@TedShifrin can you elaborate, actually, on what 'tangentially related' means here? my impression had been that Picard-Fuchs stuff was essentially a special case of Gauss-Manin
well, the physical idea of the path-integral formulation is that rather than trying to look for a particular path for a quantum particle in time, you look at the entire space of such paths
those which correspond to classical paths will have stationary phase, whereas those which are not will be oscillatory. that means that as $h\to 0$, the only paths which contribute to the solution are the classical trajectories
on the other hand, if $h$ is small but nonzero, then paths which are near to the classical trajectories will also contribute
so from that you get both the classical limit as well as semiclassical approximations to the real quantum mechanical behavior
in quantum mechanics, we instead have the time-dependent eigenvalue problem $i\hbar\partial_t \Psi = \hat{H} \Psi$ where $\hat{H}=-\frac{1}{2}D^2+\frac{1}{2}x^2$
okay. now, the time-independent equation is simply $H\psi = E\psi$
i also need to have boundary conditions, as since I want to be able to interpret $|\psi|^2$ as a probability distribution I need to require that $\psi$ be square-integrable
i'll treat that a little sloppily, but it'll work fine in this concrete case
at the very least, though, i need to be sure that $\psi$ goes to zero at infinity
to obtain that behavior, let me write out the Schrodinger equation as $$-\frac{\hbar^2}{2}\psi''(x)+\frac{1}{2}x^2 \psi = E\psi$$
to proceed, it'll be convenient for me to write $\psi = e^{iS/\hbar}$
taking derivatives and dividing out by $\psi/2$, this gives
it's a little sad how our decisions force our future. while I could certainly take the time to learn all of a physics curriculum, I don't have time to, and it's unlikely I will
there's too much out there and we'll always be unsatisfied.
anyways, i said earlier that I needed to be careful about the behavior at infinity. to that end, let me approximate things further and say $S'(x)\approx \pm i x$ (i.e. dropping the energy as irrelevant)
then $S(x)=S(0)\pm ix^2/2$ and so $\psi(x)\approx e^{i S(x)/\hbar} = A e^{\pm x^2/2}$
of these two signs, only the negative one is reasonable since otherwise it blows up at infinity
typo: should've been $x^2/2\hbar$
if I don't do this large-$x$ approximation, then instead I formally have $S(x)=\int\sqrt{E-u^2}\,du$ and $$\psi(x)\sim \exp\left(-\frac{1}{2\hbar}\int^x\sqrt{u^2-E}\,du\right)$$
consequently, we heuristically expect $\psi(x)$ to be a complex exponential if $u$ is small, and to be a damped real exponential if it's large.
and that reflects the fact that the potential $x^2$ blows up as $x\to\infty$, meaning that the particle must be found inside this well
by contrast, i could instead consider a potential like $-x^2/(1+x^2)$. formally, everything I've said still goes through
except that now, if $E>0$, the result analogous to the one I ended at above leads to the $\ln \psi(x)$ being oscillatory everywhere
in which case the meaning of my boundary conditions will have to be a lot more subtle, since $\psi$ will fail to be square-integrable
but i can't disregard them, since they are physically meaningful: If I send a particle in from the left, it'll can pass by that well unimpeded or it can be reflected
and in that case, i'll end up having to deal with oscillatory integrals in order to make calculations
(obviously, that's a ton of hand-waving)
i'm sort of losing the thread of what i was saying, so i'll stop here
how do i solve this problem (rate of change): a ball starts to fall 10 ft from the top, each .5 second the ball get's closer to hitting the ground and it is accelerating more and more, how do i find the rate of change and when will the ball hit the ground if the first .5 secs the ball covered 2 ft, the next .5 secs it covered 4 ft?
@Semiclassical I got a quick question that you can probably answer. I have the matrix $$\begin{bmatrix} 1 & 1 & 0 & 1 & 0 \\ 1 & 1 & 0 & 1 & 0 \\ 1 & 1 & 0 & 0 & 1 \\ 1 & 1 & 0 & 0 & 1 \\ 1 & 0 & 1 & 1 & 0 \\ 1 & 0 & 1 & 1 & 0 \\ 1 & 0 & 1 & 0 & 1 \\ 1 & 0 & 1 & 0 & 1 \end{bmatrix}$$ What's the easiest way to see that the rank of this matrix is $3$? Obviously the first column is redundant, so I know the rank is less than or equal to $4$.
@clarinetist main thing i notice is that the first three pairs of rows are identical
@adan my main concern is that, while its true that quadratic functions have slopes which are linear in time, that leaves open what linear function that is
@Semiclassical And you get something similar for the columns (drop the first column, you would - I would think - get a matrix of rank 4). But my program suggests the rank is 3. Hmm.
You can start with the last two columns, and observe that the third column can not be constructed from these two last columns. Thus, the column rank is at least 3. Finally, you can see that ($c_i$ being column $i$)
$$
c_4 + c_5 - c_3 = c_2
$$
thereby eliminating $c_2$. As a result, you are left ...
As a math graduate student I feel that I should have better math jokes to keep my students interested, but all I can think of is the old standard: Why is six afraid of seven? Because seven eight nine!
Any help?
