Diagonalization of total angular momentum over creation operators for an isotropic harmonic oscillator?
$begingroup$
You have an isotropic three dimensional quantum harmonic oscillator so the Hamiltonian is
$$
H=frac{p^2}{2}+frac{r^2}2
$$
If you do the creation-annihilation operator-algebra trick and define creation operator
$$
a^dagger_j=frac{x_j-ip_j}{sqrt{2hbar}}
$$
so that the Hamiltonian is
$$
H=hbar left (a_1^dagger a_1+a_2^dagger a_2+a_3^dagger a_3+frac 3 2 right)
$$
so that the Hamiltonian is diagonalized by states characterized by the excitation number in each direction.
BUT if we want to compute angular momentum, we could define $3$-rotating creation operators
$$
a_uparrow^dagger=frac{a^dagger_1+ia^dagger_2}{sqrt 2} a_downarrow^dagger=frac{a^dagger_1-ia^dagger_2}{sqrt 2}
$$
and, as $a_1^dagger a_1+a_2^dagger a_2=a_uparrow^dagger a_uparrow+a_downarrow^dagger a_downarrow$, the Hamiltonian is still nicely diagonized; moreover we also have the 3-angular momentum operator diagonal, $L_3=hbar (a_uparrow^dagger a_uparrow-a_downarrow^dagger a_downarrow)$, so $L_3 left | n_uparrow n_downarrow n_z right rangle=hbar (n_uparrow-n_downarrow) left | n_uparrow n_downarrow n_z right rangle$.
How might you similarly diagonalize $L^2$ with respect to creation-annihilation? The spherical harmonics are hellish, but Schwinger's creation operator approach is so nice and intuitive. How could we compute $L^2$ probabilities without ever integrating again?
Edit More precisely: Are there commuting operators $a,b,c$ such that $[a,a^dagger]=hbar$ (etc for $b$ and $c$) and the set of states defined by
$$left | n_a n_b n_c right rangle=(a^dagger)^{n_a}(b^dagger)^{n_b}(c^dagger)^{n_c}left | psi_0 right rangle$$
form an orthogonal basis diagonalizing $H,L^2$,and $L_3$? Here
$left | psi_0 right rangle$ is the ground state. How can you determine $a,b,c$ algebraically?
physics mathematical-physics operator-algebras quantum-mechanics
edited Dec 27 '18 at 18:00
Cosmas Zachos
1,582520
asked Sep 26 '14 at 4:50
spitespikespitespike
611316
$endgroup$
add a comment |
$begingroup$
You have an isotropic three dimensional quantum harmonic oscillator so the Hamiltonian is
$$
H=frac{p^2}{2}+frac{r^2}2
$$
If you do the creation-annihilation operator-algebra trick and define creation operator
$$
a^dagger_j=frac{x_j-ip_j}{sqrt{2hbar}}
$$
so that the Hamiltonian is
$$
H=hbar left (a_1^dagger a_1+a_2^dagger a_2+a_3^dagger a_3+frac 3 2 right)
$$
so that the Hamiltonian is diagonalized by states characterized by the excitation number in each direction.
BUT if we want to compute angular momentum, we could define $3$-rotating creation operators
$$
a_uparrow^dagger=frac{a^dagger_1+ia^dagger_2}{sqrt 2} a_downarrow^dagger=frac{a^dagger_1-ia^dagger_2}{sqrt 2}
$$
and, as $a_1^dagger a_1+a_2^dagger a_2=a_uparrow^dagger a_uparrow+a_downarrow^dagger a_downarrow$, the Hamiltonian is still nicely diagonized; moreover we also have the 3-angular momentum operator diagonal, $L_3=hbar (a_uparrow^dagger a_uparrow-a_downarrow^dagger a_downarrow)$, so $L_3 left | n_uparrow n_downarrow n_z right rangle=hbar (n_uparrow-n_downarrow) left | n_uparrow n_downarrow n_z right rangle$.
How might you similarly diagonalize $L^2$ with respect to creation-annihilation? The spherical harmonics are hellish, but Schwinger's creation operator approach is so nice and intuitive. How could we compute $L^2$ probabilities without ever integrating again?
Edit More precisely: Are there commuting operators $a,b,c$ such that $[a,a^dagger]=hbar$ (etc for $b$ and $c$) and the set of states defined by
$$left | n_a n_b n_c right rangle=(a^dagger)^{n_a}(b^dagger)^{n_b}(c^dagger)^{n_c}left | psi_0 right rangle$$
form an orthogonal basis diagonalizing $H,L^2$,and $L_3$? Here
$left | psi_0 right rangle$ is the ground state. How can you determine $a,b,c$ algebraically?
physics mathematical-physics operator-algebras quantum-mechanics
edited Dec 27 '18 at 18:00
Cosmas Zachos
1,582520
asked Sep 26 '14 at 4:50
spitespikespitespike
611316
$endgroup$
1
$begingroup$
Aren’t $H$ and $L^2$ essentially the same operator? (up to some scalar constants)
$endgroup$
– Incnis Mrsi
May 5 '15 at 20:35
$begingroup$
You probably want to read up in Messiah QM vI, Ch XII, § 15, p 456.
