TITLE: Finite expectation
QUESTION [0 upvotes]: Let $X_1,X_2,...$ be i.i.d with mean $\mu$. Let $T$ be a stopping time with respect to $X_,X_2,...$ with $E(T)<\infty$. Show that $E\left(\sum_{n=1}^\infty |X_n|I\{T\ge n\}\right)<\infty$.
My idea is  $E\left(\sum_{n=1}^\infty |X_n|I\{T\ge n\}\right)=\sum_{n=1}^\infty E(|X_n|I\{T\ge n\})$ and if I am able to prove that $|X_n| \le K$ where K is non zero constatnt, then $\sum_{n=1}^\infty E(|X_n|I\{T\ge n\})\le K\sum_{n=1}^\infty P(\{T\ge n\})=KE(T)<\infty$. But I am not sure whether my claim is true. Can anyone give me any hints in solving this question?

REPLY [2 votes]: Observe that $ I \{ T \geq n \}) =  1 - I \{ T \leq n-1 \}) $. Now, $ T $ is a stopping time, hence the event $ \{ T \leq n-1 \} \in {\cal F}_{n-1} := \sigma ( X_1, X_2, \dotsc, X_{n-1}) $. Thus, $ I \{ T \geq n \}) $ is $ {\cal F}_{n-1} $ measurable. Since $ X_n $ is independent of $ X_1, X_2, \dotsc, X_{n-1} $, we have $ X_n $ is independent of $ {\cal F}_{n-1} $. Thus, for any $ n \geq 1 $, we have
\begin{align*}
\mathbb{E} \bigl( |X_n| I \{ T \geq n\} \bigr) & =  \mathbb{E} \Bigl( \mathbb{E} \bigl( |X_n| I \{ T \geq n \} \mid {\cal F}_{n-1} \bigr) \Bigr) \\
& =  \mathbb{E} \Bigl( I \{T \geq n\} \mathbb{E} \bigl(  |X_n|  \mid {\cal F}_{n-1} \bigr) \Bigr) \\
& =  \mathbb{E} \Bigl( I \{T \geq n\} \mathbb{E} \bigl(  |X_n|  \bigl) \Bigr) \\
& =  \beta \mathbb{E} \bigl( I \{T \geq n\} \bigr) = \beta \mathbb{P} \bigl( T \geq n\bigr)
\end{align*}
where $ \beta = \mathbb{E}( |X_n| ) $.
Thus, $ \mathbb{E} \Bigl(  \sum_{n=1}^{\infty} |X_n| I \{ T \geq n \} \Bigr) = \sum_{n=1}^{\infty} \mathbb{E} \Bigl(  |X_n| I \{T \geq n \} \Bigr) = \sum_{n=1}^{\infty}  \beta \mathbb{P} ( T \geq n) = \beta \mathbb{E} (T) $.


TITLE: trigonometry application question
QUESTION [0 upvotes]: I have a trigonometry question from my textbook that I am not sure how to approach.
View from a Satellite: The figures on the next page indicate
that the higher the orbit of a satellite, the more of the earth
the satellite can “see.” Let u, s, and h be as in the figure, and
assume that the earth is a sphere of radius 3960 mi.
(a) Express the angle $\theta$ as a function of h.

I know the dotted lines measure 3960 since that is the radius but there is no right angle in the triangles.  I am not sure how to approach this question.

REPLY [1 votes]: There is, in fact, a right angle in this question, but it isn't drawn in for you. The black lines in the diagram are tangent to the circle, so they are perpendicular to its radius, forming a right triangle with sides of the circle's radius, the tangent to the circle, and the line from the satellite to the center of the earth. In this case, since you now have a right triangle, you can say that
$$\cos\theta=\frac{3960}{3960+h}$$
$$\frac{1}{\cos\theta}=\frac{3960+h}{3960}$$
$$\frac{3960}{\cos\theta}=3960+h$$
$$h=\frac{3960}{\cos\theta}-3960$$
Which should be your final answer.

