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heather
heather
12:47 AM
in a table listing polarizations, what do 'e' and 'o' mean - i.e., if something has a 'e' or 'o' polarization, what does that mean?
 
vzn
vzn
@Blue good question! really want to know the answer? think what youre asking about touches on local realism. have looked into this heavily. still looking for someone else to follow leads with & really get to the bottom of it... to put it briefly, there are a lot of things you cant find in textbooks or even from experts/ authorities (who are all confident after over a century theres nothing more to the story)... but it can be like pulling thread on sweater, or falling down rabbit hole... :|
 
heather
heather
(context being the table here)
 
vzn
vzn
... or taking the red pill ... o_O
 
heather
heather
perhaps elliptical and circular polarization?
not sure why they wouldn't do c for circular then...
 
Jaime Gallego
Jaime Gallego
1:50 AM
@Kaumudi.H Yeah, check out the explanation @Blue mentioned.
@DavidZ lol
 
vzn
vzn
2:35 AM
@heather are you starting school soon? watcha studying this year? what prjs are you working on these days?
 
Secret
Secret
2:55 AM
1
Q: Dissipative adaptation vs. principle of least entropy production: I'm confused

user42541I am interested in Jeremy England's non-equilibrium thermodynamic theory of "dissipative adaptation". See refs like https://journals.aps.org/prl/pdf/10.1103/PhysRevLett.119.038001. My understanding of his theory is that systems, when driven out of equilibrium, tend to organize themselves such t...

thermodynamics statistical-mechanics biophysics
 
Dawood ibn Kareem
Dawood ibn Kareem
@heather It's mentioned in the text. "... Typically a crystal has three axes, one or two of which have a different refractive index than the other one(s). Uniaxial crystals, for example, have a single preferred axis, called the extraordinary (e) axis, while the other two are ordinary axes (o) ..."
 
DanielSank
DanielSank
3:31 AM
@vzn wut?
Nobody is confident "there's nothing more to the story". No self-respecting physicist is ever goign to tell you that we know everything and the theory will never change.
We've tested Bell violations pretty well and haven't found anything surprising yet.
That doesn't mean we think the theory is done and will never change.
I mean, dude, for heaven's sake, one of my colleagues did a pretty famous Bell violation experiment and even she is confident that physics theories are never the final word.
 
Dawood ibn Kareem
Dawood ibn Kareem
3:49 AM
@DanielSank It would be interesting if they did. "As from today, there will be no further physics. All professional physicists must choose a new career".
 
dmckee
dmckee
@DawoodibnKareem Nawh. Even if we found the final theory there are still endless unexplored corners arising from the physics we already know. Not to mention all the obscure corners that will arise in the final theory itself.
It's not like continuum fluid dynamics of classical mechanics have changed their basic principles in the last 150 years, yet people keep finding new surprises in them.
 
Secret
Secret
o and don't forget the myriad of quasiparticles in condensed matter, or even simple and common things like the properties of water.
though to be honest, unless infinite but bounded sequences do exists physically, if you have a field of knowledge where the foundations and the unified theory is known, then perhaps it might only take a finite (however large) amount of time to fill in the gaps in between, thus making knowledge exhaustible
Another possibility is that knowledge in reality formed such a complex interrated network that expand quicker than all attempt to exhaust it, thus effectively making it endless as required
e.g. Let's suppose for a simple model that there are only 4 things need to know about the nature of water and then everything about water will become known and these physicists have to change their carrier. From what we have seen so far in the nature of knowledge, each of these 4 things are interrelated to some other things, and so on down the rabbit hole it goes, so the exponential increase in the number of pieces one need to know will mean one can never fully
know everything about water even if it seems you know all the 4 pieces of knowledge
 
John Rennie
John Rennie
4:12 AM
@0celóñe7 I must admit I hadn't realised that SSDs had got that fast. The SSD I use is only a couple of years old but is half that speed.
 
vzn
vzn
@DanielSank then maybe she should seriously look into available alternatives or "leads" etc. the fact remains the basic theory has not changed in roughly a century. ps recently got kicked from this room for basically asserting "there may be more to the story"...
 
Secret
Secret
A cautionary tale for everywhere in academia
(though the lack of tenure in australia makes it harder to happen in australia)
 
vzn
vzn
4:28 AM
@Secret scanning it, quite dramatic, it reads like/ has some impression of a persecution complex. but dont know further details or context.
 
Secret
Secret
The case in a nutshell: A US PhD being plaglarised by her supervisor
 
vzn
vzn
@Secret she. (did find some context) allisonharbin.com/bio
reminds me of sayres law en.wikipedia.org/wiki/Sayre%27s_law
 
John Rennie
John Rennie
4:52 AM
@Blue you were asking about quantum mechanics earlier?
 
Anonymous
@JohnRennie Yes. I wanted to know how to find the energy eigenvalues E in $\hat{H}\psi(x)=E\psi(x)$ experimentally in the lab.
 
