|
|
Line 38: |
Line 38: |
| <math>\hat{H}_1=\hbar\omega\left (\hat{a}\dagger\hat{a}+\tfrac{1}{2}\right ),</math> | | <math>\hat{H}_1=\hbar\omega\left (\hat{a}\dagger\hat{a}+\tfrac{1}{2}\right ),</math> |
|
| |
|
| where <math>\omega=\frac{eB}{mc}.</math> This is just the Hamiltonian for a [[Harmonic Oscillator States and Eigenvalues|harmonic oscillator]]. The contribution to the energy from this term is therefore | | where <math>\omega=\frac{eB}{mc}.</math> This is just the Hamiltonian for a [[Harmonic Oscillator Spectrum and Eigenstates|harmonic oscillator]]. The contribution to the energy from this term is therefore |
|
| |
|
| <math>E_1=\left (n+\tfrac{1}{2}\right )\hbar\omega.</math> | | <math>E_1=\left (n+\tfrac{1}{2}\right )\hbar\omega.</math> |
(a) In the symmetric gauge, 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 A_{x}=-\tfrac{1}{2}By,}
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 A_{y}=\tfrac{1}{2}Bx,}
and
(b) The Hamiltonian for the system is
If we label the first two terms as
, and the last one as
, then we may write the Hamiltonian as
Using the identity,
we may rewrite
as
If we now define the operators,
and
this becomes
where
This is just the Hamiltonian for a harmonic oscillator. The contribution to the energy from this term is therefore
The remaining part of the Hamiltonian,
is just that of a free particle in one dimension, and thus its contribution to the energy is just
The total energy is then just
Back to Charged Particles in an Electromagnetic Field.