Zero uncertainity in components of angular momentum in Hydrogen atom












6














It is given that L and Lz,Lx,Ly commute.(L is total angular momentum, Lx is angular momentum along x axis).
So, I can simultaneously know the value of let's say L and Lz. But, if I perform huge no of measurements and in a certain measurement, I get the value of L = Lz, then I know for certain that Lx and Ly are 0.
But, according to the uncertainty principle, I can't know the exact values of any two of Lx, Ly and Lz.
So, where did I go wrong?






quantum-mechanics angular-momentum atomic-physics quantum-spin heisenberg-uncertainty-principle







share|cite|improve this question



























    6














    It is given that L and Lz,Lx,Ly commute.(L is total angular momentum, Lx is angular momentum along x axis).
    So, I can simultaneously know the value of let's say L and Lz. But, if I perform huge no of measurements and in a certain measurement, I get the value of L = Lz, then I know for certain that Lx and Ly are 0.
    But, according to the uncertainty principle, I can't know the exact values of any two of Lx, Ly and Lz.
    So, where did I go wrong?






    quantum-mechanics angular-momentum atomic-physics quantum-spin heisenberg-uncertainty-principle







    share|cite|improve this question

























      6












      6








      6


      1





      It is given that L and Lz,Lx,Ly commute.(L is total angular momentum, Lx is angular momentum along x axis).
      So, I can simultaneously know the value of let's say L and Lz. But, if I perform huge no of measurements and in a certain measurement, I get the value of L = Lz, then I know for certain that Lx and Ly are 0.
      But, according to the uncertainty principle, I can't know the exact values of any two of Lx, Ly and Lz.
      So, where did I go wrong?






      quantum-mechanics angular-momentum atomic-physics quantum-spin heisenberg-uncertainty-principle







      share|cite|improve this question













      It is given that L and Lz,Lx,Ly commute.(L is total angular momentum, Lx is angular momentum along x axis).
      So, I can simultaneously know the value of let's say L and Lz. But, if I perform huge no of measurements and in a certain measurement, I get the value of L = Lz, then I know for certain that Lx and Ly are 0.
      But, according to the uncertainty principle, I can't know the exact values of any two of Lx, Ly and Lz.
      So, where did I go wrong?






      quantum-mechanics angular-momentum atomic-physics quantum-spin heisenberg-uncertainty-principle




      quantum-mechanics angular-momentum atomic-physics quantum-spin heisenberg-uncertainty-principle






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Nov 24 at 13:47









      Astik

      405




      405






















          3 Answers
          3






          active

          oldest

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          8














          You can't get



          $$leftlVertvec LrightrVert = L_z $$



          for non-zero $l$, since:



          $$leftlVertvec LrightrVert = hbarsqrt{l(l+1)} $$



          while the maximum value of $L_z$ is



          $$ (L_z)_{mathrm{max}} = hbar l $$



          Also: The maximal state is:



          $$ Y_l^l(theta, phi) propto sin^l{theta}e^{ilphi} $$



          which is not an eigenstate of $L_x$, nor $L_z$.






          share|cite|improve this answer































            2














            JEB's answer is correct: you can't do a huge number of measurements so that out of luck in one of them you will find



            $$||vec L|| = L_z$$



            You could try another way: since the reference frame is arbitrary, you could just choose it having the z-axis parallel to the angular momentum vector. This won't work either, because for doing so you would have to know where the angular momentum vector is pointing, and this would require simultaneous knowledge of its three components.






            share|cite|improve this answer

















            • 1




              This answer is wrong, since there is no single value of the vector $L$. The state is a superposition of various eigenstates, none of which has a value of $L$. Indeed, since definition of a measurable property is one for which there exists an operator and eigenstates, and there are no eigenstates of $L$, this is not a property a quantum system can have.
              – Eddy
              Nov 24 at 17:08






            • 1




              @Eddy what do you mean there is no single value of $L$? As JEB has pointed out, spherical harmonics are eigenfunctions of $L^2$ with eigenvalue $hbar l(l+1)$, so you can measure $L$. What you cannot do is measure simultaneously its three components, since they not commute and therefore do not share simultaneous eigenstates.
              – TeneT
              Nov 24 at 17:22