REPLY [0 votes]: There is, actually a right angle. The lines that form a big triangle are nothing but tangents to the surface. Thus that is actually nothing but a right triangle.
Therefore, $$\sec\theta=\frac{r+h}{r}$$
where $r=360 miles$


TITLE: In a complex metrizable topological vector space, $d(0,\alpha x)\neq |\alpha|d(0,x), \ \alpha \in \mathbb C.$
QUESTION [0 upvotes]: Let $(X,\tau)$ be a complex metrizable topological vector space with the metric $d$. Does the following hold:
$$d(0,\alpha x)=d(0,x),\ \forall \alpha \in \mathbb C, |\alpha|=1 \ ?$$
In general, the following holds: $$d(0,\alpha x)\neq |\alpha|d(0,x), \ \alpha \in \mathbb C.$$

REPLY [0 votes]: Absolutely not. One can easily find counterexamples, but it seems more important to understand why one should not expect such an equality. "Metrizable topological vector space" just means that the topology generated by the metric satisfies certain conditions (namely making the vector space operations continuous). This leaves plenty of room to break your proposed equality without changing the topology.
For a concrete example take $\mathbb{C}$ and let
$$\phi\colon\mathbb{C}\to\mathbb{C},\,z\mapsto \operatorname{Re}z+4 \mathrm{i}\operatorname{Im} z.$$
This map is clearly continuous with continuous inverse, thus
$$
d\colon\mathbb{C}\times \mathbb{C}\to [0,\infty),\,d(z,w)=|\phi(z)-\phi(w)|
$$
induces the Eucliean topology on $\mathbb{C}$ (which is of course a vector space topology). However $d(0,2)=2$, while $d(0,2i)=8$.


TITLE: Computing the trigonometric sum $ \sum_{j=1}^{n} \cos(j) $
QUESTION [2 upvotes]: I have a task to compute such a sum:
$$ \sum_{j=1}^{n} \cos(j) $$
Of course I know that the answer is $$ \frac{1}{2} (\cos(n)+\cot(\frac{1}{2}) \sin(n)-1) = \frac{\cos(n)}{2}+\frac{1}{2} \cot(\frac{1}{2}) \sin(n)-\frac{1}{2} $$
but I don't have any idea how to start proving it.

REPLY [3 votes]: You could also use $\sin(n+\frac{1}{2})-\sin(n-\frac{1}{2})=2\cos n\sin\frac{1}{2}$; $\;\;$ so if we let $a=2\sin\frac{1}{2}$,
$\cos n =\frac{1}{a}\left[\sin(n+\frac{1}{2})-\sin(n-\frac{1}{2})\right]\implies$
$\cos 1+\cos2+\cos 3+\cdots+\cos(n-1)+\cos n=$
$\frac{1}{a}\left[(\sin\frac{3}{2}-\sin\frac{1}{2})+(\sin\frac{5}{2}-\sin\frac{3}{2})+(\sin\frac{7}{2}-\sin\frac{5}{2})+\cdots+(\sin(n+\frac{1}{2})-\sin(n-\frac{1}{2}))\right]=$
$\frac{1}{a}\left[\sin(n+\frac{1}{2})-\sin\frac{1}{2}\right]=\frac{1}{2\sin\frac{1}{2}}\left[\sin n\cos\frac{1}{2}+\cos n\sin\frac{1}{2}-\sin\frac{1}{2}\right]$
$=\frac{1}{2}\left[\sin n\cot\frac{1}{2}+\cos n-1\right]$

REPLY [1 votes]: Let $$S_n = \sum_{j=1}^{n}\cos j.$$
Then:
$$S_n\cdot \sin\frac{1}{2}=\frac{1}{2}\sum_{j=1}^{n}\left(\sin(j+1/2)-\sin(j-1/2)\right)=\frac{1}{2}\left(\sin(n+1/2)-\sin(1/2)\right).$$


TITLE: Is there a natural way to give a bisimplicial structure on a 2-category?
QUESTION [6 upvotes]: I mean by the nerve functor.
Given a 2-category $\mathcal{C}$, if we forget the 2-category structure (just view $\mathcal{C}$ as a category), the nerve functor will give us a simplicial set $N\mathcal{C}$. However, $\mathcal{C}$ is a 2-category, thus for any two objects $x,y\in\mathcal{C}$, $Hom_{\mathcal{C}}(x,y)$ is a category, applying the nerve functor gives us a simplicial set $N(Hom(x,y))$.
My question is, can these two simplicial set structure compatible in some way, gives us a bisimplicial set $N_{p,q}(\mathcal{C})$, say? Or is there another way to give a bisimplicial structure on a 2-category?

