Not sure about shape of this quantum well

[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.

 

[Redbelly98]

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.

I'm having trouble figuring out your expression. If I wrap it in [noparse]

[/noparse] tags, it looks like this:

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]

[/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.

 

[Davio]

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

http://img269.imageshack.us/img269/8703/dsc009141.jpg

Thanks for your reply.

 

[Redbelly98]

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):

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.

 

[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:

    ____
____

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.

 

[Redbelly98]

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.

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

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

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.

 

[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

 

[Redbelly98]

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):

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

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.

 

[Redbelly98]

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.

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.

 

[Redbelly98]

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

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

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-?

 

[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!

 

[George Jones]

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.

 

[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 apparently. Interestingly the dirac function was bought up, apparently 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 =(

 

[George Jones]

Now I have another question:

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

 

[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..

 

[Redbelly98]

it must be smaller!

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:

  ____
 /
/

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

 /\
/  \

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

____
    \
     \

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

 

[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.

 

[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.

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?

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.

 

[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?

 

[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.

 

[Davio]

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

 

[Redbelly98]

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.

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

• E is negative. Let's say E = -E_o
• In the regions x>0 and x<0, V is ... what? You can use the definition of the delta function to answer that:
  http://en.wikipedia.org/wiki/Dirac_delta_function#Definitions

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

 

[Feldoh]

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

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).

 

[Davio]

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

• E is negative. Let's say E = -E_o
• In the regions x>0 and x<0, V is ... what? You can use the definition of the delta function to answer that:
  http://en.wikipedia.org/wiki/Dirac_delta_function#Definitions

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

"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?

 

[Redbelly98]

Yes. And also E is negative.

 

[Davio]

Just to confirm, is E negative because dirac is negative?

I solved equation to get phi =

Cexp i (Kx- (ET/ hbar))+ Dexp i (Kx- (ET/ hbar))

how do i know how to get it for x>0 and x<0? I would have thought them to be the same but I'm worried the question asks for solutionS, and not solution.

 

[Redbelly98]

Just to confirm, is E negative because dirac is negative?

No. It's negative because ... read part (a) of the problem statement.

I solved equation to get phi =

Cexp i (Kx- (ET/ hbar))+ Dexp i (Kx- (ET/ hbar))

First, since you're solving the time-independent Schrodinger equation, you don't need to include the time-dependent terms.

Second, here's a question for you: if you substitute this equation into the differential equation given in the problem, what do you get for E?

how do i know how to get it for x>0 and x<0? I would have thought them to be the same but I'm worried the question asks for solutionS, and not solution.

The problem statement asks, "in addition to being a solution of this equation, what other conditions must ψ(x) satisfy?"
Have answered that part yet? Those other conditions can be used to find C and D for the two regions (x<0 and x>0).

 

[Feldoh]

Just to confirm, is E negative because dirac is negative?

E is negative because it was specified in the problem, meaning that it is a bound state. However the (negative) dirac-delta potential can also have scattering states (E > 0)

 

[Davio]

Hi part A just says E smaller than 0.. oh

ohh I've done that part already, 3 things: must be continuous and single valued function
have continuous first derivative unless potential goes to infinity. and have a finite normalization integral.

Do i just take away the t's for the time independentness?

I have yet to substitute it in as its time for bed, however I think it would be something like E= ((hbared)^2 / 2m) . (lamda squared) as per my notes and taking into account the negative sign.

I was answering the next question which is: find the gneeral solutions to the shrodinger equation for E<0 in the two regions x <0 and x>0.

Thanks.

looking through my notes, I notice infinite square wells have V=0 and also has no time dependancy, is that what i want?

 

[Redbelly98]

Hi part A just says E smaller than 0.. oh

ohh I've done that part already, 3 things: must be continuous and single valued function
have continuous first derivative unless potential goes to infinity. and have a finite normalization integral.

Looks good. Note that the potential does go to infinity at x=0.

Do i just take away the t's for the time independentness?

Yes.

I have yet to substitute it in as its time for bed, however I think it would be something like E= ((hbared)^2 / 2m) . (lamda squared) as per my notes and taking into account the negative sign.

Please post back when you have time to figure out E.

 

[Davio]

Taking the solution with no time dependance: Cexp ikx + D exp -ikx
and then I substite into the equation: I will do left hand side then right hand side.

-(hbared)/2m d^2 phi over dx^2 = ( -k^2 Cexp ikx + k^2 Dexp - ikx) . -(hbared)/2m

= (right hand side) E. phi

E= left hand side divided by Phi

E=0

That surely is wrong?!

(dphi/dx = ik Cexp ikx - ik Dexp - ikx)

from my notes I notice i should get E= ((hbared . k)^2 / 2m)

Thanks.

 

[Redbelly98]

Taking the solution with no time dependance: Cexp ikx + D exp -ikx
and then I substite into the equation: I will do left hand side then right hand side.

-(hbared)/2m d^2 phi over dx^2 = ( -k^2 Cexp ikx + k^2 Dexp - ikx) . -(hbared)/2m

On the RHS here, the expression in parentheses should be

( -k^2 Cexp ikx - k^2 Dexp - ikx)​

= (right hand side) E. phi

E= left hand side divided by Phi

E=0

That surely is wrong?!

Yes, it's wrong, E is not 0.

(dphi/dx = ik Cexp ikx - ik Dexp - ikx)

from my notes I notice i should get E= ((hbared . k)^2 / 2m)

Thanks.

Yes, that looks right. The important point is, this wavefunction has a positive energy. So it is not the solution you want, since you want a solution with a negative energy (E<0).

 

[Davio]

Hiya, thank you for your help. I managed to meet up with some friends and we managed to do it. Your help was invaluable. Thanks :-D

 

[Redbelly98]

Great, glad it worked out.