 5:42 AM
Good Morning sir :) @JohnRennie 5:55 AM
@user8718165 hi :-) hi
@JohnRennie do you have some time for a quantum q? @psa yes I have a wavefunction defined by $\psi(x) = x(L-x)$ when $0<x<L$ and $\psi(x) = 0$ otherwise. I've already plotted + normalized it, but I'm just trying to verify (or not verify) that it solves the time independent Schrodinger equation for a particle in a box between 0 and L.
Where $V(x) = 0$ for $x \in [0,L]$ and $V(x) = \infty$ for $x \not\in [0,L]$
cause it's a particle in a box
The boundary conditions are fine, so I've considered that as well There's a quick shortcut to the answer ... I'm very confused what I'm doing with the equation though. I tried to show that LHS = RHS but the LHS just gave me $\frac{A\hbar^2}{m}$ where $A$ is the normalization constant.
OK, I like shortcuts 6:03 AM
Only the eigenfunctions of the Hamiltonian are time independent.
Superpositions of different (non-degenerate) eigenfunctions are not time independent.
And this is clearly a superposition. You'd express it as a linear combination of the eigenfunctions by Fourier analysis. can you help express it as one? The eigenfunctions of the particle in a box are sine waves, or more precisely parts of a sine wave. This function is a quadratic not a sine wave. I thought the eigenfunctions took the form $u_{n} = \sqrt{\frac{2}{a}} \sin \frac{n{\pi}x}{a}$
oh ok yes
I was slow typing lol :-) so I can just say "it's not a sinusoidal" 6:07 AM
The reason is this: "so it's not an eigenfunction of the hamiltonian ---> not time independent" The solutions to the time independent equation are:
$$\psi_{n} = \sqrt{\frac{2}{a}} \sin \frac{n{\pi}x}{a} e^{i E_n/\hbar t}$$
Where $E_n$ is the energy of the eigenfunction $\psi_n$
This is simpler than you think. If we write $E = h\nu$ then $E/\hbar = 2\pi f = \omega$ so the time part is just $e^{i\omega t}$.
So the time part is just a sinusoidal oscillation in time as well. OK so far? yes
wait, isn't there a negative there?
$e^{-iE_{n}t/\hbar}$ And if you add two different eigenfunctions you are adding oscillations with different frequencies and they interfere in a time dependent way i.e. you get a beat frequency. I might be mistaken... 6:12 AM
@psa I honestly can't remember ... haha ok fair enough
big picture... Anyway, the time dependent interference means the probability distribution becomes time dependent. right And therefore it cannot be a solution to the time independent equation.
With a single frequency $|e^{-iE/\hbar t}| = 1$ at all times so the probability distribution is time independent. right, so are we adding frequencies somehow with a functional form $x(L-x)$? 6:16 AM
Yes, because:
$$x(L-x) = \sum_n \psi_{n} = A_n \sqrt{\frac{2}{L}} \sin \frac{n{\pi}x}{L} e^{i E_n/\hbar t}$$
where $A_n$ are the coefficients you get by Fourier analysing $x(L-x)$. oh because $A_{n}$ varies Yes. ah I see
that makes a lot of sense The quadratic actually looks very similar to the ground state, so $A_1$ is going to be large and for higher values of $n$ the amplitude $A_n$ will fall off rapidly. But the higher coefficients will be non-zero.
I suppose we could Fourier analyse the function, but that seems a lot of hassle for 6 a.m. :-) 10:20 :p 6:20 AM
OK, not so late :-) interesting This is probably more than your interested in, but there's an answer on the SE explaining how a superposition of the hydrogen 1s and 2p states produces a time dependent oscillating charge, and that oscillation generates the photon with an energy equal to the difference in the 1s and 2p energies. yeah I'm down to take a look 64  In this specific instance you are correct. If you have a hydrogen atom that is completely isolated from the environment, and which has been prepared in a pure quantum state given by a superposition of the $1s$ and $2p$ states, then yes, the charge density of the electron (defined as the electron ... @JohnRennie hello sir 6:29 AM
wow the animations are beautiful
thanks! It's good isn't it :-)
It makes the point. For an eigenfunction $|\Psi|^2$ is time independent, and that's why the eigenfunctions are solutions of the time independent eqn. For any other function this is not true. @JohnRennie are you free now sir? @user8718165 yes 7:03 AM
@JohnRennie hello sir...I want to ask you a qn @user8718165 yes? @JohnRennie sir please have a look at this I need some context. Where is that graph from?
