Two different definitions of ladder operator for Harmonic Oscillators

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As it happened, I accidentally referred to two different editions of Introduction to QM by Griffiths. In the second chapter, while defining the ladder operator for harmonic oscillators, he used different terms. Now different definitions mean that their commutator change. Which of the two operators should I use? Also, why is there ambiguity in the definition of the same?



$$a_pm equiv frac 1 sqrt2mbigg( frac hbari frac ddx pm im omega xbigg)$$



$$a_pm equiv frac 1 sqrt2hbar m omegabigg(mp ip + m omega xbigg)$$










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    As it happened, I accidentally referred to two different editions of Introduction to QM by Griffiths. In the second chapter, while defining the ladder operator for harmonic oscillators, he used different terms. Now different definitions mean that their commutator change. Which of the two operators should I use? Also, why is there ambiguity in the definition of the same?



    $$a_pm equiv frac 1 sqrt2mbigg( frac hbari frac ddx pm im omega xbigg)$$



    $$a_pm equiv frac 1 sqrt2hbar m omegabigg(mp ip + m omega xbigg)$$










    share|cite|improve this question

























      up vote
      2
      down vote

      favorite









      up vote
      2
      down vote

      favorite











      As it happened, I accidentally referred to two different editions of Introduction to QM by Griffiths. In the second chapter, while defining the ladder operator for harmonic oscillators, he used different terms. Now different definitions mean that their commutator change. Which of the two operators should I use? Also, why is there ambiguity in the definition of the same?



      $$a_pm equiv frac 1 sqrt2mbigg( frac hbari frac ddx pm im omega xbigg)$$



      $$a_pm equiv frac 1 sqrt2hbar m omegabigg(mp ip + m omega xbigg)$$










      share|cite|improve this question















      As it happened, I accidentally referred to two different editions of Introduction to QM by Griffiths. In the second chapter, while defining the ladder operator for harmonic oscillators, he used different terms. Now different definitions mean that their commutator change. Which of the two operators should I use? Also, why is there ambiguity in the definition of the same?



      $$a_pm equiv frac 1 sqrt2mbigg( frac hbari frac ddx pm im omega xbigg)$$



      $$a_pm equiv frac 1 sqrt2hbar m omegabigg(mp ip + m omega xbigg)$$







      quantum-mechanics operators harmonic-oscillator






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      edited 2 hours ago









      Hugo V

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      asked 2 hours ago









      Ashutosh Singh

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          2 Answers
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          Your second equation is the standard definition for $a$ ($a_-$) and $a^dagger$ ($a_+$) as found e.g. on Wikipedia. Note that the ladder operators here are dimensionless.



          Griffiths is doing it slightly differently. I'm not sure why, but if you continue consistently with this alternative definition, you will get to the same correct results of course. For example, Griffiths (2.45) is
          $$ [a_-, a_+] = hbaromega , $$
          compare with the standard $[a, a^dagger] = 1$.






          share|cite|improve this answer



























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            In the first definition, $a_+$ and $a_-$ have been defined such that $[a_-,a_+]=hbaromega$. Here the position and momentum have been scaled as
            $X=sqrtmx$ & $P=sqrtfrac1mp$ and $a_pm= frac1sqrt2(X pmiP)$.



            Whereas, in the second definition, $a_+$ and $a_-$ have been defined such that $[a_-,a_+]=1$. Here the position and momentum have been scaled as
            $X=sqrtfracmomegahbarx$ & $P=sqrtfrac1momegahbarp$ and $a_pm=frac1sqrt2(XpmiP)$.






            share|cite|improve this answer










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            Bhavesh Valecha is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
            Check out our Code of Conduct.

















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              2 Answers
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              2 Answers
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              up vote
              4
              down vote













              Your second equation is the standard definition for $a$ ($a_-$) and $a^dagger$ ($a_+$) as found e.g. on Wikipedia. Note that the ladder operators here are dimensionless.



              Griffiths is doing it slightly differently. I'm not sure why, but if you continue consistently with this alternative definition, you will get to the same correct results of course. For example, Griffiths (2.45) is
              $$ [a_-, a_+] = hbaromega , $$
              compare with the standard $[a, a^dagger] = 1$.






              share|cite|improve this answer
























                up vote
                4
                down vote













                Your second equation is the standard definition for $a$ ($a_-$) and $a^dagger$ ($a_+$) as found e.g. on Wikipedia. Note that the ladder operators here are dimensionless.



