ladder operators

[William W chen]

Can someone explain why is it when one attempts to factor the Hamiltonian
for the harmonic oscillator, one ends up with these two operators that raise
and lower the eigenstates of the harmonic oscillator. i.e. H=hbar*w(a'a+1/2)
where a*|n>=|n-1>, and a'*|n>=|n+1>. I've neglected normalization. In other
words, what's the physical basis for these step-up and step-down operators?

In the same vein, I want to know what motivates one to combine Lx and Ly to
produce angular momentum ladder operators. i.e. why is it that L+=Lx+iLy,
and why L_=Lx-iLy?

[rocke...@my-dejanews.com]

In article <7f0hbh$1n...@itssrv1.ucsf.edu>,

wc...@itsa.ucsf.edu (William W chen) wrote:
> Can someone explain why is it when one attempts to factor the Hamiltonian
> for the harmonic oscillator, one ends up with these two operators that raise
> and lower the eigenstates of the harmonic oscillator. i.e. H=hbar*w(a'a+1/2)
> where a*|n>=|n-1>, and a'*|n>=|n+1>. I've neglected normalization. In other
> words, what's the physical basis for these step-up and step-down operators?
>

It's a convenient choice such that the commutator of these raising and
lowering operators gives you "1". I'm assuming you've got some text in
front of you, so I won't go through the details, but basically that's
where the q.m. is entering. Look at the h-bar in the commutator if you
left it as between p and q (a and a-dagger should be written in terms
of p and q there somewhere). In the classical limit of h-bar -> 0, these
a's loose their commutation structure and you'd just have classical p's
and q's, so the physical basis is rather subtle in that it's the essence
of where the q.m. enters.

>
> In the same vein, I want to know what motivates one to combine Lx and Ly to
> produce angular momentum ladder operators. i.e. why is it that L+=Lx+iLy,
> and why L_=Lx-iLy?
>

Put these L's in between up and down states, you'll easily deduce that
adding combinations of them as Lx+-Ly gives you the current linking the
two states. It helps to write it all out longhand in detail and try
every combination to get the feel for what's going on. Don't try to
put too much 'mechanical' meaning into it, remember it's q.m.

Greg Trayling
http://www.cs.uwindsor.ca/meta-index/people/traylin

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