The Schrödinger Equation in Spherical Coordinates (3D)

(image unrelated)

I’m totally screwed.

[MENTAL BREAKDOWN WARNING POST]

Let’s get started.

OK, looking back at what we’ve done so far, it seems like there wasn’t anything that special.

The main content was pretty much only in Chapter 2, and Chapters 1 & 3 were just, like, preparation??? for Chapter 2, I think,

and honestly, I think this is where we’re really taking our first steps now.

(I heard that Chapter 1 of the quantum mechanics textbook used in grad school is the hydrogen atom?)

What we’re going to do now is approach the state of the electron in the ‘hydrogen atom’ quantum-mechanically.

To do that, here we first need to conceptually extend the quantum mechanics we’ve been looking at in 1D to 3D.

The 3D Schrödinger equation!!!! Simple~

What used to be a 1D total derivative just gets bam~~ turned into a Laplacian (partial derivatives), and we’re done.

Then in the Cartesian coordinate system we’ve been using a lot,

​Alright, and since this chapter is gonna start with the hydrogen atom~~

we just need to plug in the potential V of the hydrogen atom for the potential V, right???!??!

Since the hydrogen atom is an atom with 1 proton, its electric potential is

so~~~ plugging this into the above gives (e: electron charge!!!)

Since that r also needs to be converted into the Cartesian coordinate (rectangular coordinate) system,​

Yeah… exploring the hydrogen atom further~~~ in the rectangular coordinate system stops right here.

That red guy!!! is bundled up as squares, and on top of that there’s even a square root,,,,,,,

anyway, x, y, z are all bundled together, so the differential equation can’t be solved!!

So we can’t solve it in the x,y,z coordinate system.

So the choice we can make is

[HEART ATTACK WARNING]

Well, what choice do we have… they say switching to the spherical coordinate system is at least the more convenient option….(?)

Ugh, convenient my ass

It’s just gonna make my hair fall out. lol

Just kidding. Anyway, so we have to switch to spherical coordinates,

Switching the Laplacian to spherical coordinates… is not all that simple.

I could prove switching the Laplacian to spherical coordinates here too, but it would get super long,

so I’ll just attach a file. Please refer to it.

I’ll just write the conclusion.

The Laplacian in spherical coordinates is written as

And then we plug it into the Schrödinger equation.

Oh, and for now, I’ll plug it into the time-independent Schrödinger equation first!!

Its looks are seriously outrageous lol

lol lol lol

All I did was switch the coordinate system as told and it ended up looking massively massively massively massively..heh. hehe haha

Whenever we hit a situation like this, the physics department has a special measure!?!?!!!

“Let’s assume it’s separable*^0^*”

Now ψ is no longer a function depending on x,y,z but on r,θΦ, so

it’s ψ(r,θΦ).

And since we assumed this is separable,

it would be written like this, right.

And plugging this into the Schrödinger equation above,

<From here on, I referred to modern physics (Beiser, 6th ed.), not the quantum mechanics textbook.>

Ah, and rather than partial derivatives, total derivatives would now be the more correct notation.

I’ll change it to the more mathematically correct expression.​

So that the red guy becomes a term depending only on Φ,

I’ll keep only the red guy on the right side and move everything else to the left side~~~~.

Then I’m going to say this:

The red guy: an expression in Φ

The blue guy: an expression in r & θ

Now!!!! For the equation to hold!!!

“The left and right sides must be equal as a constant!!!!”

<As for why they have to be equal as a constant…… I don’t really need to write that, right?~?~~?!!!>

(If I were to speak on behalf of the professor, it would be: “If there’s a case where they’re equal without being equal as a constant, why don’t you bring it and show me lol Let’s take a look lol”)

So let’s say the left and right sides are equal to m squared.

Writing only the right side,

like this.

Setting it as the square of some number is because there are no particular conditions on m, so we set it as m squared,

meaning m can be positive, negative, complex…. they say we’ve left ALL those possibilities open!!!

Now let’s start playing with the above equation by plugging in m-squared.

Then I’ll say this:

The green-green guy: an expression in r

The orange-orange guy: an expression in θ​

Ah, and since the potential V in the hydrogen atom is a function depending on r,

we can say the left side is f(r) and the right side is g(θ)!!!

Also, since the left and right sides are equal with different variables!!!! For the equation to hold!!

they can only be equal as a constant!!!!! Ooh!!!!

Let’s call that constant l(l+1)!!!

(Let’s think of setting the constant like this as the wisdom of our forebears or something…..

You could actually just set it as any old constant C~, but the result will end up being like this anyway, so

they say it’s better for your mental health to just,,, accept setting the constant this way… hehe)

OK, let me summarize.

OK, now we’re going to solve these one by one. Let’s start with the Φ expression, which is easy to solve.

Now, we’re not the math department, we’re the physics department~~~ Let’s look at the meaning, the meaning!!

Do you remember earlier when I said “there’s no particular condition on m”? So, there was no condition like m has to be positive~ has to be something. Right?

So whether it’s the front term or the back term, it’s the same thing, so there’s no need to bother writing it as a linear combination for the general solution.

Since they’re the same, taking a linear combination of things that are the same just gives the same thing anyway?

(Since Φ is an angle from the x-axis, (+) is rotation in the (+) direction and (-) is rotation in the (-) direction, so we just pick one of the two and use it.)