$endgroup$
– Cosmas Zachos
Dec 27 '18 at 16:06
$begingroup$
Related.
$endgroup$
– Cosmas Zachos
Dec 27 '18 at 17:20
$begingroup$
You are probably aware that the basis $|n_uparrow, n_downarrow,n_3rangle$ does not diagonalize $L^2$. You should emphasize that, to explain why you are wondering about the existence of yet another basis, a,b,c. It does not exist.
$endgroup$
– Cosmas Zachos
Dec 27 '18 at 17:37
add a comment |
$begingroup$
You have an isotropic three dimensional quantum harmonic oscillator so the Hamiltonian is
$$
H=frac{p^2}{2}+frac{r^2}2
$$
If you do the creation-annihilation operator-algebra trick and define creation operator
$$
a^dagger_j=frac{x_j-ip_j}{sqrt{2hbar}}
$$
so that the Hamiltonian is
$$
H=hbar left (a_1^dagger a_1+a_2^dagger a_2+a_3^dagger a_3+frac 3 2 right)
$$
so that the Hamiltonian is diagonalized by states characterized by the excitation number in each direction.
BUT if we want to compute angular momentum, we could define $3$-rotating creation operators
$$
a_uparrow^dagger=frac{a^dagger_1+ia^dagger_2}{sqrt 2} a_downarrow^dagger=frac{a^dagger_1-ia^dagger_2}{sqrt 2}
$$
and, as $a_1^dagger a_1+a_2^dagger a_2=a_uparrow^dagger a_uparrow+a_downarrow^dagger a_downarrow$, the Hamiltonian is still nicely diagonized; moreover we also have the 3-angular momentum operator diagonal, $L_3=hbar (a_uparrow^dagger a_uparrow-a_downarrow^dagger a_downarrow)$, so $L_3 left | n_uparrow n_downarrow n_z right rangle=hbar (n_uparrow-n_downarrow) left | n_uparrow n_downarrow n_z right rangle$.
How might you similarly diagonalize $L^2$ with respect to creation-annihilation? The spherical harmonics are hellish, but Schwinger's creation operator approach is so nice and intuitive. How could we compute $L^2$ probabilities without ever integrating again?
Edit More precisely: Are there commuting operators $a,b,c$ such that $[a,a^dagger]=hbar$ (etc for $b$ and $c$) and the set of states defined by
$$left | n_a n_b n_c right rangle=(a^dagger)^{n_a}(b^dagger)^{n_b}(c^dagger)^{n_c}left | psi_0 right rangle$$
form an orthogonal basis diagonalizing $H,L^2$,and $L_3$? Here
$left | psi_0 right rangle$ is the ground state. How can you determine $a,b,c$ algebraically?
physics mathematical-physics operator-algebras quantum-mechanics
edited Dec 27 '18 at 18:00
Cosmas Zachos
1,582520
asked Sep 26 '14 at 4:50
spitespikespitespike
611316
$endgroup$
You have an isotropic three dimensional quantum harmonic oscillator so the Hamiltonian is
$$
H=frac{p^2}{2}+frac{r^2}2
$$
If you do the creation-annihilation operator-algebra trick and define creation operator
$$
a^dagger_j=frac{x_j-ip_j}{sqrt{2hbar}}
$$
so that the Hamiltonian is
$$
H=hbar left (a_1^dagger a_1+a_2^dagger a_2+a_3^dagger a_3+frac 3 2 right)
$$
so that the Hamiltonian is diagonalized by states characterized by the excitation number in each direction.
BUT if we want to compute angular momentum, we could define $3$-rotating creation operators
$$
a_uparrow^dagger=frac{a^dagger_1+ia^dagger_2}{sqrt 2} a_downarrow^dagger=frac{a^dagger_1-ia^dagger_2}{sqrt 2}
$$
and, as $a_1^dagger a_1+a_2^dagger a_2=a_uparrow^dagger a_uparrow+a_downarrow^dagger a_downarrow$, the Hamiltonian is still nicely diagonized; moreover we also have the 3-angular momentum operator diagonal, $L_3=hbar (a_uparrow^dagger a_uparrow-a_downarrow^dagger a_downarrow)$, so $L_3 left | n_uparrow n_downarrow n_z right rangle=hbar (n_uparrow-n_downarrow) left | n_uparrow n_downarrow n_z right rangle$.
How might you similarly diagonalize $L^2$ with respect to creation-annihilation? The spherical harmonics are hellish, but Schwinger's creation operator approach is so nice and intuitive. How could we compute $L^2$ probabilities without ever integrating again?
