The h Bar

GPhys

00:00

hmm

dmckee

On the other hand, it should be possible to (a) use only that part of the question which is purely conceptual (in this case part a), or (b) extracting a conceptual core to ask about.

Jake Rose

Why does the time independent schrodinger equation have to be continuous?

and also why does its derivative have to be continuous?

Anonymous

12

A: How to know if a wave function is physically acceptable solution of a Schrödinger equation?

The very minimum that a wavefunction needs to satisfy to be physically acceptable is that it be square-integrable; that is, that its $L_2$ norm, $$ \int |\psi(x)|^2\mathrm d x, $$ be finite. This rules out functions like $\sin(x)$, which have nonzero amplitude all the way into infinity, and funct...

Jake Rose

The level to which we have done qm is not sufficient to understand them

In my lecture notes it says a discontinuous wave function implies infinite momentum and a discontinuous gradient implies infinite energy. Could you explain where those statements arise from?

user187604

00:16

@JohnRennie man I didn't watch Lord of rings. BTW good night from India......

Anonymous

@JakeRose Actually that claim your lecture note makes isn't completely true.

GPhys

@dmckee I'm still processing how I feel about this. A big part of me really wants to say to hell with the rule here, since I think the questions in (b) and (c) well motivate the conceptual back of the question and help make it an interesting problem that people could learn from and would be of general use of the site. To the extent it's asking a specific thing, it seems clear enough it's for motivational reasons. Changing it too much just feels like I'm taking the "fun" out of it.

Jake Rose

@Blue explain?

If possible that is

Anonymous

What they're saying is that the expectation value of your momentum or kinetic energy might be infinite in certain such discontinuous cases, but that isn't necessarily always the case.

Jake Rose

mhmm

Dont really understand the expectation value

We have done what I would call stupid qm where is just a time filler for next year

Anonymous

00:20

8

A: Can a physical wavefunction be non-smooth (its first derivative is discontinuous)?

There is no physical requirement for a wave function to be smooth, or even continuous. At least if we accept the common interpretation that a wavefunction is nothing else than a "probability amplitude". I.e. it represents, when multiplied with its complex conjugate, a probability density. Now a p...

Jake Rose

We spend a good term on nothing but in detail qm next year

Anonymous

@JakeRose Better to learn QM from mathematicians if you're concerned about mathematical details ;)

Jake Rose

Blasphemy in the most sacred of physics chats @Blue

Is there an even simpler to way of seeing why those statements might be implied though?

Anonymous

Those statements are clearly false...I don't know why they would include that. As for why they might have included it, see the alternative answer in the link I posted

Anonymous

At least you need to learn about expectation values before that though

Jake Rose

00:25

Okay fair enough

Anonymous

@JakeRose I think everyone here agrees that physicists are second class citizens compared to mathematicians ;) Gosh, I so much wish @0celo7 was around :P

Bob van de Voort

@Mithrandir24601 that would be awesome, I'll try to remind you

dmckee

00:51

@GPhys Don't feel like my thoughts are the end all or be all on this. Ask some other users. Or even post one and see.

As a group the moderators have been trying not to cast diamond-powered votes and things that aren't completely obvious so you'd get a chance to see what the wider user base thinks.

Mithrandir24601

@Blue I wouldn't go as far as saying 'clearly false' - discontinuities can exist in models, sure (I've got one example open right now), but they tend to arise from considering things like 'point interactions' or 'point objects' or point-something-or-other in my experience, so while a discontinuity doesn't automatically make it wrong or anything, it seems to arise from simplified models (in my relatively limited experience of this kind of thing).

What I'm trying to say is that if you get a discontinuity in your wavefunction, asking 'how did that get there?' is probably something worth asking

dmckee

@Mithrandir24601 The important thing is is knowing what happens when you relax the model condition that gives rise to the unphysical effects.

For instance, consider the infinite square well. $\dot{\psi}$ is discontinuous at the bounds of the well, but in the relaxed condition of a very deep finite well, you simple have a very high value for $\ddot{\psi}$ near the edge of the well, so all is well.

Mithrandir24601

@dmckee Maybe not in the particular paper I'm looking at where it's the long-range effects that matter though(?)

dmckee

@Mithrandir24601 Don't know right off. Developing a tool kit for understanding these questions is part of what you should be doing if you want to pursue a sub-discipline where QM is important.

Mithrandir24601

@dmckee Fair. This particular one comes from considering a point-coupling to a waveguide, which unsurprisingly gives a discontinuity in the wavefunction at the point of interaction, but I'm not really concerned about that just yet, as I've got a pile of other stuff to be worried about first...

GPhys

01:13

@dmckee this is a draft of what I intend to post

with the title "Can you sail down a moving river on a windless day?"

@dmckee physics.stackexchange.com/questions/410484/…

we'll see

GPhys

01:52

ahh I wish the downvoter left a comment :(

user187604

@CooperCape @Blue drive.google.com/file/d/1w5bv5cJPJuiGSct4RHtInGYiw25oELNO/…

Sorry guys I gave the answer key before question. Quite brilliant :p

It's bilangual so you have to count on English...

It's bilingual so you have to look for English....

It was not my intention to befool you sorry...

user187604