In a spherically symmetric central potential why do we look for eigenfunctions of the angular momentum...












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In finding the solutions to the wave equation for a spherically symmetric potential $V(r)$, we look for the eigenfunctions of $hat L^2$ and $hat L_z$ operators. However, what is the reasoning behind this? The time-independent Schrodinger equation:
$$hat H Psi = E Psi$$



This is an eigenvalue equation and any eigenfunction of $hat H$ this equation is obviously a solution for this equation. Now when there is a spherically symmetric potential $V(r)$, I understand that because of separation of variables we can look for solutions of the form:
$$Psi (vec r,theta, phi) = R(vec r) Y(theta, phi)$$



Now to find the angular part why do we look for eigenstates of $hat L^2$ and $hat L_z$ operators (I know that they commute)?






quantum-mechanics schroedinger-equation spherical-harmonics







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    $begingroup$


    In finding the solutions to the wave equation for a spherically symmetric potential $V(r)$, we look for the eigenfunctions of $hat L^2$ and $hat L_z$ operators. However, what is the reasoning behind this? The time-independent Schrodinger equation:
    $$hat H Psi = E Psi$$



    This is an eigenvalue equation and any eigenfunction of $hat H$ this equation is obviously a solution for this equation. Now when there is a spherically symmetric potential $V(r)$, I understand that because of separation of variables we can look for solutions of the form:
    $$Psi (vec r,theta, phi) = R(vec r) Y(theta, phi)$$



    Now to find the angular part why do we look for eigenstates of $hat L^2$ and $hat L_z$ operators (I know that they commute)?






    quantum-mechanics schroedinger-equation spherical-harmonics







    share|cite|improve this question









    $endgroup$















      2












      2








      2





      $begingroup$


      In finding the solutions to the wave equation for a spherically symmetric potential $V(r)$, we look for the eigenfunctions of $hat L^2$ and $hat L_z$ operators. However, what is the reasoning behind this? The time-independent Schrodinger equation:
      $$hat H Psi = E Psi$$



      This is an eigenvalue equation and any eigenfunction of $hat H$ this equation is obviously a solution for this equation. Now when there is a spherically symmetric potential $V(r)$, I understand that because of separation of variables we can look for solutions of the form:
      $$Psi (vec r,theta, phi) = R(vec r) Y(theta, phi)$$



      Now to find the angular part why do we look for eigenstates of $hat L^2$ and $hat L_z$ operators (I know that they commute)?






      quantum-mechanics schroedinger-equation spherical-harmonics







      share|cite|improve this question









      $endgroup$




      In finding the solutions to the wave equation for a spherically symmetric potential $V(r)$, we look for the eigenfunctions of $hat L^2$ and $hat L_z$ operators. However, what is the reasoning behind this? The time-independent Schrodinger equation:
      $$hat H Psi = E Psi$$



      This is an eigenvalue equation and any eigenfunction of $hat H$ this equation is obviously a solution for this equation. Now when there is a spherically symmetric potential $V(r)$, I understand that because of separation of variables we can look for solutions of the form:
      $$Psi (vec r,theta, phi) = R(vec r) Y(theta, phi)$$



      Now to find the angular part why do we look for eigenstates of $hat L^2$ and $hat L_z$ operators (I know that they commute)?






      quantum-mechanics schroedinger-equation spherical-harmonics




      quantum-mechanics schroedinger-equation spherical-harmonics






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      asked 4 hours ago









      daljit97daljit97

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          $begingroup$

          Well, to put it shortly, it's just because $L^2$ and $L_z$ are two observables that have no $r$ dependence. Since the kinetic terms of $H$ can be written as functions of $r$ and $L^2$, and the potential depends only on $r$, it is clear that the Hamiltonian commutes with the angular momenta. As such, it makes sense to write down $Y$ in terms of these two.



