Occupation (or particle) number operator. Eigenvalues and eigenvectors. [closed]












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https://homepage.univie.ac.at/reinhold.bertlmann/pdfs/T2_Skript_Ch_5.pdf



Help me please. I made a screen (below) from the article above and highlighted what I did not understand.
Why is it true?



enter image description here










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closed as off-topic by Brahadeesh, amWhy, TMM, José Carlos Santos, Daniel Moskovich Dec 13 '18 at 15:49


This question appears to be off-topic. The users who voted to close gave this specific reason:


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    0












    $begingroup$


    https://homepage.univie.ac.at/reinhold.bertlmann/pdfs/T2_Skript_Ch_5.pdf



    Help me please. I made a screen (below) from the article above and highlighted what I did not understand.
    Why is it true?



    enter image description here










    share|cite|improve this question









    $endgroup$



    closed as off-topic by Brahadeesh, amWhy, TMM, José Carlos Santos, Daniel Moskovich Dec 13 '18 at 15:49


    This question appears to be off-topic. The users who voted to close gave this specific reason:


    • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Brahadeesh, amWhy, TMM, José Carlos Santos, Daniel Moskovich

    If this question can be reworded to fit the rules in the help center, please edit the question.



















      0












      0








      0





      $begingroup$


      https://homepage.univie.ac.at/reinhold.bertlmann/pdfs/T2_Skript_Ch_5.pdf



      Help me please. I made a screen (below) from the article above and highlighted what I did not understand.
      Why is it true?



      enter image description here










      share|cite|improve this question









      $endgroup$




      https://homepage.univie.ac.at/reinhold.bertlmann/pdfs/T2_Skript_Ch_5.pdf



      Help me please. I made a screen (below) from the article above and highlighted what I did not understand.
      Why is it true?



      enter image description here







      ordinary-differential-equations eigenvalues-eigenvectors quantum-mechanics






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      asked Dec 13 '18 at 12:20









      Иван ПетровИван Петров

      112




      112




      closed as off-topic by Brahadeesh, amWhy, TMM, José Carlos Santos, Daniel Moskovich Dec 13 '18 at 15:49


      This question appears to be off-topic. The users who voted to close gave this specific reason:


      • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Brahadeesh, amWhy, TMM, José Carlos Santos, Daniel Moskovich

      If this question can be reworded to fit the rules in the help center, please edit the question.







      closed as off-topic by Brahadeesh, amWhy, TMM, José Carlos Santos, Daniel Moskovich Dec 13 '18 at 15:49


      This question appears to be off-topic. The users who voted to close gave this specific reason:


      • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Brahadeesh, amWhy, TMM, José Carlos Santos, Daniel Moskovich

      If this question can be reworded to fit the rules in the help center, please edit the question.






















          1 Answer
          1






          active

          oldest

          votes


















          1












          $begingroup$

          The operator $N$ works like this



          $$
          N psi_{😺} = 😺 psi_{😺} tag{1}
          $$



          Which you can read as: If I apply $N$ to a state, I get the label of state times the state (more technically, $psi_{😺}$ is an eigenvector of $N$ with eigenvalue $😺$).



          In your problem you have



          $$
          N(a^daggerpsi_nu) = (nu + 1)(a^dagger psi_nu) tag{2}
          $$



          If you compare this with (1) you will conclude that



          $$
          a^daggerpsi_nu sim psi_{nu + 1}
          $$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            I will ask in another way. Why can't one eigenvalue correspond to two linearly independent vectors? In this situation.
            $endgroup$
            – Иван Петров
            Dec 13 '18 at 13:08










          • $begingroup$
            @ИванПетров You mean something like $a^{dagger} psi_{nu} = (nu + 1)( psi^{(1)}_{nu+1} + psi^{(2)}_{nu+1} )$?
            $endgroup$
            – caverac
            Dec 13 '18 at 13:14












          • $begingroup$
            I mean that a^daggerpsi_nu and psi_{nu + 1} may be linearly independent. Why not?
            $endgroup$
            – Иван Петров
            Dec 13 '18 at 13:27










          • $begingroup$
            @ИванПетров And have the same eigenvalue?
            $endgroup$
            – caverac
            Dec 13 '18 at 13:36










          • $begingroup$
            Yes, the same eigenvalue
            $endgroup$
            – Иван Петров
            Dec 13 '18 at 13:49


















          1 Answer
          1






          active

          oldest

          votes








          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          1












          $begingroup$

          The operator $N$ works like this



          $$
          N psi_{😺} = 😺 psi_{😺} tag{1}
          $$



          Which you can read as: If I apply $N$ to a state, I get the label of state times the state (more technically, $psi_{😺}$ is an eigenvector of $N$ with eigenvalue $😺$).