REPLY [5 votes]: Yes. This is called the double nerve of a 2-category.
See in particular the first reference cited at that link.


TITLE: If an parabola has its focus at the (a,b) and has directrix at x=c....
QUESTION [1 upvotes]: If an parabola has its focus at the (a,b) and has directrix at x=c, what would the equation 4p(x – h) = (y – k)^2 look like in terms of a,b, and c?

REPLY [0 votes]: Hint: $|a-c|=2p$, $a+c=2h$, and $b=k.$  Try also drawing a general graph.


TITLE: Magnification - When is it negative?
QUESTION [0 upvotes]: In magnification, I keep on confusing the signs.
From what I understand currently, magnification is positive when the image is erect. An image is only erect when it is a virtual image, therefore virtual images = positive magnification. Vice versa, magnification is negative when the image is inverted, therefore a real image.
However using the equation $m = v/u$, $m$ is negative when $v$ is negative. And $v$ is only negative when the image is on the same side of the lens as the object. Therefore meaning a virtual image (According to physicsclassroom "The negative value for image distance indicates that the image is a virtual image located on the object's side of the lens."). This contradicts the statement above where virtual images = positive magnification.
Where am I going wrong?

REPLY [0 votes]: In optics, the following concepts should be kept distinct in your thinking:

where an object or image is located (e.g. on one side or another of a lens or mirror)

whether an image is real or virtual

whether an image is erect or inverted

In general all combinations of the above are possible when there are multiple lenses.You can have a real image before one lens combination, and after another lens combination. You can have a real image which is erect for one lens combination, and inverted for another, etc.
For a single lens, (idealized), the quantities $u$ (object distance) and $v$ (image distance) and $f$ (focal length), related by
$$
\frac{1}{u} + \frac{1}{v} = \frac{1}{f},
$$
are all signed quantities. That is, they can each be positive or negative.
The standard convention on these signs, for a lens, is:

$f$ is positive for a converging lens (e.g. a convex-convex one), and negative for a diverging lens (e.g. a concave-concave one).

if light is travelling left to right then $u$ is positive when the object is before, i.e. to left of, lens

$v$ is positive when the image is after lens, (i.e. to right of lens if
the light is travelling left to right)

For example, when an object is placed at distance $2f$ in front of a converging lens then we get a real inverted image at distance $2f$ after the lens. This situation is described by $f > 0$, $u = 2f$, $v=2f$ and the magnification is $M = -1$. More generally the magnification for a single lens, with $u$ and $v$ as described above, is given by
$$
M = - \frac{v}{u} .
$$
The best way to be confident of the sign in this formula is to draw the ray diagram for the case $u=v=2f$ and it is plain as day that the image is real and inverted.
When we use a magnifying glass, for example to magnify text on a page, we hold the lens a little closer to the paper than $f$, so this case is described by $u < f$. But since
$$
\frac{1}{v} = \frac{1}{f} - \frac{1}{u} = \frac{u - f}{uf}
$$
we shall then find that $v$ is negative. This means the image is on the same side of the lens as the object, and now we will find that $M$ is positive. The formula
$$
M = - \frac{v}{u} \;\;\;\; > 0  \; \mbox{ for magnifying glass used with $u < f$}
$$
is still working just fine. The image is a erect, not inverted. In this case the image is a virtual image, but the formulae do not tell you that. (The concept of a virtual image is simply that it is a location where light appears to have come from if we pretend the light arrived at its final direction of travel just by travelling in a straight line, undeviated by any lens or mirror. Of course the light did not really do that but it is a useful way to describe the final outcome.)


