Mixing
Bloch's theorem in one dimension, confusion about proof
$begingroup$
I was looking at the derivation of Bloch's theorem in Griffith's QM:
If $V(x+a)=V(x)$ for any $x$ and some $a$, and
$psi$ solves
$$
Hpsi =lambda psi
$$
for $H=-frac{hbar^2}{2pi}frac{d^2}{dx^2}+V(x)$
then
$$
psi(x+a)=e^{ika}psi(x)
$$
The proof relies on the fact that since the shift operator $Dpsi(x)=psi(x+a)$ commutes with the Hamiltonian, we may take an eigenfunction of $H$ to serve as an eigenfunction for $D$. The wikipedia article on the theorem (on a 3 dimensional crystal lattice but the principle is the same) once again asserts the existence of a simultaneous eigenbasis for the two operators.
What am I missing here? I know that this is a fact for matrices, but we are working over $L^2(mathbb{R})$ here, where I don't believe this is true for unbounded self adjoint operators, and it is not clear to me why it should be true with this particular operator (indeed I think it is not).
Thanks for any help and I apologize if I am being naive.
functional-analysis mathematical-physics quantum-mechanics unbounded-operators
asked May 21 '18 at 2:28
qbertqbert
22.1k32560
$endgroup$
|
show 6 more comments
$begingroup$
I was looking at the derivation of Bloch's theorem in Griffith's QM:
If $V(x+a)=V(x)$ for any $x$ and some $a$, and
$psi$ solves
$$
Hpsi =lambda psi
$$
for $H=-frac{hbar^2}{2pi}frac{d^2}{dx^2}+V(x)$
then
$$
psi(x+a)=e^{ika}psi(x)
$$
The proof relies on the fact that since the shift operator $Dpsi(x)=psi(x+a)$ commutes with the Hamiltonian, we may take an eigenfunction of $H$ to serve as an eigenfunction for $D$. The wikipedia article on the theorem (on a 3 dimensional crystal lattice but the principle is the same) once again asserts the existence of a simultaneous eigenbasis for the two operators.
What am I missing here? I know that this is a fact for matrices, but we are working over $L^2(mathbb{R})$ here, where I don't believe this is true for unbounded self adjoint operators, and it is not clear to me why it should be true with this particular operator (indeed I think it is not).
Thanks for any help and I apologize if I am being naive.
functional-analysis mathematical-physics quantum-mechanics unbounded-operators
asked May 21 '18 at 2:28
qbertqbert
22.1k32560
$endgroup$
2
$begingroup$
I think Griffiths is not the place you should be if you are even asking these questions. Try Brian Hall's Quantum Mechanics for Mathematicians, or Barry Simon's Methods of Mathematical Physics.
$endgroup$
– user14717
May 21 '18 at 2:50
$begingroup$
Fair enough. I am also reading through the first Volume of the Simon series and couldn't find a reference to the theorem; do you know if it is in a later volume? Thank you for the other reference as well. @user14717
$endgroup$
– qbert
May 21 '18 at 2:54
$begingroup$
You can show that $D$ commutes with the Laplacian, and by virtue of $V(x) = V(x+a)$ you can then show that both $H$ and $D$ are simultaneously diagonalizable.
$endgroup$
– Guillermo Angeris
May 21 '18 at 4:31
$begingroup$
@GuillermoAngeris diagonalizable? My point is that we are not in the finite dimensional setting.
$endgroup$
– qbert
May 21 '18 at 4:32
1
$begingroup$
@qbert: just found a nice paper which roughly addresses the problems which you are concerned about, but in a much more complicated setting: ams.org/journals/tran/1951-070-02/S0002-9947-1951-0041010-X/…, Tosio Kato, Fundamental Properties of Hamiltonian Operators of Schrodinger Type.
