# Homework Help: Not sure about shape of this quantum well

1. Aug 15, 2009

### Davio

Hiya guys, i have a quantum well in strange notation I have never seen before. I think i can do the question, i just need to know the shape of the well.

If someone would be so kind to draw me a shape (I think its a step but not sure?).

It is:

dphi dphi -2m
---- minus ----- equals ----- . U . phi(0)
dx x=0+ dx x=0- (h barred)^2

where V(x) = - U dirac delta function (x)

and U is a postivie constant.

x=0- and x= 0+ are positions immediately to the left and right of the origin.

Thanks a bunch.

2. Aug 15, 2009

### Redbelly98

Staff Emeritus
I'm having trouble figuring out your expression. If I wrap it in [noparse]
Code (Text):

[/noparse] tags, it looks like this:

Code (Text):
dphi dphi -2m
---- minus ----- equals ----- . U . phi(0)
dx x=0+ dx x=0- (h barred)^2
Could you edit this expression to show it more clearly? Use the "Quote" button, then "Preview Post" to make sure it looks the way you want before posting it. The [noparse]
Code (Text):

[/noparse] tags will allow you to use space characters for proper spacing of the terms.

Also, there is no V(x) in what you wrote.

3. Aug 17, 2009

### Davio

Whoops sorry about that. I took a picture of the question actually:

http://img269.imageshack.us/img269/8703/dsc009141.jpg [Broken]

Last edited by a moderator: May 4, 2017
4. Aug 17, 2009

### Redbelly98

Staff Emeritus
Okay.

There are two questions that you seem to be asking ...

1. What is the shape of the potential V(x)?
2. What is the shape of the wavefunction ψ(x)?

... and so I'll address both:

1. V(x) is shaped like the Dirac delta function, only it's negative. Are you familiar with that function?

2. This equation refers to ψ(x):

Code (Text):

dψ |          dψ |          -2m
-- |      -   -- |    =    -----  U ψ(0)
dx |x=0+      dx |x=0-    (h_bar)[SUP]2[/SUP]

And it is telling us that the slope of ψ is discontinuous, similar to the way the absolute value function's slope is discontinuous.

5. Aug 18, 2009

### Davio

I am not familiar with it. As far as I'm aware I haven't been taught it yet.

Does that mean the shape of the well is:

Code (Text):
____
____
ie. like a potential step? My reasoning being that the right hand side take the left hand side equals something left, therefore the right must be bigger than the left?

By discontinuous, does that mean it goes up vertically like what I drew above?

Thanks.

Last edited by a moderator: Aug 19, 2009
6. Aug 18, 2009

### Redbelly98

Staff Emeritus
Are you aware of the distinction between V(x) and ψ(x)? Sorry if this question seems too obvious, but I have the impression that you are mixing them together as the same thing, which they are not.

For an overview of the delta function, see the first three sections (through and including the "Definitions" section) at wiki:
http://en.wikipedia.org/wiki/Dirac_delta_function

And just to clarify: you seem to be asking about the shape of the potential, V(x).
The equation that was given, with the x=0+ and x=0- terms, is referring to the wavefunction ψ(x), not to the potential V(x). If it is V(x) that you are asking about, then that equation is not relevant.

7. Aug 18, 2009

### Davio

I believe i'm asking about the wavefunction? Based on my notes, I've always had a well drawn, and from there I normally work out the rest of the questions. I've read that wiki before lol, I actually have no clue what any of it is on about.

Perhaps I don't understand this question at all, it seems to ask basically a normal well question, but with that hard formula.

I'm extremely confused now.

If its shaped like the dirac delta function. Surely its just a flat plane ? Unless I don't understand the picture :-s

8. Aug 18, 2009

### Redbelly98

Staff Emeritus
Hi,

Rather than continue to jump back and forth and confusing V(x) and ψ(x), let's take a deep breath ....

I'm going to take a step back and outline 3 necessary requirements for you to understand what is going on with this HW problem. Then I'll help you with 2 of those 3 items.

1. First, it's important for you to get an understanding of what is generally meant by the potential function V(x), as well as the wavefunction ψ(x). These are two separate and distinct concepts. It's necessary for you to understand them so that you are able to solve homework and exam questions in quantum mechanics and pass (or at least get a half-decent grade in) this class

2. You'll also need to understand what the Dirac delta function is, on an intuitive level. Then when you see something like

V(x) = -U δ(x)​

you'll have a mental picture of what that looks like. It is a type of well, as you seem to vaguely suspect, but I'll explain more on this later.

3. Finally, you'll need to understand what this equation is telling us about the shape of ψ(x):

$$\frac{d\psi}{dx}\right| _{(x=0+)} - \frac{d\psi}{dx}\right| _{(x=0-)} \ = \ -\frac{2m}{\hbar^2} \ \psi(0)$$​

Now that we have these 3 items outlined, here is what I recommend. I will help with items 2 and 3, in separate posts from this one (so stay tuned).

For item 1, I recommend you talk to either your professor, the teaching assistant / grader (if there is one), OR 1 or 2 of your fellow students and ask one of them to help explain what V(x) and ψ(x) (i.e., potentials and wavefunctions) are. Or you could also start up a separate thread at Physics Forums asking about this, but I think you would understand it better if you can talk to somebody in person.

9. Aug 18, 2009

### Redbelly98

Staff Emeritus
The graph on the left is one way to represent δ(x), the Dirac delta function:

You may think of δ(x) as the limit, as a→0, of the function graphed on the left. I like to picture it as a rectangle of infinite height and infinitesimally small width, so that it has a total area=1.