            • 1




              As I stated and you also just stated, there is no single value of the vector $L$. However, in your answer, you propose setting up a coordinate system parallel to this vector, which can't be done because it cannot have a well defined value for any state.
              – Eddy
              Nov 24 at 17:25






            • 1




              There is a measurable value for the lenght of the vector $L$. That which you cannot measure is its orientation, and the impossibility of setting up such a coordinate system is precisely my point.
              – TeneT
              Nov 24 at 17:36












            • Guys, I get that mathematically the value of $L_z$ can never attain the value of $L$, but what I don't understand is what would go wrong if we got the same value for $L_z and L$. I mean will any principle of QM be violated by it?
              – Astik
              Dec 1 at 18:26





















            0














            To add to the others,



            L does not commute with $L_x$, $L^2$ does. Also, one is a scalar operator, the other is a vector.



            Angular momentum is the product of position vector and momentum vector. So this scenario can serve as prototypical example of uncertainty principle arising in a single entity.



            To elaborate on above, if you randomly pick a unit vector an infinite number of times, you'll never match the direction of the Angular momentum.






            share|cite|improve this answer





















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              3 Answers
              3






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              3 Answers
              3






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              8














              You can't get



              $$leftlVertvec LrightrVert = L_z $$



              for non-zero $l$, since:



              $$leftlVertvec LrightrVert = hbarsqrt{l(l+1)} $$



              while the maximum value of $L_z$ is



              $$ (L_z)_{mathrm{max}} = hbar l $$



              Also: The maximal state is:



              $$ Y_l^l(theta, phi) propto sin^l{theta}e^{ilphi} $$



              which is not an eigenstate of $L_x$, nor $L_z$.






              share|cite|improve this answer




























                8














                You can't get



                $$leftlVertvec LrightrVert = L_z $$



                for non-zero $l$, since:



                $$leftlVertvec LrightrVert = hbarsqrt{l(l+1)} $$



                while the maximum value of $L_z$ is



                $$ (L_z)_{mathrm{max}} = hbar l $$



                Also: The maximal state is:



                $$ Y_l^l(theta, phi) propto sin^l{theta}e^{ilphi} $$



                which is not an eigenstate of $L_x$, nor $L_z$.






                share|cite|improve this answer


























                  8












                  8








                  8






                  You can't get



                  $$leftlVertvec LrightrVert = L_z $$



                  for non-zero $l$, since:



                  $$leftlVertvec LrightrVert = hbarsqrt{l(l+1)} $$



                  while the maximum value of $L_z$ is



                  $$ (L_z)_{mathrm{max}} = hbar l $$



                  Also: The maximal state is:



                  $$ Y_l^l(theta, phi) propto sin^l{theta}e^{ilphi} $$



                  which is not an eigenstate of $L_x$, nor $L_z$.






                  share|cite|improve this answer














                  You can't get



                  $$leftlVertvec LrightrVert = L_z $$



                  for non-zero $l$, since:



                  $$leftlVertvec LrightrVert = hbarsqrt{l(l+1)} $$



                  while the maximum value of $L_z$ is



                  $$ (L_z)_{mathrm{max}} = hbar l $$



                  Also: The maximal state is:



                  $$ Y_l^l(theta, phi) propto sin^l{theta}e^{ilphi} $$



                  which is not an eigenstate of $L_x$, nor $L_z$.







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Nov 24 at 18:43









                  Ruslan

                  9,20143070




                  9,20143070










                  answered Nov 24 at 15:06









                  JEB

                  5,6431717




                  5,6431717























                      2














                      JEB's answer is correct: you can't do a huge number of measurements so that out of luck in one of them you will find



                      $$||vec L|| = L_z$$



                      You could try another way: since the reference frame is arbitrary, you could just choose it having the z-axis parallel to the angular momentum vector. This won't work either, because for doing so you would have to know where the angular momentum vector is pointing, and this would require simultaneous knowledge of its three components.






                      share|cite|improve this answer

















                      • 1




                        This answer is wrong, since there is no single value of the vector $L$. The state is a superposition of various eigenstates, none of which has a value of $L$. Indeed, since definition of a measurable property is one for which there exists an operator and eigenstates, and there are no eigenstates of $L$, this is not a property a quantum system can have.
                        – Eddy
                        Nov 24 at 17:08






