When is separating the total wavefunction into a space part and a spin part possible? The Next CEO of Stack OverflowAnti-symmetric 2 particle wave functionA conceptual question about spinAntisymmetry requirement for the total wavefunctionConnection between singlet, triplet two-electron states and the Slater determinantMeasuring total angular momentum of two electronsTwo identical particlesConfusion on good quantum numbersSpectrum of two particles system hamiltonianAbout the symmetric spatial part of a two-electron wavefunction: Can it be that $r_1= r_2$ less favoured than $|r_1-r_2|neq 0$?What is the simplest possible Hamiltonian that yields an Antisymmetric Wavefunction?
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When is separating the total wavefunction into a space part and a spin part possible?
The Next CEO of Stack OverflowAnti-symmetric 2 particle wave functionA conceptual question about spinAntisymmetry requirement for the total wavefunctionConnection between singlet, triplet two-electron states and the Slater determinantMeasuring total angular momentum of two electronsTwo identical particlesConfusion on good quantum numbersSpectrum of two particles system hamiltonianAbout the symmetric spatial part of a two-electron wavefunction: Can it be that $r_1= r_2$ less favoured than $|r_1-r_2|neq 0$?What is the simplest possible Hamiltonian that yields an Antisymmetric Wavefunction?
$begingroup$
The total wavefunction of an electron $psi(vecr,s)$ can always be written as $$psi(vecr,s)=phi(vecr)zeta_s,m_s$$ where $phi(vecr)$ is the space part and $zeta_s,m_s$ is the spin part of the total wavefunction $psi(vecr,s)$. In my notation, $s=1/2, m_s=pm 1/2$.
Question 1 Is the above statement true? I am asking about any wavefunction here. Not only about energy eigenfunctions.
Now imagine a system of two electrons. Even without any knowledge about the Hamiltonian of the system, the overall wavefunction $psi(vecr_1,vecr_2;s_1,s_2)$ is antisymmetric. I think (I have this impression) under this general conditions, it is not possible to decompose $psi(vecr_1,vecr_2;s_1,s_2)$ into a product of a space part and spin part. However, if the Hamiltonian is spin-independent, only then can we do such a decomposition into space part and spin part.
Question 2 Can someone properly argue that how this is so? Please mention about any wavefunction of the system and about energy eigenfunctions.
quantum-mechanics wavefunction quantum-spin pauli-exclusion-principle identical-particles
asked Mar 25 at 12:21
mithusengupta123mithusengupta123
1,32811539
$endgroup$
add a comment |
$begingroup$
The total wavefunction of an electron $psi(vecr,s)$ can always be written as $$psi(vecr,s)=phi(vecr)zeta_s,m_s$$ where $phi(vecr)$ is the space part and $zeta_s,m_s$ is the spin part of the total wavefunction $psi(vecr,s)$. In my notation, $s=1/2, m_s=pm 1/2$.
Question 1 Is the above statement true? I am asking about any wavefunction here. Not only about energy eigenfunctions.
Now imagine a system of two electrons. Even without any knowledge about the Hamiltonian of the system, the overall wavefunction $psi(vecr_1,vecr_2;s_1,s_2)$ is antisymmetric. I think (I have this impression) under this general conditions, it is not possible to decompose $psi(vecr_1,vecr_2;s_1,s_2)$ into a product of a space part and spin part. However, if the Hamiltonian is spin-independent, only then can we do such a decomposition into space part and spin part.
Question 2 Can someone properly argue that how this is so? Please mention about any wavefunction of the system and about energy eigenfunctions.
quantum-mechanics wavefunction quantum-spin pauli-exclusion-principle identical-particles
asked Mar 25 at 12:21
mithusengupta123mithusengupta123
1,32811539
$endgroup$
add a comment |
$begingroup$
The total wavefunction of an electron $psi(vecr,s)$ can always be written as $$psi(vecr,s)=phi(vecr)zeta_s,m_s$$ where $phi(vecr)$ is the space part and $zeta_s,m_s$ is the spin part of the total wavefunction $psi(vecr,s)$. In my notation, $s=1/2, m_s=pm 1/2$.
Question 1 Is the above statement true? I am asking about any wavefunction here. Not only about energy eigenfunctions.