Let’s go with this.

OK, also, since Φ is the angle from the x-axis,

this condition must hold.

Applying that to the function of Φ we found,

Oooh!!!!! We’ve even found the condition that m has to be an integer!!!

OK, now let’s look at the equation for θ.

I’ll multiply both sides by Θ sin²θ and move everything to one side!!!!

Whoa… this….. is a kind of differential equation……

We need to find the big-Θ that satisfies this,

and since the above equation is an associated Legendre function,

the solution to the above diffeq is

they say,

so I’ll just check whether that diffeq really is an ‘associated Legendre equation’ or not.!!

And now I’ll deal with just the pink derivative!!!!!!

OK so please check Boas, Mathematical Physics, page 597, the associated Legendre functions part!!!

So then

the solution to this equation is

this should make sense…..

Oh but, x=cosθ was what we had!

OK, now the part we need to pay attention to is right there, blue blue blue.

If m > ℓ, it becomes zero. If it becomes zero, it becomes meaningless.

Around here, I’ll now reveal the identities of the constant m and the constant ℓ.

Actually, the ‘constant ℓ’ is called the azimuthal quantum number, and

ℓ=0 → s-orbital

ℓ=1 → p-orbital

ℓ=2 → d-orbital

…

it determines these orbitals!!! (As for why~~~ they become orbitals, try running it in MATLAB later….hehehe)​

The ‘constant m’ is called the magnetic quantum number, and they say it’s the number that shows

“how many of those orbitals are possible~”.

(ℓ=1) s has 1

(ℓ=2) p has 3

…

(ℓ=n) has m///(2ℓ+1)

(Also, as comes up later, as you already know, m also indicates the direction of the L vector!??!?!)

OK, now let’s also go to R(r).

This is what we had.

Since the electric potential in the hydrogen atom is a function depending on r,

‘at last’ V(r) comes out in the equation we’re dealing with!!!!!!

Up until now, that’s why there was no V potential term.

Now, how is this handled?…

First, they say that substituting rR(r) = u(r) makes the diffeq solve nicely.

(Let’s copy off the wisdom of the forebears…. hehe)

Then now, let’s churn through the equations from top to bottom to reach the goal.

Now let’s plug the equation we got from the substitution into the expression in r~~~

This thing, this thing, this thing~~~ Haven’t you seen it somewhere before?!?!?!?! In classical mechanics, ‘Kepler’s laws’,

it looks like the ‘planetary orbit equation’!!!

(It kinda makes sense. Whether it’s the universal gravitation of planets or the r-dependent equation inside an atom lol feels like the same thing~?)

[Even calling what’s inside the square brackets the ’effective potential’ is the same!!!!~~ yahoo~ my hard work studying classical mechanics paid off hehe]

OK so, the reason we did the rR = u substitution on the r equation like that will be revealed in just a moment….

Ugh…. this post is long this time……. I’ve been writing just this for several hours too…. T_T

Now we’re almost at the end.

We have to solve this problem.

Normalization

Ha…… I’m screwed hehehe I’m going to go submit the transfer paperwork after just this…..

I want to live a long life… hehe

What we’ve done so far is assume that the wave function

will separate like this, and we’ve unraveled each one!!!!

So now, as we’ve always~ done,

we need to normalize this!!!!

(So that the statistical interpretation is possible!!! damn it)

you’ll know this.

But here, dτ in the rectangular coordinate system is

like this, but

in the spherical coordinate system

this r-squared sine-theta pops out, and as for why,

it’s something you can easily see by looking at the ‘Jacobian’ in mathematical physics, so I’ll just pass over it as if it’s obvious.

Doing the normalization,

The cases where red times blue = +1 could be insanely many with infinite possibilities…

“but they say we set red to 1 and blue to 1 too!!!”

“As for the detailed reason behind it, I don’t know if it’s because we’re still in the early quantum mechanics part that they’re hiding it or if it’s because I’m an undergrad that they’re hiding it……

I’m sorry for keep yapping on and on about the wisdom of our forebears……

Even going around asking questions, I haven’t gotten a good answer yet.. T_T”

OK, so from the red one first!!

Now the reason why we substituted Rr = u… I think you can kind of see!!!!!

But in the current situation where V(r) is not specified, they say this is the end of the information we can figure out!!!!!

I’ll push the detailed integration for the hydrogen atom to later!!! Because there’s still more to do before that!!!!

With this, for the equation in r we’ve done everything we can do!!!(yes!)

Now let’s go to the blue one too!!!!!

Before integrating this,

let’s switch it up like this~~~!!

The reason is

because we can define a function Y that satisfies exactly this~~~~~.

It’s because there’s a certain function defined as the spherical harmonics!

Anyway, its essence is also surely ‘some expression’ related to the quantum numbers m and ℓ! Right!??!?

Obviously this one also needs m and ℓ to be determined for Y to be determined exactly,

so we put the superscripts and subscripts on Y nice and neat~!~!!

and the form of the spherical harmonics is

(For reference, epsilon is a kind of ‘device’ that acts differently when m is positive vs negative.

When m is positive it oscillates + - + - + -, and when m is negative it behaves like + + + + + +; calling it such a ‘device’ would probably be the easiest way to explain it.)

Off to submit the transfer paperwork go go go go go

Ha…. lol