TITLE: Solve limit without L'Hopital
QUESTION [0 upvotes]: I'd like some help in solving this limit without using L'Hopital -
$$ \lim_{x\to 1}
\left(
\frac3{1-x^{1/2}} -
\frac2{1-x^{1/3}}
\right)
$$
(Original formula)

REPLY [1 votes]: To facilitate the analysis, we note that be making a substitution $x\to x^6$, the limit is equivalent to
$$\begin{align}
\lim_{x\to 1}\left(\frac{3}{1-x^3}-\frac{2}{1-x^2}\right)&=\lim_{x\to 1}\left(\frac{3}{(1-x)(1+x+x^2)}-\frac{2}{(1-x)(1+x)}\right)\\\\
&=\lim_{x\to 1}\left(\frac{1+x-2x^2}{(1-x)(1+x)(1+x+x^2)}\right)\\\\
&=\lim_{x\to 1}\left(\frac{(1-x)(1+2x)}{(1-x)(1+x)(1+x+x^2)}\right)\\\\
&=\lim_{x\to 1}\left(\frac{(1+2x)}{(1+x)(1+x+x^2)}\right)\\\\
&=\frac12
\end{align}$$


TITLE: A $2$-Sylow subgroup of $A_5$ has five conjugates?
QUESTION [1 upvotes]: How can I show a $2$-Sylow subgroup of $A_5$ has exactly $5$ conjugates?
Recall $A_5$ is the group of even permutation of the first five positive integers.
Thanks
Sketch: We have $|A_5|=5\cdot 3\cdot 2^2$. Suppose $P$ is a $2$-Sylow of $A_5$. It suffices showing $$|N_{A_5}(P)|=12,$$ for in this case we would have: $$[N_{A_5}(P): P]=\frac{12}{4}=3$$ so that $$[A_5:N_{A_5}(P)]=\frac{[A_5:P]}{[N_{A_5}(P):P]}=\frac{15}{[N_{A_5}(P): P]}=\frac{15}{3}=5.$$
And, $[A_5: N_{A_5}(P)]$ is exactly the number of conjugates to $P$ in $A_5$.
Here $N_G(P):=\{g\in G: gPg^{-1}=P\}$ is the normalizer of $P$ in $G$.
Now I don't know how to show $|N_{A_5}(P)|=12$, however I conjecture $N_{A_5}(P)=A_4$.

REPLY [0 votes]: $<(12)(34),(13)(24)>$ sits in $A_4$ when the latter is taken as the alternating subgroup of the permutation group on the symbols 1,2,3 and 4. Substitute 4 by 5 and you have $<(12)(35),(13)(25)>$ sitting in $A_4$ when the latter is taken as the alternating subgroup on the symbols 1,2,3 and 5. And both $A_4$s, or more precisely, both isomorphic copies of $A_4$ sit in $A_5$.


TITLE: Representations of a non-compact group are labeled by its maximal compact subgroup?
QUESTION [10 upvotes]: I don't have much of any awareness about the representation theory of non-compact Lie groups but I bumped into it for my work.
Is there some idea that the representations of a non-compact group are labeled by those of its maximal compact subgroup? If yes then I would like to know of explanations for the above and of references from where I can pick this up.

The conformal group of $3+1$ dimensional space-time is $SO(4,2)$ and apparently any representation of it can be written as a direct sum over representations of $SO(4) \times SO(2)$.
The N=2 superconformal group for $2+1$ dimensions is $SO(3,2)\times SO(2)$ and its maximal compact subgroup $SO(2) \times SO(3) \times SO(2)$ ? If yes then I would like to know how can this be proven. (In physics contexts these two $SO(2)$ factors are distinguished by physical generators of different meanings.)

I would like to know of the general framework in which the above fits in.
To quote from a paper a typical argument where such a thing seems to get used,
"Any irreducible representation of the superconformal algebra may be decomposed into a finite number of distinct irreducible representations of the conformal algebra...which are in turn labeled by their own primary states...hence the state content of an irreducible representation of the superconformal algebra is completely specified by the quantum numbers of its conformal primaries"
I would be very glad if someone can also give expository references or explanations specific to the above argument.