John Rennie
John Rennie
As a general rule we measure the difference between energy levels rather than the energy levels themselves. And that's exactly what atomic spectra measure.
 
Anonymous
@JohnRennie Does $E$ stand for the total K.E. + P.E. of say an electron in any orbit of an atom ?
 
John Rennie
John Rennie
Yes. Note however that because PE is always relative to some reference point the total energy is also relative to some rference point.
e.g. for an electron in an atom we usually take the potential energy to be zero at infinity.
 
Anonymous
Umm. But isn't the Schrodinger's equation valid for all elementary particles...even photons? What does it mean to say....say "ground state of a photon" ?
 
John Rennie
John Rennie
5:00 AM
This is a common problem students have with QM.
QM doesn't necessarily mean observables are quantised.
 
Anonymous
@JohnRennie Okay, that's why it makes sense to take the difference of the energy levels rather than finding the absolute energy
 
Anonymous
@JohnRennie What does that mean?
 
John Rennie
John Rennie
@Blue In general if we have a bound state then things like energy are quantised. So energy levels in a hydrogen atom are quantised. But a photon whizzing through space is not a bound state so its energy levels are not quantised.
A photon, or any other free particle, does obey the Schrodinger equation, but that doesn't mean its energy is quantised.
 
Anonymous
@JohnRennie Then, how do we find E for photons?
 
John Rennie
John Rennie
Photons are a bad choice because they are inherently relativistic and the SE is a non-relativistic equation. Can we consider a free electron instead?
 
Anonymous
5:03 AM
@JohnRennie Ah, sure
 
John Rennie
John Rennie
OK you know a free particle has a de Broglie wavelength?
 
Anonymous
@JohnRennie Yes
 
John Rennie
John Rennie
So we can represent a free particle by a plane wave with the coirrect de Brogliue wavelength. So its wavefunction will be something like: $$ \psi(t,x) = \exp(i(kx - \omega t))$$
 
Anonymous
@JohnRennie Yes. Then?
 
Anonymous
Got it so far
 
John Rennie
John Rennie
5:06 AM
Then apply $\hat{H}$ to that function, and out drops the energy
 
Anonymous
Yes. But that is assuming that after applying the hamiltonian operator the result will be of the form constant times the wavefunction. For some other type of wavefunction that might not be true. Am I going wrong?
 
John Rennie
John Rennie
Two points: firstly the particle obeys the SE so its wavefunction must be a solution to the SE
But the second point is more interesting.
 
0celóñe7
0celóñe7
@JohnRennie that's 600 less than advertised
 
Anonymous
@JohnRennie What's that?
 
John Rennie
John Rennie
@Blue What you wrote down above is the time independent SE, and only the eigenfunctions of the time independent SE have a precisely defined energy.
 
Anonymous
5:11 AM
@JohnRennie Umm, why so?
 
Anonymous
Why only time independent SE have precise energy?
 
John Rennie
John Rennie
Hmm, I need to be a bit careful here because my grasp of the fundamentals of QM isn't as strong as it could be and I don't want to mislead you ...
The Hamiltonian is the time evolution operator, that is when you apply it to a wavefunction it evolves that wavefunction with time.
If you have a time independent wavefunction then obviously evolving it in time leaves it unchanged, that's why we get the time independent equation: $$\hat{H}\psi = E\psi$$ for some constant $E$.
where $E$ turns out to be the energy of the state $\psi$.
 
Anonymous
Hmm....basically if the function is time dependent the result we get after applying the hamiltonian operator is also a time dependent function (usually). Am I right?
 
John Rennie
John Rennie
If the wavefunction is time dependent then we get $$\hat{H}\psi = i\hbar \frac{d\psi}{dt}$$
 
Anonymous
@JohnRennie What do you mean by "state" $\psi$ ? Isn't $\psi$ just the wavefunction of particle like electron?
 
John Rennie
John Rennie
5:16 AM
Yes, $\psi$ is the wavefunction of the particle.
 
Anonymous
Does $\psi$ have things like ground "state", excited "state"...etc?
 
John Rennie
John Rennie
@Blue Remember than as a general rule only bound states have quantised energy levels.
If you have a free electron you can certainly increase and decrease its energy, but the energy change is continuous not discrete.
 
Anonymous
@JohnRennie Ok. But I didn't understand what you meant by:
 
Anonymous
4 mins ago, by John Rennie
where $E$ turns out to be the energy of the state $\psi$.
 
Anonymous
What's "state" there?
 
John Rennie
John Rennie
5:19 AM
Sorry, by state I meant wavefunction. I was using the two words to mean the same thing.
 
0celóñe7
0celóñe7
@JohnRennie no
 
Anonymous
Oh. Got it till here! Now, coming to the bounded electron system
 
John Rennie
John Rennie
Before we do that, you know a particle can be in a superposition?
 
0celóñe7
0celóñe7
You need to exponentiate it to get the time evolution
 
John Rennie
John Rennie
@0celóñe7 oops, yes.
 