The V(x) in the problem is drawn in the graph to the right. It is a well, as is typical in many of these quantum mechanics problems. And while it should really have zero width and infinite depth, it's easier to picture it with merely a very small width and a very large depth.

That, in a nutshell, is what V(x) looks like.

10. Aug 18, 2009

### Redbelly98

Staff Emeritus
Okay, you were thinking before that this describes a step function, but that isn't right. This equation says the slope of ψ(x) changes abruptly at x=0, not that ψ(x) itself changes abruptly.

As I mentioned before, this is similar to how the absolute value function behaves near x=0. It is the slope that is discontinuous, not the function itself.

Question for you: according to the equation above, is the slope at x=0+ greater than or less than the slope at x=0-? Assume, for the sake of argument, that ψ(0) is positive.

Hint for above question: is the R.H.S. of the equation positive or negative? What does that say about dψ/dx at x=0+ vs. at x=0-?

11. Aug 19, 2009

### Davio

it must be smaller!

"What does that say about dψ/dx at x=0+ vs. at x=0-"

err rate of change is less? Surely postive/negative is just a constant or am I being dumb again :-s

I may seem kinda dumb in this post, but this question is actually not taught in our course but we're expected to do it!

12. Aug 19, 2009

### George Jones

Staff Emeritus
I'm really helping here, I'm just curious: What is the course, and what is the text? From the way the question is worded, it seems that the students are not expected to have seen the Schrodinger equation, but they are expected to have seen the Dirac delta function. Is it a mathematical methods course?

Since the Dirac delta function is used in the question, and since the delta function is not defined in the question, surely the text covers delta functions somewhere before the page on which the question appears.

13. Aug 19, 2009

### Davio

Hiya, its just a beginners course to quantum physics. We were expecting an OK paper, but put it this way, my mates who are graduates think the paper is ridicolous!

The tutors petitioned against this paper apparantly. Interestingly the dirac function was bought up, apparantly it was litually on the last page of the a prerequistite maths course, where it commented on it briefly in the space of a couple of lines! Basically not enough for us to really understand it. We think the tutor based the exam paper on previous math's syllabus.

So yeah =(

14. Aug 19, 2009

### George Jones

Staff Emeritus
Now I have another question:

Is this a question on a take-home examination on which you are currently working?

15. Aug 19, 2009

### Davio

It's now considered a past paper. Quite a few people are doing a retest, we are betting that whatever paper he will give us, is going to be VERY similar to this paper. My tactic is to do this entire paper and then the rest of the past papers and see what happens.

heh, take home examinations exist? That seems to defeat the idea..

16. Aug 19, 2009

### Redbelly98

Staff Emeritus
Correct! The slope suddenly becomes less as you go from 0- to 0+.

This can happen in a few different ways:

The slope could change from a large positive value to a smaller (or zero) positive value:

Code (Text):

____
/
/

The slope could change from a positive to a negative value:

Code (Text):

/\
/  \

Or the slope could change from a shallow negative (or zero) value to a steeper negative value:

Code (Text):

____
\
\

That is what the equation with the x=0+ and x=0- is telling us.

17. Aug 19, 2009

### Davio

Ah I see. Um.. this opens a new kettle of fish now...

how would I find a general solution to the schodinger equation for that?! I would use the right hand side picture from before? How does this new information alter my general method of solving schodinger equations?

Thanks.

18. Aug 19, 2009

### Feldoh

I'd just like to add: If you have the resources I'd look up Griffiths QM, particularly section 2.5 which covers delta-function potentials in an easy to understand way.

Generally speaking most people solve the Schrodinger equation for any given well/barrier potential by dividing up the regions based on the potential V(x) and solving the Schrodinger equations for these regions individually.

Once you've found psi for the individual regions you usually apply continuity equations because Psi(x) has to be continuous, and usually Psi'(x) is also continuous. These continuity equations (and normalization) will give you a system of equations which you can solve for unknown coefficients.

However the dirac-delta function potential, as you just learned, gives Psi'(x) a discontinuity at x=0 which means you need to find a different equation to use to solve for unknown coefficients.

19. Aug 19, 2009

### Davio

hmm, the only thing i can think of which has a "discontinuety" is the infinite well, which allows discontinueties due to the infinite potential.

I thought wavefunctions had to by defintion be continuous?

20. Aug 19, 2009

### Feldoh

The wave function is continuous for an infinite well.

The wave functions simple goes to zero at the the boundaries.

The wave function for an infinite well is all differentiable everywhere (continuous derivative)

------------------------

For the dirac potential the wave function is continuous but not differentiable everywhere.

21. Aug 19, 2009

### Davio

hmmm, the dirac potential i mentioned. Is it an infinite well? or are you just explaining why I'm wrong :-s

22. Aug 19, 2009

### Redbelly98

Staff Emeritus
Looking at the problem statement you provided, part (a) looks pretty straightforward. The key points here are:

Then use that information, plus any other constraints on the wavefunction, to solve the differential equation (the Schrodinger equation given in the problem statement).

23. Aug 19, 2009

### Feldoh

I was just explaining why an infinite well doesn't have a discontinuity. The delta-dirac potential is a different entity from the infinite well (essentially).

24. Aug 20, 2009

### Davio

"The Dirac delta can be loosely thought of as a function on the real line which is zero everywhere except at the origin, where it is infinite"

WOuld it be 0?

and if that is the case, I Would use just a normal schoinger equation without the V or rather the V set as 0 and solve as normal?

Last edited: Aug 20, 2009
25. Aug 20, 2009

### Redbelly98

Staff Emeritus
Yes. And also E is negative.