                Griffiths is doing it slightly differently. I'm not sure why, but if you continue consistently with this alternative definition, you will get to the same correct results of course. For example, Griffiths (2.45) is
                $$ [a_-, a_+] = hbaromega , $$
                compare with the standard $[a, a^dagger] = 1$.






                share|cite|improve this answer






















                  up vote
                  4
                  down vote










                  up vote
                  4
                  down vote









                  Your second equation is the standard definition for $a$ ($a_-$) and $a^dagger$ ($a_+$) as found e.g. on Wikipedia. Note that the ladder operators here are dimensionless.



                  Griffiths is doing it slightly differently. I'm not sure why, but if you continue consistently with this alternative definition, you will get to the same correct results of course. For example, Griffiths (2.45) is
                  $$ [a_-, a_+] = hbaromega , $$
                  compare with the standard $[a, a^dagger] = 1$.






                  share|cite|improve this answer












                  Your second equation is the standard definition for $a$ ($a_-$) and $a^dagger$ ($a_+$) as found e.g. on Wikipedia. Note that the ladder operators here are dimensionless.



                  Griffiths is doing it slightly differently. I'm not sure why, but if you continue consistently with this alternative definition, you will get to the same correct results of course. For example, Griffiths (2.45) is
                  $$ [a_-, a_+] = hbaromega , $$
                  compare with the standard $[a, a^dagger] = 1$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered 1 hour ago









                  Noiralef

                  3,495926




                  3,495926




















                      up vote
                      1
                      down vote













                      In the first definition, $a_+$ and $a_-$ have been defined such that $[a_-,a_+]=hbaromega$. Here the position and momentum have been scaled as
                      $X=sqrtmx$ & $P=sqrtfrac1mp$ and $a_pm= frac1sqrt2(X pmiP)$.



                      Whereas, in the second definition, $a_+$ and $a_-$ have been defined such that $[a_-,a_+]=1$. Here the position and momentum have been scaled as
                      $X=sqrtfracmomegahbarx$ & $P=sqrtfrac1momegahbarp$ and $a_pm=frac1sqrt2(XpmiP)$.






                      share|cite|improve this answer










                      New contributor




                      Bhavesh Valecha is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                      Check out our Code of Conduct.





















                        up vote
                        1
                        down vote













                        In the first definition, $a_+$ and $a_-$ have been defined such that $[a_-,a_+]=hbaromega$. Here the position and momentum have been scaled as
                        $X=sqrtmx$ & $P=sqrtfrac1mp$ and $a_pm= frac1sqrt2(X pmiP)$.



                        Whereas, in the second definition, $a_+$ and $a_-$ have been defined such that $[a_-,a_+]=1$. Here the position and momentum have been scaled as
                        $X=sqrtfracmomegahbarx$ & $P=sqrtfrac1momegahbarp$ and $a_pm=frac1sqrt2(XpmiP)$.






                        share|cite|improve this answer










                        New contributor




                        Bhavesh Valecha is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                        Check out our Code of Conduct.



















                          up vote
                          1
                          down vote










                          up vote
                          1
                          down vote









                          In the first definition, $a_+$ and $a_-$ have been defined such that $[a_-,a_+]=hbaromega$. Here the position and momentum have been scaled as
                          $X=sqrtmx$ & $P=sqrtfrac1mp$ and $a_pm= frac1sqrt2(X pmiP)$.



                          Whereas, in the second definition, $a_+$ and $a_-$ have been defined such that $[a_-,a_+]=1$. Here the position and momentum have been scaled as
                          $X=sqrtfracmomegahbarx$ & $P=sqrtfrac1momegahbarp$ and $a_pm=frac1sqrt2(XpmiP)$.






                          share|cite|improve this answer










                          New contributor




                          Bhavesh Valecha is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                          Check out our Code of Conduct.









                          In the first definition, $a_+$ and $a_-$ have been defined such that $[a_-,a_+]=hbaromega$. Here the position and momentum have been scaled as
                          $X=sqrtmx$ & $P=sqrtfrac1mp$ and $a_pm= frac1sqrt2(X pmiP)$.



                          Whereas, in the second definition, $a_+$ and $a_-$ have been defined such that $[a_-,a_+]=1$. Here the position and momentum have been scaled as
                          $X=sqrtfracmomegahbarx$ & $P=sqrtfrac1momegahbarp$ and $a_pm=frac1sqrt2(XpmiP)$.







                          share|cite|improve this answer










                          New contributor




                          Bhavesh Valecha is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                          Check out our Code of Conduct.









                          share|cite|improve this answer



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                          edited 31 mins ago





















                          New contributor




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                          answered 1 hour ago









                          Bhavesh Valecha

                          112




                          112




                          New contributor




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                          New contributor





                          Bhavesh Valecha is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                          Check out our Code of Conduct.






                          Bhavesh Valecha is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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