Anonymous
5:21 AM
@JohnRennie Nope. What do you mean by that? Superposition of two wavefunctions? Like we do for YDSE explanations ?
 
0celóñe7
0celóñe7
YD?
 
John Rennie
John Rennie
@Blue Yes, a superposition of two wavefunctions.
@0celóñe7 Young's Double Slits Expt
 
Anonymous
@JohnRennie Okay. Right. Then?
 
John Rennie
John Rennie
Suppose we have $$\psi(t,x) = \exp(i(k_1x - \omega_1t)) + \exp(i(k_2x - \omega_2 t)) $$ i.e. a superposition of two plane waves.
 
Anonymous
mmm hmm ?
 
John Rennie
John Rennie
5:24 AM
What do you think the energy of that wavefunction will be?
 
Anonymous
Summation of the E's of the two individual wavefunction?
 
Anonymous
After applying H operator
 
John Rennie
John Rennie
In fact the energy is not defined.
If we measured the energy of that state we would get either the result $E_1$ or $E_2$ i.e. the energy of one of the two parts of the superposition.
That's because the state isn't time independent.
If you're interested Emilio wrote an absolutely brilliant answer explaining exactly this.
I could with some effort find it ...
 
Anonymous
Sure. I'd like to read it
 
Anonymous
You could give the link later also
 
John Rennie
John Rennie
5:27 AM
Hang on then ...
27
Q: Is there oscillating charge in a hydrogen atom?

Marty GreenIn another post, I claimed that there was obviously an oscillating charge in a hydrogen atom when you took the superposition of a 1s and a 2p state. One of the respected members of this community (John Rennie) challenged me on this, saying: Why do you say there is an oscillating charge distri...

quantum-mechanics atomic-physics charge
 
Anonymous
I couldn't understand something about about the bound state of electron. And what E denotes in that case. Suppose we know the wavefunction of an electron in the 1st energy level. Say $\psi_1$. We apply the H operator on it and get some constant times $\psi_1$. What does that "constant" denote? (It can't denote total energy because total energy is relative as you had mentioned earlier). Also, how does measuring the difference between two energy levels give us E (i.e. the energy eigenvalue)?
 
Anonymous
@JohnRennie Checking it
 
John Rennie
John Rennie
@Blue Some of the other answers are also very good, though annoyingly the accepted answer isn't that great.
 
Fawad
Fawad
Hi @Blue where you got seat?
 
Anonymous
@JohnRennie I think I'll take some time to understand the answers. Maybe we can get back to it later in the day. Is understanding the answer to that question essential to understanding my previous question about bound state of electron (just previous message ^) ?
 
Anonymous
5:35 AM
@Fawad Jadavpur University
 
John Rennie
John Rennie
@Blue The Hamiltonian is $$\nabla^2 + V$$ Yes?
 
Anonymous
@JohnRennie Yes
 
John Rennie
John Rennie
So it contains the potential energy. But the potential energy has what we call a gauge symmetry i.e. the absolute value has no physical meaning. Potential energies are always relative to some reference point.
Typically if we are studying an electron in an atom we would say $v=0$ as $r \to \infty$ but this is just a convenient choice of reference point.
And because we use $V$ in the SE it means the energies we get out of the time independent SE also have this global gauge symmetry.
So when we say the energy of the ground state in hydrogen is 13.3eV what we mean is that it takes 13.6eV of energy to remove the electron from that ground state to infinity.
 
Anonymous
Okay. So say we know the wavefunction of an electron in the K shell i.e. $\psi$. We find $\hat{H}\psi$. Then does the $E$ come out to be 13.6 eV or not?
 
John Rennie
John Rennie
If we define $V$ to be the potential energy using $V(\infty)=0$ then yes $E$ will come out as 13.6eV.
As in $$\hat{H}\psi_{1s} = -13.6eV\psi_{1s} $$
 
Anonymous
5:43 AM
@JohnRennie Okay! Now that makes sense. I couldn't understand before as to how you could measure $E$ (i.e. the eigenvalue) by finding the difference in energy levels
 
John Rennie
John Rennie
Well that 13.6 eV is a difference.
It's the difference between the $\psi_{1s}$ energy and the energy at infinity.
 
Anonymous
Difference between $n=\infty$ and $n=0$ ? Oh, sure! :) That makes sense because we took $V(\infty)=0$
 
John Rennie
John Rennie
In fact it's the ionisation energy of the hydrogen atom
Which of course we can easily measure in the lab :-)
 
Anonymous
thanks...now before going to Emilio's answer...I have two more questions to ask about the H operator:P
 
John Rennie
John Rennie
OK?
 
Anonymous
5:48 AM
First of all, Balarka told me yesterday that the Hamiltonian operator can be thought of as an infinite dimensional vector space (since there are infinite infinitely differentiable functions). But then how does one multiply an infinite dimensional vector space with a wavefunction $\psi$ ?
 