                      • 1




                        @Eddy what do you mean there is no single value of $L$? As JEB has pointed out, spherical harmonics are eigenfunctions of $L^2$ with eigenvalue $hbar l(l+1)$, so you can measure $L$. What you cannot do is measure simultaneously its three components, since they not commute and therefore do not share simultaneous eigenstates.
                        – TeneT
                        Nov 24 at 17:22








                      • 1




                        As I stated and you also just stated, there is no single value of the vector $L$. However, in your answer, you propose setting up a coordinate system parallel to this vector, which can't be done because it cannot have a well defined value for any state.
                        – Eddy
                        Nov 24 at 17:25






                      • 1




                        There is a measurable value for the lenght of the vector $L$. That which you cannot measure is its orientation, and the impossibility of setting up such a coordinate system is precisely my point.
                        – TeneT
                        Nov 24 at 17:36












                      • Guys, I get that mathematically the value of $L_z$ can never attain the value of $L$, but what I don't understand is what would go wrong if we got the same value for $L_z and L$. I mean will any principle of QM be violated by it?
                        – Astik
                        Dec 1 at 18:26


















                      2














                      JEB's answer is correct: you can't do a huge number of measurements so that out of luck in one of them you will find



                      $$||vec L|| = L_z$$



                      You could try another way: since the reference frame is arbitrary, you could just choose it having the z-axis parallel to the angular momentum vector. This won't work either, because for doing so you would have to know where the angular momentum vector is pointing, and this would require simultaneous knowledge of its three components.






                      share|cite|improve this answer

















                      • 1




                        This answer is wrong, since there is no single value of the vector $L$. The state is a superposition of various eigenstates, none of which has a value of $L$. Indeed, since definition of a measurable property is one for which there exists an operator and eigenstates, and there are no eigenstates of $L$, this is not a property a quantum system can have.
                        – Eddy
                        Nov 24 at 17:08






                      • 1




                        @Eddy what do you mean there is no single value of $L$? As JEB has pointed out, spherical harmonics are eigenfunctions of $L^2$ with eigenvalue $hbar l(l+1)$, so you can measure $L$. What you cannot do is measure simultaneously its three components, since they not commute and therefore do not share simultaneous eigenstates.
                        – TeneT
                        Nov 24 at 17:22








                      • 1




                        As I stated and you also just stated, there is no single value of the vector $L$. However, in your answer, you propose setting up a coordinate system parallel to this vector, which can't be done because it cannot have a well defined value for any state.
                        – Eddy
                        Nov 24 at 17:25






                      • 1




                        There is a measurable value for the lenght of the vector $L$. That which you cannot measure is its orientation, and the impossibility of setting up such a coordinate system is precisely my point.
                        – TeneT
                        Nov 24 at 17:36












                      • Guys, I get that mathematically the value of $L_z$ can never attain the value of $L$, but what I don't understand is what would go wrong if we got the same value for $L_z and L$. I mean will any principle of QM be violated by it?
                        – Astik
                        Dec 1 at 18:26
















                      2












                      2








                      2






                      JEB's answer is correct: you can't do a huge number of measurements so that out of luck in one of them you will find



                      $$||vec L|| = L_z$$



                      You could try another way: since the reference frame is arbitrary, you could just choose it having the z-axis parallel to the angular momentum vector. This won't work either, because for doing so you would have to know where the angular momentum vector is pointing, and this would require simultaneous knowledge of its three components.






                      share|cite|improve this answer












                      JEB's answer is correct: you can't do a huge number of measurements so that out of luck in one of them you will find



                      $$||vec L|| = L_z$$



                      You could try another way: since the reference frame is arbitrary, you could just choose it having the z-axis parallel to the angular momentum vector. This won't work either, because for doing so you would have to know where the angular momentum vector is pointing, and this would require simultaneous knowledge of its three components.







                      share|cite|improve this answer












                      share|cite|improve this answer



                      share|cite|improve this answer










                      answered Nov 24 at 15:28









                      TeneT

                      236




                      236








                      • 1




                        This answer is wrong, since there is no single value of the vector $L$. The state is a superposition of various eigenstates, none of which has a value of $L$. Indeed, since definition of a measurable property is one for which there exists an operator and eigenstates, and there are no eigenstates of $L$, this is not a property a quantum system can have.
                        – Eddy
                        Nov 24 at 17:08






