Problem Solving Strategies

user8718165

5:42 AM

Good Morning sir :) @JohnRennie

John Rennie

5:55 AM

@user8718165 hi :-)

psa

hi

@JohnRennie do you have some time for a quantum q?

John Rennie

@psa yes

psa

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

John Rennie

There's a quick shortcut to the answer ...

psa

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

John Rennie

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.

psa

can you help express it as one?

John Rennie

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.

psa

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

John Rennie

:-)

psa

so I can just say "it's not a sinusoidal"

John Rennie

6:07 AM

The reason is this:

psa

"so it's not an eigenfunction of the hamiltonian ---> not time independent"

John Rennie

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?

psa

yes

wait, isn't there a negative there?

$e^{-iE_{n}t/\hbar}$

John Rennie

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.

psa

I might be mistaken...

John Rennie

6:12 AM

@psa I honestly can't remember ...

psa

haha ok fair enough

big picture...

John Rennie

Anyway, the time dependent interference means the probability distribution becomes time dependent.

psa

right

John Rennie

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.

psa

right, so are we adding frequencies somehow with a functional form $x(L-x)$?

adding eigenfunctions

John Rennie

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 $x(L-x)$.

psa

oh because $A_{n}$ varies

John Rennie

Yes.

psa

ah I see

that makes a lot of sense

John Rennie

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. :-)

Presumably midnight your time ...

psa

10:20 :p

John Rennie

6:20 AM

OK, not so late :-)

psa

interesting

John Rennie

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.

psa

yeah I'm down to take a look

John Rennie

64

A: Is there oscillating charge in a hydrogen atom?

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 ...

It is an awesome answer. One of those answers I wish I had written myself :-)

user8718165

@JohnRennie hello sir

psa

6:29 AM

wow the animations are beautiful

thanks!

John Rennie

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.

user8718165

@JohnRennie are you free now sir?

John Rennie

@user8718165 yes

user8718165

7:03 AM

@JohnRennie hello sir...I want to ask you a qn

John Rennie

@user8718165 yes?

user8718165

@JohnRennie sir please have a look at this

John Rennie

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

user8718165

@JohnRennie okay sir...please tell me after finishing :)

yuvraj singh

7:44 AM

@user8718165 why Don, t you put any name instead of this number.

John Rennie

@user8718165 hi, do you want to ask now?

user8718165

@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?

John Rennie

That graph shows a calculation of the electronic states when a cloud of silicon atoms is brought together to form a silicon crystal.

yuvraj singh

Sorry. @user8718165

Enjoy.........

user8718165

@yuvrajsingh Nah...No sorry ;-)

John Rennie

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.

user8718165

@JohnRennie okay sir...can I show you one more image that will clear my qn?

@JohnRennie yeah sir...got it

John Rennie

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.

user8718165

@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...

John Rennie

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.

user8718165

@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...

John Rennie

8:09 AM

That's showing something completely different.

user8718165

@JohnRennie why sir...

John Rennie

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.

user8718165

@JohnRennie okay sir...got it

John Rennie

It's a general rule that if you compress any solid enough you will make the bands overlap and you'll get metallic conduction.

user8718165

@JohnRennie even insulators sir?

John Rennie

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.

Yes. If you Google "metallic hydrogen" you'll find lots of links.

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...

user8718165

@JohnRennie thank you very much sir...I'll have a look :)

Intellex

11:40 AM

@JohnRennie, Hi sir. Are you free now?

John Rennie

@Intellex hi, yes I'm free.

Intellex

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?

John Rennie

The complication is that objects can have an angular momentum even when they aren't rotating.

Intellex

@JohnRennie But in COM frame it must be entirely rotational right sir?

John Rennie

For example if a car is moving in a straight line at constant speed $v$ we'd normally say it isn't rotating.

Intellex

11:45 AM

@JohnRennie Yes sir. But angular momentum is non zero about a point not on the road.

John Rennie

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.

Intellex

@JohnRennie Understood this point sir :)

John Rennie

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.

Intellex

@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.

John Rennie

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.

Intellex

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 :)

Transcript for

Problem Solving Strategies