Harmonic oscillator inside square well again

This does make it easier to separate. I'll try this and post more questions later.In summary, the problem involves finding the energy levels in a square well potential with a soda can-shaped potential inside. The time-independent Schrodinger equation is used to solve for the wave function, which is written as the product of separate functions for the x and y directions. The potential is simplified to 1/2 x^2 for y < |a|, making it easier to separate the equation.
  • #1
syang9
61
0

Homework Statement



Consider the SHO inside a square well, which looks like a soda can cut in half inside of a box.

[tex]

\[
V(x,y) = \left\{ \begin{array}{l}
\frac{1}{2}kx^2 ,{\rm{ }}y{\rm{ }} < {\rm{ }}|a| \\
\infty ,{\rm{ }}y{\rm{ }} \ge {\rm{ }}|a|{\rm{ }} \\
\end{array} \right\}
\]

[/tex]

a. Find the energy levels in this potential.
b. Write the wave function for the ground state.
c. Find a relation between the constants of the problem that makes the first excited state degenerate.

Homework Equations



Since the potential is time-invariant (which means the Hamiltonian is also time-invariant, right..?) then we should use the time-independent schrodinger eqn:

[tex]
\[
- \frac{{\hbar ^2 }}{{2m}}\nabla ^2 \psi + V\psi = E\psi
\]
[/tex]

The Attempt at a Solution



So (in a previous post) I indicated I would try to solve this by finding a separable solution, of the form [tex] \[\psi (x,y) = \psi _x (x) \cdot \psi _y (y)\] [/tex]. So that gives us

[tex]
\[
- \frac{{\hbar ^2 }}{{2m}}\left( {\frac{{\partial ^2 \psi _x }}{{\partial x^2 }} + \frac{{\partial ^2 \psi _y }}{{\partial y^2 }}} \right) + V\left( {\psi _x \psi _y } \right) = E(\psi_x \psi_y)
\]
[/tex]

According to my text (griffiths), we should be able to make this into two equations of x and y only, which each have to be a constant. But I'm having trouble with that, since I'm not sure how to separate the potential such that [tex] \[V(x,y) = V_x + V_y \] [/tex]. Assuming I can even do that, I'm not sure how that would help me. I'd get something like this:

[tex]
\[
\left( {\frac{{\partial ^2 \psi _x }}{{\partial x^2 }} + \frac{{\partial ^2 \psi _y }}{{\partial y^2 }}} \right) - \frac{{2m}}{{\hbar ^2 }}\left( {V_x + V_y } \right)\left( {\psi _x \psi _y } \right) = - l^2 \left( {\psi _x \psi _y } \right),{\rm{ }}l \equiv \frac{{\sqrt {2mE} }}{\hbar }
\]
[/tex]

Divide by [tex] \psi _x \psi _y [/tex] to get

[tex]
\[
\frac{1}{{\psi _x \psi _y }}\left( {\frac{{\partial ^2 \psi _x }}{{\partial x^2 }} + \frac{{\partial ^2 \psi _y }}{{\partial y^2 }}} \right) - \frac{{2m}}{{\hbar ^2 }}\left( {V_x + V_y } \right) = - l^2
\]
[/tex]

...and so I'm stuck here.. I'm not sure how to write this in order to make this more conducive to separation, so I could use some hints here.

Once I get a differential equation for just x and y though, I can solve that for [tex] \psi [/tex] and get a general solution, upon which I can impose the boundary conditions of the problem to get the energies. (..right?) The thing that concerns me, though, is that there are only boundary conditions for the y-direction; there are no constraints on the x-direction. Based on that, wouldn't I get two energies for the x-direction and y-direction? How do I resolve the two into one expression for the energy? Do I just sum them?

Once I have the expression for the energy, it should be trivial to just tack the standard time dependence [tex] \[\left( {\varphi = e^{\frac{{ - iE_n }}{\hbar }t}} \right)\][/tex] onto the general solution and plug in the corresponding ground state energy. Or is that not the right way to go about that?

