@heather Hi.

hello

Hi!

So yeah, your solution is right...

$$D^2 f = - f$$ is solved by $$f(t) = A \sin(t)$$

Note that it's also solved by $$A\cos(t) \, .$$

In fact! The most general solution is $$f(t) = A\cos(t) + B\sin(t) \, .$$

Hey, regarding quantum mechanics, can anyone explain me difference between the states |1 0> (1/sqrt(2) (up down + down up)) and |0 0> (1/sqrt(2) (up down - down up)) of an electron pair?

whoa...that's awesome

i didn't think to find other solutions =)

@heather There's a pretty general rule: the number of free constants in a differential equation is equal to the highest power of derivative.

Here we have $D^2$, so you know there will be two free constants, $A$ and $B$ in this case.

oh, okay.

good to know.

It's kinda obvious that's true because you have to integrate as many times as the highest derivative power, and each integration brings in a constant.

Anyway, this equation $D^2f = -f$ is super duper amazingly incredibly important. It's the most common differential equation in your life.

@Kelthar One of these states has total spin 1 and the other has total spin 0, as you may see by evaluating $S^2$ on both.

Why?, you ask.

Well, because it applies to physical systems at equilibrium.

Suppose I have some object in a potential energy $V(x)$.

The object moves around until it winds up somewhere with zero force, i.e. somewhere such that $(DV)(x)=0$.

@DanielSank Because it applies approximately, you mean ;P

(i.e. the derivative of the potential is zero. You can imagine $V(x)$ is a hill or something and the derivative is zero when you get to a flat spot.

@ACuriousMind Getting there. No peanut gallery.

For example, point C is an equilibrium (so are the other points but ignore them for a minute).

Here they wrote $U(x)$ instead of $V(x)$, so I'll use $U$ from now on.

@heather you with me?

@DanielSank, I believe so, yes.

@heather Ok, do you know what a Taylor series is?

@DanielSank, heard of it, don't know what it is. I can look it up real quick...

@heather Yeah, go for it.

okay

@ACuriousMind yes, I know, but I'm trying to understand it conceptually...

@Kelthar But that is the conceptual difference - the first state has spin 1 and the other has spin 0, i.e. you can attempt to raise/lower the spin-z value of the first (it's 0) and get a non-zero state, but you get nothing for the second.

"In mathematics, a Taylor series is a representation of a function as an infinite sum of terms that are calculated from the values of the function's derivatives at a single point. [...] A function can be approximated by using a finite number of terms of its Taylor series."

Correct.

Basically, this is tru:

so basically, take the derivative of a function at a point, you get an infinite number of terms, and then the function can in turn be approximated by a finite number of terms of the series?

@heather It's easier to understand if you think about building up the series step by step:

$$f(x) = f(0) + f'(0) x + \frac{1}{2} f''(0) x^2 + \cdots $$

that looks hideous

oh nvm, that's not too bad...

Is that better? Using $f'$ instead of $Df$.

is the next term $1/3(D^3f'''(0)x^3$?

Take $f(x)$, we want to approximate it around $x_0$. Then the constant function $f_0(x) = f(x_0)$ is usually a very bad approximation. But taking $f_1(x) = f(x_0) + f'(x_0)(x-x_0)$ is the tangent to the function at $x_0$, that's already a bit better.

Actually it's $1/6$ not $1/3$.

oh.

The nth term has $1/n!$.

oh, cool! I love factorials.

Right, what @ACuriousMind is saying is a great way to think about it.

so the fourth term would be $1/24f''''(0)x^4$

Now, $f_2(x) = f(x) + f'(x_0)(x-x_0) + \frac{1}{2}f''(x_0)(x-x_0)^2$ is a parabola tangent to the function at $x_0$, and this goes on for taking higher orders.

@ACuriousMind oh...that makes sense

The idea is that each order of $x$ gives you one more point that defines the polynomial - a constant function is defined by a single value, a line is defined by two points, a parabola by three, and so on

@ACuriousMind hmm, okay, this is a bit confusing. Another thing that boggles me is that, in the hydrogen atom, the fundamental state has total angular momentum 0... But if the electron is going round the proton, how is that possible?

Ok @heather so here's the point:

Suppose we're at point C of that energy diagram.

sure.

@ACuriousMind nevermind, this made no sense, the total angular momentum isn't 0

So, going to the case of "infinite" orders, you get something that's determined by infinite points - which hopefully are the points of the function (which annoyingly can fail to be true, but let's not worry about that)

Newton's law says $$F = m a \equiv m \frac{d^2 q}{dt^2} \, .$$

@Kelthar The orbital angular momentum of the ground state is indeed zero (its spin is not). The electron is not "going around the proton", that's a classical picture that has no meaning in quantum mechanics.

what's that third part there @DanielSank? I've never heard of that.

$q$ is the position of the particle.

@heather Clarify, please.

the derivative of the position so really change in position over change in time, so speed...

okay, so you're rewriting acceleration with what represents how to calculate speed/acceleration, that's what's going on there.

@DanielSank, I just misunderstood what the $\frac{d^2q}{dt^2}$ was.

Oh, ok.

@heather you might know $q$ better as $x$ (or $y$ or $z$ or $\vec{x}$ or $\vec{r}$ or so on)

Ok so what's the force? Well, force is minus the position derivative of potential energy, i.e. $$F = - \frac{dU}{dx} \, .$$

Now, if we're near an equilibrium point, we can Taylor expand $U$ like this: $$U(x) = U(0) + U'(0) x + \frac{1}{2} U''(0) x^2 + \cdots$$

er...why? @DanielSank I mean I don't quite see that intuitively. You are saying F=ma, okay, that makes sense, I am familiar with that. But this second equation doesn't make sense to me.

@heather Which second equation?

@DanielSank I think you are once again trying to explain math by physical application to someone who can't appreciate it because they're not as steeped in physics (yet) as you are

@ACuriousMind Indeed possibly true.

@DanielSank, $F=-\frac{dU}{dx}$

If we get a physics lesson out of this, that's fine too.

@heather Aha! Good question.

@ACuriousMind, but whenever I do get through what is meant, it makes a lot of sense and is just plain cool. =)

Well, the work you do to push an object from one place to another is the force you exert multiplied by the distance over which you exert it, right?

(Have you done mechanics yet?)

@DanielSank no, sorry. But that statement makes sense.

@heather Ooooooohhhh, ok that changes things.

Basically, think of it like this:

If my energy would be lower if I go to the right, then that means, kinda by definition, that there's a force pushing me to the right.

Like... if I jump off a building, my potential energy decreases if I go downward, which is the same thing as saying gravity is pulling me downward.

@ACuriousMind Okay, thanks, this made me understand the singlet/triplet question: inspirehep.net/record/1392518/files/SingletTriplet.png I was thinking of oposite total angular momentum instead of just oposite z-momentum for the |1 0> state :P

I might suggest using $U_g = mgh$ as an example

Right, so $U_g$ goes down as $h$ (my height above ground) goes down.

So the force is negative here, and in fact $F = -mg$ which you'll notice is the same as $-dU_g / dh$.

what is g there? Gravity?

Yep

It's the acceleration due to gravity of the Earth.

okay. i think that is making sense.

Wanna ask questions / for clarifications?

brb

okay, back

@DanielSank it makes sense so far.

@heather Ok.

So we have $$- \frac{dU}{dx} = m \frac{d^2 q}{dt^2}$$

Now, suppose our object only moves around near the little valley that it's sitting in.

Then we can approximate $U$ by its Taylor series near that valley.