Expressions for energy in 1d well

Yes, this is the correct way to find an expression for the bound states of the particle. Unfortunately, sometimes we have to resort to graphical solutions for transcendental equations like these. It may seem odd, but it is a common approach in physics and mathematics. If you are unable to draw the graphs during an exam, perhaps you can use a calculator or computer program to help you find the solutions.
  • #1
Brewer
212
0
Question states:

Derive an expression for the energy of the bound states of a particle in the one-dimensional well defined by:

[tex]V(x) = \infinity x<=0[/tex] (Region I)
[tex]V(x) = 0 0<x<L[/tex] (Region II)
[tex]V(x) = V_0 x>=L[/tex] (Region III)

And its not been too bad. Up until a point.

I know for I [tex]\psi(x)=0[/tex].

In the region II [tex]\psi(x) = Acos(kx) + Bsin(kx)[/tex]

In the region III [tex]\psi(x) = Ce^{-\alpha x}[/tex]

where [tex]k = \sqrt{\frac{2mE}{\hbar^2}}[/tex]
and [tex]\alpha = \sqrt{\frac{2m(V_0 - E)}{\hbar^2}}[/tex]

I also know that at the boundary of II and III (x=L) that [tex]\psi_{II}(x) = \psi_{III}(x)[/tex]
Solving for the even parity solutions I get [tex]tan(kL) = \frac{\alpha}{k}[/tex] and for the odd parity solutions I get [tex]cot(kL) = -\frac{\alpha}{k}[/tex].

Now in order to find an expression for E I want to solve these equations for k (as k is a function of E).

Equating the 2 equations I end up with [tex]tan(kl) = - cot(kl)[/tex], which on first glimpse seems ok, but after a bit of working I manage to get this to cancel to [tex]tan(kl) = i[/tex], and correct me if I'm wrong, but I'm sure that I can't do the inverse tan of i can I? Does this look like the right way to go about attempting this question? Have I just messed up the maths somewhere? Is there anything you suggest that might point me in the right direction?

Thanks guys.
 
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  • #2
Brewer said:
Question states:

Derive an expression for the energy of the bound states of a particle in the one-dimensional well defined by:

[tex]V(x) = \infinity x<=0[/tex] (Region I)
[tex]V(x) = 0 0<x<L[/tex] (Region II)
[tex]V(x) = V_0 x>=L[/tex] (Region III)

And its not been too bad. Up until a point.

I know for I [tex]\psi(x)=0[/tex].

In the region II [tex]\psi(x) = Acos(kx) + Bsin(kx)[/tex]

In the region III [tex]\psi(x) = Ce^{-\alpha x}[/tex]

where [tex]k = \sqrt{\frac{2mE}{\hbar^2}}[/tex]
and [tex]\alpha = \sqrt{\frac{2m(V_0 - E)}{\hbar^2}}[/tex]

I also know that at the boundary of II and III (x=L) that [tex]\psi_{II}(x) = \psi_{III}(x)[/tex]
Solving for the even parity solutions I get [tex]tan(kL) = \frac{\alpha}{k}[/tex] and for the odd parity solutions I get [tex]cot(kL) = -\frac{\alpha}{k}[/tex].

Now in order to find an expression for E I want to solve these equations for k (as k is a function of E).

Equating the 2 equations I end up with [tex]tan(kl) = - cot(kl)[/tex], which on first glimpse seems ok,
no! You can not set them equal! Those are conditions for the energies of the even and odd solutions..Those are *different* solutions so those two equations should not be mixed! They correspond to different solutions of the Schrodinger equation.

What you must do is to solve separately

[tex]tan(kL) = \frac{\alpha}{k}[/tex] and
[tex]cot(kL) = -\frac{\alpha}{k}[/tex].

Each is a transcendental solution. You must solve each graphically.

Hope this helps

Patrick
 
  • #3
Is there no way to solve either algebraicly?
 
  • #4
Brewer said:
Is there no way to solve either algebraicly?

Unfortunately, no.

Patrick
 
  • #5
But this is the way to go about finding an expression for the bound states of the particle, correct?

It just seems odd to me that you have to do it this way, because while it is now a homework question, it is just a direct lift from an exam paper, where I wouldn't have time (or the capabilities) to draw these graphs.
 

Related to Expressions for energy in 1d well

1. What is an energy expression in a 1-dimensional well?

The energy expression in a 1-dimensional well refers to the mathematical equation that describes the energy levels of a particle confined within a 1-dimensional potential well. It takes into account the size and shape of the well, as well as the properties of the particle, such as its mass and charge.

2. How is the energy expression derived?

The energy expression is derived using the Schrödinger equation, which is a fundamental equation in quantum mechanics. This equation takes into account the potential energy of the well and the kinetic energy of the particle to determine the allowed energy levels.

3. What are the factors that affect the energy expression in a 1-dimensional well?

The energy expression is affected by the width and depth of the potential well, as well as the mass and charge of the particle. Additionally, the energy levels are also dependent on the boundary conditions set by the potential well.

4. What is the significance of the energy expression in 1-dimensional well?

The energy expression in a 1-dimensional well is significant because it helps us understand the behavior of particles confined in a potential well. It allows us to predict the energy levels and the corresponding wave functions of the particle, which in turn, can provide valuable insights into the physical properties and behavior of the system.

5. How does the energy expression change for different types of potential wells?

The energy expression will vary depending on the type of potential well. For example, in a square well, the energy levels are discrete and equally spaced, while in a triangular well, the energy levels are continuous. The shape and size of the well also play a crucial role in determining the energy expression.

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