REPLY [12 votes]: Suppose that we have (to fix ideas) a unitary irreducible representation of a semi-simple Lie group $G$ (such as the non-compact Lie groups that you write down) on a Hilbert space.  (Here I mean irreducible in the Hilbert space sense, i.e. there are no proper invariant closed subspaces.)  Let's call the Hilbert space $V$ (just to give it a name).
A theorem of Harish-Chandra then says that $V$ is admissible, which means the following: fix a maximal compact subgroup $K$ of $G$.  Then each irreducible representation $W$ of $K$ appears with finite multiplicity as a subrepresentation
of $V$.  If we call this multiplicity $m_W$, then we may write
$V = \hat{\oplus}_W W^{m_W}$, i.e. as the Hilbert space direct sum (i.e. the completed direct sum) of the various $W$, each appearing with multiplicity $m_W$.  (This is a consequence of the Peter--Weyl theorem, and is true
for any unitary representation of a compact group in which each irrep. appears
with finite multiplicity.)
Now inside $\hat{\oplus} W^{m_W}$ we have the actual algebraic direct sum
$\oplus_W W^{m_W}$, and this has an intrinsic characterization as a subspace
of $V$, as the $K$-finite vectors.  (A vector $v$ is called $K$-finite if the
linear span of all its translates by elements of $K$ is finite-dimensional.)
Let's denote it by $V_K \subset V$.
It turns out that $V_K$, although it is not invariant under the action of $G$
(typically, unless $V$ happens to be finite-dimensional, which it usually won't be), is invariant under $\mathfrak g$, the Lie algebra of $G$.
One calls $V_K$ a $(\mathfrak{g},K)$-module, or also a Harish-Chandra module
(because it has actions of $\mathfrak{g}$ and $K$).  It turns out that
$V_K$ determines $V$, and the basis of Harish-Chandra's approach to the study
of unitary reps. of $G$ is to work instead with the underlying $(\mathfrak g,K)$-modules.
Now in principle, to recover $V$, one really needs $V_K$ as a $(\mathfrak g, K)$-module; i.e. forgetting $\mathfrak g$ and just remembering the $K$-action is throwing away a lot of information.
But in practice (at least in the examples that I know) non-isomorphic irreducible $V$ have different list of multiplicities $m_W$, and so just knowing
$V_K$ as a $K$-rep. may well already pin down $V$.
In fact, often one doesn't even have to know all the $m_W$, but just
the first non-zero vanishing value. (If I think of the reps. $W$ as being
labelled by their highest weights lying in some choice of dominant Weyl chamber for $K$.)
A good place to read about this (it is not short, but I found it very good for dipping into) is Knapp's book Representation theory of semisimple groups: an
overview based on examples.  He gives the basic definitions, a lot of examples, and goes on to develop various aspects of the theory (e.g. the theory of
the relationship between $V$ and the multiplicities $m_W$: this is known as
the theory of $K$-types).
Incidentally, Harish-Chandra was a student of Dirac, and (as far as I know) his study of unitary reps. of semisimple groups was inspired by Bargmann's treatment of the special case of $SL_2(\mathbb R)$, which was in turn inspired in part by the role of this group in physics.
On the other hand, I don't know of a treatment of the theory which directly relates it to the physics literature, and I can't parse the physics argument that you wrote down in detail.


TITLE: how come hot bodies emit radiations of several wavelengths?
QUESTION [0 upvotes]: Atoms in a rod of iron has discrete energy levels. When we heat it, its electrons after absorbing photons rise to next energy levels and then fall back emitting radiations. As all iron atoms are alike so radiations of same wavelength should be given off by iron atoms. Am I missing something?

REPLY [1 votes]: Iron atoms, as a dilute gas, have discrete energy levels.  An iron rod, however, has broad energy bands instead, has oxides and other scale compounds on its surface,
and at high temperatures, even Doppler shifts and internal vibrations
can significantly wash out the atomic spectrum features.
Impurities also give rise to extra energy levels.
One can expect a few gaps, but many features of the blackbody curve will
show up in the hot-solid-iron spectrum.   All the 'lines' of iron, all the
hybridized orbitals that bond the metal together (bands), all the spreading
of those lines due to pressure, shifts due to fields, and just plain randomness
of temperature, work against a sharp spectral-lines result.


