Triangular potential Well

In summary, using the chain rule and the equation for w, the TISE expression found in part (a) can be simplified to a simpler form, \frac{d^{2} \psi}{dw^{2}} - w \psi = 0, by replacing the denominator in \frac{d^2 \psi}{dx^2} with the expression for dw/dx and dividing out the resulting factor in front of the new 2nd derivative. This leads to an ODE in terms of w only and no other terms. The cube roots are necessary to cancel out coefficients in the final form of the equation.
  • #1
QuantumJG
32
0

Homework Statement


Show that the TISE expression you found in part (a):

I found it (a) to be:

[tex]\frac{d^{2} \psi}{dx^{2}} - \frac{2m}{\hbar ^{2}}(e \xi x - E) \psi (x) = 0[/tex]

Show it (a) can be simplified to:

[tex]\frac{d^{2} \psi}{dw^{2}} - w \psi = 0[/tex]

Homework Equations



w = z - z0

where:

[tex]z = \left( \frac{2me \xi }{ \hbar ^{2} } \right)^{ \frac{1}{3} } x[/tex]

[tex]z_{0} = \frac{2mE}{\hbar ^{2}} \left( \frac{ \hbar ^{2} }{ 2me \xi } \right)^{ \frac{2}{3} } [/tex]

The Attempt at a Solution



So far I have found:

[tex]w = \left( \frac{2m}{ \hbar ^{2} e^{2} \xi ^{2} } \right)^{ \frac{1}{3} } (e \xi x - E) [/tex]

But my problem is:

how do I convert:

[tex] \frac{d^{2} \psi}{dx^{2}} [/tex] to [tex] \frac{d^{2} \psi}{dw^{2}} [/tex], because I'm totally stuck on that.
 
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  • #2
Well, you know w in terms of x. So what is dw in terms of dx? Then solve for dx^2, and plug it into your Schrodinger Eq.
 
  • #3
well,

[tex] dw = \left( \frac{2me \xi }{ \hbar ^{2} } \right)^{ \frac{1}{3} } dx[/tex]

[tex] \therefore dw = z dx [/tex]

But I'm confused as to how that's going to help.
 
  • #4
QuantumJG said:

Homework Statement


Show that the TISE expression you found in part (a):

I found it (a) to be:

[tex]\frac{d^{2} \psi}{dx^{2}} - \frac{2m}{\hbar ^{2}}(e \xi x - E) \psi (x) = 0[/tex]

Show it (a) can be simplified to:

[tex]\frac{d^{2} \psi}{dw^{2}} - w \psi = 0[/tex]

Homework Equations



w = z - z0

where:

[tex]z = \left( \frac{2me \xi }{ \hbar ^{2} } \right)^{ \frac{1}{3} } [/tex]

[tex]z_{0} = \frac{2mE}{\hbar ^{2}} \left( \frac{ \hbar ^{2} }{ 2me \xi } \right)^{ \frac{2}{3} } [/tex]
Are you sure this is right? Where did x go? And where did the cube roots come from?

The Attempt at a Solution



So far I have found:

[tex]w = \left( \frac{2m}{ \hbar ^{2} e^{2} \xi ^{2} } \right)^{ \frac{1}{3} } (e \xi x - E) [/tex]
This seems unnecessarily complicated. Perhaps I'm missing something obvious, but if you compare the original and simplified equations, can't you see what w is equal to by inspection?
But my problem is:

how do I convert:

[tex] \frac{d^{2} \psi}{dx^{2}} [/tex] to [tex] \frac{d^{2} \psi}{dw^{2}} [/tex], because I'm totally stuck on that.
Use the chain rule:

[tex]\frac{df}{dx} = \frac{dw}{dx}\frac{df}{dw}[/tex]
 
  • #5
QuantumJG said:
well,

[tex] dw = \left( \frac{2me \xi }{ \hbar ^{2} } \right)^{ \frac{1}{3} } dx[/tex]

[tex] \therefore dw = z dx [/tex]

But I'm confused as to how that's going to help.