Now imagine a system of two electrons. Even without any knowledge about the Hamiltonian of the system, the overall wavefunction $psi(vecr_1,vecr_2;s_1,s_2)$ is antisymmetric. I think (I have this impression) under this general conditions, it is not possible to decompose $psi(vecr_1,vecr_2;s_1,s_2)$ into a product of a space part and spin part. However, if the Hamiltonian is spin-independent, only then can we do such a decomposition into space part and spin part.
Question 2 Can someone properly argue that how this is so? Please mention about any wavefunction of the system and about energy eigenfunctions.
quantum-mechanics wavefunction quantum-spin pauli-exclusion-principle identical-particles
asked Mar 25 at 12:21
mithusengupta123mithusengupta123
1,32811539
$endgroup$
The total wavefunction of an electron $psi(vecr,s)$ can always be written as $$psi(vecr,s)=phi(vecr)zeta_s,m_s$$ where $phi(vecr)$ is the space part and $zeta_s,m_s$ is the spin part of the total wavefunction $psi(vecr,s)$. In my notation, $s=1/2, m_s=pm 1/2$.
Question 1 Is the above statement true? I am asking about any wavefunction here. Not only about energy eigenfunctions.
Now imagine a system of two electrons. Even without any knowledge about the Hamiltonian of the system, the overall wavefunction $psi(vecr_1,vecr_2;s_1,s_2)$ is antisymmetric. I think (I have this impression) under this general conditions, it is not possible to decompose $psi(vecr_1,vecr_2;s_1,s_2)$ into a product of a space part and spin part. However, if the Hamiltonian is spin-independent, only then can we do such a decomposition into space part and spin part.
Question 2 Can someone properly argue that how this is so? Please mention about any wavefunction of the system and about energy eigenfunctions.
quantum-mechanics wavefunction quantum-spin pauli-exclusion-principle identical-particles
quantum-mechanics wavefunction quantum-spin pauli-exclusion-principle identical-particles
asked Mar 25 at 12:21
mithusengupta123mithusengupta123
1,32811539
asked Mar 25 at 12:21
mithusengupta123mithusengupta123
1,32811539
edited Mar 25 at 12:51
mithusengupta123
asked Mar 25 at 12:21
mithusengupta123mithusengupta123
1,32811539
asked Mar 25 at 12:21
mithusengupta123mithusengupta123
1,32811539
asked Mar 25 at 12:21
mithusengupta123mithusengupta123
1,32811539
1,32811539
add a comment |
add a comment |
1 Answer
1
active
oldest
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$begingroup$
Your claim
[any arbitrary] wavefunction of an electron $psi(vecr,s)$ can always be written as $$psi(vecr,s)=phi(vecr)zeta_s,m_s tag 1$$ where $phi(vecr)$ is the space part and $zeta_s,m_s$ is the spin part of the total wavefunction $psi(vecr,s)$
is false. It is perfectly possible to produce wavefunctions which cannot be written in that separable form - for a simple example, just take two orthogonal spatial wavefunctions, $phi_1$ and $phi_2$, and two orthogonal spin states, $zeta_1$ and $zeta_2$, and define
$$
psi = frac1sqrt2bigg[phi_1zeta_1+phi_2zeta_2 bigg].
$$
Moreover, to be clear: the hamiltonian of a system has absolutely no effect on the allowed wavefunctions for that system. The only thing that depends on the hamiltonian is the energy eigenstates.
The result you want is the following:
If the hamiltonian is separable into spatial and spin components as $$ H = H_mathrmspaceotimes mathbb I+ mathbb I otimes H_mathrmspin,$$ with $H_mathrmspaceotimes mathbb I$ commuting with all spin operators and $mathbb I otimes H_mathrmspin$ commuting with all space operators, then there exists an eigenbasis for $H$ of the separable form $(1)$.
To build that eigenbasis, simply diagonalize $H_mathrmspace$ and $H_mathrmspin$ independently, and form tensor products of their eigenstates. (Note also that the quantifiers here are crucial, particularly the "If" in the hypotheses and the "there exists" in the results.)
answered Mar 25 at 13:08
Emilio PisantyEmilio Pisanty
86.3k23214434
$endgroup$
1
$begingroup$
@SRS The claim is specifically that there exists a separable eigenbasis. There is no claim that all eigenbases for such a hamiltonian are separable, because that claim is false. Please read more carefully.
