The h Bar

12:02 AM

What is a momentum operator?

user54412

@StanShunpike $g(\vec{e}_{(i)}, \vec{e}_{(j)}) = g_{kl} (\mathrm{d}x^k \otimes \mathrm{d}x^l)(\vec{e}_{(i)}, \vec{e}_{(j)}) = g_{kl} \mathrm{d}x^k(\vec{e}_{(i)}) \mathrm{d}x^l(\vec{e}_{(j)}) = g_{kl} \delta^k_i \delta^l_j = g_{ij}$

Stan Shunpike

Lol hang on I'm getting my laptop lol

Chajax

user54412

i've got to go, but digest that equation :)

user54412

also, am I the only one left in the room who doesn't use chatjax?

Stan Shunpike

Doubt it, but it is useful

user54412

12:05 AM

also also, I was inconsistent with that notation

user54412

I usually place indices inside parentheses to indicate which vector/tensorial/form object I'm using, and indices without parentheses mean the component of the object

user54412

I did that for the vectors, but maybe I should have for the forms too

user54412

meh

0celo7

12:32 AM

@ChrisWhite Eh, I don't like that notation. It should be clear from context if the index is a component index or a coordinate label.

Parentheses look ugly.

Stan Shunpike

@0celo7 Can every wave function be written as $\Psi = A e^{i(kx-\omega t)}%?

0celo7

@StanShunpike Naw.

Stan Shunpike

Didn't think so.

0celo7

That would make for a very boring world.

Stan Shunpike

But I don't get what the operators in the SE do. They make sense to me if the wave function is of that form. But the simple statement "operators map from one Hilbert space vector to another" doesn't.

@0celo7

Like how do I know which vector it maps to?

It could be any one.

0celo7

12:40 AM

@StanShunpike Uh, it maps to the one you get when you apply the operator...

Stan Shunpike

How do we know which one that is? It just sounds like a vague thing.

@0celo7 like usually when I use an operator, I am given a matrix for it. So I know how to operate with it. I don't know what do do with for instant the Hamiltonian operator for the Coulomb potential

Like are they matrices?

0celo7

Infinite dimensional matrices, although the mathematical physicists might call me out on that.

alarge

@StanShunpike Maybe start think about simpler problems first. Like take the Laplacian operator in some simple geometry and find its eigenvectors. I suppose you could then discretize the operator to a matrix and numerically find its eigenvectors, too, just to see how they compare.

Stan Shunpike

@alarge What do you mean?

Like isn't the 3D wave function just the Laplacian?

0celo7

@StanShunpike The Laplacian is an operator, not a function.

Stan Shunpike

12:55 AM

Sorry, I bungled my words

For the 3D wave function, isn't the momentum operator the Laplacian.

alarge

@StanShunpike Laplacian is the operator, $\nabla^2$ or however you want to write it. So you can trivially approximate that by a matrix, for example (with finite differences, say; but do remember to include the BCs).

Just forget QM for a bit, and just concentrate on computing the eigenvectors (eigenfunctions) of the Laplacian as a start (also numerically, if you think that is something that might interest you).

Stan Shunpike

Is the collection of all the eigenvalues for an operator called the spectrum?

dmckee

@Sean I had a staff scientist acting as a mentor when I first started full time research who used to introduce grad students as "the future doctor <name>". A lot of our high rep users are future doctors.

Mark Mitchison

@StanShunpike Thinking about the Coulomb potential or the Laplacian as a matrix might be a bit over the top. Does it make sense to you if I say that differentiation maps the function $f_1(x) = x^2/2$ to the new function $f_2(x) = df_1/dx = x$? Then you should be able to accept that $d/dx$ is an operator that maps from one function to another.

The only difference from what you wrote about operators mapping from one vector to another is that here we are talking about functions, not vectors.

So the only conceptual leap now is that functions $\approx$ infinite-dimensional vectors

alarge

@MarkMitchison When we're talking linear operators, I think there's some intuition in noting that they are in some sense infinite matrices (so some things, like the spectrum as pointed out by Stan, generalize nicely). But yes, there are some intricacies with infinite-dimensional spaces, so in that sense the matrix analogy is a bit "over the top". But again, a trivial numerical implementation (of say, solving the Poisson equation to take a simple example) might use matrices.