Edit More precisely: Are there commuting operators $a,b,c$ such that $[a,a^dagger]=hbar$ (etc for $b$ and $c$) and the set of states defined by
$$left | n_a n_b n_c right rangle=(a^dagger)^{n_a}(b^dagger)^{n_b}(c^dagger)^{n_c}left | psi_0 right rangle$$
form an orthogonal basis diagonalizing $H,L^2$,and $L_3$? Here
$left | psi_0 right rangle$ is the ground state. How can you determine $a,b,c$ algebraically?
physics mathematical-physics operator-algebras quantum-mechanics
physics mathematical-physics operator-algebras quantum-mechanics
edited Dec 27 '18 at 18:00
Cosmas Zachos
1,582520
asked Sep 26 '14 at 4:50
spitespikespitespike
611316
edited Dec 27 '18 at 18:00
Cosmas Zachos
1,582520
asked Sep 26 '14 at 4:50
spitespikespitespike
611316
edited Dec 27 '18 at 18:00
Cosmas Zachos
1,582520
edited Dec 27 '18 at 18:00
Cosmas Zachos
1,582520
edited Dec 27 '18 at 18:00
Cosmas Zachos
1,582520
1,582520
asked Sep 26 '14 at 4:50
spitespikespitespike
611316
asked Sep 26 '14 at 4:50
spitespikespitespike
611316
asked Sep 26 '14 at 4:50
spitespikespitespike
611316
611316
1
$begingroup$
Aren’t $H$ and $L^2$ essentially the same operator? (up to some scalar constants)
$endgroup$
– Incnis Mrsi
May 5 '15 at 20:35
$begingroup$
You probably want to read up in Messiah QM vI, Ch XII, § 15, p 456.
$endgroup$
– Cosmas Zachos
Dec 27 '18 at 16:06
$begingroup$
Related.
$endgroup$
– Cosmas Zachos
Dec 27 '18 at 17:20
$begingroup$
You are probably aware that the basis $|n_uparrow, n_downarrow,n_3rangle$ does not diagonalize $L^2$. You should emphasize that, to explain why you are wondering about the existence of yet another basis, a,b,c. It does not exist.
$endgroup$
– Cosmas Zachos
Dec 27 '18 at 17:37
add a comment |
1
$begingroup$
Aren’t $H$ and $L^2$ essentially the same operator? (up to some scalar constants)
$endgroup$
– Incnis Mrsi
May 5 '15 at 20:35
$begingroup$
You probably want to read up in Messiah QM vI, Ch XII, § 15, p 456.
$endgroup$
– Cosmas Zachos
Dec 27 '18 at 16:06
$begingroup$
Related.
$endgroup$
– Cosmas Zachos
Dec 27 '18 at 17:20
$begingroup$
You are probably aware that the basis $|n_uparrow, n_downarrow,n_3rangle$ does not diagonalize $L^2$. You should emphasize that, to explain why you are wondering about the existence of yet another basis, a,b,c. It does not exist.
$endgroup$
– Cosmas Zachos
Dec 27 '18 at 17:37
1
1
$begingroup$
Aren’t $H$ and $L^2$ essentially the same operator? (up to some scalar constants)
$endgroup$
– Incnis Mrsi
May 5 '15 at 20:35
$begingroup$
Aren’t $H$ and $L^2$ essentially the same operator? (up to some scalar constants)
$endgroup$
– Incnis Mrsi
May 5 '15 at 20:35
$begingroup$
You probably want to read up in Messiah QM vI, Ch XII, § 15, p 456.
$endgroup$
– Cosmas Zachos
Dec 27 '18 at 16:06
$begingroup$
You probably want to read up in Messiah QM vI, Ch XII, § 15, p 456.
$endgroup$
– Cosmas Zachos
Dec 27 '18 at 16:06
$begingroup$
Related.
$endgroup$
– Cosmas Zachos
Dec 27 '18 at 17:20
$begingroup$
Related.
$endgroup$
– Cosmas Zachos
Dec 27 '18 at 17:20
$begingroup$
You are probably aware that the basis $|n_uparrow, n_downarrow,n_3rangle$ does not diagonalize $L^2$. You should emphasize that, to explain why you are wondering about the existence of yet another basis, a,b,c. It does not exist.
$endgroup$
– Cosmas Zachos
Dec 27 '18 at 17:37
$begingroup$
You are probably aware that the basis $|n_uparrow, n_downarrow,n_3rangle$ does not diagonalize $L^2$. You should emphasize that, to explain why you are wondering about the existence of yet another basis, a,b,c. It does not exist.
$endgroup$
– Cosmas Zachos
Dec 27 '18 at 17:37
add a comment |
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1
$begingroup$
Aren’t $H$ and $L^2$ essentially the same operator? (up to some scalar constants)
$endgroup$
– Incnis Mrsi
May 5 '15 at 20:35
$begingroup$
You probably want to read up in Messiah QM vI, Ch XII, § 15, p 456.
$endgroup$
– Cosmas Zachos
Dec 27 '18 at 16:06
$begingroup$
Related.
$endgroup$
– Cosmas Zachos
Dec 27 '18 at 17:20
$begingroup$
You are probably aware that the basis $|n_uparrow, n_downarrow,n_3rangle$ does not diagonalize $L^2$. You should emphasize that, to explain why you are wondering about the existence of yet another basis, a,b,c. It does not exist.
$endgroup$
– Cosmas Zachos
Dec 27 '18 at 17:37