Knowing this and squaring it to get dx^2, you can replace the denominator in

[tex]\frac{d^2 \psi}{dx^2}[/tex]

with that and get the 2nd derivative in terms of w. Then you will want to divide out the factor in front of your new 2nd derivative to get the 2nd equation you listed in your first post.

EDIT: The chain rule will also get you the same thing. But since w is linearly proportional to x, I just take the proportionality constant out and square it.

EDIT2: Also, the cube roots look right. They are needed to cancel out all the coefficients in the final form of the equation. If QuantumJG finishes this last step, then he should get a simple ODE in terms of w only and no other terms.
 
  • #6
vela said:
Are you sure this is right? Where did x go? And where did the cube roots come from?

I just realized I forgot to put x into the equation for z

vela said:
This seems unnecessarily complicated. Perhaps I'm missing something obvious, but if you compare the original and simplified equations, can't you see what w is equal to by inspection?

Use the chain rule:

[tex]\frac{df}{dx} = \frac{dw}{dx}\frac{df}{dw}[/tex]

but then wouldn't:

[tex] \frac{d}{dx} \left( \frac{df}{dx} \right) = \frac{d}{dx} \left( \frac{dw}{dx}\frac{df}{dw} \right) [/tex]

[tex] = \frac{df}{dw} \frac{d}{dx} \left( \frac{dw}{dx} \right) + \frac{dw}{dx} \frac{d}{dx} \left( \frac{df}{dw} \right) [/tex]
 
  • #7
Use the chain rule on the last term:

[tex]\frac{dw}{dx}\frac{dw}{dx}\frac{d}{dw}\left(\frac{df}{dw}\right)[/tex]
 
  • #8
nickjer said:
Use the chain rule on the last term:

[tex]\frac{dw}{dx}\frac{dw}{dx}\frac{d}{dw}\left(\frac{df}{dw}\right)[/tex]

Wait that's confusing, why did you put in that extra [tex] \frac{dw}{dx} [/tex]?
 
  • #9
nickjer said:
EDIT2: Also, the cube roots look right. They are needed to cancel out all the coefficients in the final form of the equation. If QuantumJG finishes this last step, then he should get a simple ODE in terms of w only and no other terms.
Oh, okay, I did miss something obvious. :smile:
 
  • #10
QuantumJG said:
Wait that's confusing, why did you put in that extra [tex] \frac{dw}{dx} [/tex]?
That's the one from the first differentiation. Because dw/dx is a constant, you get

[tex]\left(\frac{d}{dx}\right)^2 = \frac{dw}{dx} \frac{d}{dw}\left(\frac{dw}{dx} \frac{d}{dw}\right) = \left(\frac{dw}{dx}\right)^2 \frac{d^2}{dw^2}[/tex]
 
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1. What is a triangular potential well?

A triangular potential well is a theoretical concept used in quantum mechanics to describe the behavior of a particle in a confined space. It refers to a potential energy function that resembles a triangle, with the sides representing the boundaries of the well.

2. How does a particle behave in a triangular potential well?

The behavior of a particle in a triangular potential well depends on its energy. If the particle has less energy than the height of the potential well, it will be confined within the well and will oscillate back and forth between the boundaries. If the particle has more energy than the height of the well, it can escape from the well and move freely.

3. What is the significance of the triangular potential well in quantum mechanics?

The triangular potential well is an important concept in quantum mechanics as it helps us understand the behavior of particles in confined spaces. It is also used to model various physical systems, such as atoms and molecules, and can provide insights into their properties and behavior.

4. How is the triangular potential well related to the Schrödinger equation?

The Schrödinger equation is a fundamental equation in quantum mechanics that describes the evolution of a particle's wave function over time. In the case of a particle in a triangular potential well, the Schrödinger equation can be solved to determine the allowed energy levels and corresponding wave functions of the particle within the well.

5. Can the triangular potential well be applied to real-world systems?

While the triangular potential well is a theoretical concept, it can be used to model real-world systems, such as quantum dots, nanowires, and other confined structures. It has also been used in various fields, including condensed matter physics, chemistry, and material science, to study the behavior of particles in confined environments.

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