Billy&#x27;s (modestly sized) fourth term &quot;help!&quot; thread

OP

Oops. I forgot to subscribe to this thread, so I didn't see it got updated. :o:

No worries, though - I hadn't gone back to linear algebra since then. Thanks for your help, you two.

My quantum mechanics lecture looks like he's being intentionally difficult here:

"Time-dependent Schrödinger equation,

\displaystyle i\hbar \frac{\partial \psi}{\partial t} = -\frac{\hbar^2}{2m} \nabla^2 \psi + U(\mathbf{x}) \psi.

Separation of variables,

\displaystyle \psi(\mathbf{x},t) = \chi(\mathbf{x}) e^{-i\omega t}.

"

How on earth does the second line follow from the first?

Reply 4

generalebriety

Oops. I forgot to subscribe to this thread, so I didn't see it got updated. :o:

No worries, though - I hadn't gone back to linear algebra since then. Thanks for your help, you two.

My quantum mechanics lecture looks like he's being intentionally difficult here:

"Time-dependent Schrödinger equation,

\displaystyle i\hbar \frac{\partial \psi}{\partial t} = -\frac{\hbar^2}{2m} \nabla^2 \psi + U(\mathbf{x}) \psi.

Separation of variables,

\displaystyle \psi(\mathbf{x},t) = \chi(\mathbf{x}) e^{-i\omega t}.

"

How on earth does the second line follow from the first?

What has he defined omega as?

Reply 5

OP

yusufu

What has he defined omega as?

I don't think he has. He refers to it as "angular frequency", though of what I don't know.

Reply 6

JohnnySPal

2

What I found when I studied PDEs was that they often try a separation ansatz and work from the assumption that there is a solution of that form.

So the second line *might* not follow as such. It might just be a matter of saying "throw this into the equation and find what

\chi

must be if there is a solution of this form".

I'm probably wrong though :p:

EDIT: This would make sense, actually, since a typical ansatz would be to assume a solution of the form

\psi (x,t) = \alpha (x) \beta (t)

, which is what s/he's done here except s/he's given you a specific function of t here. Unless

\omega

is somehow a function of x. *shrugs*

Reply 7

generalebriety

I don't think he has. He refers to it as "angular frequency", though of what I don't know.

hmm. I didn't spot that last year.

I'll have a think.

Reply 8

Right, I have an idea that seems to work. I'll type it up as soon as I get back from the kitchen. Give me 5 mins

Reply 9

generalebriety

Oops. I forgot to subscribe to this thread, so I didn't see it got updated. :o:

No worries, though - I hadn't gone back to linear algebra since then. Thanks for your help, you two.

My quantum mechanics lecture looks like he's being intentionally difficult here:

"Time-dependent Schrödinger equation,

\displaystyle i\hbar \frac{\partial \psi}{\partial t} = -\frac{\hbar^2}{2m} \nabla^2 \psi + U(\mathbf{x}) \psi.

Separation of variables,

\displaystyle \psi(\mathbf{x},t) = \chi(\mathbf{x}) e^{-i\omega t}.

"

How on earth does the second line follow from the first?

Do you want my full explanation or hints and stuff?

Reply 10

MC REN

14

generalebriety

Oops. I forgot to subscribe to this thread, so I didn't see it got updated. :o:

No worries, though - I hadn't gone back to linear algebra since then. Thanks for your help, you two.

My quantum mechanics lecture looks like he's being intentionally difficult here:

"Time-dependent Schrödinger equation,

\displaystyle i\hbar \frac{\partial \psi}{\partial t} = -\frac{\hbar^2}{2m} \nabla^2 \psi + U(\mathbf{x}) \psi.

Separation of variables,

\displaystyle \psi(\mathbf{x},t) = \chi(\mathbf{x}) e^{-i\omega t}.

"

How on earth does the second line follow from the first?

He's almost certainly postulating a solution of that form, and the actual form of the time part is from solving the equation

\displaystyle i\hbar \frac{\partial \psi}{\partial t} = E \psi

i.e. the stationary state eigensolution of the Hamiltonian (from the spatial part of the wavefunction

H\psi(x) = E \psi(x)

where there is no time dependence in the wavefunction)
and

\omega

is easily identifiable

Reply 11

generalebriety

Oops. I forgot to subscribe to this thread, so I didn't see it got updated. :o:

No worries, though - I hadn't gone back to linear algebra since then. Thanks for your help, you two.

My quantum mechanics lecture looks like he's being intentionally difficult here:

"Time-dependent Schrödinger equation,

\displaystyle i\hbar \frac{\partial \psi}{\partial t} = -\frac{\hbar^2}{2m} \nabla^2 \psi + U(\mathbf{x}) \psi.

Separation of variables,

\displaystyle \psi(\mathbf{x},t) = \chi(\mathbf{x}) e^{-i\omega t}.

"

How on earth does the second line follow from the first?

Dammit. Stupid new combi ovens burnt my dinner! :mad:

I'm gonna do some work now, so I'll just spoilerify each step.

\displaystyle i\hbar \frac{\partial \psi}{\partial t} = -\frac{\hbar^2}{2m} \nabla^2 \psi + U(\mathbf{x}) \psi.

Consider a function

\psi = \chi (x) f(t)

Spoiler

Reply 12

Is it just me or are the spoiler tags in my post not working?

Reply 13

Kevin_B

1

They're not working. You forgot to put them on the other end of the sentence.

It should be

Spoiler

Reply 14

MC REN

14

yusufu

Is it just me or are the spoiler tags in my post not working?

you've got the slashes the wrong way round

Reply 15

OP

That took me quite a while to get through, but I think I got it in the end. Cheers. :smile:

(Edit: I had a question, but I solved it. :p:

)

Reply 16

OP

"Show that the Green's function for the initial value problem y'''' + k^2 y'' = f(t), y(0) = y'(0) = y''(0) = y'''(0) = 0, is given by

G(t, T) = 0 for 0 < t < T,
G(t, T) = (1/k^2)(t - T) - (1/k^3) sin k(t - T) for t > T.

Hence write down the solution for f(t) = e^-t and verify that it satisfies the equation and initial conditions. (Hint: make life easy by noting G(T, T) = 0 for an IVP (initial value problem?) Green's function and so use the time invariance of the equation to take G(t, T) = f(t - T) for t > T.)"

I've found G, which was easy enough. I don't see how I'm meant to "write down" the solution for f = e^-t (unless he just wants me to write down a big messy integral?) - the hint doesn't mean much to me, and I really don't think he means G(t, T) = f(t - T), does he?

Thanks.

Reply 17

I'd write the integral. I think the hint is more for the verification part of the wuestion.

Reply 18

OP