John Rennie
John Rennie
This is actually why I pinged you this morning.
There are two ways to start learning QM. The traditional way is to start with the SE and learning how to solve it. Then you learn the wavefunctions for free particles, hydrogen atoms and so on.
That is usually the way QM is taught to undergraduates.
 
Anonymous
By vector space I'd think of something like $\{\partial_x,\partial_y,\partial_z\}$ (i.e. a 3D vector space)or $\{1,2,3\}$ which could be equivalent to 1 i + 2 j + 3 k ...(maybe I can't understand because I don't know linear algebra)
 
John Rennie
John Rennie
But it turns out QM has an amazingly interesting mathematical structure, and the second way to learn QM is to start with the mathematical structure.
That is what Balarka is talking about.
You remember I said a particle can be in a superposition?
 
Anonymous
@JohnRennie Yup
 
John Rennie
John Rennie
OK suppose we have some system, free particle hydrogen atom or whatever, and the time independent SE has some solutions $\psi_1$, $\psi_2$, etc. There are an infinite number of these solutions. OK so far?
 
Anonymous
5:54 AM
@JohnRennie Solution to the Differential equation? Yup, sure
 
John Rennie
John Rennie
OK. So a general wavefunction can be an arbitrary superposition of these: $$\psi = a_1 \psi_1 + a_2 \psi_2 + ... $$ where $a_1$, $a_2$, etc are just numbers. OK so far? This is where the vector space creeps in so you need to make sure you're OK with what I've said.
 
Anonymous
@JohnRennie I get it. Linear superpositions of wavefunction solution also gives a wavefunction solution. Adding all independent wavefunction solutions gives a general solution like the one you gave above. OK :)
 
John Rennie
John Rennie
Yes. We're nearly there! :-)
We can drop the $\psi$s and write the wavefunction as: $$ \psi = (a_1, a_2, a_3, ...)$$ where there are an infinite number of the $a$s. Now, doesn't that look a bit like a vector written as its components?
 
Anonymous
@JohnRennie Yup, it does!
 
John Rennie
John Rennie
Bingo!
 
Anonymous
6:00 AM
One sec. Then how come the $\hat{H}$ operator an infinite dimensional vector space? I could understand upto what you said
 
Balarka Sen
Balarka Sen
oh wow you're still doing this shit
much dedication
 
John Rennie
John Rennie
@BalarkaSen stay out of this you ... you ... mathematician!!! :-)
 
Anonymous
@BalarkaSen Overthinking got me confused again =P
 
John Rennie
John Rennie
@Blue remember I said there are an infinite number of the $\psi$s? That means our vector has an infinite number of components.
 
Anonymous
@JohnRennie Yes
 
John Rennie
John Rennie
6:03 AM
So our vector space has an infinite number of dimensions.
In effect each of the $\psi_i$s is one dimension.
And our superposition $\psi$ is a vector in this infinite dimensional vector space.
But don't let the idea of infinite dimensions blow your mind. It's a lot simpler than it seems.
 
Anonymous
Yes. I understand that. I wanted to know how we can say that the Hamiltonian operator can be written like $\hat{H}=\{...,....,....,....,....,\}$ (i.e. an infinite dimensional vector space). I have understood that $\psi = (a_1, a_2, a_3, ...)$....
 
John Rennie
John Rennie
Operators are infinite dimensional matrices not vectors
 
Balarka Sen
Balarka Sen
Hamiltonian is an operator slash infinite dimensional matrix. It is not an "infinite dimensional vector space"
 
John Rennie
John Rennie
^
I can explain why operators are matrices ...
 
Anonymous
@JohnRennie Okay. How can you write the $\hat{H}$ as an infinite dimensional matrix then?
 
Balarka Sen
Balarka Sen
6:07 AM
If $V$ is your "vector space of wavefunctions", Hamiltonian is a map $V \to V$ (or at least, I hope so).
 
John Rennie
John Rennie
Let's go back to a simple system we can all understand intuitively. A 2D vector space $x,y$ i.e. just a piece of graph paper. OK so far?
 
Balarka Sen
Balarka Sen
But I'll slither back to my cave.
 
John Rennie
John Rennie
@BalarkaSen :-)
 
Anonymous
@JohnRennie Sure. Then?
 
John Rennie
John Rennie
So an arbitrary vector is $$\hat{r} = a_1 \hat{x} + a_2 \hat{y} $$ and we simply write $$ r = (a_1, a_2) $$ Yes?
 
Anonymous
6:10 AM
@JohnRennie Yes
 
John Rennie
John Rennie
Or let me write $$r = (x,y)$$ to avoid all those subscripts.
Anyhow suppose I have an operator that multiplies the length of vectors by a scale factor $c$
 
Anonymous
Wait. $r=(a_1,a_2)$
 
Anonymous
We don't write $r=(x,y)$
 
Anonymous
Or did you take $x=a_1$
 
Anonymous
:P
 
John Rennie
John Rennie
6:11 AM
I was just using the number (not vector) $x$ to mean $a_1$ and $y$ to mean $a_2$.
But I'll continue to use the $a$s if you prefer ...
 