$endgroup$
– Emilio Pisanty
Mar 25 at 13:23
$begingroup$
@SRS "H atom problem" is undefined. (I.e.: are you including fine structure and spin-orbit coupling?) If you're only talking about the Keplerian hamiltonian (i.e. kinetic energy plus electrostatic potential energy, with a frozen proton) then yes, a separable eigenbasis exists; there the $zeta_j$ are arbitrary (as there is no spin hamiltonian). If you're including spin-orbit coupling, then the hamiltonian does not satisfy the hypotheses I laid out, and the result is false.
$endgroup$
– Emilio Pisanty
Mar 25 at 13:37
$begingroup$
Comments are not for back-and-forth - particularly about another user's question. If you have further queries, take them to chat or ask separately.
$endgroup$
– Emilio Pisanty
Mar 25 at 13:38
$begingroup$
Is the statement that all the $H_spaceotimes I$ must commute with all the $Iotimes H_spin$ not redundant? From the way you have written them, it seems like they must commute, no?
$endgroup$
– user1936752
Mar 25 at 14:55
$begingroup$
@user1936752 Yes, this is redundant, but I don't think it hurts.
$endgroup$
– Emilio Pisanty
Mar 25 at 14:56
add a comment |
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1 Answer
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1 Answer
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$begingroup$
Your claim
[any arbitrary] wavefunction of an electron $psi(vecr,s)$ can always be written as $$psi(vecr,s)=phi(vecr)zeta_s,m_s tag 1$$ where $phi(vecr)$ is the space part and $zeta_s,m_s$ is the spin part of the total wavefunction $psi(vecr,s)$
is false. It is perfectly possible to produce wavefunctions which cannot be written in that separable form - for a simple example, just take two orthogonal spatial wavefunctions, $phi_1$ and $phi_2$, and two orthogonal spin states, $zeta_1$ and $zeta_2$, and define
$$
psi = frac1sqrt2bigg[phi_1zeta_1+phi_2zeta_2 bigg].
$$
Moreover, to be clear: the hamiltonian of a system has absolutely no effect on the allowed wavefunctions for that system. The only thing that depends on the hamiltonian is the energy eigenstates.
The result you want is the following:
If the hamiltonian is separable into spatial and spin components as $$ H = H_mathrmspaceotimes mathbb I+ mathbb I otimes H_mathrmspin,$$ with $H_mathrmspaceotimes mathbb I$ commuting with all spin operators and $mathbb I otimes H_mathrmspin$ commuting with all space operators, then there exists an eigenbasis for $H$ of the separable form $(1)$.
To build that eigenbasis, simply diagonalize $H_mathrmspace$ and $H_mathrmspin$ independently, and form tensor products of their eigenstates. (Note also that the quantifiers here are crucial, particularly the "If" in the hypotheses and the "there exists" in the results.)
answered Mar 25 at 13:08
Emilio PisantyEmilio Pisanty
86.3k23214434
$endgroup$
1
$begingroup$
@SRS The claim is specifically that there exists a separable eigenbasis. There is no claim that all eigenbases for such a hamiltonian are separable, because that claim is false. Please read more carefully.
$endgroup$
– Emilio Pisanty
Mar 25 at 13:23
$begingroup$
@SRS "H atom problem" is undefined. (I.e.: are you including fine structure and spin-orbit coupling?) If you're only talking about the Keplerian hamiltonian (i.e. kinetic energy plus electrostatic potential energy, with a frozen proton) then yes, a separable eigenbasis exists; there the $zeta_j$ are arbitrary (as there is no spin hamiltonian). If you're including spin-orbit coupling, then the hamiltonian does not satisfy the hypotheses I laid out, and the result is false.
$endgroup$
– Emilio Pisanty
Mar 25 at 13:37
$begingroup$
Comments are not for back-and-forth - particularly about another user's question. If you have further queries, take them to chat or ask separately.
$endgroup$
– Emilio Pisanty
Mar 25 at 13:38
$begingroup$
Is the statement that all the $H_spaceotimes I$ must commute with all the $Iotimes H_spin$ not redundant? From the way you have written them, it seems like they must commute, no?