          In your problem you have



          $$
          N(a^daggerpsi_nu) = (nu + 1)(a^dagger psi_nu) tag{2}
          $$



          If you compare this with (1) you will conclude that



          $$
          a^daggerpsi_nu sim psi_{nu + 1}
          $$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            I will ask in another way. Why can't one eigenvalue correspond to two linearly independent vectors? In this situation.
            $endgroup$
            – Иван Петров
            Dec 13 '18 at 13:08










          • $begingroup$
            @ИванПетров You mean something like $a^{dagger} psi_{nu} = (nu + 1)( psi^{(1)}_{nu+1} + psi^{(2)}_{nu+1} )$?
            $endgroup$
            – caverac
            Dec 13 '18 at 13:14












          • $begingroup$
            I mean that a^daggerpsi_nu and psi_{nu + 1} may be linearly independent. Why not?
            $endgroup$
            – Иван Петров
            Dec 13 '18 at 13:27










          • $begingroup$
            @ИванПетров And have the same eigenvalue?
            $endgroup$
            – caverac
            Dec 13 '18 at 13:36










          • $begingroup$
            Yes, the same eigenvalue
            $endgroup$
            – Иван Петров
            Dec 13 '18 at 13:49
















          1












          $begingroup$

          The operator $N$ works like this



          $$
          N psi_{😺} = 😺 psi_{😺} tag{1}
          $$



          Which you can read as: If I apply $N$ to a state, I get the label of state times the state (more technically, $psi_{😺}$ is an eigenvector of $N$ with eigenvalue $😺$).



          In your problem you have



          $$
          N(a^daggerpsi_nu) = (nu + 1)(a^dagger psi_nu) tag{2}
          $$



          If you compare this with (1) you will conclude that



          $$
          a^daggerpsi_nu sim psi_{nu + 1}
          $$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            I will ask in another way. Why can't one eigenvalue correspond to two linearly independent vectors? In this situation.
            $endgroup$
            – Иван Петров
            Dec 13 '18 at 13:08










          • $begingroup$
            @ИванПетров You mean something like $a^{dagger} psi_{nu} = (nu + 1)( psi^{(1)}_{nu+1} + psi^{(2)}_{nu+1} )$?
            $endgroup$
            – caverac
            Dec 13 '18 at 13:14












          • $begingroup$
            I mean that a^daggerpsi_nu and psi_{nu + 1} may be linearly independent. Why not?
            $endgroup$
            – Иван Петров
            Dec 13 '18 at 13:27










          • $begingroup$
            @ИванПетров And have the same eigenvalue?
            $endgroup$
            – caverac
            Dec 13 '18 at 13:36










          • $begingroup$
            Yes, the same eigenvalue
            $endgroup$
            – Иван Петров
            Dec 13 '18 at 13:49














          1












          1








          1





          $begingroup$

          The operator $N$ works like this



          $$
          N psi_{😺} = 😺 psi_{😺} tag{1}
          $$



          Which you can read as: If I apply $N$ to a state, I get the label of state times the state (more technically, $psi_{😺}$ is an eigenvector of $N$ with eigenvalue $😺$).



          In your problem you have



          $$
          N(a^daggerpsi_nu) = (nu + 1)(a^dagger psi_nu) tag{2}
          $$



          If you compare this with (1) you will conclude that



          $$
          a^daggerpsi_nu sim psi_{nu + 1}
          $$






          share|cite|improve this answer









          $endgroup$



          The operator $N$ works like this



          $$
          N psi_{😺} = 😺 psi_{😺} tag{1}
          $$



          Which you can read as: If I apply $N$ to a state, I get the label of state times the state (more technically, $psi_{😺}$ is an eigenvector of $N$ with eigenvalue $😺$).



