Proving HO Eigenvalues Using Ladder Operators

[Loro]

This has already been adressed here: https://www.physicsforums.com/showthread.php?t=173896 , but I still didn't get the answer.

The Harmonic Oscillator is fully described (according to my favourite QM book) by the HO Hamiltonian, and the commutation relations between the position and momentum operators.

So starting there, we define the ladder operators and show that the energy eigenvalues are:

[itex] E_n = (n + 1/2) \hbar ω [/itex]
where n is a non-negative integer

I know how to do this.

But how do we know there are no other eigenvalues? Or equivalently - that the corresponding eigenkets form a complete set? Can we show it without any extra assumptions?

I know it can be shown from solving the pde, that all other solutions blow up, but I'd like to do it purely by ladder operators.

 

[The_Duck]

Here's a way of doing this that only requires solving a single simple differential equation.

Suppose there was an eigenket that was not generated by repeatedly applying the raising operator to the ground state. Repeatedly apply the lowering operator to this new eigenket to get a sequence of more new eigenkets, until you get one that's annihilated by the lowering operator. Write out the differential equation corresponding to a|psi> = 0 and show that it only has one solution--the familiar ground state. Therefore when we repeatedly applied the lowering operator to our "new" eigenket, we eventually got the regular ground state. Therefore we actually could have generated the original eigenket by applying raising operators repeatedly to the ground state.

 

[Loro]

Indeed! Now I'm totally convinced!

Thanks a lot!

 

[dextercioby]

That the spectrum of the Hamiltonian can be set into a bijective correspondence with [itex] \mathbb{N} [/itex] one can show in a purely algebraic fashion (that is without assuming an a apriori topology on the linear space in which the spectral equation for H is solved). One automatically derives that the linear space contains a countable set of eigenvectors of the Hamiltonian, each of them corresponding to a spectral value. Without resorting to topology, one shows that the spectrum is non-degenerate.

However, QM is made with topological spaces and, as one endows the linear space with a scalar product and makes the completion of it with respect to its associated strong topology, one obtains the following result:

The Hamiltonian when defined on the initial space (the non-completed one) endowed with the strong topology and realized as the space of Schwartz test functions (this realization automatically brings in a new topology which makes the Schwartz space itself a countable Hilbert space) is an essentially self adjoint operator. Its closure with respect to the strong topology is an unbounded self-adjoint operator with the following property deriving from the spectral theorem (Hilbert, von Neumann, Stone): the set of eigenvectors is a countable basis in the Hilbert space, therefore it's maximal.

It's interesting to note that the spectral problem for the Hamiltonian can be solved without resorting to the full Schwartz space rigging of the Hilbert space (thus no need for tempered distributions). The latter is necessary only when the spectral problem for the position and momentum operators is solved.

 

[Loro]

You know you just did a horrible thing to me? Wrote an explanation referring to a few mathematical notions that I don't understand, and I wish I could. And it won't leave me alone until I eventually sit down and study them!

Thanks though...

 

[vanhees71]

A good source to study QM at this level of mathematical sophistication is the two-volume textbook by Galindo and Pascual.

 

[Bill_K]

Awesome!