Mark Mitchison

1:11 AM

@alarge Sorry, I didn't mean to tread on your toes there, and I agree that the idea of infinite-dimensional matrices is enormously useful. I just mean that one does not need to appeal to the idea of matrix representations in order to understand why people call $-\frac{\hbar^2\nabla^2}{2m}$ an operator.

It is just a thing that takes in a function and spits out another one.

That seemed to be the core of @StanShunpike's confusion, although perhaps I'm wrong

Stan Shunpike

1:29 AM

@MarkMitchison I like your example. It works well. But how do I know what specific operation the operators in the SE perform on the wave function.

Mark Mitchison

@StanShunpike In exactly the same way. The Coulomb potential $1/r$ takes a wavefunction (vector) $\psi(r)$ and maps it onto a new wavefunction (vector) $\psi^\prime(r) = \psi(r)/r$.

You literally just take the function and divide it by $r$.

Stan Shunpike

Is $\psi'(r)$ a derivative or just denoting function after being operated on?

Mark Mitchison

Sorry, it's not a derivative

I just mean $\psi_{new} = \psi_{old}/r$

Stan Shunpike

What is the time derivative of the wave function? Like when I have taken partial derivatives they are usually of the form $f(x,y) = xy$ or something like that. So it's obvious how to take a partial derivative. But I thought the probability amplitude yielded a complex number not another function.

@MarkMitchison ^

Mark Mitchison

@StanShunpike I'm not sure I understand the question. A probability amplitude is a complex number.

If you want to find the time derivative you just differentiate with respect to time.

Stan Shunpike

1:43 AM

That's it. But isn't that always zero?

Mark Mitchison

Are you asking about the physical interpretation of $\partial\psi/\partial t$?

No.

Stan Shunpike

Why?

Oh, because there is no time element?

Mark Mitchison

Well, if the wave function depends on time then the derivative is not zero

In almost all cases this is true

user105491

If there's no variable in time $t$, then the partial derivative is $0$. However, there are many interesting wavefunctions whose time derivative isn't zero.

Stan Shunpike

Okay, so how do we know the time dependency of $\psi$?

Mark Mitchison

1:44 AM

That's what the Schrodinger equation is for!!

user105491

Yup. Look at the Hamiltonian operator. (I think.)

Mark Mitchison

@StanShunpike You have to solve it to find the time dependence.

Finding the time dependence of the wave function is the quantum mechanical problem. If you can solve that then you can calculate anything you want.

Stan Shunpike

So for particle in a box, what is the wave function? If you don't know off hand, that's okay I'll try to figure it out. I'm just curious because I don't think I fully understand the SE purpose. So the SE defines a PDE whose solutions are the wave function for the particle and allow us to find the probability amplitude at different times?

So really, it isn't nearly as complicated as I'm thinking it is.

I probably just haven't seen enough examples and therefore have never really seen what a solution looks like.

user105491

You might want to take the 8.05x course online at edX offered by MIT, @StanShunpike.

user105491

I'm taking it as well, and it's challenging enough to be fun.

Stan Shunpike

1:57 AM

Great suggestion. It looks hard enough to be interesting. Thanks!

user105491

Sure!

Stan Shunpike

Are you a chat regular?

user105491

Me?

user105491

I come to the homotopy theory chat room often.

user105491

I'm an aspiring mathematician.

Stan Shunpike

1:59 AM

Nice, so is my cousin. My uncle is a mathematician at Drexler.

user105491

Nice!

user105491

I used to do some physics some years ago but then got bored of manipulating equations. I found that I loved topology and algebra more, and got into it. So now I'm just trying to study some physics, recall a few things.

Mark Mitchison

@StanShunpike The wave function at some time depends on the initial conditions. Just like most problems in physics!

But the usual trick is to first find the energy eigenfunctions.

Kyle Kanos

@SanathDevalapurkar Then you probably weren't really doing physics

Mark Mitchison

Then any wave function can be represented as a superposition of these, each oscillating with a phase.

user105491

2:01 AM

Yes, @Kyle, I probably wasn't! :-P

Mark Mitchison

(This is basically equivalent to solving the differential equation in the Laplace domain rather than the time domain.)

Stan Shunpike

@MarkMitchison What is an example of "initial conditions"?

