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I Determining if an operator is degenerate

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  1. Nov 24, 2016 #1
    Hi,

    I was wondering how you can formally determine whether a given operator is degenerate. I undertand you can produce the 'usual equation' det(Q-##\lambda ##)=0 and solve for ##\lambda ##, where Q is our operator. But if Q is a differential (for example ##\frac{p^2}{2m}= - \bar{h} \frac{d^2}{dx^2}## ) how can one write such an equation.

    p.s. I only used this example as I know that the operator is degenerate with sin and cos of the same argument. Other examples are welcome :)

    Many thanks in advance!!

    Also- dont know why I cant get my equations to format properly? ^^
    < Moderator's note: fixed. You had a ( instead of { >
     
    Last edited by a moderator: Nov 24, 2016
  2. jcsd
  3. Nov 24, 2016 #2

    hilbert2

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    Degeneracy of an operator's eigenvalue spectrum is usually a result of some symmetry. In your example the operator ##\frac{d^2}{dx^2}## is translation invariant which means that the set of all displaced sine functions of same wavelength, ##sin(a x + b)##, is a set of mutually degenerate states (a is kept constant and b can have any real value).
     
    Last edited: Nov 24, 2016
  4. Nov 24, 2016 #3
    Ahh okay that makes sense in this case. In general, do you know how this can be formally determined for any given operator (other than looking for a symmetry)?

    Thanks :)
     
  5. Nov 24, 2016 #4

    hilbert2

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    ^ There are probably some methods to systematically find all the symmetries of a differential operator, but that's quite advanced group theory. There are also so called "accidental" degeneracies that aren't caused by symmetries, but there's probably no efficient way to find them if they exist.
     
  6. Nov 25, 2016 #5

    PeterDonis

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    As a differential equation. For example, in your case we would have, for a function ##f(x)##,

    $$
    - \hbar^2 \frac{d^2}{dx^2} f(x) - \lambda f(x) = 0
    $$

    which obviously has solutions for ##f(x)## that are sines and cosines. This still doesn't guarantee finding a solution, since solving differential equations is not a completely mechanical process. But that's how you would start.
     
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