# Eigenstates of a free electron in a uniform magnetic field

• Zhuangzi
In summary, the conversation discusses the process of deriving the Hamiltonian for a charged particle in a magnetic field and using it to find the wavefunction for the particle. The relevant equations are substituted and expanded to yield the Hamiltonian, which includes terms for kinetic energy, magnetic moment, and potential energy. The conversation then mentions difficulties in finding the eigenvalue for the wavefunction, using a suggested form and the Schrödinger equation.
Zhuangzi
Homework Statement
Consider the (non-relativistic) Hamiltonian of a particle of charge -e in the presence of an external magnetic field B=B_0*ẑ, in the symmetric gauge A=(1/2)B x r.

a) Explicitly write the Hamiltonian described and show that p_z is a constant of motion.
b) Using your reasoning from (a), show that the problem admits a separation of variables with eigenfunctions of the form ψ(r)=exp(i*k_z*z)φ(x,y).
Relevant Equations
H = (1/2m)(p-qA)^2 + qV
L_z = -iħ(x∂_y - y∂_x)
B=B_0*ẑ
A=(1/2)B x r
ψ(r)=exp(i*k_z*z)φ(x,y)
I started with the first of the relevant equations, replacing the p with the operator -iħand expanding the squared term to yield:

H = (-ħ^2 / 2m)^2 + (iqħ/m)A·∇ + (q^2 / 2m)A^2 + qV

But since A = (1/2)B x r

(iqħ/m)A·∇ = (iqħ / 2m)(r x )·B = -(q / 2m)L·B = -(qB_0 / 2m)L_z

and A^2 = (1/4)(r^2*B^2 - (r·B)^2) = (B_0^2 / 4)(x^2 + y^2)

and V = 0

which gives a total Hamiltonian of

H = (-ħ^2 / 2m)^2 + (eB_0 / 2m) L_z + (e^2*B_0^2 / 8m)(x^2 + y^2).

At this point, however, I get stuck. I tried plugging in the wavefunction suggested in the problem, but I couldn't get an eigenvalue to pop out (I've attached a picture of my work). I want to know if I've made a mistake in calculating the Hamiltonian or in applying it to the wavefunction.

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The idea is to assume a solution of the form ##\psi(x,y,z) = \phi(x,y)Z(z)##, substitute it into the Schrödinger equation, and show it separates into a differential equation for ##\phi## and one for ##Z##.

## 1. What is an eigenstate of a free electron in a uniform magnetic field?

An eigenstate of a free electron in a uniform magnetic field is a state in which the electron's wavefunction is a solution to the Schrödinger equation for the system. This means that the electron's energy and position are well-defined and do not change over time.

## 2. How does a uniform magnetic field affect the energy of a free electron?

A uniform magnetic field causes the energy levels of a free electron to split into multiple discrete levels, known as Landau levels. The energy of an electron in a uniform magnetic field is quantized, meaning it can only take on certain discrete values.

## 3. What is the significance of Landau levels in the context of a free electron in a uniform magnetic field?

Landau levels are important because they explain the phenomenon of magnetoresistance, where the resistance of a material changes in the presence of a magnetic field. They also play a crucial role in the Hall effect, which is the generation of a voltage perpendicular to both the direction of current flow and the direction of the magnetic field.

## 4. How do the eigenstates of a free electron in a uniform magnetic field differ from those of a free electron in a vacuum?

The eigenstates of a free electron in a uniform magnetic field are different from those in a vacuum because the magnetic field causes the energy levels to split into discrete Landau levels. In a vacuum, the energy levels are continuous and not affected by a magnetic field.

## 5. Can an electron in a uniform magnetic field have a superposition of eigenstates?

Yes, an electron in a uniform magnetic field can have a superposition of eigenstates, just like any other quantum system. This means that the electron can exist in a combination of different energy levels, with each level having a certain probability of being measured. However, the energy of the electron will still be quantized and will only take on discrete values.

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