Is the First Excited State of a Particle in a Rectangular Box Degenerate?

In summary, the energy of the first excited state of a particle confined in a rectangular box with sides of length a,a, and 2a is determined by the equation E=Ex+Ey+Ez, where Ex, Ey, and Ez are determined by the values of nx, ny, and nz respectively. The smallest energy state is the ground state, with nx, ny, and nz all equal to 1. The first excited state has the next lowest energy, and nx, ny, and nz must all be greater than or equal to 1. This is determined by the boundary conditions and the requirement that the wave function is normalizable.
  • #1
Squid138
3
0
A particle is confined inside a rectangular box with sides of length a,a, and 2a. What is the energy of the first excited state? Is this state degenerate? If so, determine how many different wave fuctions have this energy.

The relevant equations are

E=Ex+Ey+Ez
=((hbar^2*pi^2)/2m) +(nx^2)/a^2+(ny^2)/b^2+(nz^2)/c^2

So far I have plugged in the a,a, and 2a into the equation and factored out the a. I am confused at what it means by first excited state. In a hydrogen atom n=2. In this case should each component =2 or should the energy add up to 2?
 
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  • #2
Ignore
 
Last edited:
  • #3
Wouldn't that imply that there is zero energy in the ground state and is that possible?
 
  • #4
Squid138 said:
Wouldn't that imply that there is zero energy in the ground state and is that possible?

Ooops sorry I meant (1,1,1) is the ground state and then (1,1,2), (1,2,1) and (2,1,1) will be the first excited state.
 
  • #5
How do you know that? Is there something I am not seeing?
 
  • #6
Squid138 said:
How do you know that? Is there something I am not seeing?

Just look at the energy, if nx,ny,nz are all >=1 then the smallest the energy can be is when they're all equal to 1, that's the ground state. The first excited state is the state of next lowest energy so just look at the expression for energy and decide what nx,ny,nz all have to be, the choice will of course depend on a,b,c.
 
  • #7
Squid138 said:
How do you know that? Is there something I am not seeing?

[What Tangent means] From the boundary conditions you get that n is an integer. You reject negative integers because of they give you no new information. You reject zero because that makes your wave function zero everywhere, and hence makes it non-normalizable. Only things left are n>=1.
 

Related to Is the First Excited State of a Particle in a Rectangular Box Degenerate?

1. What is degeneracy in a rectangular box?

Degeneracy in a rectangular box refers to the phenomenon in quantum mechanics where multiple quantum states have the same energy level. These states are called degenerate states.

2. How does degeneracy occur in a rectangular box?

Degeneracy in a rectangular box occurs when the dimensions of the box are such that they allow for multiple quantum states to exist with the same energy. This typically happens when the dimensions of the box are significantly larger than the wavelength of the particle.

3. What is the significance of degeneracy in a rectangular box?

Degeneracy in a rectangular box is important because it affects the energy levels of particles within the box. It can also lead to unique properties, such as the splitting of degenerate states under certain conditions.

4. How does degeneracy in a rectangular box affect the behavior of particles?

Degeneracy in a rectangular box can affect the behavior of particles by changing their energy levels and therefore influencing their interactions with other particles and external fields. It can also result in special quantum effects, such as Bose-Einstein condensation.

5. Can degeneracy in a rectangular box be manipulated or controlled?

Yes, degeneracy in a rectangular box can be controlled by changing the dimensions of the box or by applying external fields. This allows for the manipulation of particle behavior and the exploration of various quantum effects.

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