Zero uncertainity in components of angular momentum in Hydrogen atom
$begingroup$
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
asked Nov 24 '18 at 13:47
AstikAstik
405
$endgroup$
add a comment |
$begingroup$
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
asked Nov 24 '18 at 13:47
AstikAstik
405
$endgroup$
add a comment |
$begingroup$
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
asked Nov 24 '18 at 13:47
AstikAstik
405
$endgroup$
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
asked Nov 24 '18 at 13:47
AstikAstik
405
asked Nov 24 '18 at 13:47
AstikAstik
405
asked Nov 24 '18 at 13:47
AstikAstik
405
asked Nov 24 '18 at 13:47
AstikAstik
405
asked Nov 24 '18 at 13:47
AstikAstik
405
405
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
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 $L_z$.
edited Nov 24 '18 at 18:43
Ruslan
9,22643070
answered Nov 24 '18 at 15:06
JEBJEB
5,8031717
$endgroup$
add a comment |
$begingroup$
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.
answered Nov 24 '18 at 15:28
TeneTTeneT
236
$endgroup$
1
$begingroup$
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.
$endgroup$
– Eddy
Nov 24 '18 at 17:08
1
$begingroup$
@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.
$endgroup$
– TeneT
Nov 24 '18 at 17:22
1
$begingroup$
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.
$endgroup$
– Eddy
Nov 24 '18 at 17:25
1
$begingroup$
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.
$endgroup$
– TeneT
Nov 24 '18 at 17:36
$begingroup$
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?
$endgroup$
– Astik
Dec 1 '18 at 18:26
|
show 1 more comment
$begingroup$
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.
answered Nov 24 '18 at 18:28
R. RomeroR. Romero
3136
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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 $L_z$.
edited Nov 24 '18 at 18:43
Ruslan
9,22643070
answered Nov 24 '18 at 15:06
JEBJEB
5,8031717
$endgroup$
add a comment |
$begingroup$
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 $L_z$.
edited Nov 24 '18 at 18:43
Ruslan
9,22643070
answered Nov 24 '18 at 15:06
JEBJEB
5,8031717
$endgroup$
add a comment |
$begingroup$
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 $L_z$.
edited Nov 24 '18 at 18:43
Ruslan
9,22643070
answered Nov 24 '18 at 15:06
JEBJEB
5,8031717
$endgroup$
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 $L_z$.
edited Nov 24 '18 at 18:43
Ruslan
9,22643070
answered Nov 24 '18 at 15:06
JEBJEB
5,8031717
edited Nov 24 '18 at 18:43
Ruslan
9,22643070
edited Nov 24 '18 at 18:43
Ruslan
9,22643070
edited Nov 24 '18 at 18:43
Ruslan
9,22643070
9,22643070
answered Nov 24 '18 at 15:06
JEBJEB
5,8031717
answered Nov 24 '18 at 15:06
JEBJEB
5,8031717
answered Nov 24 '18 at 15:06
JEBJEB
5,8031717
5,8031717
add a comment |
add a comment |
$begingroup$
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.
answered Nov 24 '18 at 15:28
TeneTTeneT
236
$endgroup$
1
$begingroup$
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.
$endgroup$
– Eddy
Nov 24 '18 at 17:08
1
$begingroup$
@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.
$endgroup$
– TeneT
Nov 24 '18 at 17:22
1
$begingroup$
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.
$endgroup$
– Eddy
Nov 24 '18 at 17:25
1
$begingroup$
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.
$endgroup$
– TeneT
Nov 24 '18 at 17:36
$begingroup$
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?
$endgroup$
– Astik
Dec 1 '18 at 18:26
|
show 1 more comment
$begingroup$
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.
answered Nov 24 '18 at 15:28
TeneTTeneT
236
$endgroup$
1
$begingroup$
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.
$endgroup$
– Eddy
Nov 24 '18 at 17:08
1
$begingroup$
@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.
$endgroup$
– TeneT
Nov 24 '18 at 17:22
1
$begingroup$
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.
$endgroup$
– Eddy
Nov 24 '18 at 17:25
1
$begingroup$
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.
$endgroup$
– TeneT
Nov 24 '18 at 17:36
$begingroup$
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?
$endgroup$
– Astik
Dec 1 '18 at 18:26
|
show 1 more comment
$begingroup$
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.
answered Nov 24 '18 at 15:28
TeneTTeneT
236
$endgroup$
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.
answered Nov 24 '18 at 15:28
TeneTTeneT
236
answered Nov 24 '18 at 15:28
TeneTTeneT
236
answered Nov 24 '18 at 15:28
TeneTTeneT
236
answered Nov 24 '18 at 15:28
TeneTTeneT
236
236
1
$begingroup$
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.
$endgroup$
– Eddy
Nov 24 '18 at 17:08
1
$begingroup$
@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.
$endgroup$
– TeneT
Nov 24 '18 at 17:22
1
$begingroup$
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.
$endgroup$
– Eddy
Nov 24 '18 at 17:25
1
$begingroup$
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.
$endgroup$
– TeneT
Nov 24 '18 at 17:36
$begingroup$
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?
$endgroup$
– Astik
Dec 1 '18 at 18:26
|
show 1 more comment
1
$begingroup$
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.
$endgroup$
– Eddy
Nov 24 '18 at 17:08
1
$begingroup$
@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.
$endgroup$
– TeneT
Nov 24 '18 at 17:22
1
$begingroup$
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.
$endgroup$
– Eddy
Nov 24 '18 at 17:25
1
$begingroup$
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.
$endgroup$
– TeneT
Nov 24 '18 at 17:36
$begingroup$
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?
$endgroup$
– Astik
Dec 1 '18 at 18:26
1
1
$begingroup$
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.
$endgroup$
– Eddy
Nov 24 '18 at 17:08
$begingroup$
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.
$endgroup$
– Eddy
Nov 24 '18 at 17:08
1
1
$begingroup$
@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.
$endgroup$
– TeneT
Nov 24 '18 at 17:22
$begingroup$
@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.
$endgroup$
– TeneT
Nov 24 '18 at 17:22
1
1
$begingroup$
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.
$endgroup$
– Eddy
Nov 24 '18 at 17:25
$begingroup$
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.
$endgroup$
– Eddy
Nov 24 '18 at 17:25
1
1
$begingroup$
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.
$endgroup$
– TeneT
Nov 24 '18 at 17:36
$begingroup$
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.
$endgroup$
– TeneT
Nov 24 '18 at 17:36
$begingroup$
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?
$endgroup$
– Astik
Dec 1 '18 at 18:26
$begingroup$
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?
$endgroup$
– Astik
Dec 1 '18 at 18:26
|
show 1 more comment
$begingroup$
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.
answered Nov 24 '18 at 18:28
R. RomeroR. Romero
3136
$endgroup$
add a comment |
$begingroup$
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.
answered Nov 24 '18 at 18:28
R. RomeroR. Romero
3136
$endgroup$
add a comment |
$begingroup$
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.
answered Nov 24 '18 at 18:28
R. RomeroR. Romero
3136
$endgroup$
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.
answered Nov 24 '18 at 18:28
R. RomeroR. Romero
3136
answered Nov 24 '18 at 18:28
R. RomeroR. Romero
3136
answered Nov 24 '18 at 18:28
R. RomeroR. Romero
3136
answered Nov 24 '18 at 18:28
R. RomeroR. Romero
3136
3136
add a comment |
add a comment |
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Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