$endgroup$
– user14717
Jun 8 '18 at 2:05
|
show 6 more comments
$begingroup$
I was looking at the derivation of Bloch's theorem in Griffith's QM:
If $V(x+a)=V(x)$ for any $x$ and some $a$, and
$psi$ solves
$$
Hpsi =lambda psi
$$
for $H=-frac{hbar^2}{2pi}frac{d^2}{dx^2}+V(x)$
then
$$
psi(x+a)=e^{ika}psi(x)
$$
The proof relies on the fact that since the shift operator $Dpsi(x)=psi(x+a)$ commutes with the Hamiltonian, we may take an eigenfunction of $H$ to serve as an eigenfunction for $D$. The wikipedia article on the theorem (on a 3 dimensional crystal lattice but the principle is the same) once again asserts the existence of a simultaneous eigenbasis for the two operators.
What am I missing here? I know that this is a fact for matrices, but we are working over $L^2(mathbb{R})$ here, where I don't believe this is true for unbounded self adjoint operators, and it is not clear to me why it should be true with this particular operator (indeed I think it is not).
Thanks for any help and I apologize if I am being naive.
functional-analysis mathematical-physics quantum-mechanics unbounded-operators
asked May 21 '18 at 2:28
qbertqbert
22.1k32560
$endgroup$
I was looking at the derivation of Bloch's theorem in Griffith's QM:
If $V(x+a)=V(x)$ for any $x$ and some $a$, and
$psi$ solves
$$
Hpsi =lambda psi
$$
for $H=-frac{hbar^2}{2pi}frac{d^2}{dx^2}+V(x)$
then
$$
psi(x+a)=e^{ika}psi(x)
$$
The proof relies on the fact that since the shift operator $Dpsi(x)=psi(x+a)$ commutes with the Hamiltonian, we may take an eigenfunction of $H$ to serve as an eigenfunction for $D$. The wikipedia article on the theorem (on a 3 dimensional crystal lattice but the principle is the same) once again asserts the existence of a simultaneous eigenbasis for the two operators.
What am I missing here? I know that this is a fact for matrices, but we are working over $L^2(mathbb{R})$ here, where I don't believe this is true for unbounded self adjoint operators, and it is not clear to me why it should be true with this particular operator (indeed I think it is not).
Thanks for any help and I apologize if I am being naive.
functional-analysis mathematical-physics quantum-mechanics unbounded-operators
functional-analysis mathematical-physics quantum-mechanics unbounded-operators
asked May 21 '18 at 2:28
qbertqbert
22.1k32560
asked May 21 '18 at 2:28
qbertqbert
22.1k32560
edited May 22 '18 at 2:09
qbert
asked May 21 '18 at 2:28
qbertqbert
22.1k32560
asked May 21 '18 at 2:28
qbertqbert
22.1k32560
asked May 21 '18 at 2:28
qbertqbert
22.1k32560
22.1k32560
2
$begingroup$
I think Griffiths is not the place you should be if you are even asking these questions. Try Brian Hall's Quantum Mechanics for Mathematicians, or Barry Simon's Methods of Mathematical Physics.
$endgroup$
– user14717
May 21 '18 at 2:50
$begingroup$
Fair enough. I am also reading through the first Volume of the Simon series and couldn't find a reference to the theorem; do you know if it is in a later volume? Thank you for the other reference as well. @user14717
$endgroup$
– qbert
May 21 '18 at 2:54
$begingroup$
You can show that $D$ commutes with the Laplacian, and by virtue of $V(x) = V(x+a)$ you can then show that both $H$ and $D$ are simultaneously diagonalizable.
$endgroup$
– Guillermo Angeris
May 21 '18 at 4:31
$begingroup$
@GuillermoAngeris diagonalizable? My point is that we are not in the finite dimensional setting.
$endgroup$
– qbert
May 21 '18 at 4:32
1
$begingroup$
@qbert: just found a nice paper which roughly addresses the problems which you are concerned about, but in a much more complicated setting: ams.org/journals/tran/1951-070-02/S0002-9947-1951-0041010-X/…, Tosio Kato, Fundamental Properties of Hamiltonian Operators of Schrodinger Type.
$endgroup$
– user14717
Jun 8 '18 at 2:05
|
show 6 more comments
2
$begingroup$
I think Griffiths is not the place you should be if you are even asking these questions. Try Brian Hall's Quantum Mechanics for Mathematicians, or Barry Simon's Methods of Mathematical Physics.