Mark Mitchison

2:34 AM

@StanShunpike For example, suppose I have a particle in 1D perfectly localised at the position $x_0$ at time $t = 0$. Then we have that $\psi(x,0) = \delta(x-x_0)$. Using the Schroedinger equation, you can then solve for the wavefunction $\psi(x,t)$ at future times. From this you can calculate the position-space probablity distribution or any other observable quantity you choose.

Generally speaking, you would expect the wave function to initially spread out. If there is a confining potential, this will tend to inhibit this spreading. For example if you have a square well you would expect reflection at the boundaries and so on. All of this information is captured by the Hamiltonian and its eigenfunctions.

Stan Shunpike

@MarkMitchison is the SE like the EFEs in that it doesn't just work for any old function...you have to specify the wave function in question and show that this equation solves for it.

And that may be different for different cases then.

0celo7

@StanShunpike Of course it doesn't work for every function...it's a differential equation to be solved for the function.

It seems I left a pot out in the garage when it was colder and it was OK to do so.

Now I have to get rid of a month's old half-full pot of pasta sauce.

Stan Shunpike

2:50 AM

@0celo7 Well, smarty pants, I'm still learning! :p But yeah, now that you mention it, that was a very obvious statement. I just thought there was a general wave function it worked for. I didn't realize that you had to find specific wave functions for different cases. No wonder I have been confused!

@0celo7 Why would you leave a pot in the garage? Is your garage near your kitchen?

0celo7

@StanShunpike Well with $V=0$ it always works with plane waves.

@StanShunpike When it's colder the garage is like a really big fridge.

But it's been 70 degrees for like a week, so I'm scared what I'll find.

Stan Shunpike

For a week? Dude, lol. Who know what might be growing there....

@0

0celo7

Oh man, chance of 1-3 inches of snow

Pray for a snow day

::does the snow dance::

Stan Shunpike

We almost never got snowdays when I was in K-12

I went to the same school and we always had school.

Didn't matter how much it snowed

0celo7

State?

Stan Shunpike

2:56 AM

IL

0celo7

In VA we melt in the snow

We're the northernmost southerners

Stan Shunpike

So you how much do you get on average? A few inches?

0celo7

Although some MD people consider themselves southern.

A few inches.

Kyle Kanos

@0celo7 They're below the canonical M-D line

0celo7

The problem is that the chilluns don't know how to drive.

@KyleKanos Some texts don't abide by that rule.

Mine did, but we learned both conventions for the AP test.

They will never ask if MD is a southern state because not everyone agrees.

Stan Shunpike

2:59 AM

Are all the possible solutions to the SE known?

Kyle Kanos

It probably depends on the where in the state they're from

Like the more southern they are, the more likely they say they're Southerners and vice versa

0celo7

The county doesn't want to be responsible for damages to student vehicles. The first snow we had this year was 1 inch...300 accidents in the tri-county area.

~70 by 8 am in our county alone.

I got to school 2 hours late...I live 15 minutes away.

Kyle Kanos

That's awful

I was born & raised in NJ, I'm perfectly capable of driving in the snow

The idiots in SC cry for fear when we get 0.5" of snow

We literally closed campus this year with that much snow

Stan Shunpike

Both of those comments are crazy lol

70 accidents by 8am...wtf

closing a school for .5 inch of snow...

0celo7

Yo that pot is fucking toxic.

Most vile thing I've smelled this year.

user54412

3:03 AM

Snow closings are great. I get to drive with no traffic into work, and I get to park wherever I want on campus.

Stan Shunpike

lol yeah there are definitely perks to them

user54412

(Yes, for grad students snow day = more work day)

NeuroFuzzy

Southern california...

sometimes the sky gets a bit grey?

:P

that can make driving sooo hard

Kyle Kanos

Well to be fair, I also went about 2 or 3 years without snow down here

NeuroFuzzy

And the moment I take a road trip I'm going to get into an accident haha

road trip in winter*

0celo7

3:05 AM

@NeuroFuzzy My sister lives in Studio City and she sends out mass texts to everyone in the family when the temperature drops below 50.

"It's a Christmas miracle"

Kyle Kanos

My wife was talking to her mom (who still is in NJ) this afternoon and said it was a "bit chilly here." I commented loud enough (so the MIL could hear) that she meant 60 degrees F.

0celo7

That was horrifying.

There were gray tomato chunks in there.

