Charged Particles in an Electromagnetic Field: Difference between revisions
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We recognize this as the cyclotron frequency of the particle. We may immediately write down the full eigenfunctions and energy levels of the system. The wave functions are | We recognize this as the cyclotron frequency of the particle. We may immediately write down the full eigenfunctions and energy levels of the system. The wave functions are | ||
<math>\psi(x,y,z)=\frac{1}{\sqrt{L_yL_z}e^{i(k_yy+k_zz)}</math> | <math>\psi(x,y,z)=\frac{1}{\sqrt{L_yL_z}}e^{i(k_yy+k_zz)}</math> | ||
== Problems about Motion in electromagnetic field == | == Problems about Motion in electromagnetic field == |
Revision as of 16:20, 12 August 2013
A problem with some relation to the harmonic oscillator is that of the motion of a charged particle in a constant and uniform magnetic field. In classical mechanics, we know that the Hamiltonian for this system is
Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle H=\frac{1}{2m}\left (\mathbf{p}-\frac{e}{c}\mathbf{A}\right )^2,}
where Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle e\!} is the charge of the particle and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathbf{A}} is the vector potential. In fact, to obtain the Hamiltonian for any system in the presence of a magnetic field, we simply make the replacement, Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathbf{p}\rightarrow\mathbf{p}-\frac{e}{c}\mathbf{A}.} In quantum mechanics, we introduce the magnetic field in the same way; this process is referred to as minimal coupling.
Gauge Invariance in Quantum Mechanics
We know from Maxwell's equations that the classical physics of a charged particle in an electromagnetic field is invariant under a gauge transformation, Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Phi\rightarrow\Phi-\frac{1}{c}\frac{\partial\chi}{\partial t}} and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathbf{A}\rightarrow\mathbf{A}+\nabla\chi,} where Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Phi\!} is the scalar potential and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \chi(\mathbf{r},t)\!} is a single-valued real function. We will now show how this is expressed in quantum mechanics.
In the position basis, the Schrödinger equation for a charged particle in an electromagnetic field is
Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle -i\hbar\frac{\partial\Psi}{\partial t}-e\Phi\Psi=-\frac{\hbar^2}{2m}\left (\nabla+\frac{ie}{\hbar c}\mathbf{A}\right )^2\Psi.}
If we now perform the above gauge transformation on the electromagnetic field, then this equation becomes
Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle -i\hbar\frac{\partial\Psi}{\partial t}-e\Phi\Psi+\frac{e}{c}\frac{\partial\chi}{\partial t}\Psi-\frac{\hbar^2}{2m}\left (\nabla+\frac{ie}{\hbar c}\mathbf{A}+\frac{ie}{\hbar c}\nabla\chi\right )^2\Psi.}
If we make the substitution, then we recover the original equation. Therefore, a gauge transformation of the magnetic field effectively introduces a phase factor to the wave function. This does result in a change in the canonical momentum, but it will have no effect on, for example, the probability density for finding the particle at a given position or, as we will see later, on the expectation value of the position or velocity of the particle.
We see that, in quantum mechanics, gauge invariance is expressed as follows. If one introduces a single-valued phase factor into the wave function, then it may be "canceled out" by a corresponding change in the electromagnetic potentials that the particle is subject to.
For a constant and uniform magnetic field we typically work with one of two gauges. One of these is the Laudau gauge,
or
The other is the symmetric gauge,
Eigenstates of a Charged Particle in a Static and Uniform Magnetic Field
Let us now find the eigenstates of a charged particle in a static and uniform magnetic field. We will be working in the Landau gauge,
The Schrödinger equation for this system is
In this gauge, the Hamiltonian is translationally invariant along the and axes. Therefore, our wave function will have the form,
Substituting this form into the equation, we obtain
Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle -\frac{\hbar^2}{2m}\frac{d^2f}{dx^2}+\frac{\hbar^2}{2m}\left (k_y+\frac{e}{\hbar c}Bx\right )^2f+\frac{\hbar^2k_z^2}{2m}f=Ef,}
or
Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle -\frac{\hbar^2}{2m}\frac{d^2f}{dx^2}+\frac{e^2B^2}{2mc^2}\left (x+\frac{\hbar c}{eB}k_y\right )^2f=\left (E-\frac{\hbar^2k_z^2}{2m}\right )f.}
If we now introduce the shifted position coordinate Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x'=x+\frac{\hbar c}{eB}k_y} and the shifted energy Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle E'=E-\frac{\hbar^2k_z^2}{2m},} this becomes
Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle -\frac{\hbar^2}{2m}\frac{d^2f}{dx'^2}+\frac{e^2B^2}{2mc^2}x'^2f=E'f.}
This is just the equation for a harmonic oscillator with frequency
Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \omega=\frac{eB}{mc}.}
We recognize this as the cyclotron frequency of the particle. We may immediately write down the full eigenfunctions and energy levels of the system. The wave functions are
Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \psi(x,y,z)=\frac{1}{\sqrt{L_yL_z}}e^{i(k_yy+k_zz)}}