                      • 1




                        @Eddy what do you mean there is no single value of $L$? As JEB has pointed out, spherical harmonics are eigenfunctions of $L^2$ with eigenvalue $hbar l(l+1)$, so you can measure $L$. What you cannot do is measure simultaneously its three components, since they not commute and therefore do not share simultaneous eigenstates.
                        – TeneT
                        Nov 24 at 17:22








                      • 1




                        As I stated and you also just stated, there is no single value of the vector $L$. However, in your answer, you propose setting up a coordinate system parallel to this vector, which can't be done because it cannot have a well defined value for any state.
                        – Eddy
                        Nov 24 at 17:25






                      • 1




                        There is a measurable value for the lenght of the vector $L$. That which you cannot measure is its orientation, and the impossibility of setting up such a coordinate system is precisely my point.
                        – TeneT
                        Nov 24 at 17:36












                      • Guys, I get that mathematically the value of $L_z$ can never attain the value of $L$, but what I don't understand is what would go wrong if we got the same value for $L_z and L$. I mean will any principle of QM be violated by it?
                        – Astik
                        Dec 1 at 18:26
















                      • 1




                        This answer is wrong, since there is no single value of the vector $L$. The state is a superposition of various eigenstates, none of which has a value of $L$. Indeed, since definition of a measurable property is one for which there exists an operator and eigenstates, and there are no eigenstates of $L$, this is not a property a quantum system can have.
                        – Eddy
                        Nov 24 at 17:08






                      • 1




                        @Eddy what do you mean there is no single value of $L$? As JEB has pointed out, spherical harmonics are eigenfunctions of $L^2$ with eigenvalue $hbar l(l+1)$, so you can measure $L$. What you cannot do is measure simultaneously its three components, since they not commute and therefore do not share simultaneous eigenstates.
                        – TeneT
                        Nov 24 at 17:22








                      • 1




                        As I stated and you also just stated, there is no single value of the vector $L$. However, in your answer, you propose setting up a coordinate system parallel to this vector, which can't be done because it cannot have a well defined value for any state.
                        – Eddy
                        Nov 24 at 17:25






                      • 1




                        There is a measurable value for the lenght of the vector $L$. That which you cannot measure is its orientation, and the impossibility of setting up such a coordinate system is precisely my point.
                        – TeneT
                        Nov 24 at 17:36












                      • Guys, I get that mathematically the value of $L_z$ can never attain the value of $L$, but what I don't understand is what would go wrong if we got the same value for $L_z and L$. I mean will any principle of QM be violated by it?
                        – Astik
                        Dec 1 at 18:26










                      1




                      1




                      This answer is wrong, since there is no single value of the vector $L$. The state is a superposition of various eigenstates, none of which has a value of $L$. Indeed, since definition of a measurable property is one for which there exists an operator and eigenstates, and there are no eigenstates of $L$, this is not a property a quantum system can have.
                      – Eddy
                      Nov 24 at 17:08




                      This answer is wrong, since there is no single value of the vector $L$. The state is a superposition of various eigenstates, none of which has a value of $L$. Indeed, since definition of a measurable property is one for which there exists an operator and eigenstates, and there are no eigenstates of $L$, this is not a property a quantum system can have.
                      – Eddy
                      Nov 24 at 17:08




                      1




                      1




                      @Eddy what do you mean there is no single value of $L$? As JEB has pointed out, spherical harmonics are eigenfunctions of $L^2$ with eigenvalue $hbar l(l+1)$, so you can measure $L$. What you cannot do is measure simultaneously its three components, since they not commute and therefore do not share simultaneous eigenstates.
                      – TeneT
                      Nov 24 at 17:22






                      @Eddy what do you mean there is no single value of $L$? As JEB has pointed out, spherical harmonics are eigenfunctions of $L^2$ with eigenvalue $hbar l(l+1)$, so you can measure $L$. What you cannot do is measure simultaneously its three components, since they not commute and therefore do not share simultaneous eigenstates.
                      – TeneT
                      Nov 24 at 17:22