Anonymous
@JohnRennie OK.!
 
Anonymous
Then?
 
John Rennie
John Rennie
Anyway, my operator $\hat{O}$ doubles the length of vectors so $$\hat{O}(x,y) = (2x,2y)$$ Make sense so far?
 
Anonymous
@JohnRennie Sure. It does
 
John Rennie
John Rennie
Now how can I write down my operator $\hat{O}$? Well one way is to write it as a matrix. This'll take a moment while I try to remember how to write matrices using MathJax ...
 
Anonymous
6:16 AM
Operator $\hat{O}$ is simply a scalar $2$...isn't it?
 
John Rennie
John Rennie
Bugger, I can't remember how to do this. I'll have to go look it up, unless you can get the idea from my incorrect MathJax ...
 
Anonymous
Mathjax error. You can just write a matrix as $\{...,...,...\}$
 
Anonymous
for now
 
John Rennie
John Rennie
@Blue Suppose I wanted different factors in the $x$ and $y$ directions e.g. $$\hat{O}(x,y) = (2x,3y)$$ then $\hat{O}$ can't simply be a scalar.
 
Anonymous
OK. So you basically mean $\hat{O}=\{\{2,0\},\{0,2\}\}$ ?
 
Anonymous
6:20 AM
Right?
 
Anonymous
In row-major order
 
Anonymous
I think I get it
 
John Rennie
John Rennie
$$ \left(\begin{matrix} 2 & 0 \\ 0 & 3 \end{matrix}\right) \left(\begin{matrix} x \\ y \end{matrix}\right) = \left(\begin{matrix} 2x \\ 3y \end{matrix}\right) $$
 
Anonymous
$\hat{O}=\{\{2,0\},\{0,3\}\} . \{x,y\}=(2x,3y)$
 
John Rennie
John Rennie
It works! :-)
Anyway ...
 
Anonymous
6:22 AM
@JohnRennie Yes!!!
 
John Rennie
John Rennie
The point is that I can represent my operator $\hat{O}$ as a matrix.
 
Anonymous
Got it got it got it :D Yay....phew....finally!!!!!
 
Anonymous
@JohnRennie Yes
 
John Rennie
John Rennie
That doesn't mean it is a matrix, that means in this vector space I can write it as a matrix.
 
Anonymous
@JohnRennie Right
 
Anonymous
6:24 AM
Now just the "infinite" dimensions stuff is left
 
John Rennie
John Rennie
And obviously, we now let the vector space increase to infinite size and our operator becomes an infinite matrix. That's why $\hat{H}$ can be reprtesnted as an infinite dimensional matrix.
 
Anonymous
How does $\hat{H}$ have infinite dimension
 
Balarka Sen
Balarka Sen
@Blue It's a linear transformation of an infinite dimensional vector space. Is that much clear?
 
John Rennie
John Rennie
@Blue Because our vector space has as its axes the eigenfunctions $\psi_i$ and there are an infinite number of those eigenfunctions.
So our vector $\psi = (a_1, a_2, ...)$ is infinite dimensional
And when you multiply a vector by a matrix the sizes of the vector and matrix must match.
→ 1 message moved to trash
 
Anonymous
@JohnRennie Yes. Understood that. My confusion at is this point is:
 
Anonymous
6:27 AM
How does the $\hat{H}$ look like when written as a matrix?
 
Anonymous
Could you write out a couple of rows of the matrix?
 
John Rennie
John Rennie
Err ... I'll have to think about that
 
Balarka Sen
Balarka Sen
@Blue If $T : \Bbb R^n \to \Bbb R^n$ is a linear transformation, the matrix corresponding to $T$ has rows $T(1, 0, \cdots, 0)$, $T(0, 1, 0, \cdots 0)$, etc etc etc $T(0, \cdots, 0, 1)$.
Do you agree?
Well, rows/columns depending on your convention. I prefer to have those as my columns.
 
Anonymous
@BalarkaSen Yes. Agreed
 
John Rennie
John Rennie
Incidentally, we have a special name for the type of vector space made up from the $\psi_i$s. It's called a Hilbert space.
 
Balarka Sen
Balarka Sen
6:30 AM
@Blue So those vectors are called the standard basis vectors, $e_i$. They represent the unit vector along each coordinate axis.
What are the "standard basis vectors" for the space of wavefunctions?
 
Anonymous
@BalarkaSen "those vectors"....which vectors?
 
Balarka Sen
Balarka Sen
$(1, 0, 0, \cdots, 0)$, $(0, 1, 0 \cdots, 0)$ etc
$e_i$ is the vector (0, 0, ..., 0, 1, 0, ..., 0) with 1 at the i-th slot
It's the unit vector pointing along the positive i-th coordinate axis.
Think about $(1, 0)$ and $(0, 1)$ in $\Bbb R^2$. Those are $e_1, e_2$.
 
John Rennie
John Rennie
@Blue speaking as a mere physicist, I wouldn't worry too much about exactly what $\hat{H}$ looks like as a matrix. The idea of approaching the subject this way is to get a feel for the overall mathematical structure without worrying too much about the detail.
 
