Zero uncertainity in components of angular momentum in Hydrogen atom

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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?










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    up vote
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    favorite
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    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?










    share|cite|improve this question























      up vote
      6
      down vote

      favorite
      1









      up vote
      6
      down vote

      favorite
      1






      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?










      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






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      asked Nov 24 at 13:47









      Astik

      384




      384




















          3 Answers
          3






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          You can't get



          $$leftlVertvec LrightrVert = L_z $$



          for non-zero $l$, since:



          $$leftlVertvec LrightrVert = hbarsqrtl(l+1) $$



          while the maximum value of $L_z$ is



          $$ (L_z)_mathrmmax = hbar l $$



          Also: The maximal state is:



          $$ Y_l^l(theta, phi) propto sin^lthetae^ilphi $$



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






          share|cite|improve this answer





























            up vote
            2
            down vote













            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
              6 hours ago


















            up vote
            0
            down vote













            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




















              Your Answer





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              3 Answers
              3






              active

              oldest

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              3 Answers
              3






              active

              oldest

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              active

              oldest

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              active

              oldest

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              up vote
              8
              down vote













              You can't get



              $$leftlVertvec LrightrVert = L_z $$



              for non-zero $l$, since:



              $$leftlVertvec LrightrVert = hbarsqrtl(l+1) $$



              while the maximum value of $L_z$ is



              $$ (L_z)_mathrmmax = hbar l $$



              Also: The maximal state is:



              $$ Y_l^l(theta, phi) propto sin^lthetae^ilphi $$



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






              share|cite|improve this answer


























                up vote
                8
                down vote













                You can't get



                $$leftlVertvec LrightrVert = L_z $$



                for non-zero $l$, since:



                $$leftlVertvec LrightrVert = hbarsqrtl(l+1) $$



                while the maximum value of $L_z$ is



                $$ (L_z)_mathrmmax = hbar l $$



                Also: The maximal state is:



                $$ Y_l^l(theta, phi) propto sin^lthetae^ilphi $$



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






                share|cite|improve this answer
























                  up vote
                  8
                  down vote










                  up vote
                  8
                  down vote









                  You can't get



                  $$leftlVertvec LrightrVert = L_z $$



                  for non-zero $l$, since:



                  $$leftlVertvec LrightrVert = hbarsqrtl(l+1) $$



                  while the maximum value of $L_z$ is



                  $$ (L_z)_mathrmmax = hbar l $$



                  Also: The maximal state is:



                  $$ Y_l^l(theta, phi) propto sin^lthetae^ilphi $$



                  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:



                  $$leftlVertvec LrightrVert = hbarsqrtl(l+1) $$



                  while the maximum value of $L_z$ is



                  $$ (L_z)_mathrmmax = hbar l $$



                  Also: The maximal state is:



                  $$ Y_l^l(theta, phi) propto sin^lthetae^ilphi $$



                  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

                  8,24342864




                  8,24342864










                  answered Nov 24 at 15:06









                  JEB

                  5,4321717




                  5,4321717




















                      up vote
                      2
                      down vote













                      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
                        6 hours ago















                      up vote
                      2
                      down vote













                      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
                        6 hours ago













                      up vote
                      2
                      down vote










                      up vote
                      2
                      down vote









                      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

                      215




                      215







                      • 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
                        6 hours ago













                      • 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
                        6 hours ago








                      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
                      6 hours ago





                      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
                      6 hours ago











                      up vote
                      0
                      down vote













                      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
























                        up vote
                        0
                        down vote













                        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






















                          up vote
                          0
                          down vote










                          up vote
                          0
                          down vote









                          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

                          2536




                          2536



























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