Regarding part c, I have no idea where to even start. I know that degeneracy means a measurement of the total energy of two energy eigenstates returns the same eigenvalue, but I'm not sure what the problem means when it says that the first excited state itself is degenerate. Usually I think of degenerate (bound) states as two separate states that somehow have the same energy.. I guess what I'm saying is I would have expected the problem to say something like 'fiddle with the constants such that the first _and third_ excited states are degenerate'. Any ideas here?

Thanks very much for any help at all.
 
Last edited:
Physics news on Phys.org
  • #2
syang9 said:

Homework Statement



Consider the SHO inside a square well, which looks like a soda can cut in half inside of a box.

[tex]

\[
V(x,y) = \left\{ \begin{array}{l}
\frac{1}{2}x^2 ,{\rm{ }}y{\rm{ }} < {\rm{ }}|a| \\
\infty ,{\rm{ }}y{\rm{ }} \ge {\rm{ }}|a|{\rm{ }} \\
\end{array} \right\}
\]

[/tex]

a. Find the energy levels in this potential.
b. Write the wave function for the ground state.
c. Find a relation between the constants of the problem that makes the first excited state degenerate.

Homework Equations



Since the potential is time-invariant (which means the Hamiltonian is also time-invariant, right..?) then we should use the time-independent schrodinger eqn:

[tex]
\[
- \frac{{\hbar ^2 }}{{2m}}\nabla ^2 \psi + V\psi = E\psi
\]
[/tex]

The Attempt at a Solution



So (in a previous post) I indicated I would try to solve this by finding a separable solution, of the form [tex] \[\psi (x,y) = \psi _x (x) \cdot \psi _y (y)\] [/tex]. So that gives us

[tex]
\[
- \frac{{\hbar ^2 }}{{2m}}\left( {\frac{{\partial ^2 \psi _x }}{{\partial x^2 }} + \frac{{\partial ^2 \psi _y }}{{\partial y^2 }}} \right) + V\left( {\psi _x \psi _y } \right) = E(\psi_x \psi_y)
\]
[/tex]ll.

for y greater than |a| the wavefunction is obviously zero. So consider only y < |a|.

Then your potential is simply 1/2 x^2 and the separation will be easy.
 
  • #3
syang9 said:
[tex]
\[
- \frac{{\hbar ^2 }}{{2m}}\left( {\frac{{\partial ^2 \psi _x }}{{\partial x^2 }} + \frac{{\partial ^2 \psi _y }}{{\partial y^2 }}} \right) + V\left( {\psi _x \psi _y } \right) = E(\psi_x \psi_y)
\]
[/tex]]

That doesn't look right. Since the wavefunction is written as the product of [tex]\psi_x[/tex] and [tex]\psi_y[/tex], it should be

[tex]
- \frac{{\hbar ^2 }}{{2m}}\left( \psi_y{\frac{{\partial ^2 \psi _x }}{{\partial x^2 }} + \psi_x\frac{{\partial ^2 \psi _y }}{{\partial y^2 }}} \right) + V\left( {\psi _x \psi _y } \right) = E(\psi_x \psi_y)
[/tex]
 
  • #4
Okay yeah you're right. I didn't write it correctly. So now we have

[tex]
\[
\left( {\frac{1}{{\psi _x }}\frac{{d^2 \psi _x }}{{dx^2 }} + \frac{1}{{\psi _y }}\frac{{d^2 \psi _y }}{{dy^2 }}} \right) + \frac{{2m}}{{\hbar ^2 }}\left( {\frac{1}{2}kx^2 } \right) - l^2 = 0
\]

[/tex]

This does make it easier to separate. I'll try this and post more questions later.
 
  • #5
syang9 said:
Okay yeah you're right. I didn't write it correctly. So now we have

[tex]
\[
\left( {\frac{1}{{\psi _x }}\frac{{d^2 \psi _x }}{{dx^2 }} + \frac{1}{{\psi _y }}\frac{{d^2 \psi _y }}{{dy^2 }}} \right) + \frac{{2m}}{{\hbar ^2 }}\left( {\frac{1}{2}kx^2 } \right) - l^2 = 0
\]

[/tex]

This does make it easier to separate. I'll try this and post more questions later.