In the discrete topology, a singleton $\{x\}$ is open, so $X-\{x\}$ is closed. Since the discrete topology is (in this case) equal to the finite-closed topology, $X-\{x\}$ is finite.
@Semiclassical: OK, so the theme is that in working with solutions of the Schroedinger equation, you get an extra term: the oscillatory integral, and you'd really like to ignore it; so frwuently you want to know precisely how small it is. Yes?
in the case of the harmonic oscillator potential, the energy levels are quantized $E_n = n+\frac12$. this can be expressed in terms of the classical action as
here, the integration is to be understood over one classical oscillation (i.e. negative amplitude to positive back to negative, with positive momentum on first part and negative on the second)
to get an idea of where complications arise, note that if there's two minima for the potential and one maximum between them, then there are two cases. if the energy is above the maximum, you get precisely the same result as above
if you put $E$ between the minimum and the maximum, though, then things are different because there's a region in the middle where $K<0$
to a first approximation, it's enough to just consider the two regions as 'separate' potentials and use the same formula above
on the other hand, if the potential is even, then that prescription would give the same eigenvalue twice
I don't know how to write the second equation for a systems of equation.
Mr.Salvatore wants to mix 2 kinds of nuts that cost $0.90 per pound and $1.30 per pound to make 50 pounds of a combination that cost $1.20 per pound. How many of each should he use?
For the first equation, I got x +y = 50. For the second it's 0.9x + 1.3y = 1.2?
If a statement says that there is a set of all elements whose orders are powers of $p$ (where $p$ is a prime) then if I call that set $H$ does that mean that $H$={$e,p,p^2, ....., p^k$}?
Can anyone explain that what does a group that is abelian have anything to do with proving a subgroup is a set of elements whose orders are powers of primes?
@BalarkaSen By the way, I made a mistake earlier; the Euler characteristic of a disk with a handle is -1, not the genus. (I don't know how one defines genus for surfaces with boundary.)
@XTImpossible Don't use 2 variables. There are X pounds of the $1.30 nut, and (50-X) pounds of the $.90 nut. The question is fine as the result is between the two extremes. You can mix to 100% of cheap nut for .90/lb or higher, $1.30/lb, or anything in between.
@BalarkaSen You probably just glue a circle onto each boundary component and take the genus of that, in which case the genus of a disk with a handle would be 1.
Mr.Salvatore wants to mix $2$ kinds of nuts that cost $0.90$ per pound and $1.30$ per pound to make $50$ pounds of a combination that costs $1.20$ per pound. How many of each should he use? So I did $0.90x + 1.30y = 1.20$ and $x + y = 50$. When I try to solve it, it gives me some very long decima...
@Sᴋᴜʟʟᴘᴇᴛʀᴏʟ - thx - I see it was closed. I know this isn't the site for this kind of question. We might consider referring to algebra.com , this question is welcome there. I have no affiliation, FYI.
@DanielFischer If you are given that $$\int^{\infty}_{-\infty}|f(x)|^2dx=1,$$ is it clear to you why $f(x)$ must go to zero faster than $\frac{1}{\sqrt{|x|}}$?
How would describe the space of polynomials $ l(t) = \left\{at^4+bt^2+c: a,b,c \in \mathbb{R}\right\}$? The set of all polynomials of degree $0,2,4$ sounds wrong.
Maybe the set of all polynomials of degree less than $4$ that contain no terms of degree $1$ and $3$?
Never mind, probably the best way to describe it is the space of all biquadratic functions, I think.
$\left\{1+t,(1+t)^2\right\}$ spans the set $ l(t) = \left\{a_1(1+t)+a_2(1+t)^2: a_1, a_2 \in \mathbb{R}\right\}$. It is also a basis for this space because $a_1(1+t)+a_2(1+t)^2 = (a_1+a_2)+(2a_2+a_1)t+a_2t^2 = 0 $ for all $t$ implies $a_1 = a_2 = 0$. Therefore $\text{dim}(l(t)) = 2$. Is this correct?
Let $\Omega \neq \emptyset$ and $\mathcal{F}$ a $\sigma$-algebra of subsets of $\Omega$.
Let $\mu_1, \mu_2: \mathcal{F} \to [0, \infty)$ be measures.
Theorem. Let $\mathcal{C} \subset \mathcal{F}$ be a $\pi$-system such that $\mathcal{F} = \sigma\langle \mathcal{C}\rangle$. If
$\mu_1(C) =...
@J.M. I mean, you usually show up for a day or two, then we don't see you for a long time. Or perhaps you have been on Mathematica and I hadn't noticed.
Perhaps it has been I who have not been on Mathematica in a while :-)
As a math graduate student I feel that I should have better math jokes to keep my students interested, but all I can think of is the old standard: Why is six afraid of seven? Because seven eight nine!
Any help?
Let $\Omega \neq \emptyset$ and $\mathcal{F}$ a $\sigma$-algebra of subsets of $\Omega$.
Let $\mu_1, \mu_2: \mathcal{F} \to [0, \infty)$ be measures.
Theorem. Let $\mathcal{C} \subset \mathcal{F}$ be a $\pi$-system such that $\mathcal{F} = \sigma\langle \mathcal{C}\rangle$. If
$\mu_1(C) =...