$endgroup$
– user1936752
Mar 25 at 14:55
$begingroup$
@user1936752 Yes, this is redundant, but I don't think it hurts.
$endgroup$
– Emilio Pisanty
Mar 25 at 14:56
add a comment |
$begingroup$
Your claim
[any arbitrary] wavefunction of an electron $psi(vecr,s)$ can always be written as $$psi(vecr,s)=phi(vecr)zeta_s,m_s tag 1$$ where $phi(vecr)$ is the space part and $zeta_s,m_s$ is the spin part of the total wavefunction $psi(vecr,s)$
is false. It is perfectly possible to produce wavefunctions which cannot be written in that separable form - for a simple example, just take two orthogonal spatial wavefunctions, $phi_1$ and $phi_2$, and two orthogonal spin states, $zeta_1$ and $zeta_2$, and define
$$
psi = frac1sqrt2bigg[phi_1zeta_1+phi_2zeta_2 bigg].
$$
Moreover, to be clear: the hamiltonian of a system has absolutely no effect on the allowed wavefunctions for that system. The only thing that depends on the hamiltonian is the energy eigenstates.
The result you want is the following:
If the hamiltonian is separable into spatial and spin components as $$ H = H_mathrmspaceotimes mathbb I+ mathbb I otimes H_mathrmspin,$$ with $H_mathrmspaceotimes mathbb I$ commuting with all spin operators and $mathbb I otimes H_mathrmspin$ commuting with all space operators, then there exists an eigenbasis for $H$ of the separable form $(1)$.
To build that eigenbasis, simply diagonalize $H_mathrmspace$ and $H_mathrmspin$ independently, and form tensor products of their eigenstates. (Note also that the quantifiers here are crucial, particularly the "If" in the hypotheses and the "there exists" in the results.)
answered Mar 25 at 13:08
Emilio PisantyEmilio Pisanty
86.3k23214434
$endgroup$
1
$begingroup$
@SRS The claim is specifically that there exists a separable eigenbasis. There is no claim that all eigenbases for such a hamiltonian are separable, because that claim is false. Please read more carefully.
$endgroup$
– Emilio Pisanty
Mar 25 at 13:23
$begingroup$
@SRS "H atom problem" is undefined. (I.e.: are you including fine structure and spin-orbit coupling?) If you're only talking about the Keplerian hamiltonian (i.e. kinetic energy plus electrostatic potential energy, with a frozen proton) then yes, a separable eigenbasis exists; there the $zeta_j$ are arbitrary (as there is no spin hamiltonian). If you're including spin-orbit coupling, then the hamiltonian does not satisfy the hypotheses I laid out, and the result is false.
$endgroup$
– Emilio Pisanty
Mar 25 at 13:37
$begingroup$
Comments are not for back-and-forth - particularly about another user's question. If you have further queries, take them to chat or ask separately.
$endgroup$
– Emilio Pisanty
Mar 25 at 13:38
$begingroup$
Is the statement that all the $H_spaceotimes I$ must commute with all the $Iotimes H_spin$ not redundant? From the way you have written them, it seems like they must commute, no?
$endgroup$
– user1936752
Mar 25 at 14:55
$begingroup$
@user1936752 Yes, this is redundant, but I don't think it hurts.
$endgroup$
– Emilio Pisanty
Mar 25 at 14:56
add a comment |
$begingroup$
Your claim
[any arbitrary] wavefunction of an electron $psi(vecr,s)$ can always be written as $$psi(vecr,s)=phi(vecr)zeta_s,m_s tag 1$$ where $phi(vecr)$ is the space part and $zeta_s,m_s$ is the spin part of the total wavefunction $psi(vecr,s)$
is false. It is perfectly possible to produce wavefunctions which cannot be written in that separable form - for a simple example, just take two orthogonal spatial wavefunctions, $phi_1$ and $phi_2$, and two orthogonal spin states, $zeta_1$ and $zeta_2$, and define
$$
psi = frac1sqrt2bigg[phi_1zeta_1+phi_2zeta_2 bigg].
$$
Moreover, to be clear: the hamiltonian of a system has absolutely no effect on the allowed wavefunctions for that system. The only thing that depends on the hamiltonian is the energy eigenstates.