$endgroup$
– user14717
May 21 '18 at 2:50
$begingroup$
Fair enough. I am also reading through the first Volume of the Simon series and couldn't find a reference to the theorem; do you know if it is in a later volume? Thank you for the other reference as well. @user14717
$endgroup$
– qbert
May 21 '18 at 2:54
$begingroup$
You can show that $D$ commutes with the Laplacian, and by virtue of $V(x) = V(x+a)$ you can then show that both $H$ and $D$ are simultaneously diagonalizable.
$endgroup$
– Guillermo Angeris
May 21 '18 at 4:31
$begingroup$
@GuillermoAngeris diagonalizable? My point is that we are not in the finite dimensional setting.
$endgroup$
– qbert
May 21 '18 at 4:32
1
$begingroup$
@qbert: just found a nice paper which roughly addresses the problems which you are concerned about, but in a much more complicated setting: ams.org/journals/tran/1951-070-02/S0002-9947-1951-0041010-X/…, Tosio Kato, Fundamental Properties of Hamiltonian Operators of Schrodinger Type.
$endgroup$
– user14717
Jun 8 '18 at 2:05
2
2
$begingroup$
I think Griffiths is not the place you should be if you are even asking these questions. Try Brian Hall's Quantum Mechanics for Mathematicians, or Barry Simon's Methods of Mathematical Physics.
$endgroup$
– user14717
May 21 '18 at 2:50
$begingroup$
I think Griffiths is not the place you should be if you are even asking these questions. Try Brian Hall's Quantum Mechanics for Mathematicians, or Barry Simon's Methods of Mathematical Physics.
$endgroup$
– user14717
May 21 '18 at 2:50
$begingroup$
Fair enough. I am also reading through the first Volume of the Simon series and couldn't find a reference to the theorem; do you know if it is in a later volume? Thank you for the other reference as well. @user14717
$endgroup$
– qbert
May 21 '18 at 2:54
$begingroup$
Fair enough. I am also reading through the first Volume of the Simon series and couldn't find a reference to the theorem; do you know if it is in a later volume? Thank you for the other reference as well. @user14717
$endgroup$
– qbert
May 21 '18 at 2:54
$begingroup$
You can show that $D$ commutes with the Laplacian, and by virtue of $V(x) = V(x+a)$ you can then show that both $H$ and $D$ are simultaneously diagonalizable.
$endgroup$
– Guillermo Angeris
May 21 '18 at 4:31
$begingroup$
You can show that $D$ commutes with the Laplacian, and by virtue of $V(x) = V(x+a)$ you can then show that both $H$ and $D$ are simultaneously diagonalizable.
$endgroup$
– Guillermo Angeris
May 21 '18 at 4:31
$begingroup$
@GuillermoAngeris diagonalizable? My point is that we are not in the finite dimensional setting.
$endgroup$
– qbert
May 21 '18 at 4:32
$begingroup$
@GuillermoAngeris diagonalizable? My point is that we are not in the finite dimensional setting.
$endgroup$
– qbert
May 21 '18 at 4:32
1
1
$begingroup$
@qbert: just found a nice paper which roughly addresses the problems which you are concerned about, but in a much more complicated setting: ams.org/journals/tran/1951-070-02/S0002-9947-1951-0041010-X/…, Tosio Kato, Fundamental Properties of Hamiltonian Operators of Schrodinger Type.
$endgroup$
– user14717
Jun 8 '18 at 2:05
$begingroup$
@qbert: just found a nice paper which roughly addresses the problems which you are concerned about, but in a much more complicated setting: ams.org/journals/tran/1951-070-02/S0002-9947-1951-0041010-X/…, Tosio Kato, Fundamental Properties of Hamiltonian Operators of Schrodinger Type.
$endgroup$
– user14717
Jun 8 '18 at 2:05
|
show 6 more comments
1 Answer
1
active
oldest
votes
$begingroup$
The statement, if considered for a Hamiltonian with periodic potential which acts as a densely defined selfadjoint operator on an L^2 of the full space R^n, is wrong. Therefore, you won't find "Bloch's theorem" in this form in Reed/Simon.
