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. Feb 14, 2014 at 0:43
• I'm not sure that they duplicates... The links talk about non-degenerate vacua and non-integer quantum numbers. Feb 14, 2014 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. Feb 14, 2014 at 3:27
• What else is necessary to show that? Feb 14, 2014 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. Feb 14, 2014 at 3:36

1. The potential of the system is bounded from below. Thus there is no way to have a particle with energy below this point. This would require negative kinetic energy. The way to see that is $$E_{tot}= E_kin + V \ge V$$ Thus the total energy must always be greater than or equal to the potential energy.
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.