Mixing
In a spherically symmetric central potential why do we look for eigenfunctions of the angular momentum...
$begingroup$
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 hilbert-space wavefunction schroedinger-equation spherical-harmonics
edited Jan 29 at 4:39
Qmechanic♦
107k121981229
asked Jan 28 at 21:16
daljit97daljit97
17713
$endgroup$
add a comment |
$begingroup$
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 hilbert-space wavefunction schroedinger-equation spherical-harmonics
edited Jan 29 at 4:39
Qmechanic♦
107k121981229
asked Jan 28 at 21:16
daljit97daljit97
17713
$endgroup$
add a comment |
$begingroup$
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 hilbert-space wavefunction schroedinger-equation spherical-harmonics
edited Jan 29 at 4:39
Qmechanic♦
107k121981229
asked Jan 28 at 21:16
daljit97daljit97
17713
$endgroup$
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 hilbert-space wavefunction schroedinger-equation spherical-harmonics
quantum-mechanics hilbert-space wavefunction schroedinger-equation spherical-harmonics
edited Jan 29 at 4:39
Qmechanic♦
107k121981229
asked Jan 28 at 21:16
daljit97daljit97
17713
edited Jan 29 at 4:39
Qmechanic♦
107k121981229
asked Jan 28 at 21:16
daljit97daljit97
17713
edited Jan 29 at 4:39
Qmechanic♦
107k121981229
edited Jan 29 at 4:39
Qmechanic♦
107k121981229
edited Jan 29 at 4:39
Qmechanic♦
107k121981229
107k121981229
asked Jan 28 at 21:16
daljit97daljit97
17713
asked Jan 28 at 21:16
daljit97daljit97
17713
asked Jan 28 at 21:16
daljit97daljit97
17713
17713
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
One objective of quantum mechanics is to fully labelled states of the system. In some cases energy is not enough: for instance in the hydrogen atom some energy levels occur more than once; the same happens in the 3d harmonic oscillator. In such cases, simply giving the energy is not enough to completely identify the state so we look for other ``quantum numbers'' to refine the identification.
Of course one would prefer to label states with constants of motion, since they are constant so what we call state $nell m$ at the start we can use the same name at the end. Hence the idea of using a complete set of commuting operators - the Hamiltonian and others that commute with it - so as to uniquely identify the quantum states with constants of motion.
Of course when the potential is spherical $V(r)$, what enters in the radial part of the Schrödinger equation is actually the effective potential
$$
V_{eff }=V(r)+frac{hbar^2}{2m}frac{ell(ell+1)}{r^2} tag{1}
$$
and so you see that the radial Schrödinger equation actually explicitly contains an angular momentum part so it is quite natural to look for functions $psi(r,theta,phi)$ which are also eigenfunctions $Y^ell_m(theta,phi)$ of $hat L^2$ as these will provide the correct extra factor in Eq.(1) when separating the variables. Since there are many functions $Y^ell_m(theta,phi)$ with the same $hbar^2ell(ell+1)$, one uses the eigenvalue $hbar m$ of $hat L_z$ to distinguish them.
answered Jan 29 at 1:58
ZeroTheHeroZeroTheHero
21.1k53364
$endgroup$
$begingroup$
What do you mean by "these will provide the correct extra factor"?
$endgroup$
– daljit97
Jan 29 at 22:33
1
$begingroup$
@daljit97 Even if the "core" potential depends only on $r$, the effective potential has a centrifugal part in $ell(ell+1)$, so it's natural to look for eigenfunctions $Y^ell_m(theta,phi)$ of the angular part which are eigenstates of $L^2$ with eigenvalues $ell(ell+1)$.
$endgroup$
– ZeroTheHero
Jan 30 at 0:19
add a comment |
$begingroup$
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.
After reading @ZeroTheHero 's comment, i figured it would be instructive to provide some intuition as to why angular momentum does not depend on radius. Remember the classical definition
$$mathbf L=mathbf rtimesmathbf p$$
Let's then give and estimate of $L$. Suppose an atom or something has some characteristic radius $R$, and the electron moves with some characteristic momentum $p$. Then we estimate
$$Lsim Rp$$
But the de Broigle relation gives us an estimate of momentum in terms of the characteristic length
$$psimfrac{h}{R}$$
Where $h$ is Planck's constant. As such, out estimate for angular momentum amounts to
$$Lsim h,$$
Which has no explicit dependence on characteristic lengths of the system. As such it is reasonable to say it has no $r$ dependence, even though I've not given a rigorous proof.
answered Jan 28 at 21:25
Gabriel GolfettiGabriel Golfetti
1,3341713
$endgroup$
$begingroup$
and then how do we know that their eigenfunctions will be eigenfunctions of $hat H$?