TITLE: What does $\vee$ mean in set theory?
QUESTION [2 upvotes]: The following proof is from Probability by Davar Khoshnevisan. There is a symbol $\vee$ in the third sentence of the proof. What does this symbol mean, please? There seems no definition about it in the book and I certainly do not recall seeing this symbol in previous studies. Thank you!
Kolmogorov's Zero-One Law. If $\{ X_i\}^\infty_{i=1}$ are independent
random variables, then their tail $\sigma$-algebra $\mathscr{T}$ is
trivial in the sense that for all $E\in\mathscr{T}$, $P(E)=0$ or $1$.
Consequently, any $\mathscr{T}$-measurable random variable is a constant
almost surely.
Proof. Our strategy is to prove that every $E\in\mathscr{T}$ is
independent of itself, so that $P(E)=P(E\cap E)=P(E)P(E)$. Since
$E\in\mathscr{T}$, it follows that $E$ is independent of $\sigma(\{
X_i\}^{n-1}_{i=1})$. Because this is true for each $n$, Lemma 6.15 (iii)
ensures that $E$ is independent of $\vee^\infty_{n=1}\sigma(\{
X_i\}^{n-1}_{i=1})$, which is defined to be the smallest
$\sigma$-algebra that contains $\cup^\infty_{n=1}\sigma(\{
X_i\}^{n-1}_{i=1})$. In other words, $E$ is independent of all the
$X_i$'s, and hence of itself.

REPLY [2 votes]: The context is not quite set theory, but Boolean algebras, in particular $\sigma$-algebras. The symbol $\lor$ in that context is the supremum of two elements, or any collection of elements when defined.
If we think about these algebras as algebras of sets then this is exactly the union of the two sets, or in the case of a $\sigma$-algebra countably many sets.


TITLE: Magnetic field generated by an electron in an electric field
QUESTION [1 upvotes]: While studying the corrections of fine structure in the hydrogen atom my book tries to justify the spin-orbit term with some arguments about magnetic energy an spin. I'm fine with the calculations but I don't understand where his starting point comes from. It says that the magnetic field generated by an electron
moving in an eletric field $\vec{E}$ with velocity $\vec{v}$ is $$\vec{B}=-\frac{1}{c^2}\vec{E}\times\vec{v}$$
I've studied electromagnetism and a bit of special relativity, but I don't understand where this formula comes from.

REPLY [1 votes]: Here's how I would show it very briefly using Special Relativity:
Let's consider two frames $S$ and $S'$, with $S'$ moving with respect to $S$ at a velocity $v$. There is an electric field $\mathbf{E}$ in $S$, and the charge is at rest in $S'$. According to an observer in $S$, therefore, we have a charge moving in the electric field $\mathbf{E}$.
Now let's look at it from the charge's point of view: the charge only sees a static electric field in the universe, and describes it using a scalar potential. The four-potential the charge sees is therefore:
$$A'^\mu = \begin{pmatrix}\frac{\phi'}{c}\\0\\0\\0\end{pmatrix}$$
(note that primed quantities represent quantities measured in the $S'$ frame). How can we relate this to the four-potential in the $S$ (or "lab") frame? Well, using a Lorentz Transformation it's easy to see that (if the $S'$ is moving along the positive $\hat{\mathbf{x}}$ direction),
$$A^\mu = \begin{pmatrix}\gamma \frac{\phi'}{c}\\\gamma \frac{v}{c^2}\phi'\\0\\0\end{pmatrix},$$
which means that the new electric and magnetic potentials are
\begin{aligned}\phi &= \gamma \phi'\\ A_x &= \frac{v}{c^2}\gamma\phi',\end{aligned}
and substituting for $\phi'$, you can show that $$A_x = \frac{v}{c^2}\phi.$$
Of course, what we have done so far assumes the velocity to be only along $\mathbf{\hat{x}}$ which is just to simplify our work. It's possible (but for more tedious) to show that for an arbitrary velocity $\mathbf{v}$,
$$\mathbf{A} = \frac{\mathbf{v}}{c^2} \phi.$$
From here, it's a very short step to calculate the magnetic field, since
$$\mathbf{B} = \mathbf{\nabla \times A} = \frac{1}{c^2}\mathbf{\nabla \times }(\mathbf{v}\phi).$$
Since $\mathbf{v}$ is a constant, we can use a standard relation from vector calculus, that for a scalar function $\psi$ and a constant vector $\mathbf{C}$:
$$\mathbf{\nabla \times}(\psi \mathbf{C}) = \psi (\mathbf{\nabla \times C}) + (\nabla\psi)\times\mathbf{C}.$$
Using this relation, it can be shown very simply (since $\mathbf{v}$ is constant) that
$$\mathbf{B} = \frac{1}{c^2} (\nabla \phi) \times \mathbf{v},$$
and the definition of the electric field tells us that $$\mathbf{E} = - \nabla\phi,$$
and so $$\mathbf{B} = -\frac{1}{c^2} \left(\mathbf{E} \times \mathbf{v}\right)$$


