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Eigenvector proof from Dirac's QM

  1. Aug 28, 2013 #1
    Hi everyone,

    I'm currently working my way through Dirac's Quantum Mechanics, and I found this proof really irritating.

    We're trying to demonstrate that any eigenket can be expressed as a sum of eigenkets of a real linear function [itex]\xi[/itex] which satisfies the equation [itex]\varphi[/itex]([itex]\xi[/itex]) = a[itex]_{1}[/itex][itex]\xi[/itex][itex]^{n}[/itex]+a[itex]_{2}[/itex][itex]\xi[/itex][itex]^{n-1}[/itex]...+a[itex]_{n}[/itex]

    I attach Dirac's proof. I'm confused by how 22 vanishing for [itex]\chi (\xi)[/itex] in general follows from the substitution.


    Attached Files:

  2. jcsd
  3. Aug 28, 2013 #2
    It would help me if you can attach previous two pages too. I dont have access to this book right now.
  4. Aug 28, 2013 #3
    Before I answer your question, let me point out that your version of eqn (17) has a small problem. If [itex]a_1[/itex] is the coefficient on [itex]\xi^n[/itex], then the constant term should be [itex]a_{n+1}[/itex]. Also, in eqn (17), [itex]\phi[/itex] is set to zero. Here is the eqn as it appears in the book.
    [tex]\phi(\xi) = \xi^n+a_1\xi^{n-1}+a_2\xi^{n-2}+\cdots+a_{n} = 0[/tex]

    However, this has nothing to do with the problem you are facing. As for your problem, in the text below expression (21), there is an explanation of why that expression is equal to zero. Equation (22) is then the result of setting (21) to zero and applying (21) to the ket [itex]|P>[/itex].
    Last edited: Aug 28, 2013
  5. Aug 28, 2013 #4
    Sorry about copying that equation incorrectly - I think my brain must have given up on the formatting.

    The thing I don't understand is why 21 is zero for all ξ given that it is zero for the substitution Cs (s = 1, 2, 3...n). Dirac justifies it by saying that the expression is of degree n-1 in ξ. This is obviously the case but I don't understand why it means you can go from the substitution of complex numbers to a generalized real linear operator. I'm sure there's something really simple here I haven't grasped.
  6. Aug 28, 2013 #5
    Because if an n-1 degree polynomial has n distinct zeros, then the polynomial itself is zero. For instance, a quadratic has no more than 2 zeros, and a cubic has no more than 3. etc.
  7. Aug 29, 2013 #6
    Yes, that would be the case. I think I was just confused by ξ being an operator.
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