$endgroup$
– daljit97
Jan 28 at 21:29
$begingroup$
@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
Jan 28 at 21:33
1
$begingroup$
@daljit97 Commuting operators have simultaneous eigenstates
$endgroup$
– Aaron Stevens
Jan 28 at 21:47
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
Jan 28 at 23:44
1
$begingroup$
given that classically $vec L=vec rtimes vec p$ you might want to expand on why $L^2$ has no $r$-dependence.
$endgroup$
– ZeroTheHero
Jan 29 at 3:55
|
show 3 more comments
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2 Answers
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active
oldest
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2 Answers
2
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
One objective of quantum mechanics is to fully labelled states of the system. In some cases energy is not enough: for instance in the hydrogen atom some energy levels occur more than once; the same happens in the 3d harmonic oscillator. In such cases, simply giving the energy is not enough to completely identify the state so we look for other ``quantum numbers'' to refine the identification.
Of course one would prefer to label states with constants of motion, since they are constant so what we call state $nell m$ at the start we can use the same name at the end. Hence the idea of using a complete set of commuting operators - the Hamiltonian and others that commute with it - so as to uniquely identify the quantum states with constants of motion.
Of course when the potential is spherical $V(r)$, what enters in the radial part of the Schrödinger equation is actually the effective potential
$$
V_{eff }=V(r)+frac{hbar^2}{2m}frac{ell(ell+1)}{r^2} tag{1}
$$
and so you see that the radial Schrödinger equation actually explicitly contains an angular momentum part so it is quite natural to look for functions $psi(r,theta,phi)$ which are also eigenfunctions $Y^ell_m(theta,phi)$ of $hat L^2$ as these will provide the correct extra factor in Eq.(1) when separating the variables. Since there are many functions $Y^ell_m(theta,phi)$ with the same $hbar^2ell(ell+1)$, one uses the eigenvalue $hbar m$ of $hat L_z$ to distinguish them.
answered Jan 29 at 1:58
ZeroTheHeroZeroTheHero
21.1k53364
$endgroup$
$begingroup$
What do you mean by "these will provide the correct extra factor"?
$endgroup$
– daljit97
Jan 29 at 22:33
1
$begingroup$
@daljit97 Even if the "core" potential depends only on $r$, the effective potential has a centrifugal part in $ell(ell+1)$, so it's natural to look for eigenfunctions $Y^ell_m(theta,phi)$ of the angular part which are eigenstates of $L^2$ with eigenvalues $ell(ell+1)$.
$endgroup$
– ZeroTheHero
Jan 30 at 0:19
add a comment |
$begingroup$
One objective of quantum mechanics is to fully labelled states of the system. In some cases energy is not enough: for instance in the hydrogen atom some energy levels occur more than once; the same happens in the 3d harmonic oscillator. In such cases, simply giving the energy is not enough to completely identify the state so we look for other ``quantum numbers'' to refine the identification.
Of course one would prefer to label states with constants of motion, since they are constant so what we call state $nell m$ at the start we can use the same name at the end. Hence the idea of using a complete set of commuting operators - the Hamiltonian and others that commute with it - so as to uniquely identify the quantum states with constants of motion.
Of course when the potential is spherical $V(r)$, what enters in the radial part of the Schrödinger equation is actually the effective potential
$$
V_{eff }=V(r)+frac{hbar^2}{2m}frac{ell(ell+1)}{r^2} tag{1}
$$
and so you see that the radial Schrödinger equation actually explicitly contains an angular momentum part so it is quite natural to look for functions $psi(r,theta,phi)$ which are also eigenfunctions $Y^ell_m(theta,phi)$ of $hat L^2$ as these will provide the correct extra factor in Eq.(1) when separating the variables. Since there are many functions $Y^ell_m(theta,phi)$ with the same $hbar^2ell(ell+1)$, one uses the eigenvalue $hbar m$ of $hat L_z$ to distinguish them.
answered Jan 29 at 1:58
ZeroTheHeroZeroTheHero
21.1k53364
$endgroup$
$begingroup$
What do you mean by "these will provide the correct extra factor"?
