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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?

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yankeefan11
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  • Essentially a duplicate of http://physics.stackexchange.com/q/23028/2451 , http://physics.stackexchange.com/q/54691/2451 and links therein. – Qmechanic Feb 14 '14 at 00:43
  • I'm not sure that they duplicates... The links talk about non-degenerate vacua and non-integer quantum numbers. – JeffDror Feb 14 '14 at 01:02
  • 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 03:27
  • What else is necessary to show that? – yankeefan11 Feb 14 '14 at 03:34
  • @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 03:36
  • ny recommendations? Ive been going off of wikipedia a bit, which is doing better than any notes I have from class – yankeefan11 Feb 14 '14 at 03:41
  • @yankeefan11 My favorite resource for this is Cohen-Tannoudji volume 1, chapter 5. – joshphysics Feb 14 '14 at 06:05
  • Thanks! I will take a look. In the meantime, I found that $[K_+,K_-]=-2 J_3$ So does my J act as the number operator? – yankeefan11 Feb 14 '14 at 06:22
  • @yankeefan11 I wrote another answer, yesterday, giving the final explicit expression of your operators $K_\pm$ and their action on eigenstates of $J_3$. Unfortunately, as far as I understand, it was not in agreement with the homework policy of this site and it was deleted by the moderator. As your system relies upon standard $q$, $p$ operators in $L^2(\mathbb R)$, my suggestion is: Try to write your $K_\pm$ in terms of $a,a^\dagger$ and the eigenstates/eigenvalues of $J_3$ in terms of the ones of the Hamiltonian standard harmonic oscillator. – Valter Moretti Feb 14 '14 at 07:23
  • $|m_0\rangle$ does exists just because the ground state of the harmonic oscillator in $L^2(\mathbb R)$ exists, when translating your model into the standard harmonic oscillator. Actually, you have here TWO vectors verifying $K_-|a\rangle =0$. – Valter Moretti Feb 14 '14 at 09:11

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One can show that the energy spectrum is bounded from below using a few ways:

  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 \begin{equation} E_{tot}= E_kin + V \ge V \end{equation} 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.
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