TITLE: Second order relativistic corrections to Pauli equation from Dirac equation
QUESTION [4 upvotes]: I'm trying to derive the full and correct Hamiltonian for spin$\frac{1}{2}$ particles from Dirac equation up to second order in $v/c$. For a potential and magnetic field constant in time.
In particular, there should be Zeeman term, relativistic kinetic term, spin-orbit term and Darwin term:
$$\hat{H}=\frac{\mathbf{P}^2}{2m}+U-e \hbar \mathbf{ \sigma} \mathbf{B}-\frac{\mathbf{P}^4}{8m^3 c^2}- \frac{e \hbar}{4 m^2 c^2} \mathbf{ \sigma} (\mathbf{F} \times \mathbf{P})- \frac{e \hbar^2}{8 m^2 c^2} (\nabla \mathbf{F}) \tag{1}$$
Here:
$$\mathbf{P}=- i \hbar \nabla-e \mathbf{A} \\ \mathbf{B}= \nabla \times \mathbf{A} \\ \mathbf{F}=- \frac{1}{e} \nabla U$$
This is taken from various sources, for example, Berestetsky, Lifshitz, Pitaevsky "Quantum electrodynamics", (33.12) though I have added the Zeeman term and changed some notation.
Now, assuming the above is correct, I want to derive this expression from Dirac equation directly.
I use the following form of the equation as a starting point:
$$\begin{cases} \displaystyle \Psi_+ = - \frac{\sigma \mathbf{P}}{mc} \Psi_-+\frac{E}{mc^2} \Psi_+ \\ \displaystyle \Psi_- =  \frac{\sigma \mathbf{P}}{mc} \Psi_+ -\frac{E}{mc^2} \Psi_- \end{cases} \tag{2}$$
Where $\Psi_{\pm}$ are two component spinors and $\sigma$ is the vector of Pauli matrices.
Introducing a new energy:
$$E=mc^2+\mathcal{E} \\  \mathcal{E}=i \hbar \frac{\partial}{\partial t} -U $$
We obtain:
$$\begin{cases} \displaystyle  \frac{\mathcal{E}}{mc^2} \Psi_+= \frac{\sigma \mathbf{P}}{mc} \Psi_- \\ \displaystyle \left(2+ \frac{\mathcal{E}}{mc^2} \right)\Psi_- =  \frac{\sigma \mathbf{P}}{mc} \Psi_+  \end{cases} \tag{3}$$
Assuming $$\Vert \mathcal{E} \Vert \ll mc^2$$
We obtain, up to first order in $\Vert \mathcal{E} \Vert / mc^2$:

$$\mathcal{E} \Psi_+ = \frac{\sigma \mathbf{P}}{2m} \left(1- \frac{\mathcal{E}}{2mc^2} \right) \sigma \mathbf{P} \Psi_+ \tag{4}$$

I assume that this should directly give us (1).
However, after some simplifications, while I get most of the terms above right (including the most important SO term), I get an incorrect Darwin term (twice as large) and also a bunch of extra terms.
My full derivation is lengthy, so I wanted to ask first if I missed something? Is (4) correct (up to second order in $v/c$) and should it lead to (1) without any extra terms?
I have extensively used the identity:
$$(\sigma \mathbf{a}) (\sigma \mathbf{b})= \mathbf{a} \mathbf{b}+i \sigma (\mathbf{a} \times \mathbf{b})$$
though it's not always clear how to apply it to operators. In any case, it gives a lot of extra terms not present in (1) when applied to (4).

