The Harmonic Oscillator and Ladder Operators

Alright, as I announced earlier, we’re now handling V(x) case by case.

The first case of potential was

exactly the infinite square well, the infinitely deep square potential well.

So now, the second case~

When the potential is that of the harmonic oscillator!!

Huh??! But what was it again that I said I’d solve case by case for V(x)..T_T

Finding the ψ(x) that satisfies this ODE!!!!

We’re handling the case where this holds!!! Let me talk a bit more.

That’s the potential of a spring system. That is, what I want to say is

the fact that the potential is

means we’re in a force system described by Hooke’s law of force!

Force following Hooke’s law means x and F are in a first-order linear relationship!

So that means

The x(t) that satisfies this is

But,,,,, does a system that experiences this kind of force even exist in nature..??…..heh heh heh heh

There won’t be many relationships that are exactly linear like that,

why is it that we physics majors, from the moment we enroll it’s spring spring, up until graduation it’s spring spring, all day long spring spring why is it like this

Let me explain why the spring system is important,

If we first look at the Carbon-Carbon potential and get a feel for it,

ahh first of all

it’s not this!

No it’s not, no it’s not

hmm~~ but!!

We do a Taylor series expansion, and where we’ll do it is

at the x where V(x) is minimum!

At that point the slope is 0, so the red one is 0 !!

And we’ll make this approximation!!

Ah also, the blue one~~~ since it’s the potential at the minimum point, it should be fine to set it as 0.

And what’s left, that V double-prime, what it means is the same as the spring constant k.

So like this, even when there’s no actual spring attached,

in our natural world there are many situations where we can interpret things as if a spring were attached, so studying spring systems is important!!!!!!that’s the story.

I feel like I’ve scribbled the same story, this exact story, in many different places…so I’ll stop here and

go back to the main topic

plugging

into the (time-independent) Schrödinger equation

.

It becomes like this

and now we just need to find the solution to this second-order differential equation…………………ha…………this is daunting……….

There’s that x-squared term, so finding a solution is a little tricky!

For this problem, our ancestors came up with countless methods for solving that equation,

and among them, we’ll be learning the 2 most opposing methods.

The first one is, so-called, 1. Super-Crafty method!!!!

We’ll cleverly~~~~~ use trickery to untangle the problem,

the second method is the so-called 2. Super Armored Tank method!!!!!

Just mindlessly ram through everything, break it all down and go in!!!

If you ram and break things down, I’ll probably get hurt too….?hehehehe

(There’s no royal road in life, let’s walk the right path (正道)!!! I’ll recommend the 2nd method.)

Alright, now entering 1. Super-Crafty method!!!!!

I’ll jump right in. Let’s start with an equation.

Coming down this far, no problem!!! Now we’re going to factor that!

cf.) Rather than specifically just that one case, any of these three would give the same result, right

Wait a minute!!!!!!!!!!!!!!!!!!!!!!!!!!!@@!@!@!@!@!@!!!!@@!

Is this factoring even possible?!?!?!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

To see whether it’s possible or not, let me expand it again

Now this is where the problem is……

The problem we’re facing is, here p and x are not just ‘values! (numbers)’

they’re ‘operators’, so

this holds.

This kind of relationship between operators, xp-px, is mathematically called a commutator,

and the commutator is expressed as xp-px = [x , p],

and what it means physically is

for a state function psi, “what do you measure first, and what remainder do you measure after that” — this changes the result.

That’s roughly the meaning,

so then, not for arbitrary operators Q, T,

but specifically!!!! for the operators p and x,

let’s see what (xp-px) reduces to after it’s done all its operator duty.

If we cloak ψ away~~~

Hmm~~ so our equation,

since it’s about x and p,

I guess we can write it like this!!! Since it’ll end up like that anyway,

now finally, the problem we’re facing in the Schrödinger equation —

how do we solve it~~

I don’t really know..T_T

An ordinary commoner-peasant-civilian like me has no idea how the equation on the next page came to be, so

just…

let’s say that some ’new operator’ made of operators fell from the sky — plop!

And I picked it up off the street lolololol

Ohoh what is this, so you’re saying these two operators give us the above equation?

Alright let’s find out right now!

Oh-hoooo~~~~~ the term that came out looks just like before, doesn’t it?!?!!?!!

Hmm yes, now what we’re going to do is take the calculation that couldn’t proceed because of the order of the operators,

and fix it piece by piece so it can proceed,

and we’ll do that with the newly-picked-up operators a+, a_.

So now let’s slowly head to the Schrödinger equation with the New Operator!!!

Okay good!!!! Using the nameless a_ & a+ new operator we picked up, made of x, p operators!

We’re able to express the Hamiltonian!!!!!!!

That means we can express the Time-Independent Schrödinger Equation in terms of the New Operation

Done!hehehehehe

We can now express the Schrödinger equation with the New Operator like this.

Now we need to look into that new operator.

To say the conclusion first, those two operators are called ’ladder operators’,

and let me explain why we call these ladder operators!!!

Energy-measuring guy

Let me call this guy over and request an energy measurement.

What am I going to request,

I’m going to request the energy measurement for this state!!!!! -> (let’s see what eigen-value comes out)

By the same method

if we request the energy measurement of this,

we can see that an equation like this comes out

and if I explain the above situation with my cartoon again, which looks like I was on drugs,

&

Let me explain why these are called ladder operators:

is called the raising operator (or Creation operator),

((

is the energy of a single photon, so (: meaning it creates one photon’s worth of energy) ))

is called the lowering operator (or

Annihilation operator (: because it removes one photon’s worth of energy,)

and lumping

and

together, they are called ladder operators…hehe

So then, when we’ve found some solution,

if we keep~~~~~ applying the lowering operator to that solution ψ(x), at some point the energy will become 0.

The moment will come when E+V = 0,

But! For any substance, there is no such thing as E+V = 0!!!!! (You’ll find out why as we go further! Or you might already know,)

Aaah so

the ground state of some particle, the bottom state

if we apply the lowering operator just once more here, it will become 0

we have to accept this!! ….

Yep yep! No problem accepting this!!

Describing the ground state by the above logic,

since it was this,

The psi-zero wavefunction we found at the ground state

also, its square has to have meaning as a probability, so

let’s normalize it!!!!! Time to determine the undetermined coefficient A via normalization!

The above integral must be done via the Gaussian integral!!!!!!!

Actually we used the Gaussian integral before too, but let me introduce once more the most common use of the Gaussian integral,

More generally,

Hitting the above integral with this principle,

Okay, (state function,) now that we know the complete shape of the wavefunction, let’s measure the energy?!!!!?!!

Hellooo~~~~~

You can slap either one of the two in,

but which one to slap in????????? — the latter one is more convenient hehehehe