States with minimum energy for electrons in mag field with nonzero Lz

In summary, we have an electron in a uniform magnetic field with no angular momentum. The Hamiltonian can be expressed as two separate harmonic oscillators and the ground state is Ψ=Ne^{-\frac{mω}{ℏ}r^2}. There are other states with nonzero angular momentum L_z that yield the same energy, such as Ψ_n=Nr^ne^{inφ}e^{-\frac{mω}{ℏ}r^2}. The question is to prove that \hat{H}Ψ_n=\frac{\hbarω_B}{2}Ψ_n since Ψ_n is also an eigenstate of the Hamiltonian. By using polar coordinates and the Laplace operator, we can
  • #1
vdweller
2
0

Homework Statement



For an electron in a uniform magnetic field, say [itex]B\hat{z}[/itex] with no angular momentum, the Hamiltonian can be expressed as [itex]\hat{H}=\frac{1}{2m}\Big(\hat{p}_x^2+\frac{mω^2}{2}\hat{x}^2\Big)+\frac{1}{2m}\Big(\hat{p}_y^2+\frac{mω^2}{2}\hat{y}^2\Big)[/itex]

Which is equivalent to two separate harmonic oscillators.

Now the ground state is [itex]Ψ=Ne^{\frac{mω}{\hbar}r^2}[/itex]

where [itex]r=\sqrt{x^2+y^2}[/itex] and [itex]ω=ω_B/2=eB/2mc[/itex]

which yields energy equal to [itex]\hbarω[/itex]

Now there are other states with nonzero angular momentum [itex]L_z[/itex] which yield the same energy. Those states are [itex]Ψ_n=Nr^ne^{inφ}e^{\frac{mω}{\hbar}r^2}[/itex]

(n is not the quantum number)

The question is to prove that [itex]\hat{H}Ψ_n=\frac{\hbarω_B}{2}Ψ_n[/itex]

since [itex]Ψ_n[/itex] is also an eigenstate of the Hamiltonian.

My question is, how do we prove that? I tried using the Hamiltonian with polar coordinates but I can't seem to get to the result.

Is there any other way (using the creation/annihilation operators)?
 
Physics news on Phys.org
  • #2
I'm sorry you are not finding help at the moment. Is there any additional information you can share with us?
 
  • #3
vdweller said:

Homework Statement



For an electron in a uniform magnetic field, say [itex]B\hat{z}[/itex] with no angular momentum, the Hamiltonian can be expressed as [itex]\hat{H}=\frac{1}{2m}\Big(\hat{p}_x^2+\frac{mω^2}{2}\hat{x}^2\Big)+\frac{1}{2m}\Big(\hat{p}_y^2+\frac{mω^2}{2}\hat{y}^2\Big)[/itex]

Which is equivalent to two separate harmonic oscillators.

Now the ground state is [itex]Ψ=Ne^{\frac{mω}{\hbar}r^2}[/itex]

where [itex]r=\sqrt{x^2+y^2}[/itex] and [itex]ω=ω_B/2=eB/2mc[/itex]

which yields energy equal to [itex]\hbarω[/itex]

Now there are other states with nonzero angular momentum [itex]L_z[/itex] which yield the same energy. Those states are [itex]Ψ_n=Nr^ne^{inφ}e^{\frac{mω}{\hbar}r^2}[/itex]

(n is not the quantum number)

The question is to prove that [itex]\hat{H}Ψ_n=\frac{\hbarω_B}{2}Ψ_n[/itex]

since [itex]Ψ_n[/itex] is also an eigenstate of the Hamiltonian.

My question is, how do we prove that? I tried using the Hamiltonian with polar coordinates but I can't seem to get to the result.

Is there any other way (using the creation/annihilation operators)?

Did you try simply staying in cartesian coordinates and applying the hamiltonian?
 
  • #4
Yes, it's still a mess.

By the way, the correct ground state is [itex]Ψ=Ne^{-\frac{mω}{ℏ}r^2}[/itex] (forgot a minus). I can't see the edit button any more...
 
  • #5
vdweller said:
Yes, it's still a mess.

By the way, the correct ground state is [itex]Ψ=Ne^{-\frac{mω}{ℏ}r^2}[/itex] (forgot a minus). I can't see the edit button any more...

Sorry, I did not notice your ## \phi## dependence on the one you must check.
Then you must use polar coordinates (you are working in two dimensions, right?).
You need ## \nabla^2 ## in polar coordinates. Once you reexpress the hamltonian in polar coordinates, it should be easy to check.
 

1. What is the significance of states with minimum energy for electrons in a magnetic field with nonzero Lz?

The states with minimum energy for electrons in a magnetic field with nonzero Lz are important because they represent the lowest energy levels that an electron can occupy when subjected to a magnetic field. These states are also known as the ground state and play a crucial role in understanding the behavior and properties of materials in a magnetic field.

2. How do states with minimum energy for electrons in a magnetic field with nonzero Lz affect the electronic structure of atoms?

The states with minimum energy for electrons in a magnetic field with nonzero Lz can have a significant impact on the electronic structure of atoms. In particular, they can cause the splitting of energy levels, known as the Zeeman effect, which can affect the spectral lines of atoms and provide insight into their electronic configurations.

3. Can states with minimum energy for electrons in a magnetic field with nonzero Lz be observed in real-life situations?

Yes, states with minimum energy for electrons in a magnetic field with nonzero Lz can be observed in various real-life situations. For example, they are responsible for the formation of magnetic domains in ferromagnetic materials and play a role in the behavior of electrons in the Earth's magnetic field.

4. How do the values of Lz and the magnetic field strength affect the states with minimum energy for electrons?

The values of Lz and the magnetic field strength have a direct impact on the states with minimum energy for electrons. The energy of these states is directly proportional to the magnetic field strength and inversely proportional to the square of the value of Lz. This means that higher values of Lz or stronger magnetic fields will result in higher energy states.

5. Do states with minimum energy for electrons in a magnetic field with nonzero Lz have any practical applications?

Yes, states with minimum energy for electrons in a magnetic field with nonzero Lz have practical applications in various fields, including materials science, quantum computing, and medical imaging. Understanding and manipulating these states can lead to the development of new technologies and advancements in these areas.

Similar threads

  • Advanced Physics Homework Help
Replies
1
Views
2K
  • Advanced Physics Homework Help
Replies
24
Views
2K
  • Advanced Physics Homework Help
Replies
4
Views
3K
  • Advanced Physics Homework Help
Replies
11
Views
2K
  • Advanced Physics Homework Help
Replies
1
Views
919
  • Advanced Physics Homework Help
Replies
4
Views
2K
  • Advanced Physics Homework Help
Replies
1
Views
2K
Replies
7
Views
2K
  • Advanced Physics Homework Help
Replies
5
Views
2K
  • Advanced Physics Homework Help
Replies
6
Views
1K
Back
Top