Quantum motion of a charged particle in a magnetic field

In summary: The professor has since added:##H = H_{2d} + H_z##, the writing isn't entirely clear so I'm unsure if the "2d" subscript is correct, however ##H_z = \frac {p_z^2} {2m} + \frac {kz^2} {2}## and ##H_{2d} = \frac {\hbar^2} {2m} \left[ \left( -i \frac {\partial} {\partial x} - \frac q {c\hbar} \frac {By} 2 \right)^2 + \left( -i \frac {\partial} {\partial y}+ \
  • #1
EightBells
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1
Homework Statement
Find eigenenergies for a charged particle in magnetic field B and in a parabolic central potential, ##U(r) = \frac {kr^2} 2##. The particle has mass m and charge q.
Relevant Equations
Hamiltonian for charged particle in magnetic field ##H = \frac 1 {2m} \left( p- \frac {qA} {c^2} \right)^2 +q \phi##
Once I know the Hamiltonian, I know to take the determinant ##\left| \vec H-\lambda \vec I \right| = 0 ## and solve for ##\lambda## which are the eigenvalues/eigenenergies.

My problem is, I'm unsure how to formulate the Hamiltonian. Is my potential ##U(r)## my scalar field ##\phi##? I've seen equations relating B and E fields to the scalar and vector fields, ##\vec B = \nabla {} \times \vec A## and ##\vec E = -\nabla \phi - \frac 1 c \frac {d \vec A} {dt}##, but I'm not sure how to incorporate them or if I even need to.

Do I need an equation for the vector potential ##\vec A##? What is the momentum ##p## in this case?

Thanks!
 
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  • #2
First, ##U(r)## replaces the ##q\phi##.

Now, my issue with your problem is that you are not giving us the actual magnetic field (or the vector potential). Without that the problem is ill stated and nothing can be done about it. It is an static, uniform Magnetic field?
 
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  • #3
andresB said:
First, ##U(r)## replaces the ##q\phi##.

Now, my issue with your problem is that you are not giving us the actual magnetic field (or the vector potential). Without that the problem is ill stated and nothing can be done about it. It is an static, uniform Magnetic field?

That's all the information provided, however the professor has since posted additional notes:

##\vec A = \left( \frac {By} 2 ; \frac {-Bx} 2 ; 0 \right)##
##H = H_{2d} + H_z##, the writing isn't entirely clear so I'm unsure if the "2d" subscript is correct, however ##H_z = \frac {p_z^2} {2m} + \frac {kz^2} {2}## and ##H_{2d} = \frac {\hbar^2} {2m} \left[ \left( -i \frac {\partial} {\partial x} - \frac q {c\hbar} \frac {By} 2 \right)^2 + \left( -i \frac {\partial} {\partial y}+ \frac q {c\hbar} \frac {Bx} 2 \right)^2 \right] + \frac {k(x^2+y^2)} 2##.

From there I find ##H = \frac {\hbar^2} {2m} \left[ \left( -i \frac {\partial} {\partial x} - \frac q {c\hbar} \frac {By} 2 \right)^2 + \left( -i \frac {\partial} {\partial y}+ \frac q {c\hbar} \frac {Bx} 2 \right)^2 + \left( -i \frac {\partial} {\partial z} \right)^2 \right] + \frac {k(x^2+y^2+z^2)} 2##

The professor however finds some expression ## - \frac {\hbar^2} {2m} \frac {\partial^2} {\partial r^2} + \left[ \frac {\hbar^2} 2 \left( \frac {B^2} {4mc^2} + 1 \right) \left(x^2+y^2 \right) \right] + \frac {\hbar^2} {2m} \frac {qB} {c \hbar} \vec l_z## where ##\vec l_z = x \frac {\partial} {\partial y} - y \frac {\partial} {\partial x}## is the angular momentum along the z-axis. We then look for wave functions ##\Psi = e^{i {\varphi} {l_z}} \varphi(r)## and ##l_z = 0, \pm 1, \pm 2## and ##\phi(r)## is the radial component.

I'm used to finding eigenenergies in matrix form using ##|\vec H - E \vec I| = 0##, however given a wave function could I still use ##H |\Psi \rangle = E |\Psi \rangle## and work through the partial derivatives of ##H \Psi## to solve for the eigenenergies E? How do I find the appropriate wave function to use then?
 
  • #4
andresB said:
First, ##U(r)## replaces the ##q\phi##.

Now, my issue with your problem is that you are not giving us the actual magnetic field (or the vector potential). Without that the problem is ill stated and nothing can be done about it. It is an static, uniform Magnetic field?
[itex] \vec{A} = \approx\frac{\vec{B}\times\vec{r}}{2} [/itex]
I can never remember what the sign is...
 
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Related to Quantum motion of a charged particle in a magnetic field

1. What is quantum motion of a charged particle in a magnetic field?

The quantum motion of a charged particle in a magnetic field refers to the behavior of a charged particle, such as an electron, when it is subjected to both a magnetic field and the principles of quantum mechanics. This results in the particle exhibiting wave-like behavior and having quantized energy levels.

2. How does a magnetic field affect the motion of a charged particle?

A magnetic field exerts a force on a charged particle, causing it to experience a Lorentz force that is perpendicular to both the direction of the magnetic field and the velocity of the particle. This force alters the trajectory of the particle and can lead to circular or helical motion.

3. What is the significance of quantum motion in a magnetic field?

The study of quantum motion in a magnetic field has important implications in various fields, such as condensed matter physics, materials science, and quantum computing. Understanding how charged particles behave in a magnetic field is crucial for developing new technologies and advancing our understanding of fundamental physics.

4. How is the motion of a charged particle in a magnetic field described mathematically?

The motion of a charged particle in a magnetic field is described by the Schrödinger equation, which is the fundamental equation of quantum mechanics. This equation takes into account the particle's wave function, the magnetic field strength, and the particle's mass and charge.

5. Can the motion of a charged particle in a magnetic field be observed in real life?

Yes, the effects of quantum motion in a magnetic field have been observed in various experiments, such as the Stern-Gerlach experiment and the quantum Hall effect. These experiments have provided evidence for the quantization of energy levels and the wave-like behavior of charged particles in a magnetic field.

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