          But if we'd like to go a little deeper, it is a bit related to representation theory. The fact that $L^2$ and $L_z$ form a complete description of vector rotations (they represent the Lie algebra $mathfrak{so}(3)$ of the Lie group $mathrm{SO}(3)$) and therefore are the most natural way to express the angular dependence of a rotation-invariant space in terms of orthonormalizable functions.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            and then how do we know that their eigenfunctions will be eigenfunctions of $hat H$?
            $endgroup$
            – daljit97
            4 hours ago










          • $begingroup$
            @daljit97 see the edit. Rotational invariance of the space (i.e. the potential) guarantees we can find eigenstates of $H$ in this manner. Or you could simply argue that, given that $L^2$ and $L_z$ do not depend on $r$, and $H$ has potential terms that depend only on $r$, the angular momenta have to commute with $H$ (they have a common eigenbasis)
            $endgroup$
            – Gabriel Golfetti
            4 hours ago






          • 1




            $begingroup$
            @daljit97 Commuting operators have simultaneous eigenstates
            $endgroup$
            – Aaron Stevens
            3 hours ago










          • $begingroup$
            @AaronStevens yes, that's true. However, that doesn't imply that any eigenstate of $L^2$ will be an eigenstate of $H$. Commuting operators have simultaneous states, but not the same eigenstates. Am I right?
            $endgroup$
            – daljit97
            3 hours ago






          • 1




            $begingroup$
            @dajit97 right, you may have an eigenstate of one that is not an eigenstate of the other because of degeneracy. but then you just change your basis and everything works out fine.
            $endgroup$
            – Gabriel Golfetti
            2 hours ago











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          1 Answer
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          1 Answer
          1






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          active

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          active

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          3












          $begingroup$

          Well, to put it shortly, it's just because $L^2$ and $L_z$ are two observables that have no $r$ dependence. Since the kinetic terms of $H$ can be written as functions of $r$ and $L^2$, and the potential depends only on $r$, it is clear that the Hamiltonian commutes with the angular momenta. As such, it makes sense to write down $Y$ in terms of these two.



          But if we'd like to go a little deeper, it is a bit related to representation theory. The fact that $L^2$ and $L_z$ form a complete description of vector rotations (they represent the Lie algebra $mathfrak{so}(3)$ of the Lie group $mathrm{SO}(3)$) and therefore are the most natural way to express the angular dependence of a rotation-invariant space in terms of orthonormalizable functions.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            and then how do we know that their eigenfunctions will be eigenfunctions of $hat H$?
            $endgroup$
            – daljit97
            4 hours ago










          • $begingroup$
            @daljit97 see the edit. Rotational invariance of the space (i.e. the potential) guarantees we can find eigenstates of $H$ in this manner. Or you could simply argue that, given that $L^2$ and $L_z$ do not depend on $r$, and $H$ has potential terms that depend only on $r$, the angular momenta have to commute with $H$ (they have a common eigenbasis)
            $endgroup$
            – Gabriel Golfetti
            4 hours ago






          • 1




            $begingroup$
            @daljit97 Commuting operators have simultaneous eigenstates
            $endgroup$
            – Aaron Stevens
            3 hours ago










          • $begingroup$
            @AaronStevens yes, that's true. However, that doesn't imply that any eigenstate of $L^2$ will be an eigenstate of $H$. Commuting operators have simultaneous states, but not the same eigenstates. Am I right?
            $endgroup$
            – daljit97
            3 hours ago






          • 1




            $begingroup$
            @dajit97 right, you may have an eigenstate of one that is not an eigenstate of the other because of degeneracy. but then you just change your basis and everything works out fine.
            $endgroup$
            – Gabriel Golfetti
            2 hours ago
















          3












          $begingroup$

          Well, to put it shortly, it's just because $L^2$ and $L_z$ are two observables that have no $r$ dependence. Since the kinetic terms of $H$ can be written as functions of $r$ and $L^2$, and the potential depends only on $r$, it is clear that the Hamiltonian commutes with the angular momenta. As such, it makes sense to write down $Y$ in terms of these two.