Of course it's easier to separate! Move everything that depends on x on one side and on y on the other side. Since x and y are independent variables, both sides must be equal to a common constant
 
  • #6
Okay, so now I have

[tex]

\[
\begin{array}{l}
\frac{{d^2 \psi _x }}{{dx^2 }} = \psi _x \left( {\frac{m}{{\hbar ^2 }}kx^2 - l^2 + n} \right),{\rm{ }}l \equiv \frac{{\sqrt {2mE} }}{\hbar } \\
\frac{{d^2 \psi _y }}{{dy^2 }} = n\psi _y \\
\end{array}
\]

[/tex]

Does this look reasonable? What is the significance of the separation constant? How do I even solve that first equation?
 
  • #7
Does the way I've proposed to solve the rest of the problem seem reasonable?
 
  • #8
I have no idea how to solve that first diff. eq. Am I even going about it in the right way?
 
  • #9
everyone hates me
 
  • #10
i like mangoes
 
  • #11
alright, so my friend told me that i can resolve the potential into an x and a y component, giving me two one dimensional potentials that i have already solved (aka inf square well in the y direction, SHO in the x direction). what I don't understand is why I didn't get the differential equations (i.e., why my separated equations have that stupid separation constant). shouldn't i just get the equations for the one dimensional cases?
 
  • #12
i think you made a mistake. If you group the terms, you get

[tex]\frac{-\hbar ^2}{2m}\left(\frac{\psi_x^{''}}{\psi_x}\right) + 1/2kx^2 = E + \frac{\hbar ^2}{2m} \left(\frac{\psi_y^{''}}{\psi_y}\right)[/tex]

The LHS is only a function of x, the RHS only a function of y, and they are equal. This must imply that both is equal to some constant. (Say [tex]\lambda[/tex])

So, you have
[tex]\frac{-\hbar ^2}{2m}\left(\frac{\psi_x^{''}}{\psi_x}\right) + 1/2kx^2 = \lambda[/tex]

[tex]E+\frac{\hbar ^2}{2m} \left(\frac{\psi_y^{''}}{\psi_y}\right)=\lambda[/tex]

Can you take it from here?
 

1. What is a harmonic oscillator inside a square well?

A harmonic oscillator refers to a system where the restoring force is proportional to the displacement from the equilibrium position, such as a mass attached to a spring. A square well is a potential energy function that resembles a square-shaped box, and it can be used to describe a confined region in space.

2. How does a harmonic oscillator inside a square well behave?

The behavior of a harmonic oscillator inside a square well is characterized by oscillatory motion, where the particle moves back and forth within the well. The frequency of the oscillations is determined by the mass of the particle and the strength of the restoring force.

3. What are the applications of a harmonic oscillator inside a square well?

A harmonic oscillator inside a square well is commonly used as a model in quantum mechanics to describe the behavior of particles in a confined region. It has applications in understanding the behavior of atoms in a crystal lattice, as well as in studying the energy levels of quantum systems.

4. How is the energy of a harmonic oscillator inside a square well quantized?

In quantum mechanics, the energy of a harmonic oscillator inside a square well is quantized, meaning it can only take on certain discrete values. The allowed energy levels are determined by the size of the well and the mass of the particle, and they are described by the Schrödinger equation.

5. Can a harmonic oscillator inside a square well have negative energy levels?

Yes, a harmonic oscillator inside a square well can have negative energy levels. In quantum mechanics, negative energy levels represent bound states, where the particle is confined within the well. The lowest energy level, also known as the ground state, has the most negative energy and is the most stable state for the system.

Similar threads

  • Advanced Physics Homework Help
Replies
10
Views
461
  • Advanced Physics Homework Help
Replies
1
Views
823
  • Advanced Physics Homework Help
Replies
1
Views
2K
  • Advanced Physics Homework Help
Replies
1
Views
207
  • Advanced Physics Homework Help
Replies
0
Views
618
  • Advanced Physics Homework Help
Replies
1
Views
1K
  • Advanced Physics Homework Help
Replies
30
Views
1K
Replies
7
Views
1K
Replies
16
Views
402
  • Advanced Physics Homework Help
Replies
4
Views
2K
Back
Top