The result you want is the following:
If the hamiltonian is separable into spatial and spin components as $$ H = H_mathrmspaceotimes mathbb I+ mathbb I otimes H_mathrmspin,$$ with $H_mathrmspaceotimes mathbb I$ commuting with all spin operators and $mathbb I otimes H_mathrmspin$ commuting with all space operators, then there exists an eigenbasis for $H$ of the separable form $(1)$.
To build that eigenbasis, simply diagonalize $H_mathrmspace$ and $H_mathrmspin$ independently, and form tensor products of their eigenstates. (Note also that the quantifiers here are crucial, particularly the "If" in the hypotheses and the "there exists" in the results.)
answered Mar 25 at 13:08
Emilio PisantyEmilio Pisanty
86.3k23214434
$endgroup$
Your claim
[any arbitrary] wavefunction of an electron $psi(vecr,s)$ can always be written as $$psi(vecr,s)=phi(vecr)zeta_s,m_s tag 1$$ where $phi(vecr)$ is the space part and $zeta_s,m_s$ is the spin part of the total wavefunction $psi(vecr,s)$
is false. It is perfectly possible to produce wavefunctions which cannot be written in that separable form - for a simple example, just take two orthogonal spatial wavefunctions, $phi_1$ and $phi_2$, and two orthogonal spin states, $zeta_1$ and $zeta_2$, and define
$$
psi = frac1sqrt2bigg[phi_1zeta_1+phi_2zeta_2 bigg].
$$
Moreover, to be clear: the hamiltonian of a system has absolutely no effect on the allowed wavefunctions for that system. The only thing that depends on the hamiltonian is the energy eigenstates.
The result you want is the following:
If the hamiltonian is separable into spatial and spin components as $$ H = H_mathrmspaceotimes mathbb I+ mathbb I otimes H_mathrmspin,$$ with $H_mathrmspaceotimes mathbb I$ commuting with all spin operators and $mathbb I otimes H_mathrmspin$ commuting with all space operators, then there exists an eigenbasis for $H$ of the separable form $(1)$.
To build that eigenbasis, simply diagonalize $H_mathrmspace$ and $H_mathrmspin$ independently, and form tensor products of their eigenstates. (Note also that the quantifiers here are crucial, particularly the "If" in the hypotheses and the "there exists" in the results.)
answered Mar 25 at 13:08
Emilio PisantyEmilio Pisanty
86.3k23214434
edited Mar 25 at 13:39
answered Mar 25 at 13:08
Emilio PisantyEmilio Pisanty
86.3k23214434
answered Mar 25 at 13:08
Emilio PisantyEmilio Pisanty
86.3k23214434
answered Mar 25 at 13:08
Emilio PisantyEmilio Pisanty
86.3k23214434
86.3k23214434
1
$begingroup$
@SRS The claim is specifically that there exists a separable eigenbasis. There is no claim that all eigenbases for such a hamiltonian are separable, because that claim is false. Please read more carefully.
$endgroup$
– Emilio Pisanty
Mar 25 at 13:23
$begingroup$
@SRS "H atom problem" is undefined. (I.e.: are you including fine structure and spin-orbit coupling?) If you're only talking about the Keplerian hamiltonian (i.e. kinetic energy plus electrostatic potential energy, with a frozen proton) then yes, a separable eigenbasis exists; there the $zeta_j$ are arbitrary (as there is no spin hamiltonian). If you're including spin-orbit coupling, then the hamiltonian does not satisfy the hypotheses I laid out, and the result is false.
$endgroup$
– Emilio Pisanty
Mar 25 at 13:37
$begingroup$
Comments are not for back-and-forth - particularly about another user's question. If you have further queries, take them to chat or ask separately.
$endgroup$
– Emilio Pisanty
Mar 25 at 13:38
$begingroup$
Is the statement that all the $H_spaceotimes I$ must commute with all the $Iotimes H_spin$ not redundant? From the way you have written them, it seems like they must commute, no?
$endgroup$
– user1936752
Mar 25 at 14:55
$begingroup$
@user1936752 Yes, this is redundant, but I don't think it hurts.