          In your problem you have



          $$
          N(a^daggerpsi_nu) = (nu + 1)(a^dagger psi_nu) tag{2}
          $$



          If you compare this with (1) you will conclude that



          $$
          a^daggerpsi_nu sim psi_{nu + 1}
          $$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 13 '18 at 12:45









          caveraccaverac

          14.6k31130




          14.6k31130












          • $begingroup$
            I will ask in another way. Why can't one eigenvalue correspond to two linearly independent vectors? In this situation.
            $endgroup$
            – Иван Петров
            Dec 13 '18 at 13:08










          • $begingroup$
            @ИванПетров You mean something like $a^{dagger} psi_{nu} = (nu + 1)( psi^{(1)}_{nu+1} + psi^{(2)}_{nu+1} )$?
            $endgroup$
            – caverac
            Dec 13 '18 at 13:14












          • $begingroup$
            I mean that a^daggerpsi_nu and psi_{nu + 1} may be linearly independent. Why not?
            $endgroup$
            – Иван Петров
            Dec 13 '18 at 13:27










          • $begingroup$
            @ИванПетров And have the same eigenvalue?
            $endgroup$
            – caverac
            Dec 13 '18 at 13:36










          • $begingroup$
            Yes, the same eigenvalue
            $endgroup$
            – Иван Петров
            Dec 13 '18 at 13:49


















          • $begingroup$
            I will ask in another way. Why can't one eigenvalue correspond to two linearly independent vectors? In this situation.
            $endgroup$
            – Иван Петров
            Dec 13 '18 at 13:08










          • $begingroup$
            @ИванПетров You mean something like $a^{dagger} psi_{nu} = (nu + 1)( psi^{(1)}_{nu+1} + psi^{(2)}_{nu+1} )$?
            $endgroup$
            – caverac
            Dec 13 '18 at 13:14












          • $begingroup$
            I mean that a^daggerpsi_nu and psi_{nu + 1} may be linearly independent. Why not?
            $endgroup$
            – Иван Петров
            Dec 13 '18 at 13:27










          • $begingroup$
            @ИванПетров And have the same eigenvalue?
            $endgroup$
            – caverac
            Dec 13 '18 at 13:36










          • $begingroup$
            Yes, the same eigenvalue
            $endgroup$
            – Иван Петров
            Dec 13 '18 at 13:49
















          $begingroup$
          I will ask in another way. Why can't one eigenvalue correspond to two linearly independent vectors? In this situation.
          $endgroup$
          – Иван Петров
          Dec 13 '18 at 13:08




          $begingroup$
          I will ask in another way. Why can't one eigenvalue correspond to two linearly independent vectors? In this situation.
          $endgroup$
          – Иван Петров
          Dec 13 '18 at 13:08












          $begingroup$
          @ИванПетров You mean something like $a^{dagger} psi_{nu} = (nu + 1)( psi^{(1)}_{nu+1} + psi^{(2)}_{nu+1} )$?
          $endgroup$
          – caverac
          Dec 13 '18 at 13:14






          $begingroup$
          @ИванПетров You mean something like $a^{dagger} psi_{nu} = (nu + 1)( psi^{(1)}_{nu+1} + psi^{(2)}_{nu+1} )$?
          $endgroup$
          – caverac
          Dec 13 '18 at 13:14














          $begingroup$
          I mean that a^daggerpsi_nu and psi_{nu + 1} may be linearly independent. Why not?
          $endgroup$
          – Иван Петров
          Dec 13 '18 at 13:27




          $begingroup$
          I mean that a^daggerpsi_nu and psi_{nu + 1} may be linearly independent. Why not?
          $endgroup$
          – Иван Петров
          Dec 13 '18 at 13:27












          $begingroup$
          @ИванПетров And have the same eigenvalue?
          $endgroup$
          – caverac
          Dec 13 '18 at 13:36




          $begingroup$
          @ИванПетров And have the same eigenvalue?
          $endgroup$
          – caverac
          Dec 13 '18 at 13:36












          $begingroup$
          Yes, the same eigenvalue
          $endgroup$
          – Иван Петров
          Dec 13 '18 at 13:49




          $begingroup$
          Yes, the same eigenvalue
          $endgroup$
          – Иван Петров
          Dec 13 '18 at 13:49



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