Anonymous
@BalarkaSen Why should they be "unit" vectors? That matrix should be to just extend/contract the vector formed by $\{a_1,a_2,a_3,....\}$. Like the example JR gave above. Here the $\psi_1,\psi_2,...$ should act as the independent vectors....
 
Balarka Sen
Balarka Sen
@Blue what is the magnitude of (1, 0) in R^2? ...
 
John Rennie
John Rennie
6:36 AM
@Blue careful! The example I gave of scaling up a vector was meant just to be an easily understandable example. Remember that wavefunctions have to be normalised.
So the length of a vector in the Hilbert space has to be unity.
 
Balarka Sen
Balarka Sen
Right now I don't care about wavefunctions. I am working with honest to god vectors in an honest to god Euclidean finite dimensional space everyone sane calls $\Bbb R^n$.
Just to clarify where I am at
 
Anonymous
@JohnRennie Why? ;;
 
John Rennie
John Rennie
@Blue Why are wavefunctions normalised?
Because the probability of finding the particle is related to the magnitude of the wavefunction. Specifically the probability of finding the particle in some infinitesimal volume $dV$ is $\psi^*\psi dV$.
 
Balarka Sen
Balarka Sen
(IMO It's dangerous to think about finite dimensional objects and infinite dimensional objects at the same time. Do one of the two things: Either finish with thinking about wavefunctions and then learn linear algebra, or finish learning linear algebra and then think about infinite dimensional objects.)
 
John Rennie
John Rennie
If we integrate this over all space we get the total probability of finding the particle, and the total probability has to be unity because we know the particle is there somewhere. So $$\int \psi^* \psi dV = 1$$
This places a restriction on the wavefunction $\psi$. We can't just go multiplying $\psi$ by random numbers or it won't be normalised any more.
It all goes quiet while Blue thinks this all made sense when we started ... :-)
 
Anonymous
6:46 AM
OK. I understand that the integral of $|\psi|^2dV$ is 1. Still I didn't get why the the length of Hilbert space vector i.e. $\{a_1,a_2,a_3,...\}$ has to be unity.....If that were true then $|\psi|^2$ would be $1$ rather than the integral of $|\psi|^2dV$
 
Anonymous
i.e. because $\psi = a_1 \psi_1 +a_2\psi_2+....$
 
John Rennie
John Rennie
@Blue I can prove that easily if you want ...
 
Balarka Sen
Balarka Sen
That's the catch. Norm (buzzword for magnitude) of a wavefunction is not what you think it is.
 
Anonymous
@JohnRennie So, proof? :P
 
Balarka Sen
Balarka Sen
It's called the $L^2$ norm. By definition, $\|\psi\|_{L^2} = \int |\psi|^2 dV$. That is 1.
 
Anonymous
6:49 AM
@BalarkaSen I think that makes more sense :P
 
John Rennie
John Rennie
OK. I'm going to use Dirac notation to keep things simple. This is: $$\int \psi^* \psi dV = <\psi|\psi>$$ It's just notation. That OK?
 
Anonymous
Okay!
 
John Rennie
John Rennie
I'm also going to assume the wavefunction is real to save writing complex conjugates everywhere.
And lastly our eigenfunctions are orthonormal i.e. $<\psi_i|\psi_j>$ is zero if $i$ and $j$ are different and unity if they are the same. Still OK?
 
Anonymous
$<\psi_i|\psi_j>$...means?
 
John Rennie
John Rennie
$$\int \psi_1^* \psi_2 dV = <\psi_1|\psi_2> $$
 
Anonymous
6:54 AM
@JohnRennie What is the meaning of $\int \psi_i^* \psi_j dV$? It doesn't mean anything unless $\psi_i=\psi_j$....which basically gives the probability amplitude $|\psi|^2$...
 
John Rennie
John Rennie
@Blue Do you mean what is the physical meaning?
 
Anonymous
@JohnRennie yes
 
John Rennie
John Rennie
It's called the overlap integral. What it means physically depends on the context.
In this case it has to be zero because eigenfunctions always have an overlap integral of zero. So the physical meaning is just that our $\psi_i$s are eigenfunctions.
 
Balarka Sen
Balarka Sen
@Blue If you want to think about it symbolically, it's the "dot product of the vectors $\psi_i$ and $\psi_j$"
 
John Rennie
John Rennie
^ Yes, that's a good way of thinking about it.
 
Anonymous
6:58 AM
Okay. Got it till here.
 
John Rennie
John Rennie
The rest is just algebra ...
$$<\psi|\psi> = <a_1\psi_1 + a_2\psi_2 + ... | a_1\psi_1 + a_2\psi_2 + ... > $$
where I've just expanded $\psi$
$$=<a_1\psi_1|a_1\psi_1> + <a_1\psi_1|a_2\psi_2> + ... + <a_2\psi_2|a_1\psi_1> + ... $$
 
Anonymous
mmm hmmm got it this far
 
John Rennie
John Rennie
OK. And we know that $<\psi_i|\psi_j>$ is zero if $i \ne j$ and one if $i=j$. Yes?
 
