Question

Free particle! @kainui

Answer 1

just an explanation on what exactly we're looking for would suffice!

Answer 2

\[k = \frac{ \sqrt{2mE} }{ \hbar }\] yeah k>0 traveling tot he right, k<0 traveling to the left

Answer 3

Or maybe I'll take the derivative and stuff mess around knowing no idea what I'm doing and then hopefully the answer will make some sense?

Answer 4

Ahh ok so like I'm just debating in my mind how far back to go, do you have a lot of time or do you not have much time

Answer 5

Uh well we can figure that out by using time dilation and seeing relative to you?

Answer 6

Haha no, I got some time

Answer 7

The main thing is, every system is different so it has to have its own Hamiltonian. For the free particle that system is 1 particle and no potential energy anywhere so when you write it you just have the kinetic energy term from the electron and no potential energy so:

\[\hat H = \hat T + \hat V = \frac{\hat p^2}{2m} + 0\]

Remember in classical mechanics, \[T = \frac{1}{2} mv^2 = \frac{p^2}{2m}\] It's exactly the same form of thing, just ends up being more convenient for operators... whatever these are.

Any solution to this equation are wavefunctions that describe a single electron travelling in free space.

Answer 8

Yeah so what whatever, this is still nonsense, but this is exactly where we get these plane wave solutions.

You've probably solved this like once or twice but this is where they come from but the wave function is not the particle it just holds probability information about it.

I guess I'm not sure where to continue from here so I'll let you type something to help orient me haha.

Answer 9

Yeah that's good haha, I'm just not really sure what exactly these solutions mean, like ok it's describing the wavefunction in travelling in free space, it's probably because I've never seen solutions like this before in physics so it's a bit weird

Answer 10

So those A, B, etc are normalization constants in a sense? And the exponentials represent the travelling speed uhh

Answer 11

So if we worked backwards that would give us the hamiltonian of the system?!

Answer 12

The constants out front play two roles, they normalize at first to form your basis vectors and then second they get a new constant which is like the component of the vector. think of it like \(a i+bj+ck\) is a vector, the basis vectors are normalized and now they have coefficients.

Answer 13

Nah don't think about working backwards, the Hamiltonian is the whole thing it's where this all comes from. Everything is a consequence of our choice of Hamiltonian.

Answer 14

Haha yeah, it's like classical mechanics

Answer 15

So what are the solutions to this Hamiltonian's associated Schrodinger equation? They're all of this form:

\[\psi(x) = e^{ikx}\]

So this is one of infinitely many solutions, we can plug in any real number for k to get a solution. A single one of these is called an Energy Eigenfunction. Why? Cause it's an Eigenfunction of the Hamiltonian. But it's also an Eigenfunction of the momentum operator.

How do you find the momentum of an electron represented by that wave function \(\psi(x)\)?

Answer 16

<p>!

Answer 17

You say, "Is the speed in the exponent?" Hey you have operators. That's what they do, they are what give you the observables.

Answer 18

What's the speed operator? Just make it yourself and find out since you know the momentum operator, just divide by mass to get it.

\[\hat v = \frac{\hat p }{m}\]

Now let's use it

\[\hat v \psi(x) = \frac{\hat p}{m} e^{ikx} = \frac{-i \hbar }{m} \frac{d}{dx} e^{ikx} = \frac{\hbar k}{m} e^{ikx} = \frac{\hbar k}{m} \psi(x)\]

Because \(\psi\) is an eigenfunction of the momentum operator, this works (more on that later possibly, there are tons of things to explain lol)

So the point is we are saying the velocity is \(\frac{\hbar k}{m} \) but so what?

Well plug this in then:

\[k = \frac{\sqrt{2mE}}{\hbar}\]

Now solve for E.

Answer 19

Specifically I'm saying take your equation for velocity:

\[v = \frac{\hbar k}{m}\]

and plug in that value of k and make sure you get the value of E you expect.

Answer 20

Haha this is so damn slow to type out my fingers feel like broken sausages or something lol

Answer 21

Lol it's cool, I was just reading and absorbing it all as I think this is a very intuivative way to learn it

Answer 22

So what energy do you get, I gotta test your algebra skills

Answer 23

loll ok

Answer 24

\[E = \frac{ k^2 \hbar^2 }{ 2m }\]

Answer 25

Ok now plug in the value of k from our velocity operator

Answer 26

Then that becomes

Answer 27

That looks a bit weird

Answer 28

Yeah plug in something for k, there shouldn't be any k left haha

Answer 29

Yeah hold on its because I keep scrolling up and down lol