Exp}(-\frac{n(\frac{1}{\sigma^2} + \frac{1}{\sigma_0^2})}{2}[(\mu-\frac{(\frac{2\bar{X}}{\sigma^2}-\frac{2\mu_0}{\sigma_0^2})}{2(\frac{1}{\sigma^2} + \frac{1}{\sigma_0^2})})^2])
\end{split}
\end{equation}
Which is a $\textrm{N}(\frac{(\frac{2\bar{X}}{\sigma^2}-\frac{2\mu_0}{\sigma_0^2})}{2(\frac{1}{\sigma^2} + \frac{1}{\sigma_0^2})}, (\frac{1}{n}(\frac{1}{\sigma^2} + \frac{1}{\sigma_0^2})^{-1})$.
Which corresponds to the result given here up to some canceling of values. The conjugate prior specified in the problem also has the $n$ factor, which the Wikipedia page doesn't have. You can just collapse that into $\sigma_0$, though, to get the same result they have.


TITLE: Should the Grothendieck ring of varieties be K_0 of numerical motives?
QUESTION [6 upvotes]: Assuming the Standard Conjectures, should the Grothendieck ring of varieties be the $K_0$ of the abelian category of numerical motives?

REPLY [8 votes]: As explained by Dan Petersen, the answer is no. There are additional issues though. For example, multiplication by the class $\mathbf L$ of the affine line $\mathbf A^1_k$ corresponds to Tate twist on the motivic level, hence is “motivically regular”. However, it has been proved recently by Lev Borisov that $\mathbf L$ is a zero-divisor in the Grothendieck group of varieties.


TITLE: What is the algebraic form of the momentum eigenstate?
QUESTION [0 upvotes]: I'm asking this in the context of trying to verify the equation $a^{\dagger}_{p} \vert 0 \rangle = \frac{1}{\sqrt{2\omega_p}} \vert p \rangle$.
So far I have calculated

$\vert 0 \rangle = e^{-\frac{1}{2}x^2m\omega_p}$ (omitting normalization)
$a^{\dagger}_{p} \vert 0 \rangle = \sqrt{2m\omega_p}x e^{-\frac{1}{2}x^2m\omega_p}$

Then when I try to apply $\hat{p} = -i \frac{d}{dx}$ onto $\sqrt{2\omega_p} a^{\dagger}_{p} \vert 0 \rangle$ to check that it is a momentum eigenstate, I get an eigenvalue that does not look anything like momentum. Is this the right way to check the equation? What does the momentum eigenstate look like? The eigen-equation $\hat{p} \vert p \rangle = p \vert p \rangle$ says that the solution is $\vert p \rangle = Ae^{ipx}$, but this looks very different from the algebraic form of $\sqrt{2\omega_p} a^{\dagger}_{p} \vert 0 \rangle$

REPLY [1 votes]: So the problem here is that you are confusing quantum mechanics of a single harmonic oscillator with quantum field theory, which is quantum mechanics of a field and can also be considered as quantum mechanics of an infinite number of harmonic oscillators. In quantum mechanics of a single oscillator, the ground state $|0\rangle$ can be represented as:
$$|0\rangle \dot{=} \langle x|0\rangle=\biggr(\frac{m\omega}{\pi\hbar}\biggr)^{1/4}e^{-\frac{m\omega}{2\hbar}x^2}$$
If we define the raising operator
$$a^\dagger = \sqrt{\frac{m\omega}{2\hbar}}\biggr(x-\frac{i}{m\omega}p\biggr)$$
then the first excited state of the harmonic oscillator is
$$a^\dagger |0\rangle=|1\rangle$$
Indeed, this is not a momentum eigenstate. Due to the lack of translational invariance of the harmonic oscillator potential, you wouldn't expect a stationary state of the Hamiltonian to have a definite momentum.
Now, if we move away from the harmonic oscillator and consider instead quantum mechanics of a field, the ground state $|0\rangle$ means something different. It refers to the ground state of a field. Likewise, the raising and lowering operators mean something different. They act on the states of the entire field and create particles with definite momenta. Also, the momentum operator that acts on a field is not $-i\frac{d}{dx}$. This is the momentum operator for a single harmonic oscillator. Therefore, you were confusing two different physical systems in your question. To learn more about the quantum mechanics of fields, read Schwartz again or read Peskin and Schroeder.