So in my last question, @joshphysics showed me how to prove $$K_\pm$$ were ladder operators. Now I need to show that there is a lowest state, i.e $$\langle m_0|K_+=K_-|m_0\rangle=0$$ I am not completely sure how I should approach this. I saw that $$J_3(K_-|m\rangle)=(m-1)(K_-|m\rangle$$ But I do not see how this shows that there has to be a minimum eigenvalue.

EDIT::: So I think I may have figured something out

$$\langle m_0|K_+K_-|m_0\rangle = (K_-|m_0\rangle)^\dagger K_-|m_0\rangle$$ This last quantity should be greater than 0, so we have a minimum state m_0 that we act on and get 0. Does this make sense?

• Essentially a duplicate of physics.stackexchange.com/q/23028/2451 , physics.stackexchange.com/q/54691/2451 and links therein. – Qmechanic Feb 14 '14 at 0:43
• I'm not sure that they duplicates... The links talk about non-degenerate vacua and non-integer quantum numbers. – JeffDror Feb 14 '14 at 1:02
• Your edit is exactly the right idea. There is more argumentation necessary to get to the point where you can assert that there is a state annihilated by $K_-$, but that's the start. – joshphysics Feb 14 '14 at 3:27
• What else is necessary to show that? – yankeefan11 Feb 14 '14 at 3:34
• @yankeefan11 I don't want to spoil it for you; think about it some more! Also, I'd recommend reading a good treatment of the one-dimensional harmonic oscillator that uses ladder operators. – joshphysics Feb 14 '14 at 3:36

2. To see the same effect using Quantum Mechanics you can solve the eigenvalue equation using brute force. You will find that the eigenstates are only normalizable for integer $n\ge 0$.
3. Using ladder operators one can look for the expectation value of the Hamiltonian, $\hbar \omega \left(K _+ K_-+ \frac{1 }{2} \right)$, and show that it's greater than zero.