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The ladder operator method is used to solve the one-particle Schrodinger equation with a harmonic potential.

What other potentials for the one-particle Schrodinger equation may be solved with the ladder operator method?

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  • $\begingroup$ This question (v2) seems like a list question. See also this related Wikipedia list. $\endgroup$ – Qmechanic Oct 21 '16 at 20:22
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The hydrogen atom is an example where ladder operators can be used. There is a hidden SO(4) symmetry that explains the degeneracy for the prinicpal quantum number and one can use algebraic methods to get the eigenvalues. Here is a paper that does central force problems in general. . The full symmetry group for the hydrogen atom is SO(4,2). Here is another resource.

Some people think ladder operators are only for the harmonic oscillator or equally spaced eigenvalues but this is because they are restricting themselves to the Heisenberg Lie algebra which works for the harmonic oscillator but there are other problems with other Lie algebras and their own representation theory.

For a trivial counter example for the belief you need equally spaced eigenvalues: Suppose your Hamiltonian was just $L^2$ . This is hermitian so legitimate, there is an algebraic method to solve for eigenvalues it depends on the representation theory of $su(2)$ or equivalently $sl(2,C)$ but as we all know the eigenvalues of this operator are not equally spaced.

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Given a non-degenerate observable $A$, its ladder operator is an operator $L$ such that $[A, L] = \xi L $ for some $\xi \in \mathbb{R}$.

We can then write $A (L \lvert a \rangle) = (LA + [A,L]) \lvert a \rangle = (a+\xi)(L \lvert a \rangle )$.

Which implies by non-degeneracy that $L \lvert a \rangle = \lvert a + \xi \rangle $

Also note that since $A$ is Hermitian that $[A, L^\dagger] = -\xi L^\dagger $. So $L^\dagger \lvert a \rangle = \lvert a - \xi \rangle $ similarly.

The Jacobi identity then implies that $[A, [L, L^\dagger]] = 0$.

You said you were interested in solving the 1-D Schrodinger Equation. In this case your observable $A$ is the Hamiltonian $H$. One dimensional Hamiltonians are always non-degenerate when if comes to bound states so this condition is always satisfied (see proof here). So if you can find an operator $L$ such that $[H, L] = \xi L $ for some $\xi \in \mathbb{R}$, then the problem can be solved by ladder operators. The energy of an 1-D system will also always be bounded below since $P^2$ is positive definite and $V(X)$ has a minimum since you interested in bound states.

One essential part of systems where ladder operators apply is that their spectrum (energy levels in the case of $A=H$, ${a}$ in general) is linearly spaced. If this makes qualitative sense then trying ladder operators may be the way to go.

All that said, I'm fairly certain that you could prove that linearly spaced energy levels implies a harmonic potential. If this is true, then the harmonic oscillator is the only 1-D system whose Schrodinger Equation can be solved in this way.

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