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Eigenvalue of lowering operator

  1. Jun 27, 2015 #1
    How to prove that eigenvalue of lowering operator is zero?
     
  2. jcsd
  3. Jun 27, 2015 #2

    ShayanJ

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    Eigenvalue of lowering operator is zero only on the ground state and this is so by the definition of ground state.
    For other eigenvectors of the lowering operator(coherent states), eigenvalues are non-zero.
     
  4. Jun 27, 2015 #3
    Ground state or the absolute minimum state of the system is "defined" as the state which gives zero when operated by the
    lowering operator.

    It is easy to consider the H.OSc. and operate with the lowering operator till it gets to the min = h-bar omega/2 and if you apply furthe the lowering operator then you get zero. Of course the system maybe different in different situation and hence the minimum state may vary.

    watch out that minimum state or the vacuum state is not the zero state, a zero state does not exists in nature.
    The proof is kind of trival. google on Fock space and there you get your proof.
     
  5. Jun 30, 2015 #4

    radium

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    Ladder operators (angular momentum or harmonic oscillators for example) act on states to go from one state to another state with an increased or decreased eigenvalue up to some normalization included. They are constructed so when you act on zero you get zero because that is the vacuum state by definition.

    If you are talking about an eigenvalue of the lowering operator you must consider coherent states which are constructed to be eigenstates of the annihilation operator. The eigenvalue is some complex number. Coherent states are used in path integrals. If you look at the coherent states in the harmonic oscillator, they behave the most classically in terms of minimized uncertainty.
     
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