Mixing
Harmonic oscillator in spherical coordinates
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It is probably the most well-known result in quantum mechanics that the harmonic oscillator can be solved by supersymmetry.
More precisely, the operator
$$-fracd^2dx^2+x^2$$
can be decomposed as
$$-fracd^2dx^2+x^2 = left(-fracddx-xright)left(fracddx-xright)=:a^*a.$$
When I tried to see whether the same holds true in spherical coordinates I noticed a complete suprise
$$left(-fracddr-fracn-1r-rright)left(fracddr-rright) = -fracd^2dr^2 -fracn-1r fracddr +r^2+(n-1).$$
The first three terms comprise the harmonic oscillator in spherical coordinates that is
$$-Delta_r + r^2.$$
For convenience, I discarded the angular part in this calculation.
But for some reasons I seem to get this superficial term $(n-1)$. Why is that? And can I do anything smarter to avoid this term?
fa.functional-analysis real-analysis ap.analysis-of-pdes mp.mathematical-physics quantum-mechanics
asked 3 hours ago
ErwinSchr
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up vote
4
down vote
favorite
It is probably the most well-known result in quantum mechanics that the harmonic oscillator can be solved by supersymmetry.
More precisely, the operator
$$-fracd^2dx^2+x^2$$
can be decomposed as
$$-fracd^2dx^2+x^2 = left(-fracddx-xright)left(fracddx-xright)=:a^*a.$$
When I tried to see whether the same holds true in spherical coordinates I noticed a complete suprise
$$left(-fracddr-fracn-1r-rright)left(fracddr-rright) = -fracd^2dr^2 -fracn-1r fracddr +r^2+(n-1).$$
The first three terms comprise the harmonic oscillator in spherical coordinates that is
$$-Delta_r + r^2.$$
For convenience, I discarded the angular part in this calculation.
But for some reasons I seem to get this superficial term $(n-1)$. Why is that? And can I do anything smarter to avoid this term?
fa.functional-analysis real-analysis ap.analysis-of-pdes mp.mathematical-physics quantum-mechanics
asked 3 hours ago
ErwinSchr
232
New contributor
ErwinSchr is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |Â
up vote
4
down vote
favorite
up vote
4
down vote
favorite
It is probably the most well-known result in quantum mechanics that the harmonic oscillator can be solved by supersymmetry.
More precisely, the operator
$$-fracd^2dx^2+x^2$$
can be decomposed as
$$-fracd^2dx^2+x^2 = left(-fracddx-xright)left(fracddx-xright)=:a^*a.$$
When I tried to see whether the same holds true in spherical coordinates I noticed a complete suprise
$$left(-fracddr-fracn-1r-rright)left(fracddr-rright) = -fracd^2dr^2 -fracn-1r fracddr +r^2+(n-1).$$
The first three terms comprise the harmonic oscillator in spherical coordinates that is
$$-Delta_r + r^2.$$
For convenience, I discarded the angular part in this calculation.
But for some reasons I seem to get this superficial term $(n-1)$. Why is that? And can I do anything smarter to avoid this term?
fa.functional-analysis real-analysis ap.analysis-of-pdes mp.mathematical-physics quantum-mechanics
asked 3 hours ago
ErwinSchr
232
New contributor
ErwinSchr is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
It is probably the most well-known result in quantum mechanics that the harmonic oscillator can be solved by supersymmetry.
More precisely, the operator
$$-fracd^2dx^2+x^2$$
can be decomposed as
$$-fracd^2dx^2+x^2 = left(-fracddx-xright)left(fracddx-xright)=:a^*a.$$
When I tried to see whether the same holds true in spherical coordinates I noticed a complete suprise
$$left(-fracddr-fracn-1r-rright)left(fracddr-rright) = -fracd^2dr^2 -fracn-1r fracddr +r^2+(n-1).$$
The first three terms comprise the harmonic oscillator in spherical coordinates that is
$$-Delta_r + r^2.$$
For convenience, I discarded the angular part in this calculation.
But for some reasons I seem to get this superficial term $(n-1)$. Why is that? And can I do anything smarter to avoid this term?
fa.functional-analysis real-analysis ap.analysis-of-pdes mp.mathematical-physics quantum-mechanics
fa.functional-analysis real-analysis ap.analysis-of-pdes mp.mathematical-physics quantum-mechanics
asked 3 hours ago
ErwinSchr
232
New contributor
ErwinSchr is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 3 hours ago
ErwinSchr
232
New contributor
ErwinSchr is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 3 hours ago
ErwinSchr
232
New contributor
ErwinSchr is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 3 hours ago
ErwinSchr
232
asked 3 hours ago
ErwinSchr
232
232
New contributor
ErwinSchr is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
ErwinSchr is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
ErwinSchr is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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2 Answers
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Indeed, the supersymmetric operators do not factorize the Hamiltonian of the three-dimensional harmonic oscillator, there is an additional term. See Creation and annihilation operators, symmetry and supersymmetry of the 3D isotropic harmonic oscillator, equation 16.
answered 2 hours ago
Carlo Beenakker
70.8k9157263
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Supplementing @CarloBeenakker's answer: a different sort of "factorization" of the Laplacian can be done in terms of the Dirac operator, which in some regards better preserves the nature of the situation, but/and has scalar coefficients in an appropriate Clifford algebra, which in general is not commutative. E.g., on $mathbb R$, it is $mathbb C$, which is convenient. But already in two dimensions the relevant Clifford algebra is the Hamiltonian quaternions...