          But if we'd like to go a little deeper, it is a bit related to representation theory. The fact that $L^2$ and $L_z$ form a complete description of vector rotations (they represent the Lie algebra $mathfrak{so}(3)$ of the Lie group $mathrm{SO}(3)$) and therefore are the most natural way to express the angular dependence of a rotation-invariant space in terms of orthonormalizable functions.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            and then how do we know that their eigenfunctions will be eigenfunctions of $hat H$?
            $endgroup$
            – daljit97
            4 hours ago










          • $begingroup$
            @daljit97 see the edit. Rotational invariance of the space (i.e. the potential) guarantees we can find eigenstates of $H$ in this manner. Or you could simply argue that, given that $L^2$ and $L_z$ do not depend on $r$, and $H$ has potential terms that depend only on $r$, the angular momenta have to commute with $H$ (they have a common eigenbasis)
            $endgroup$
            – Gabriel Golfetti
            4 hours ago






          • 1




            $begingroup$
            @daljit97 Commuting operators have simultaneous eigenstates
            $endgroup$
            – Aaron Stevens
            3 hours ago










          • $begingroup$
            @AaronStevens yes, that's true. However, that doesn't imply that any eigenstate of $L^2$ will be an eigenstate of $H$. Commuting operators have simultaneous states, but not the same eigenstates. Am I right?
            $endgroup$
            – daljit97
            3 hours ago






          • 1




            $begingroup$
            @dajit97 right, you may have an eigenstate of one that is not an eigenstate of the other because of degeneracy. but then you just change your basis and everything works out fine.
            $endgroup$
            – Gabriel Golfetti
            2 hours ago














          3












          3








          3





          $begingroup$

          Well, to put it shortly, it's just because $L^2$ and $L_z$ are two observables that have no $r$ dependence. Since the kinetic terms of $H$ can be written as functions of $r$ and $L^2$, and the potential depends only on $r$, it is clear that the Hamiltonian commutes with the angular momenta. As such, it makes sense to write down $Y$ in terms of these two.



          But if we'd like to go a little deeper, it is a bit related to representation theory. The fact that $L^2$ and $L_z$ form a complete description of vector rotations (they represent the Lie algebra $mathfrak{so}(3)$ of the Lie group $mathrm{SO}(3)$) and therefore are the most natural way to express the angular dependence of a rotation-invariant space in terms of orthonormalizable functions.






          share|cite|improve this answer











          $endgroup$



          Well, to put it shortly, it's just because $L^2$ and $L_z$ are two observables that have no $r$ dependence. Since the kinetic terms of $H$ can be written as functions of $r$ and $L^2$, and the potential depends only on $r$, it is clear that the Hamiltonian commutes with the angular momenta. As such, it makes sense to write down $Y$ in terms of these two.



          But if we'd like to go a little deeper, it is a bit related to representation theory. The fact that $L^2$ and $L_z$ form a complete description of vector rotations (they represent the Lie algebra $mathfrak{so}(3)$ of the Lie group $mathrm{SO}(3)$) and therefore are the most natural way to express the angular dependence of a rotation-invariant space in terms of orthonormalizable functions.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited 4 hours ago

























          answered 4 hours ago









          Gabriel GolfettiGabriel Golfetti

          1,2671713




          1,2671713












          • $begingroup$
            and then how do we know that their eigenfunctions will be eigenfunctions of $hat H$?
            $endgroup$
            – daljit97
            4 hours ago










          • $begingroup$
            @daljit97 see the edit. Rotational invariance of the space (i.e. the potential) guarantees we can find eigenstates of $H$ in this manner. Or you could simply argue that, given that $L^2$ and $L_z$ do not depend on $r$, and $H$ has potential terms that depend only on $r$, the angular momenta have to commute with $H$ (they have a common eigenbasis)
            $endgroup$
            – Gabriel Golfetti
            4 hours ago






          • 1




            $begingroup$
            @daljit97 Commuting operators have simultaneous eigenstates
            $endgroup$
            – Aaron Stevens
            3 hours ago










          • $begingroup$
            @AaronStevens yes, that's true. However, that doesn't imply that any eigenstate of $L^2$ will be an eigenstate of $H$. Commuting operators have simultaneous states, but not the same eigenstates. Am I right?
            $endgroup$
            – daljit97
            3 hours ago