                      1




                      1




                      As I stated and you also just stated, there is no single value of the vector $L$. However, in your answer, you propose setting up a coordinate system parallel to this vector, which can't be done because it cannot have a well defined value for any state.
                      – Eddy
                      Nov 24 at 17:25




                      As I stated and you also just stated, there is no single value of the vector $L$. However, in your answer, you propose setting up a coordinate system parallel to this vector, which can't be done because it cannot have a well defined value for any state.
                      – Eddy
                      Nov 24 at 17:25




                      1




                      1




                      There is a measurable value for the lenght of the vector $L$. That which you cannot measure is its orientation, and the impossibility of setting up such a coordinate system is precisely my point.
                      – TeneT
                      Nov 24 at 17:36






                      There is a measurable value for the lenght of the vector $L$. That which you cannot measure is its orientation, and the impossibility of setting up such a coordinate system is precisely my point.
                      – TeneT
                      Nov 24 at 17:36














                      Guys, I get that mathematically the value of $L_z$ can never attain the value of $L$, but what I don't understand is what would go wrong if we got the same value for $L_z and L$. I mean will any principle of QM be violated by it?
                      – Astik
                      Dec 1 at 18:26






                      Guys, I get that mathematically the value of $L_z$ can never attain the value of $L$, but what I don't understand is what would go wrong if we got the same value for $L_z and L$. I mean will any principle of QM be violated by it?
                      – Astik
                      Dec 1 at 18:26













                      0














                      To add to the others,



                      L does not commute with $L_x$, $L^2$ does. Also, one is a scalar operator, the other is a vector.



                      Angular momentum is the product of position vector and momentum vector. So this scenario can serve as prototypical example of uncertainty principle arising in a single entity.



                      To elaborate on above, if you randomly pick a unit vector an infinite number of times, you'll never match the direction of the Angular momentum.






                      share|cite|improve this answer


























                        0














                        To add to the others,



                        L does not commute with $L_x$, $L^2$ does. Also, one is a scalar operator, the other is a vector.



                        Angular momentum is the product of position vector and momentum vector. So this scenario can serve as prototypical example of uncertainty principle arising in a single entity.



                        To elaborate on above, if you randomly pick a unit vector an infinite number of times, you'll never match the direction of the Angular momentum.






                        share|cite|improve this answer
























                          0












                          0








                          0






                          To add to the others,



                          L does not commute with $L_x$, $L^2$ does. Also, one is a scalar operator, the other is a vector.



                          Angular momentum is the product of position vector and momentum vector. So this scenario can serve as prototypical example of uncertainty principle arising in a single entity.



                          To elaborate on above, if you randomly pick a unit vector an infinite number of times, you'll never match the direction of the Angular momentum.






                          share|cite|improve this answer












                          To add to the others,



                          L does not commute with $L_x$, $L^2$ does. Also, one is a scalar operator, the other is a vector.



                          Angular momentum is the product of position vector and momentum vector. So this scenario can serve as prototypical example of uncertainty principle arising in a single entity.



                          To elaborate on above, if you randomly pick a unit vector an infinite number of times, you'll never match the direction of the Angular momentum.







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Nov 24 at 18:28









                          R. Romero

                          3036




                          3036






























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                              Musa

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                              Para el género de plantas, véase Musa (planta). Apolo y la Nueve Musas En la mitología griega, las musas (en griego antiguo μοῦσαι « mousai» ) son, según los escritores más antiguos, las divinidades inspiradoras de las Artes: cada una de ellas está relacionada con ramas artísticas y del conocimiento. Son hijas de Zeus y de Mnemósine, compañeras del séquito de Apolo, dios olímpico de la música y patrón de las bellas artes, quien tuvo romances con cada una de ellas, dejando descendientes. Bajaban a la tierra a susurrar ideas e inspirar a aquellos mortales que las invocaran. En la época más arcaica eran las ninfas inspiradoras de las fuentes, en las cuales eran adoradas. Finalmente, alrededor de los siglos VIII-VII a. C. [ 1 ] ​ prevaleció en todo el territorio de la Hélade la adoración de las nueve Musas, que son Calíope, Clío, Erato, Euterpe, Melpómene, Polimnia, Talía, Terpsícore y Urania. El culto a las musas era originalmente de Tracia y Beocia, y fueron de vital impo
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