In vol 4., Reed and Simon treat Schroedinger operators with periodic potentials in chapter XIII.16.
You find there a theorem saying that under reasonable assumptions about the regularity of the periodic potential, the Hamiltonian operator has purely absolutely continuous spectrum (which gives rise to the physicist's "band structure") and therefore no eigenvalues or eigenfunctions would exist as elements in the Hilbert space. (Theorem XII.90 for 1 dimension and Theorem XIII.100 for dimensions at least 2.)
Physicists switch repeatedly between a treatment of finitely many stacked/translated primitive cells and the case of the full space with infinitely many translations of a primitive cell. Reed and Simon comment on this situation when they explain the "density of states measure" near the end of this chapter in Theorem XIII.101.
Finitely many cells give an L^2-space over a compact domain. In this case, the Hamiltonian usually will have compact resolvents and a complete orthonormal set of eigenstates. There will be no continuous "band structure", but only discrete spectrum.
In the case of the full space, the direct integral decomposition of Reed/Simon explains how the continuous band-structure is obtained from combining the discrete spectra of operators of a family H(q) (in the "p-space version") or H(theta) (in the "x-space version").
A sort of "eigenfunction expansion" is provided in Thereom XIII.98 (b).
As for Floquet's theorem for ODEs (i.e. one space dimension, no L^2-space...) you can find a nice explanation of the monodromy matrix and of the charateristic exponents in Vladimir Arnol'd's book about ordinary differential equations, and a "modern" proof in Hale's or Amann's or Walter's book about ordinary differential equations.
answered Jan 6 at 12:58
ASlateffASlateff
263
$endgroup$
$begingroup$
This is a really great answer. I will read what you suggest as soon as I can get back to the library. A question about the finite number of cells version: it seems like here the argument above could potentially be salvaged except that the shift operator wont be compact. Any ideas on this?
$endgroup$
– qbert
Jan 7 at 17:55
1
$begingroup$
In 1 dimension in the infinitely extended case, you will have eg. a closed interval [-1/2,1/2] as a primitive cell and a 1-periodic potential V (smooth, say). The finitely stacked situation consists of a finite number (say N) of such intervals arranged in a chain to give [-N/2,N/2]. Physicists start with such a finite chain and take the Hilbert space $L^2[-N/2, N/2]$. The kinetic energy operator is taking (-1) times a second derivative, you have to provide boundary conditions to obtain a selfadjoint operator. Then implement the shift translations (not faithful) and find common eigenfunctions.
$endgroup$
– ASlateff
Jan 8 at 13:09
$begingroup$
Got it. I was overthinking the shift operator. Thanks again
$endgroup$
– qbert
Jan 8 at 13:19
$begingroup$
Now, that you got it, go for Reed/Simon vol. 4 and try to understand, how physicists' Bloch functions appear in connection with the direct integral decompositions. Unfortunately, Reed/Simon did not comment explicitely on this relation, so one has to do this last step oneself.
$endgroup$
– ASlateff
Jan 8 at 17:02
add a comment |
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$begingroup$
The statement, if considered for a Hamiltonian with periodic potential which acts as a densely defined selfadjoint operator on an L^2 of the full space R^n, is wrong. Therefore, you won't find "Bloch's theorem" in this form in Reed/Simon.
In vol 4., Reed and Simon treat Schroedinger operators with periodic potentials in chapter XIII.16.
You find there a theorem saying that under reasonable assumptions about the regularity of the periodic potential, the Hamiltonian operator has purely absolutely continuous spectrum (which gives rise to the physicist's "band structure") and therefore no eigenvalues or eigenfunctions would exist as elements in the Hilbert space. (Theorem XII.90 for 1 dimension and Theorem XIII.100 for dimensions at least 2.)
Physicists switch repeatedly between a treatment of finitely many stacked/translated primitive cells and the case of the full space with infinitely many translations of a primitive cell. Reed and Simon comment on this situation when they explain the "density of states measure" near the end of this chapter in Theorem XIII.101.