          • 1




            $begingroup$
            @dajit97 right, you may have an eigenstate of one that is not an eigenstate of the other because of degeneracy. but then you just change your basis and everything works out fine.
            $endgroup$
            – Gabriel Golfetti
            2 hours ago


















          • $begingroup$
            and then how do we know that their eigenfunctions will be eigenfunctions of $hat H$?
            $endgroup$
            – daljit97
            4 hours ago










          • $begingroup$
            @daljit97 see the edit. Rotational invariance of the space (i.e. the potential) guarantees we can find eigenstates of $H$ in this manner. Or you could simply argue that, given that $L^2$ and $L_z$ do not depend on $r$, and $H$ has potential terms that depend only on $r$, the angular momenta have to commute with $H$ (they have a common eigenbasis)
            $endgroup$
            – Gabriel Golfetti
            4 hours ago






          • 1




            $begingroup$
            @daljit97 Commuting operators have simultaneous eigenstates
            $endgroup$
            – Aaron Stevens
            3 hours ago










          • $begingroup$
            @AaronStevens yes, that's true. However, that doesn't imply that any eigenstate of $L^2$ will be an eigenstate of $H$. Commuting operators have simultaneous states, but not the same eigenstates. Am I right?
            $endgroup$
            – daljit97
            3 hours ago






          • 1




            $begingroup$
            @dajit97 right, you may have an eigenstate of one that is not an eigenstate of the other because of degeneracy. but then you just change your basis and everything works out fine.
            $endgroup$
            – Gabriel Golfetti
            2 hours ago
















          $begingroup$
          and then how do we know that their eigenfunctions will be eigenfunctions of $hat H$?
          $endgroup$
          – daljit97
          4 hours ago




          $begingroup$
          and then how do we know that their eigenfunctions will be eigenfunctions of $hat H$?
          $endgroup$
          – daljit97
          4 hours ago












          $begingroup$
          @daljit97 see the edit. Rotational invariance of the space (i.e. the potential) guarantees we can find eigenstates of $H$ in this manner. Or you could simply argue that, given that $L^2$ and $L_z$ do not depend on $r$, and $H$ has potential terms that depend only on $r$, the angular momenta have to commute with $H$ (they have a common eigenbasis)
          $endgroup$
          – Gabriel Golfetti
          4 hours ago




          $begingroup$
          @daljit97 see the edit. Rotational invariance of the space (i.e. the potential) guarantees we can find eigenstates of $H$ in this manner. Or you could simply argue that, given that $L^2$ and $L_z$ do not depend on $r$, and $H$ has potential terms that depend only on $r$, the angular momenta have to commute with $H$ (they have a common eigenbasis)
          $endgroup$
          – Gabriel Golfetti
          4 hours ago




          1




          1




          $begingroup$
          @daljit97 Commuting operators have simultaneous eigenstates
          $endgroup$
          – Aaron Stevens
          3 hours ago




          $begingroup$
          @daljit97 Commuting operators have simultaneous eigenstates
          $endgroup$
          – Aaron Stevens
          3 hours ago












          $begingroup$
          @AaronStevens yes, that's true. However, that doesn't imply that any eigenstate of $L^2$ will be an eigenstate of $H$. Commuting operators have simultaneous states, but not the same eigenstates. Am I right?
          $endgroup$
          – daljit97
          3 hours ago




          $begingroup$
          @AaronStevens yes, that's true. However, that doesn't imply that any eigenstate of $L^2$ will be an eigenstate of $H$. Commuting operators have simultaneous states, but not the same eigenstates. Am I right?
          $endgroup$
          – daljit97
          3 hours ago




          1




          1




          $begingroup$
          @dajit97 right, you may have an eigenstate of one that is not an eigenstate of the other because of degeneracy. but then you just change your basis and everything works out fine.
          $endgroup$
          – Gabriel Golfetti
          2 hours ago




          $begingroup$
          @dajit97 right, you may have an eigenstate of one that is not an eigenstate of the other because of degeneracy. but then you just change your basis and everything works out fine.
          $endgroup$
          – Gabriel Golfetti
          2 hours ago


















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