Depending what you want to do, this might be a useful direction to consider, in addition to the "scalarization" of immediate reduction to the spherically symmetric case.
answered 2 hours ago
paul garrett
17.5k25298
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
Indeed, the supersymmetric operators do not factorize the Hamiltonian of the three-dimensional harmonic oscillator, there is an additional term. See Creation and annihilation operators, symmetry and supersymmetry of the 3D isotropic harmonic oscillator, equation 16.
answered 2 hours ago
Carlo Beenakker
70.8k9157263
add a comment |Â
up vote
2
down vote
accepted
Indeed, the supersymmetric operators do not factorize the Hamiltonian of the three-dimensional harmonic oscillator, there is an additional term. See Creation and annihilation operators, symmetry and supersymmetry of the 3D isotropic harmonic oscillator, equation 16.
answered 2 hours ago
Carlo Beenakker
70.8k9157263
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
Indeed, the supersymmetric operators do not factorize the Hamiltonian of the three-dimensional harmonic oscillator, there is an additional term. See Creation and annihilation operators, symmetry and supersymmetry of the 3D isotropic harmonic oscillator, equation 16.
answered 2 hours ago
Carlo Beenakker
70.8k9157263
Indeed, the supersymmetric operators do not factorize the Hamiltonian of the three-dimensional harmonic oscillator, there is an additional term. See Creation and annihilation operators, symmetry and supersymmetry of the 3D isotropic harmonic oscillator, equation 16.
answered 2 hours ago
Carlo Beenakker
70.8k9157263
answered 2 hours ago
Carlo Beenakker
70.8k9157263
answered 2 hours ago
Carlo Beenakker
70.8k9157263
answered 2 hours ago
Carlo Beenakker
70.8k9157263
70.8k9157263
add a comment |Â
add a comment |Â
up vote
0
down vote
Supplementing @CarloBeenakker's answer: a different sort of "factorization" of the Laplacian can be done in terms of the Dirac operator, which in some regards better preserves the nature of the situation, but/and has scalar coefficients in an appropriate Clifford algebra, which in general is not commutative. E.g., on $mathbb R$, it is $mathbb C$, which is convenient. But already in two dimensions the relevant Clifford algebra is the Hamiltonian quaternions...
Depending what you want to do, this might be a useful direction to consider, in addition to the "scalarization" of immediate reduction to the spherically symmetric case.
answered 2 hours ago
paul garrett
17.5k25298
add a comment |Â
up vote
0
down vote
Supplementing @CarloBeenakker's answer: a different sort of "factorization" of the Laplacian can be done in terms of the Dirac operator, which in some regards better preserves the nature of the situation, but/and has scalar coefficients in an appropriate Clifford algebra, which in general is not commutative. E.g., on $mathbb R$, it is $mathbb C$, which is convenient. But already in two dimensions the relevant Clifford algebra is the Hamiltonian quaternions...
Depending what you want to do, this might be a useful direction to consider, in addition to the "scalarization" of immediate reduction to the spherically symmetric case.
answered 2 hours ago
paul garrett
17.5k25298
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Supplementing @CarloBeenakker's answer: a different sort of "factorization" of the Laplacian can be done in terms of the Dirac operator, which in some regards better preserves the nature of the situation, but/and has scalar coefficients in an appropriate Clifford algebra, which in general is not commutative. E.g., on $mathbb R$, it is $mathbb C$, which is convenient. But already in two dimensions the relevant Clifford algebra is the Hamiltonian quaternions...
Depending what you want to do, this might be a useful direction to consider, in addition to the "scalarization" of immediate reduction to the spherically symmetric case.
answered 2 hours ago
paul garrett
17.5k25298
Supplementing @CarloBeenakker's answer: a different sort of "factorization" of the Laplacian can be done in terms of the Dirac operator, which in some regards better preserves the nature of the situation, but/and has scalar coefficients in an appropriate Clifford algebra, which in general is not commutative. E.g., on $mathbb R$, it is $mathbb C$, which is convenient. But already in two dimensions the relevant Clifford algebra is the Hamiltonian quaternions...
Depending what you want to do, this might be a useful direction to consider, in addition to the "scalarization" of immediate reduction to the spherically symmetric case.
answered 2 hours ago
paul garrett
17.5k25298
answered 2 hours ago
paul garrett
17.5k25298
answered 2 hours ago
paul garrett
17.5k25298
answered 2 hours ago
paul garrett
17.5k25298
17.5k25298
add a comment |Â
add a comment |Â
ErwinSchr is a new contributor. Be nice, and check out our Code of Conduct.
ErwinSchr is a new contributor. Be nice, and check out our Code of Conduct.
ErwinSchr is a new contributor. Be nice, and check out our Code of Conduct.
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