Finitely many cells give an L^2-space over a compact domain. In this case, the Hamiltonian usually will have compact resolvents and a complete orthonormal set of eigenstates. There will be no continuous "band structure", but only discrete spectrum.
In the case of the full space, the direct integral decomposition of Reed/Simon explains how the continuous band-structure is obtained from combining the discrete spectra of operators of a family H(q) (in the "p-space version") or H(theta) (in the "x-space version").
A sort of "eigenfunction expansion" is provided in Thereom XIII.98 (b).
As for Floquet's theorem for ODEs (i.e. one space dimension, no L^2-space...) you can find a nice explanation of the monodromy matrix and of the charateristic exponents in Vladimir Arnol'd's book about ordinary differential equations, and a "modern" proof in Hale's or Amann's or Walter's book about ordinary differential equations.
answered Jan 6 at 12:58
ASlateffASlateff
263
$endgroup$
$begingroup$
This is a really great answer. I will read what you suggest as soon as I can get back to the library. A question about the finite number of cells version: it seems like here the argument above could potentially be salvaged except that the shift operator wont be compact. Any ideas on this?
$endgroup$
– qbert
Jan 7 at 17:55
1
$begingroup$
In 1 dimension in the infinitely extended case, you will have eg. a closed interval [-1/2,1/2] as a primitive cell and a 1-periodic potential V (smooth, say). The finitely stacked situation consists of a finite number (say N) of such intervals arranged in a chain to give [-N/2,N/2]. Physicists start with such a finite chain and take the Hilbert space $L^2[-N/2, N/2]$. The kinetic energy operator is taking (-1) times a second derivative, you have to provide boundary conditions to obtain a selfadjoint operator. Then implement the shift translations (not faithful) and find common eigenfunctions.
$endgroup$
– ASlateff
Jan 8 at 13:09
$begingroup$
Got it. I was overthinking the shift operator. Thanks again
$endgroup$
– qbert
Jan 8 at 13:19
$begingroup$
Now, that you got it, go for Reed/Simon vol. 4 and try to understand, how physicists' Bloch functions appear in connection with the direct integral decompositions. Unfortunately, Reed/Simon did not comment explicitely on this relation, so one has to do this last step oneself.
$endgroup$
– ASlateff
Jan 8 at 17:02
add a comment |
$begingroup$
The statement, if considered for a Hamiltonian with periodic potential which acts as a densely defined selfadjoint operator on an L^2 of the full space R^n, is wrong. Therefore, you won't find "Bloch's theorem" in this form in Reed/Simon.
In vol 4., Reed and Simon treat Schroedinger operators with periodic potentials in chapter XIII.16.
You find there a theorem saying that under reasonable assumptions about the regularity of the periodic potential, the Hamiltonian operator has purely absolutely continuous spectrum (which gives rise to the physicist's "band structure") and therefore no eigenvalues or eigenfunctions would exist as elements in the Hilbert space. (Theorem XII.90 for 1 dimension and Theorem XIII.100 for dimensions at least 2.)
Physicists switch repeatedly between a treatment of finitely many stacked/translated primitive cells and the case of the full space with infinitely many translations of a primitive cell. Reed and Simon comment on this situation when they explain the "density of states measure" near the end of this chapter in Theorem XIII.101.
Finitely many cells give an L^2-space over a compact domain. In this case, the Hamiltonian usually will have compact resolvents and a complete orthonormal set of eigenstates. There will be no continuous "band structure", but only discrete spectrum.
In the case of the full space, the direct integral decomposition of Reed/Simon explains how the continuous band-structure is obtained from combining the discrete spectra of operators of a family H(q) (in the "p-space version") or H(theta) (in the "x-space version").
A sort of "eigenfunction expansion" is provided in Thereom XIII.98 (b).
