What are the boundary conditions for the finite square-well potential at x=0?

In summary: In that case, D would be positive.In summary, your professor wrote the equations that correspond to the wave equation and its derivative being continuous and taking the value 1 at some point. In addition, he did a different procedure where he solved for B and C in terms of A and D. He then applied the boundary conditions to find the relationships between the coefficients A, C, and D and the ratio C/D.
  • #1
Shackleford
1,656
2
24. Apply the boundary conditions to the finite square-well potential at x=0 to find the relationships between the coefficients A, C, and D and the ratio C/D.

I understand the wave equations in the three separate regions. For this question I need to only consider I, II. The wave equations need to decrease to zero as x approaches positive or negative infinity. The wave equation and its derivative need to be continuous as well. Thus, the wave equation of I equals II.

My professor did a similar problem last semester, but I can't make sense of his procedure. I think the delta function is in it, etc.

http://i111.photobucket.com/albums/n149/camarolt4z28/6t24.jpg?t=1283054054 [Broken]
 
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  • #2
Your professor wrote the two equations that correspond to
Shackleford said:
The wave equation and its derivative need to be continuous as well.
 
  • #3
vela said:
Your professor wrote the two equations that correspond to

He did it a bit differently, though.

I also re-worked my part, too.

http://i111.photobucket.com/albums/n149/camarolt4z28/6t24.jpg?t=1283054054 [Broken]

He has

1 = A + B
K = ik (A - B

ik/K = (A + B)/(A - B)

(ik + K)/(ik - K) = A/B = delta

I understand the manipulation up until here. I still don't know how in the heck this helps me related A, C, and D. Do I do the same thing?
 
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  • #4
I think you're getting your coefficients mixed up. Your equations should be

A = C+D
αA = ik(C-D)

I'm not sure what your professor is doing. It looks like his A and B are your C and D and his K is your alpha. He took (your) A to be equal to 1 for some reason. The delta is not the delta function. It's just the quantity which equals A/B.
 
  • #5
vela said:
I think you're getting your coefficients mixed up. Your equations should be

A = C+D
αA = ik(C-D)

I'm not sure what your professor is doing. It looks like his A and B are your C and D and his K is your alpha. He took (your) A to be equal to 1 for some reason. The delta is not the delta function. It's just the quantity which equals A/B.

Oops. I accidentally wrote down B + C for some reason.

Here's what the professor did last week. I assume the book is looking for something like this. I don't know how he got this.

http://i111.photobucket.com/albums/n149/camarolt4z28/2010-08-29000403.jpg?t=1283058414 [Broken]
 
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  • #6
You have two equations and three unknowns, so you can solve for two of them, say C and D, in terms of the other, A.

That's what your professor did except in his case, there were four unknowns, so he solved for B and C in terms of A and D.
 
  • #7
vela said:
You have two equations and three unknowns, so you can solve for two of them, say C and D, in terms of the other, A.

That's what your professor did except in his case, there were four unknowns, so he solved for B and C in terms of A and D.

Okay. I'll play around with the equations. Maybe I'll get partial credit. lol.

Oh, for the 1 = A + B, I think he used the free particle solution for the wave heading from the negative x direction towards the potential.
 
  • #8
vela said:
You have two equations and three unknowns, so you can solve for two of them, say C and D, in terms of the other, A.

That's what your professor did except in his case, there were four unknowns, so he solved for B and C in terms of A and D.

Well, I tried to play around with what the professor did, but I couldn't get his equations. Forgive my lack of algebraic-manipulation skills.
 
  • #9
Rewriting the equations a bit, you get

[tex]\begin{align*}
B - C & = -A + D \\
k_0 B + kC &= k_0 A + kD
\end{align*}
[/tex]

Multiply the first equation by k, add it the second, and solve for B.
 
  • #10
vela said:
Rewriting the equations a bit, you get

[tex]\begin{align*}
B - C & = -A + D \\
k_0 B + kC &= k_0 A + kD
\end{align*}
[/tex]

Multiply the first equation by k, add it the second, and solve for B.