NeuroFuzzy

Huh?

0celo7

@NeuroFuzzy Really old pot of pasta sauce had to be disposed of.

user54412

@0celo7 Oh so you've been to NJ I see?

Mark Mitchison

3:11 AM

@StanShunpike Definitely not. There are lots of potentials for which the solution is not known. When you add more particles into the mix, it becomes even more complicated. Note that the Schroedinger equation is the most fundamental equation of Nature that we know how to write down. If you knew all the solutions you'd know a hell of a lot more than could reasonably be expected.

Kyle Kanos

@0celo7 Ok, that totally clarifies the Yo that pot is fucking toxic statement from earlier

0celo7

@KyleKanos Lol.

Stan Shunpike

@MarkMitchison Why does adding more particles create problems?

Mark Mitchison

We only know how to solve problems with two particles.

In general

No one knows how to solve the many-body problem

You want to know why physically?

In quantum mechanics it is especially complex because of the way quantum correlations work. The correlations between quantum mechanical particles are so strong that one requires an exponential amount of resources to even write down a general state.

So even using computers to solve by brute force fails once you have more than, say, 20 spin-half particles on a lattice.

Stan Shunpike

I have heard we only have exact solutions to the hydrogen atom. Is this why?

Mark Mitchison

3:17 AM

@StanShunpike There are quite a few exact solutions known.

Kyle Kanos

Ooo: V=0

Mark Mitchison

The hydrogen atom is one of the most interesting, since it describes a real physical system (the Hydrogen atom) to a very good approximation.

0celo7

List of quantum-mechanical systems with analytical solutions

Much insight in quantum mechanics can be gained from understanding the solutions to the time-dependent non-relativistic Schrödinger equation in an appropriate configuration space. In vector Cartesian coordinates , the equation takes the form in which is the wavefunction of the system, H is the Hamiltonian operator, and T and V are the operators for the kinetic energy and potential energy, respectively. (Common forms of these operators appear in the square brackets.) The quantity t is the time. Stationary states of this equation are found by solving the eigenvalue-eigenfunction (time-independent...

Mark Mitchison

The harmonic oscillator is another good 'un

Stan Shunpike

I must be mixing up my words lol. So those are "exact" solutions. So there's nothing special about the hydrogen atom in terms of what we know about it relative to other atoms?

Mark Mitchison

3:19 AM

No, I would say that hydrogen is pretty special.

Kyle Kanos

Hydrogen is the best element

Mark Mitchison

You can calculate most of its measurable properties to good accuracy using analytical techniques

0celo7

@KyleKanos You're not biased at all...

Mark Mitchison

No other atom has that property.

Although helium obviously is close.

0celo7

It is the simplest one.

Kyle Kanos

3:20 AM

@0celo7 I'm a carbon-based life form, so I should be biased towards that

Stan Shunpike

There ya go! that's what I was trying to comment on. So that's what makes hydrogen special.

Mark Mitchison

And also Rydberg atoms.

Kyle Kanos

But I am (supposedly) an astrophysicist, so I should care about H a bit more than He and the metals

Mark Mitchison

Yeah hydrogen is the simplest element involving only two particles, and it is therefore special.

If you neglect the quantum fluctuations of the electromagnetic field, then the spectrum of hydrogen can be calculated exactly, unlike any other atom.

Ah, sorry. You were asking why we can only solve one atom exactly.

Yes, this is exactly why hydrogen is special in that sense.

i very much doubt that Wiki article gives a complete list of known solutions, by the way.

Those are probably just some of the most famous.

Stan Shunpike

Yeah, but it's till a lot. There were 20-30

0celo7

3:24 AM

@MarkMitchison I'd imagine one could find a more complete catalogue on arXiv.

Mark Mitchison

If you wanted to parse a lot of data, sure.

I can already think of several important ones that are not on there, mostly many-body problems.

user54412

@StanShunpike I'll just remind you that "exact solution" and "closed form" and "analytic" are pretty poorly defined

Mark Mitchison

But it doesn't matter too much, the important point is that we know a few solutions but most interesting problems cannot be solved exactly.

And yeah, what ^Chris White said.

Stan Shunpike

@ChrisWhite Yeah, they seem to be. And I don't really accept or use a word until I have a specific definition.

user54412

It's @Waffle'sCrazyPeanut !

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