As for Floquet's theorem for ODEs (i.e. one space dimension, no L^2-space...) you can find a nice explanation of the monodromy matrix and of the charateristic exponents in Vladimir Arnol'd's book about ordinary differential equations, and a "modern" proof in Hale's or Amann's or Walter's book about ordinary differential equations.
answered Jan 6 at 12:58
ASlateffASlateff
263
$endgroup$
$begingroup$
This is a really great answer. I will read what you suggest as soon as I can get back to the library. A question about the finite number of cells version: it seems like here the argument above could potentially be salvaged except that the shift operator wont be compact. Any ideas on this?
$endgroup$
– qbert
Jan 7 at 17:55
1
$begingroup$
In 1 dimension in the infinitely extended case, you will have eg. a closed interval [-1/2,1/2] as a primitive cell and a 1-periodic potential V (smooth, say). The finitely stacked situation consists of a finite number (say N) of such intervals arranged in a chain to give [-N/2,N/2]. Physicists start with such a finite chain and take the Hilbert space $L^2[-N/2, N/2]$. The kinetic energy operator is taking (-1) times a second derivative, you have to provide boundary conditions to obtain a selfadjoint operator. Then implement the shift translations (not faithful) and find common eigenfunctions.
$endgroup$
– ASlateff
Jan 8 at 13:09
$begingroup$
Got it. I was overthinking the shift operator. Thanks again
$endgroup$
– qbert
Jan 8 at 13:19
$begingroup$
Now, that you got it, go for Reed/Simon vol. 4 and try to understand, how physicists' Bloch functions appear in connection with the direct integral decompositions. Unfortunately, Reed/Simon did not comment explicitely on this relation, so one has to do this last step oneself.
$endgroup$
– ASlateff
Jan 8 at 17:02
add a comment |
$begingroup$
The statement, if considered for a Hamiltonian with periodic potential which acts as a densely defined selfadjoint operator on an L^2 of the full space R^n, is wrong. Therefore, you won't find "Bloch's theorem" in this form in Reed/Simon.
In vol 4., Reed and Simon treat Schroedinger operators with periodic potentials in chapter XIII.16.
You find there a theorem saying that under reasonable assumptions about the regularity of the periodic potential, the Hamiltonian operator has purely absolutely continuous spectrum (which gives rise to the physicist's "band structure") and therefore no eigenvalues or eigenfunctions would exist as elements in the Hilbert space. (Theorem XII.90 for 1 dimension and Theorem XIII.100 for dimensions at least 2.)
Physicists switch repeatedly between a treatment of finitely many stacked/translated primitive cells and the case of the full space with infinitely many translations of a primitive cell. Reed and Simon comment on this situation when they explain the "density of states measure" near the end of this chapter in Theorem XIII.101.
Finitely many cells give an L^2-space over a compact domain. In this case, the Hamiltonian usually will have compact resolvents and a complete orthonormal set of eigenstates. There will be no continuous "band structure", but only discrete spectrum.
In the case of the full space, the direct integral decomposition of Reed/Simon explains how the continuous band-structure is obtained from combining the discrete spectra of operators of a family H(q) (in the "p-space version") or H(theta) (in the "x-space version").
A sort of "eigenfunction expansion" is provided in Thereom XIII.98 (b).
As for Floquet's theorem for ODEs (i.e. one space dimension, no L^2-space...) you can find a nice explanation of the monodromy matrix and of the charateristic exponents in Vladimir Arnol'd's book about ordinary differential equations, and a "modern" proof in Hale's or Amann's or Walter's book about ordinary differential equations.
answered Jan 6 at 12:58
ASlateffASlateff
263
$endgroup$
The statement, if considered for a Hamiltonian with periodic potential which acts as a densely defined selfadjoint operator on an L^2 of the full space R^n, is wrong. Therefore, you won't find "Bloch's theorem" in this form in Reed/Simon.
In vol 4., Reed and Simon treat Schroedinger operators with periodic potentials in chapter XIII.16.
You find there a theorem saying that under reasonable assumptions about the regularity of the periodic potential, the Hamiltonian operator has purely absolutely continuous spectrum (which gives rise to the physicist's "band structure") and therefore no eigenvalues or eigenfunctions would exist as elements in the Hilbert space. (Theorem XII.90 for 1 dimension and Theorem XIII.100 for dimensions at least 2.)