There's a related question, but only about Pauli equation, which is simple enough to derive.

REPLY [1 votes]: Let me provide some steps of the derivation here to see where the problem had been.
$$\mathcal{E} \Psi_+ = \frac{\sigma \mathbf{P}}{2m} \left(1- \frac{\mathcal{E}}{2mc^2} \right) \sigma \mathbf{P} \Psi_+$$
Let's change the notation a little:
$$\mathcal{E}=E_c-U$$
$$E_c \Psi_+ = \frac{(\sigma \mathbf{P})^2}{2m}  \Psi_+ +U \Psi_+ -\frac{E_c}{4m^2 c^2} (\sigma \mathbf{P})^2 \Psi_+ + \frac{1}{4m^2 c^2} ( \sigma \mathbf{P} U \sigma \mathbf{P} )\Psi_+ $$
$$E_c \Psi_+ = \left(1- \frac{(\sigma \mathbf{P})^2}{4m^2 c^2}\right) \left( \frac{(\sigma \mathbf{P})^2}{2m}  +U  + \frac{1}{4m^2 c^2} ( \sigma \mathbf{P} U \sigma \mathbf{P} ) \right)\Psi_+ $$
$$E_c \Psi_+ = \left( \frac{(\sigma \mathbf{P})^2}{2m}  +U - \frac{(\sigma \mathbf{P})^4}{8m^3 c^2}- \frac{(\sigma \mathbf{P})^2 U}{4m^2 c^2} + \frac{1}{4m^2 c^2} ( \sigma \mathbf{P} U \sigma \mathbf{P} ) \right)\Psi_+ $$
It's easy to identify several correct terms from (1) immediately.
One important note: Berestetsky et al derive the expression (1) sans the Zeeman term for the case when no vector potential is present. Which is probably why I initially got all the extra terms.
In this simple case we can write (defining $\mathbf{P}=\mathbf{p}-e \mathbf{A}$):
$$(\sigma \mathbf{P})^2=\mathbf{p}^2$$
Another important note: the probability density up to $v^2/c^2$ has the following form:
$$\rho=|\Psi_+|^2+|\Psi_-|^2=|\Psi_+|^2+ \frac{1}{4m^2 c^2} |\sigma \mathbf{p} \Psi_+|^2$$
To preserve the usual definition, according to Berestetsky et al, we have to definte another wavefunction:
$$\Psi_+= \left(1- \frac{\mathbf{p}^2}{8m^2 c^2}\right) \Psi$$
Which leaves us with:
$$E_c \Psi =\left(1+ \frac{\mathbf{p}^2}{8m^2 c^2}\right) \left( \frac{\mathbf{p}^2}{2m}  +U - \frac{\mathbf{p}^4}{8m^3 c^2}- \frac{\mathbf{p}^2 U}{4m^2 c^2} + \frac{1}{4m^2 c^2} ( \sigma \mathbf{p} U \sigma \mathbf{p} ) \right) \left(1- \frac{\mathbf{p}^2}{8m^2 c^2}\right)\Psi  $$
Up to second order we have:
$$E_c \Psi = \left( \frac{\mathbf{p}^2}{2m}  +U - \frac{\mathbf{p}^4}{8m^3 c^2}- \frac{\mathbf{p}^2 U}{8m^2 c^2}- \frac{ U \mathbf{p}^2}{8m^2 c^2} + \frac{1}{4m^2 c^2} ( \sigma \mathbf{p} U \sigma \mathbf{p} ) \right) \Psi  $$
Which coincides with the expression provided in Berestetsky et al before they derive (1) (without the Zeeman term I added).
So I guess my question is answered.

If someone adds a more in depth and/or intuitive explanation for all this stuff, I will gladly award them the bounty.