Timeline for Finding the energy levels of an electron in a plane perpendicular to a uniform magnetic field
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Nov 22, 2012 at 22:58 | vote | accept | Freeman | ||
| Nov 22, 2012 at 22:58 | comment | added | Freeman | Thank you so much! I really appreciate the help explaining this to me. | |
| Nov 20, 2012 at 17:29 | comment | added | David Bar Moshe | If I may add, if you write $a = x+ip$ and $a^{\dagger} = x-ip$, you get the harmonic oscillator Hamiltonian in the usual representation, but the harmonic osillator coordinates $x$ and $p$ are not the original coordinates that you started with. | |
| Nov 20, 2012 at 17:16 | comment | added | David Bar Moshe | If you substitute the expressions of $a$ and $a^{\dagger}$ given in the answer in the Hamiltonian $H=\hbar\omega(a a^{\dagger}+\frac{1}{2})$ you get the Hamiltonian you started with expressed in terms of $X$ and $P$. | |
| Nov 20, 2012 at 17:04 | comment | added | Freeman | These creation annihilation operators, are they $a,a^{\dagger}$, I don't see how $H =\hbar \omega (a a^{\dagger}+\frac{1}{2})$? don't we want to find it in terms of $X$ and $P$ to compare to the harmonic oscillator? Sorry for being slow.. it's been a very long day! | |
| Nov 20, 2012 at 16:58 | comment | added | David Bar Moshe | The Hamiltonian expressed in terms of the creation and annihilation operators has exactly the form of the harmonic oscillator Hamiltonian. Thus the energy levels are exactly equal to those of the harmonic oscillator, however with infinite degeneracy per level (The harmonic oscillator energy levels are nondegenerate). The advantage of using this method is that it allows an algebraic solution of the energy levels (i.e. without solving differential equations), please see the quantum harmonic oscillator Wikipedia page: en.wikipedia.org/wiki/Quantum_harmonic_oscillator | |
| Nov 20, 2012 at 16:47 | comment | added | Freeman | What is the signifiance of the canonical commutation relation? | |
| Nov 20, 2012 at 16:40 | comment | added | Freeman | I'm sorry, this is very good, but would you mind explaining in a little more detail how this gives us the energy levels? | |
| Nov 20, 2012 at 15:16 | history | answered | David Bar Moshe | CC BY-SA 3.0 |