Physicists switch repeatedly between a treatment of finitely many stacked/translated primitive cells and the case of the full space with infinitely many translations of a primitive cell. Reed and Simon comment on this situation when they explain the "density of states measure" near the end of this chapter in Theorem XIII.101.
Finitely many cells give an L^2-space over a compact domain. In this case, the Hamiltonian usually will have compact resolvents and a complete orthonormal set of eigenstates. There will be no continuous "band structure", but only discrete spectrum.
In the case of the full space, the direct integral decomposition of Reed/Simon explains how the continuous band-structure is obtained from combining the discrete spectra of operators of a family H(q) (in the "p-space version") or H(theta) (in the "x-space version").
A sort of "eigenfunction expansion" is provided in Thereom XIII.98 (b).
As for Floquet's theorem for ODEs (i.e. one space dimension, no L^2-space...) you can find a nice explanation of the monodromy matrix and of the charateristic exponents in Vladimir Arnol'd's book about ordinary differential equations, and a "modern" proof in Hale's or Amann's or Walter's book about ordinary differential equations.
answered Jan 6 at 12:58
ASlateffASlateff
263
answered Jan 6 at 12:58
ASlateffASlateff
263
answered Jan 6 at 12:58
ASlateffASlateff
263
answered Jan 6 at 12:58
ASlateffASlateff
263
263
$begingroup$
This is a really great answer. I will read what you suggest as soon as I can get back to the library. A question about the finite number of cells version: it seems like here the argument above could potentially be salvaged except that the shift operator wont be compact. Any ideas on this?
$endgroup$
– qbert
Jan 7 at 17:55
1
$begingroup$
In 1 dimension in the infinitely extended case, you will have eg. a closed interval [-1/2,1/2] as a primitive cell and a 1-periodic potential V (smooth, say). The finitely stacked situation consists of a finite number (say N) of such intervals arranged in a chain to give [-N/2,N/2]. Physicists start with such a finite chain and take the Hilbert space $L^2[-N/2, N/2]$. The kinetic energy operator is taking (-1) times a second derivative, you have to provide boundary conditions to obtain a selfadjoint operator. Then implement the shift translations (not faithful) and find common eigenfunctions.
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– ASlateff
Jan 8 at 13:09
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Got it. I was overthinking the shift operator. Thanks again
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– qbert
Jan 8 at 13:19
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Now, that you got it, go for Reed/Simon vol. 4 and try to understand, how physicists' Bloch functions appear in connection with the direct integral decompositions. Unfortunately, Reed/Simon did not comment explicitely on this relation, so one has to do this last step oneself.
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– ASlateff
Jan 8 at 17:02
add a comment |
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This is a really great answer. I will read what you suggest as soon as I can get back to the library. A question about the finite number of cells version: it seems like here the argument above could potentially be salvaged except that the shift operator wont be compact. Any ideas on this?
$endgroup$
– qbert
Jan 7 at 17:55
1
$begingroup$
In 1 dimension in the infinitely extended case, you will have eg. a closed interval [-1/2,1/2] as a primitive cell and a 1-periodic potential V (smooth, say). The finitely stacked situation consists of a finite number (say N) of such intervals arranged in a chain to give [-N/2,N/2]. Physicists start with such a finite chain and take the Hilbert space $L^2[-N/2, N/2]$. The kinetic energy operator is taking (-1) times a second derivative, you have to provide boundary conditions to obtain a selfadjoint operator. Then implement the shift translations (not faithful) and find common eigenfunctions.
$endgroup$
– ASlateff
Jan 8 at 13:09
$begingroup$
Got it. I was overthinking the shift operator. Thanks again
$endgroup$
– qbert
Jan 8 at 13:19
$begingroup$
Now, that you got it, go for Reed/Simon vol. 4 and try to understand, how physicists' Bloch functions appear in connection with the direct integral decompositions. Unfortunately, Reed/Simon did not comment explicitely on this relation, so one has to do this last step oneself.
