Tuesday 1 December 2015

quantum mechanics - Why do the ladder operators in harmonic oscillators work?


The Hamiltonian can be diagonalized by transforming $x$ and $p$ to $a$ and $a^\dagger$. I understand how one proceeds from there to find the spectrum of $a^\dagger a$, the ground state $|0\rangle$ and so on. But I have trouble understanding why the simple choice $[a, a^\dagger] = 1$ is everything one needs in order to diagonalize the Hamiltonian.


In SU(2) groups one can perform the highest-weight construction for the $2j+1$ dimensional irreducible representation (spin $j$ irrep). But there one has the Cartan-Weyl basis consisting of $\sigma_3$ and then uses $\sigma_1$ and $\sigma_2$ to find $\sigma_\pm$ such that this is special with $\sigma_3$ such that $\sigma_\pm$ raises and lowers the eigenvalue of $\sigma_3$.



The harmonic oscillator feels simpler than the SU(2) group as we only have excitations of one kind. With angular momentum or spin, there seem to be much more degrees of freedom. On the other hand the basis for the harmonic oscillator is infinite and that makes all the matrix representations of $a$ and $a^\dagger$ a tad more complicated.


Why does the algebraic method work for the harmonic oscillator?




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