Okay. That makes sense. I'm still not sure about my problem. Playing with it, I got

A = [(ik + α)C + (α - ik)D] / 2α

Is that what the book is looking for?
 
  • #11
Probably not. Try solving for C and D in terms of A. Then you can calculate the ratio C/D.
 
  • #12
vela said:
Probably not. Try solving for C and D in terms of A. Then you can calculate the ratio C/D.

Crap. I forgot about the ratio C/D. Well, maybe I'll get partial credit. The homework was due today.

Just now, I got A = [-2ik/(α-ik)] D.

I suspect C would be something similar.
 
  • #13
I think that's right, and C comes out similarly.
 
  • #14
vela said:
I think that's right, and C comes out similarly.

Well, that's just great. I do something right after it's due. I'm not going to do well on this problem set. Pretty much the rest of the problem sets will come from this book. I don't expect to do well on any of them.

https://www.amazon.com/dp/0471057002/?tag=pfamazon01-20
 
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  • #15
If you don't like that text, you might want to see if your library has Griffith's book on quantum mechanics. I'll admit I've never seen it, but I used his particle physics book as an undergrad. He was very good at explaining concepts and showing how to apply them in problems.

https://www.amazon.com/dp/0131118927/?tag=pfamazon01-20

(It's the top-selling book on quantum mechanics at Amazon.)
 
  • #16
vela said:
If you don't like that text, you might want to see if your library has Griffith's book on quantum mechanics. I'll admit I've never seen it, but I used his particle physics book as an undergrad. He was very good at explaining concepts and showing how to apply them in problems.

https://www.amazon.com/dp/0131118927/?tag=pfamazon01-20

(It's the top-selling book on quantum mechanics at Amazon.)

You think it'll help me do the Gasiorowicz problems? Even the professor said Gasiorowicz isn't an ideal textbook. Also, I saw one of my friend's particle physics book, and it's Griffiths, too.
 
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  • #17
Yes, I think it would help. It couldn't hurt. Textbooks are so expensive, though, so I'd try to look through a copy in a bookstore or at the library first.
 
  • #18
I have all three Griffiths (unless there are more) books. The EM is best textbook of any kind that you will ever find, in my opinion, the particle physics is great, and the quantum book is decent. As for the quantum book: I remember he doesn't even use kets until spin and he still tries to avoid them like the plague, he dumbs down the math formalism too much, and for the most part just tries spoon feed the reader way too much. That being said, he is still as clear and concise as he usually is. Maybe it's good if you don't have a strong math background, particularly in linear algebra and PDEs, but it wasn't good for me.
 
  • #19
Oh, that's too bad. I forgot I had his E&M book as well as the particle physics book. Both were excellent, so I hoped his QM book would be too.
 

1. What is a finite square-well potential?

A finite square-well potential is a mathematical model used in quantum mechanics to describe the potential energy of a particle in a specific region. It is a potential energy function that has a finite range and is constant within that range, but drops to zero outside of it.

2. How is a finite square-well potential different from an infinite square-well potential?

An infinite square-well potential has infinite potential energy outside of its defined range, while a finite square-well potential drops to zero outside of its range. This means that particles in an infinite square-well potential are completely confined to a specific region, while particles in a finite square-well potential have the potential to escape.

3. What are the applications of a finite square-well potential?

A finite square-well potential is commonly used to model various physical systems, such as atoms, molecules, and quantum dots. It is also used to study particle interactions and bound states in quantum mechanics.

4. How is a finite square-well potential solved in quantum mechanics?

The Schrödinger equation is used to solve for the energy levels and wavefunctions of a particle in a finite square-well potential. The boundary conditions of the potential are taken into account to determine the allowed energy levels and corresponding wavefunctions.

5. What is the significance of the depth and width of a finite square-well potential?

The depth and width of a finite square-well potential determine the behavior of the particle within the potential. The depth affects the energy levels, while the width affects the localization of the particle. Changing these parameters can result in different energy levels and wavefunctions, leading to different physical properties of the system.

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