$endgroup$
– Emilio Pisanty
Mar 25 at 14:56
add a comment |
1
$begingroup$
@SRS The claim is specifically that there exists a separable eigenbasis. There is no claim that all eigenbases for such a hamiltonian are separable, because that claim is false. Please read more carefully.
$endgroup$
– Emilio Pisanty
Mar 25 at 13:23
$begingroup$
@SRS "H atom problem" is undefined. (I.e.: are you including fine structure and spin-orbit coupling?) If you're only talking about the Keplerian hamiltonian (i.e. kinetic energy plus electrostatic potential energy, with a frozen proton) then yes, a separable eigenbasis exists; there the $zeta_j$ are arbitrary (as there is no spin hamiltonian). If you're including spin-orbit coupling, then the hamiltonian does not satisfy the hypotheses I laid out, and the result is false.
$endgroup$
– Emilio Pisanty
Mar 25 at 13:37
$begingroup$
Comments are not for back-and-forth - particularly about another user's question. If you have further queries, take them to chat or ask separately.
$endgroup$
– Emilio Pisanty
Mar 25 at 13:38
$begingroup$
Is the statement that all the $H_spaceotimes I$ must commute with all the $Iotimes H_spin$ not redundant? From the way you have written them, it seems like they must commute, no?
$endgroup$
– user1936752
Mar 25 at 14:55
$begingroup$
@user1936752 Yes, this is redundant, but I don't think it hurts.
$endgroup$
– Emilio Pisanty
Mar 25 at 14:56
1
1
$begingroup$
@SRS The claim is specifically that there exists a separable eigenbasis. There is no claim that all eigenbases for such a hamiltonian are separable, because that claim is false. Please read more carefully.
$endgroup$
– Emilio Pisanty
Mar 25 at 13:23
$begingroup$
@SRS The claim is specifically that there exists a separable eigenbasis. There is no claim that all eigenbases for such a hamiltonian are separable, because that claim is false. Please read more carefully.
$endgroup$
– Emilio Pisanty
Mar 25 at 13:23
$begingroup$
@SRS "H atom problem" is undefined. (I.e.: are you including fine structure and spin-orbit coupling?) If you're only talking about the Keplerian hamiltonian (i.e. kinetic energy plus electrostatic potential energy, with a frozen proton) then yes, a separable eigenbasis exists; there the $zeta_j$ are arbitrary (as there is no spin hamiltonian). If you're including spin-orbit coupling, then the hamiltonian does not satisfy the hypotheses I laid out, and the result is false.
$endgroup$
– Emilio Pisanty
Mar 25 at 13:37
$begingroup$
@SRS "H atom problem" is undefined. (I.e.: are you including fine structure and spin-orbit coupling?) If you're only talking about the Keplerian hamiltonian (i.e. kinetic energy plus electrostatic potential energy, with a frozen proton) then yes, a separable eigenbasis exists; there the $zeta_j$ are arbitrary (as there is no spin hamiltonian). If you're including spin-orbit coupling, then the hamiltonian does not satisfy the hypotheses I laid out, and the result is false.
$endgroup$
– Emilio Pisanty
Mar 25 at 13:37
$begingroup$
Comments are not for back-and-forth - particularly about another user's question. If you have further queries, take them to chat or ask separately.
$endgroup$
– Emilio Pisanty
Mar 25 at 13:38
$begingroup$
Comments are not for back-and-forth - particularly about another user's question. If you have further queries, take them to chat or ask separately.
$endgroup$
– Emilio Pisanty
Mar 25 at 13:38
$begingroup$
Is the statement that all the $H_spaceotimes I$ must commute with all the $Iotimes H_spin$ not redundant? From the way you have written them, it seems like they must commute, no?
$endgroup$
– user1936752
Mar 25 at 14:55
$begingroup$
Is the statement that all the $H_spaceotimes I$ must commute with all the $Iotimes H_spin$ not redundant? From the way you have written them, it seems like they must commute, no?
$endgroup$
– user1936752
Mar 25 at 14:55
$begingroup$
@user1936752 Yes, this is redundant, but I don't think it hurts.
$endgroup$
– Emilio Pisanty
Mar 25 at 14:56
$begingroup$
@user1936752 Yes, this is redundant, but I don't think it hurts.
$endgroup$
– Emilio Pisanty
Mar 25 at 14:56
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