$endgroup$
– daljit97
Jan 29 at 22:33
1
$begingroup$
@daljit97 Even if the "core" potential depends only on $r$, the effective potential has a centrifugal part in $ell(ell+1)$, so it's natural to look for eigenfunctions $Y^ell_m(theta,phi)$ of the angular part which are eigenstates of $L^2$ with eigenvalues $ell(ell+1)$.
$endgroup$
– ZeroTheHero
Jan 30 at 0:19
add a comment |
$begingroup$
One objective of quantum mechanics is to fully labelled states of the system. In some cases energy is not enough: for instance in the hydrogen atom some energy levels occur more than once; the same happens in the 3d harmonic oscillator. In such cases, simply giving the energy is not enough to completely identify the state so we look for other ``quantum numbers'' to refine the identification.
Of course one would prefer to label states with constants of motion, since they are constant so what we call state $nell m$ at the start we can use the same name at the end. Hence the idea of using a complete set of commuting operators - the Hamiltonian and others that commute with it - so as to uniquely identify the quantum states with constants of motion.
Of course when the potential is spherical $V(r)$, what enters in the radial part of the Schrödinger equation is actually the effective potential
$$
V_{eff }=V(r)+frac{hbar^2}{2m}frac{ell(ell+1)}{r^2} tag{1}
$$
and so you see that the radial Schrödinger equation actually explicitly contains an angular momentum part so it is quite natural to look for functions $psi(r,theta,phi)$ which are also eigenfunctions $Y^ell_m(theta,phi)$ of $hat L^2$ as these will provide the correct extra factor in Eq.(1) when separating the variables. Since there are many functions $Y^ell_m(theta,phi)$ with the same $hbar^2ell(ell+1)$, one uses the eigenvalue $hbar m$ of $hat L_z$ to distinguish them.
answered Jan 29 at 1:58
ZeroTheHeroZeroTheHero
21.1k53364
$endgroup$
One objective of quantum mechanics is to fully labelled states of the system. In some cases energy is not enough: for instance in the hydrogen atom some energy levels occur more than once; the same happens in the 3d harmonic oscillator. In such cases, simply giving the energy is not enough to completely identify the state so we look for other ``quantum numbers'' to refine the identification.
Of course one would prefer to label states with constants of motion, since they are constant so what we call state $nell m$ at the start we can use the same name at the end. Hence the idea of using a complete set of commuting operators - the Hamiltonian and others that commute with it - so as to uniquely identify the quantum states with constants of motion.
Of course when the potential is spherical $V(r)$, what enters in the radial part of the Schrödinger equation is actually the effective potential
$$
V_{eff }=V(r)+frac{hbar^2}{2m}frac{ell(ell+1)}{r^2} tag{1}
$$
and so you see that the radial Schrödinger equation actually explicitly contains an angular momentum part so it is quite natural to look for functions $psi(r,theta,phi)$ which are also eigenfunctions $Y^ell_m(theta,phi)$ of $hat L^2$ as these will provide the correct extra factor in Eq.(1) when separating the variables. Since there are many functions $Y^ell_m(theta,phi)$ with the same $hbar^2ell(ell+1)$, one uses the eigenvalue $hbar m$ of $hat L_z$ to distinguish them.
answered Jan 29 at 1:58
ZeroTheHeroZeroTheHero
21.1k53364
answered Jan 29 at 1:58
ZeroTheHeroZeroTheHero
21.1k53364
answered Jan 29 at 1:58
ZeroTheHeroZeroTheHero
21.1k53364
answered Jan 29 at 1:58
ZeroTheHeroZeroTheHero
21.1k53364
21.1k53364
$begingroup$
What do you mean by "these will provide the correct extra factor"?
$endgroup$
– daljit97
Jan 29 at 22:33
1
$begingroup$
@daljit97 Even if the "core" potential depends only on $r$, the effective potential has a centrifugal part in $ell(ell+1)$, so it's natural to look for eigenfunctions $Y^ell_m(theta,phi)$ of the angular part which are eigenstates of $L^2$ with eigenvalues $ell(ell+1)$.
$endgroup$
– ZeroTheHero
Jan 30 at 0:19
add a comment |
$begingroup$
What do you mean by "these will provide the correct extra factor"?
$endgroup$
– daljit97
Jan 29 at 22:33
1
$begingroup$
@daljit97 Even if the "core" potential depends only on $r$, the effective potential has a centrifugal part in $ell(ell+1)$, so it's natural to look for eigenfunctions $Y^ell_m(theta,phi)$ of the angular part which are eigenstates of $L^2$ with eigenvalues $ell(ell+1)$.