$endgroup$
– ASlateff
Jan 8 at 17:02
$begingroup$
This is a really great answer. I will read what you suggest as soon as I can get back to the library. A question about the finite number of cells version: it seems like here the argument above could potentially be salvaged except that the shift operator wont be compact. Any ideas on this?
$endgroup$
– qbert
Jan 7 at 17:55
$begingroup$
This is a really great answer. I will read what you suggest as soon as I can get back to the library. A question about the finite number of cells version: it seems like here the argument above could potentially be salvaged except that the shift operator wont be compact. Any ideas on this?
$endgroup$
– qbert
Jan 7 at 17:55
1
1
$begingroup$
In 1 dimension in the infinitely extended case, you will have eg. a closed interval [-1/2,1/2] as a primitive cell and a 1-periodic potential V (smooth, say). The finitely stacked situation consists of a finite number (say N) of such intervals arranged in a chain to give [-N/2,N/2]. Physicists start with such a finite chain and take the Hilbert space $L^2[-N/2, N/2]$. The kinetic energy operator is taking (-1) times a second derivative, you have to provide boundary conditions to obtain a selfadjoint operator. Then implement the shift translations (not faithful) and find common eigenfunctions.
$endgroup$
– ASlateff
Jan 8 at 13:09
$begingroup$
In 1 dimension in the infinitely extended case, you will have eg. a closed interval [-1/2,1/2] as a primitive cell and a 1-periodic potential V (smooth, say). The finitely stacked situation consists of a finite number (say N) of such intervals arranged in a chain to give [-N/2,N/2]. Physicists start with such a finite chain and take the Hilbert space $L^2[-N/2, N/2]$. The kinetic energy operator is taking (-1) times a second derivative, you have to provide boundary conditions to obtain a selfadjoint operator. Then implement the shift translations (not faithful) and find common eigenfunctions.
$endgroup$
– ASlateff
Jan 8 at 13:09
$begingroup$
Got it. I was overthinking the shift operator. Thanks again
$endgroup$
– qbert
Jan 8 at 13:19
$begingroup$
Got it. I was overthinking the shift operator. Thanks again
$endgroup$
– qbert
Jan 8 at 13:19
$begingroup$
Now, that you got it, go for Reed/Simon vol. 4 and try to understand, how physicists' Bloch functions appear in connection with the direct integral decompositions. Unfortunately, Reed/Simon did not comment explicitely on this relation, so one has to do this last step oneself.
$endgroup$
– ASlateff
Jan 8 at 17:02
$begingroup$
Now, that you got it, go for Reed/Simon vol. 4 and try to understand, how physicists' Bloch functions appear in connection with the direct integral decompositions. Unfortunately, Reed/Simon did not comment explicitely on this relation, so one has to do this last step oneself.
$endgroup$
– ASlateff
Jan 8 at 17:02
add a comment |
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2
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I think Griffiths is not the place you should be if you are even asking these questions. Try Brian Hall's Quantum Mechanics for Mathematicians, or Barry Simon's Methods of Mathematical Physics.
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– user14717
May 21 '18 at 2:50
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Fair enough. I am also reading through the first Volume of the Simon series and couldn't find a reference to the theorem; do you know if it is in a later volume? Thank you for the other reference as well. @user14717
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– qbert
May 21 '18 at 2:54
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You can show that $D$ commutes with the Laplacian, and by virtue of $V(x) = V(x+a)$ you can then show that both $H$ and $D$ are simultaneously diagonalizable.
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– Guillermo Angeris
May 21 '18 at 4:31
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@GuillermoAngeris diagonalizable? My point is that we are not in the finite dimensional setting.
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– qbert
May 21 '18 at 4:32
1
$begingroup$
@qbert: just found a nice paper which roughly addresses the problems which you are concerned about, but in a much more complicated setting: ams.org/journals/tran/1951-070-02/S0002-9947-1951-0041010-X/…, Tosio Kato, Fundamental Properties of Hamiltonian Operators of Schrodinger Type.
$endgroup$
– user14717
Jun 8 '18 at 2:05