Anonymous
@JohnRennie Yes
 
John Rennie
John Rennie
So that big long expression simplifies to: $$ <\psi|\psi> = a_1^2 + a_2^2 + a_3^2 + ...$$
Can you see that or do I need to go into more detail?
 
Anonymous
7:04 AM
One sec. If $i \neq j$, then does it mean we can think of the functions are "perpendicular"?
 
John Rennie
John Rennie
@Blue basically yes.
Though orthonormal is the technical term.
Orthonormality
In linear algebra, two vectors in an inner product space are orthonormal if they are orthogonal and unit vectors. A set of vectors form an orthonormal set if all vectors in the set are mutually orthogonal and all of unit length. An orthonormal set which forms a basis is called an orthonormal basis. == Intuitive overview == The construction of orthogonality of vectors is motivated by a desire to extend the intuitive notion of perpendicular vectors to higher-dimensional spaces. In the Cartesian plane, two vectors are said to be perpendicular if the angle between them is 90° (i.e. if they form a right...
 
Anonymous
Are independent vectors/wave-functions always "orthonormal" to one another?
 
John Rennie
John Rennie
They are always orthogonal. It's an extra restriction in QM they they be normalised as well.
It's like Balarka said. Their dot product is zero.
 
Anonymous
18 mins ago, by Balarka Sen
It's called the $L^2$ norm. By definition, $\|\psi\|_{L^2} = \int |\psi|^2 dV$. That is 1.
 
Anonymous
If that comes out to be 1 only then it is normalized?
 
John Rennie
John Rennie
7:06 AM
Yes
 
Anonymous
Getting it
 
Anonymous
okay
 
Balarka Sen
Balarka Sen
Tagline being "normalized wavefunctions are unit vectors in the Hilbert space of wavefunctions"
 
John Rennie
John Rennie
^
 
Balarka Sen
Balarka Sen
I mostly learn mathematics by using taglines and analogies :p sorry if that sounds unmathematical
 
Anonymous
7:08 AM
 
Anonymous
So the H operator looks like the one on the left
 
Balarka Sen
Balarka Sen
Yup
 
Anonymous
And the one on the right contains $\{a_1,a_2,a_3,...\}$
 
Anonymous
Such that normalization condition holds
 
Anonymous
Why did they write $\{\psi_1,\psi_2,...\}$ in the right matrix then?
 
Balarka Sen
Balarka Sen
7:10 AM
They are not using notation consistent with JR's
 
Anonymous
Also...I still don't know what $O_{11},O_{22},O_{33}$, stand for...in the left matrix operator (in the Hamiltonian operator)
 
Balarka Sen
Balarka Sen
The matrix of $\hat{H}$ has column vectors (row vectors, depending on convention) equal to $\{a_{1i}, a_{2i}, a_{3i}, \cdots\}$ where these $a_{ni}$'s are coefficients when I write $\hat{H} \psi_i$ as $a_{1i} \psi_1 + a_{2i} \psi_2 + \cdots$
Here $\psi_k$'s are, according to JR's notation, the basis wavefunctions
This is the infinite dimensional analogue of having column vectors of the matrix of $T : \Bbb R^n \to \Bbb R^n$ as $T(e_i)$, where $e_i$ = (0, 0, ..., 0, 1, 0, ..., 0) with 1 on i-th slot.
 
Anonymous
@BalarkaSen Got it. In the example $\hat{O}=\{\{2,0\}\{0,3\}\}$ gave above...that was a diagonalized matrix. The $\hat{H}$ also may or may not be diagonalized.
 
Anonymous
It could have finite elements in each row like ${a_{1i},a_{2i},a_{3i},...}$
 
Anonymous
As you showed
 
Balarka Sen
Balarka Sen
7:16 AM
Yes. But it turns out to be diagonalizable. And the diagonal elements after we diagonalize it are precisely it's eigenvalues.
Which are solutions to the Schroedinger's equation (@JohnRennie is this right?)
 
John Rennie
John Rennie
Aha, Daniel's here!
@Blue Daniel can explain this stuff a thousand times better than I can
@BalarkaSen Yes
 
DanielSank
DanielSank
Bonjour, Monsieur LeRennie.
@JohnRennie I can?
 
Balarka Sen
Balarka Sen
Physicists are coming in hoards. Time for me to depart.
 
DanielSank
DanielSank
What are we explaining?
 
John Rennie
John Rennie
In fact I didn't understand any of this until I started reading Daniel's posts here
 
Balarka Sen
Balarka Sen
7:18 AM
Too much physics aura around here
 
John Rennie
John Rennie
@DanielSank why QM is linear algebra
 
DanielSank
DanielSank
@JohnRennie 'cuz it gives the right predictions for experiments.
Bam.
:-D
 
John Rennie
John Rennie
I think we've narrowly saved @Blue from learning QM via the Griffith's route! :-)
 
DanielSank
DanielSank
Thank the muses!
Polish the seven wonders of the world!
...and the peasants did feast upon the fruits and the vegetables, and the lambs and the breakfast cereals...
 
