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Quantum mechanics Hermitian operator

  1. Jul 2, 2017 #1
    1. The problem statement, all variables and given/known data

    I have the criteria:

    ## <p'| L_{n} |p>=0 ##,for all ##n \in Z ##

    ##L## some operator and ## |p> ##, ## |p'> ##some different physical states

    I want to show that given ## L^{+}=L_{-n} ## this criteria reduces to only needing to show that:

    ##L_n |p>=0 ## for ##n>0 ##

    2. Relevant equations
    look up , look down,

    3. The attempt at a solution

    ##<p'|L_n|p>^{+}=<p|L_n^{+}|p'>=<p|L_{-n}|p'>##

    So from this I can deduce that showing :

    ##<p'|L_n|p>^{+}## is satisfied is equivalent to showing that ##<p|L_{-n}|p'>## is satisfied.

    I.e I can reduce the criteria from showing for all ##n \in Z## to showing for ##n>0## only, BUT I am still sandwiched between two different physical states, I don't understand how this means you can reduce further to showing only that ##L_n |p>=0##, ##n >0 ## acting on solely a ket/bra...

    Many thanks in advance.
     
  2. jcsd
  3. Jul 6, 2017 #2
    So I don't understand some of your notation. Do you mean [itex]L_n^\dagger = L_{-n}[/itex]? Also are [itex]|p>[/itex] and [itex]|p'>[/itex] arbitrary? It seems like having an operator act on a ket and then taking the inner product always giving zero would mean the operator gives zero on the ket. All the extra parts about [itex]n[/itex] aren't really relevant. You can consider one [itex]n[/itex]. If [itex]L_n=0[/itex] the you can show [itex]L_{-n}[/itex] =0 if they are related by conjugation.
     
  4. Jul 10, 2017 #3
    yes
    mm I thought the idea for using two different states ## |p'> ## and ## |p> ## is to show that this result holds for general arbitrary states, rather than showing that ##<p|L_n|p> = 0 ## . I agree that if we wanted to show that this was zero then ##L_n|p> = 0## is enough for the inner product to be zero.


    yup
     
  5. Jul 10, 2017 #4
    I just can't imagine a situation where if [itex]<p|L_n|p'>=0[/itex] for all [itex]|p>[/itex] and [itex]|p'>[/itex] and they aren't equal that [itex]L_n \neq 0 [/itex]. For example say [itex] |p'> = L_n|p> [/itex]. I mean it must be some state right? Then the condition implies [itex]<p'|p'> =0 [/itex] which by the properties of vector implies [itex]|p'> = L_n |p> =0[/itex].
     
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