$endgroup$
– ZeroTheHero
Jan 30 at 0:19
$begingroup$
What do you mean by "these will provide the correct extra factor"?
$endgroup$
– daljit97
Jan 29 at 22:33
$begingroup$
What do you mean by "these will provide the correct extra factor"?
$endgroup$
– daljit97
Jan 29 at 22:33
1
1
$begingroup$
@daljit97 Even if the "core" potential depends only on $r$, the effective potential has a centrifugal part in $ell(ell+1)$, so it's natural to look for eigenfunctions $Y^ell_m(theta,phi)$ of the angular part which are eigenstates of $L^2$ with eigenvalues $ell(ell+1)$.
$endgroup$
– ZeroTheHero
Jan 30 at 0:19
$begingroup$
@daljit97 Even if the "core" potential depends only on $r$, the effective potential has a centrifugal part in $ell(ell+1)$, so it's natural to look for eigenfunctions $Y^ell_m(theta,phi)$ of the angular part which are eigenstates of $L^2$ with eigenvalues $ell(ell+1)$.
$endgroup$
– ZeroTheHero
Jan 30 at 0:19
add a comment |
$begingroup$
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.
After reading @ZeroTheHero 's comment, i figured it would be instructive to provide some intuition as to why angular momentum does not depend on radius. Remember the classical definition
$$mathbf L=mathbf rtimesmathbf p$$
Let's then give and estimate of $L$. Suppose an atom or something has some characteristic radius $R$, and the electron moves with some characteristic momentum $p$. Then we estimate
$$Lsim Rp$$
But the de Broigle relation gives us an estimate of momentum in terms of the characteristic length
$$psimfrac{h}{R}$$
Where $h$ is Planck's constant. As such, out estimate for angular momentum amounts to
$$Lsim h,$$
Which has no explicit dependence on characteristic lengths of the system. As such it is reasonable to say it has no $r$ dependence, even though I've not given a rigorous proof.
answered Jan 28 at 21:25
Gabriel GolfettiGabriel Golfetti
1,3341713
$endgroup$
$begingroup$
and then how do we know that their eigenfunctions will be eigenfunctions of $hat H$?
$endgroup$
– daljit97
Jan 28 at 21:29
$begingroup$
@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
Jan 28 at 21:33
1
$begingroup$
@daljit97 Commuting operators have simultaneous eigenstates
$endgroup$
– Aaron Stevens
Jan 28 at 21:47
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
Jan 28 at 23:44
1
$begingroup$
given that classically $vec L=vec rtimes vec p$ you might want to expand on why $L^2$ has no $r$-dependence.
$endgroup$
– ZeroTheHero
Jan 29 at 3:55
|
show 3 more comments
$begingroup$
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.
After reading @ZeroTheHero 's comment, i figured it would be instructive to provide some intuition as to why angular momentum does not depend on radius. Remember the classical definition
$$mathbf L=mathbf rtimesmathbf p$$
Let's then give and estimate of $L$. Suppose an atom or something has some characteristic radius $R$, and the electron moves with some characteristic momentum $p$. Then we estimate
$$Lsim Rp$$
But the de Broigle relation gives us an estimate of momentum in terms of the characteristic length
$$psimfrac{h}{R}$$
Where $h$ is Planck's constant. As such, out estimate for angular momentum amounts to
$$Lsim h,$$
Which has no explicit dependence on characteristic lengths of the system. As such it is reasonable to say it has no $r$ dependence, even though I've not given a rigorous proof.
answered Jan 28 at 21:25
Gabriel GolfettiGabriel Golfetti
1,3341713
$endgroup$
$begingroup$
and then how do we know that their eigenfunctions will be eigenfunctions of $hat H$?
$endgroup$
– daljit97
Jan 28 at 21:29
$begingroup$
@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
Jan 28 at 21:33
1
$begingroup$
@daljit97 Commuting operators have simultaneous eigenstates
$endgroup$
– Aaron Stevens
Jan 28 at 21:47
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
Jan 28 at 23:44
1
$begingroup$
given that classically $vec L=vec rtimes vec p$ you might want to expand on why $L^2$ has no $r$-dependence.
$endgroup$
– ZeroTheHero
Jan 29 at 3:55
|
show 3 more comments
$begingroup$
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.