Balarka Sen
Balarka Sen
bourgeois capitalists
 
Anonymous
7:21 AM
Brb in 2 mins!
 
DanielSank
DanielSank
1 min ago, by Blue
Brb in 2 mins!
damn
 
Anonymous
OK.
 
Anonymous
I'm back
 
Balarka Sen
Balarka Sen
clock keeps ticking
that was quick
 
DanielSank
DanielSank
I was hoping it would be two minutes and I could be snarky.
 
Anonymous
7:23 AM
So....let me sum up what I understood and what I didn't
 
Anonymous
@DanielSank Nah, never. Time is relative. =P
 
Balarka Sen
Balarka Sen
plot twist: Blue is a time traveler and actually went afk for a year because of his frustration with our explanations of QM
but he's just making it look like he's gone for a minute and just chillin
 
Anonymous
 
Anonymous
So. $\hat{H}$ can be considered to be a matrix operator like ^
 
Anonymous
And,
 
Anonymous
7:26 AM
 
Anonymous
$\psi$ can be written as ^
 
Anonymous
Nah....that doesn't mean anything
 
Anonymous
$\psi$ in itself could be any linear combination of independent wavefunctions $\psi_1,\psi_2,\psi_3,...$
 
DanielSank
DanielSank
What are we trying to explain here?
 
Anonymous
So basically my question is:
 
Anonymous
7:28 AM
We have a wavefunction say: $\psi = Ae^{i(wt-kx)}$ and we apply the Hamiltonian operator on it
 
Anonymous
I want to know how would you represent the above procedure in the form of a matrix operator multiplied to a vector space
 
DanielSank
DanielSank
Oh my goodness this is hilarious
 
Anonymous
 
Anonymous
Basically how do you write $\hat{H}\psi$ in the above format where $\psi$ is $ Ae^{i(wt-kx)}$ ?
 
Anonymous
@DanielSank halp
 
DanielSank
DanielSank
7:33 AM
ah
This becomes rather easy if you think about it the right way.
What is the Hamiltonian?
 
Anonymous
 
Anonymous
Replace $\frac{1}{2} kx^2$
 
Anonymous
With $V(x)$
 
DanielSank
DanielSank
Ah.
Ok
Do you understand what the elements of a matrix mean?
Suppose we have a basis for our vector space $e \equiv \{e_1, e_2, \ldots \}$.
Suppose you also have a linear transformation $T$.
How do you get the matrix for $T$ in the $e$ basis?
@Blue
 
Anonymous
@DanielSank If I understood what Balarka & JR explained then it should be the coefficients of $\psi_1,\psi_2,...$ when we write $a_1\psi_1+a_2\psi_2+....$
 
Anonymous
7:41 AM
(Considering it to be a diagonal matrix with one non-zero element in each row)
 
DanielSank
DanielSank
uh
Who said $T$ was diagonal?
 
Anonymous
@DanielSank Just assumed for simplicity. Anyway, it might be non-diagonal also
 
DanielSank
DanielSank
Ok, suppose we have this matrix $$\left[ \begin{array}{ccc} a & b & c \\ d & e & f \\ g & h & i \end{array} \right]$$
Suppose I multiply it on $$\left[ \begin{array}{c} 1 \\ 0 \\ 0 \end{array} \right]$$
What do I get?
 
Anonymous
$\{\{a\},\{d\},\{g\}\}$
 
Anonymous
$\left[ \begin{array}{c} a \\ d \\ g \end{array} \right]$
 
Anonymous
7:47 AM
@DanielSank ?
 
DanielSank
DanielSank
yes
@Blue so as you can see, multiplying a matrix by the first basis vector gives you the first column.
Same goes for the $i^\text{th}$ basis vector and the $i^\text{th}$ column.
ok?
 
Anonymous
What does "basis vector" mean?
 
DanielSank
DanielSank
do you know what a basis is?
 
Anonymous
13 hours ago, by Avantgarde
@Blue Basis is what the word implies. For example, in Cartesian coordinate system, $\hat{x},\hat{y}$ and $\hat{z}$ form the basis for any vector in 3 dimensional space. Basically, a basis is the fundamental building block, using which you can construct any other vector in the vector space in question
 
DanielSank
DanielSank
yep
 
Anonymous
7:56 AM
$\psi_1,\psi_2..$ can be considered as basis?
 
Anonymous
Since they are independent wavefunctions?
 
DanielSank
DanielSank
uhhh, I don't know what you mean by those symbols.
oh
 
Anonymous
What was the basis vector in your question btw? {1,0,0} ?
 
DanielSank
DanielSank
In a three-dimensional vector space, there are three basis vectors.
You can write them as (1,0,0), (0,1,0) and (0,0,1).
 
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