OK, so what is the question?
I'll be a few moments while I answer a question in another room @JohnRennie okay sir...please tell me after finishing :) 7:44 AM
@user8718165 why Don, t you put any name instead of this number. @user8718165 hi, do you want to ask now? @yuvrajsingh yeah :) Don't worry...I'll change it some day ;-)
@JohnRennie yeah sir... why is there the band gap when the bands are overlapping? That graph shows a calculation of the electronic states when a cloud of silicon atoms is brought together to form a silicon crystal. Sorry. @user8718165
Enjoy......... @yuvrajsingh Nah...No sorry ;-) 7:54 AM
The x axis is the separation between the atoms, so large x means we have basically isolated atoms with sharp energy levels. Then as the spacing is decreased the atomic levels broaden out to form bands. @JohnRennie okay sir...can I show you one more image that will clear my qn?
@JohnRennie yeah sir...got it The bands are the areas I have shaded in grey:
So at the actual spacing there is a gap between the valence and conduction bands. @JohnRennie yeah sir...got it...But the region where the gap is located is where the bands are overlapping sir
@JohnRennie sir this
@JohnRennie In the gap region...the two bands overlap but there is a gap... I'm not sure what you mean. At large spacings the bands overlap. i.e. here:
But at the actual spacing in crystalline silicon the bands do not overlap. @JohnRennie sir I got it...I'm not able to tell exactly what I'm thinking
@JohnRennie sir this
@JohnRennie Its not fully accurate but I'm telling about green area... 8:09 AM
That's showing something completely different. @JohnRennie why sir... If you look at small values of x the valence band energy rises rapidly. That's the area I've outlined in blue.
If the graph was continued to smaller x the valence band would rise to overlap the conduction band. @JohnRennie okay sir...got it It's a general rule that if you compress any solid enough you will make the bands overlap and you'll get metallic conduction. @JohnRennie even insulators sir? 8:13 AM
Yes, even insulators.
One of the active research areas at the moment is looking for the point at which solid hydrogen becomes a metal.
Normally solid hydrogen is an insulator. Metallic hydrogen
Metallic hydrogen is a phase of hydrogen in which it behaves like an electrical conductor. This phase was predicted in 1935 on theoretical grounds by Eugene Wigner and Hillard Bell Huntington.At high pressure and temperatures, metallic hydrogen can exist as a liquid rather than a solid, and researchers think it might be present in large quantities in the hot and gravitationally compressed interiors of Jupiter, Saturn, and in some exoplanets. == Theoretical Predictions == === Hydrogen under pressure === Though often placed at the top of the alkali metal column in the periodic table, hydrogen does... @JohnRennie thank you very much sir...I'll have a look :)

3 hours later… 11:40 AM
@JohnRennie, Hi sir. Are you free now? @Intellex hi, yes I'm free. Fine :) - Angular momentum of a body with combined translation and rotation is given to be the sum of angular momentum of the body about the centre of mass and angular momentum of the centre of mass about the origin. I understood the derivation for this. Is there any intuitive way of understanding this sir?
@JohnRennie, Further, can we say the angular momentum of the body about COM is the rotational angular momentum? The complication is that objects can have an angular momentum even when they aren't rotating. @JohnRennie But in COM frame it must be entirely rotational right sir? For example if a car is moving in a straight line at constant speed $v$ we'd normally say it isn't rotating. 11:45 AM
@JohnRennie Yes sir. But angular momentum is non zero about a point not on the road. But if I'm standing a distance $d$ from the road then from my position the angular velocity is $v/d$ so it is rotating and therefore has an angular momentum. @JohnRennie Understood this point sir :) To me this has never seemed intuitive, so if you're asking for an intuitive way of calculating the total angular momentum then I don't have one. @JohnRennie Ok sir. Thank you.
If I'm planning to ask a question about this on SE, how to explain what is meant by "intuitive way" @JohnRennie sir? Kindly guide me in this regard. I think you'll struggle to formulate that question in a clear way.
I suspect it would get closed as unlcear what you are asking. 11:49 AM
@JohnRennie Yes sir :) I understand
Ok sir. I'll proceed with this itself. Hope something strikes in my mind regarding this. Thank you :)