After reading @ZeroTheHero 's comment, i figured it would be instructive to provide some intuition as to why angular momentum does not depend on radius. Remember the classical definition
$$mathbf L=mathbf rtimesmathbf p$$
Let's then give and estimate of $L$. Suppose an atom or something has some characteristic radius $R$, and the electron moves with some characteristic momentum $p$. Then we estimate
$$Lsim Rp$$
But the de Broigle relation gives us an estimate of momentum in terms of the characteristic length
$$psimfrac{h}{R}$$
Where $h$ is Planck's constant. As such, out estimate for angular momentum amounts to
$$Lsim h,$$
Which has no explicit dependence on characteristic lengths of the system. As such it is reasonable to say it has no $r$ dependence, even though I've not given a rigorous proof.
answered Jan 28 at 21:25
Gabriel GolfettiGabriel Golfetti
1,3341713
$endgroup$
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.
After reading @ZeroTheHero 's comment, i figured it would be instructive to provide some intuition as to why angular momentum does not depend on radius. Remember the classical definition
$$mathbf L=mathbf rtimesmathbf p$$
Let's then give and estimate of $L$. Suppose an atom or something has some characteristic radius $R$, and the electron moves with some characteristic momentum $p$. Then we estimate
$$Lsim Rp$$
But the de Broigle relation gives us an estimate of momentum in terms of the characteristic length
$$psimfrac{h}{R}$$
Where $h$ is Planck's constant. As such, out estimate for angular momentum amounts to
$$Lsim h,$$
Which has no explicit dependence on characteristic lengths of the system. As such it is reasonable to say it has no $r$ dependence, even though I've not given a rigorous proof.
answered Jan 28 at 21:25
Gabriel GolfettiGabriel Golfetti
1,3341713
edited Jan 29 at 21:53
answered Jan 28 at 21:25
Gabriel GolfettiGabriel Golfetti
1,3341713
answered Jan 28 at 21:25
Gabriel GolfettiGabriel Golfetti
1,3341713
answered Jan 28 at 21:25
Gabriel GolfettiGabriel Golfetti
1,3341713
1,3341713
$begingroup$
and then how do we know that their eigenfunctions will be eigenfunctions of $hat H$?
$endgroup$
– daljit97
Jan 28 at 21:29
$begingroup$
@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
Jan 28 at 21:33
1
$begingroup$
@daljit97 Commuting operators have simultaneous eigenstates
$endgroup$
– Aaron Stevens
Jan 28 at 21:47
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
Jan 28 at 23:44
1
$begingroup$
given that classically $vec L=vec rtimes vec p$ you might want to expand on why $L^2$ has no $r$-dependence.
$endgroup$
– ZeroTheHero
Jan 29 at 3:55
|
show 3 more comments
$begingroup$
and then how do we know that their eigenfunctions will be eigenfunctions of $hat H$?
$endgroup$
– daljit97
Jan 28 at 21:29
$begingroup$
@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
Jan 28 at 21:33
1
$begingroup$
@daljit97 Commuting operators have simultaneous eigenstates
$endgroup$
– Aaron Stevens
Jan 28 at 21:47
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
Jan 28 at 23:44
1
$begingroup$
given that classically $vec L=vec rtimes vec p$ you might want to expand on why $L^2$ has no $r$-dependence.
$endgroup$
– ZeroTheHero
Jan 29 at 3:55
$begingroup$
and then how do we know that their eigenfunctions will be eigenfunctions of $hat H$?
$endgroup$
– daljit97
Jan 28 at 21:29
$begingroup$
and then how do we know that their eigenfunctions will be eigenfunctions of $hat H$?
$endgroup$
– daljit97
Jan 28 at 21:29
$begingroup$
@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
Jan 28 at 21:33
$begingroup$
@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
Jan 28 at 21:33
1
1
$begingroup$
@daljit97 Commuting operators have simultaneous eigenstates
$endgroup$
– Aaron Stevens
Jan 28 at 21:47
$begingroup$
@daljit97 Commuting operators have simultaneous eigenstates
$endgroup$
– Aaron Stevens
Jan 28 at 21:47
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
Jan 28 at 23:44
$begingroup$
@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
Jan 28 at 23:44
1
1
$begingroup$
given that classically $vec L=vec rtimes vec p$ you might want to expand on why $L^2$ has no $r$-dependence.
$endgroup$
– ZeroTheHero
Jan 29 at 3:55
$begingroup$
given that classically $vec L=vec rtimes vec p$ you might want to expand on why $L^2$ has no $r$-dependence.
$endgroup$
– ZeroTheHero
Jan 29 at 3:55
|
show 3 more comments
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