Application of Schwartz Kernel Theorem to Quantum MechanicsApplication of Fubini Theorem in Quantum MechanicsHeat operator formalism via spectral projections and Dirac measureBiorthogonality of vectorsUse of Nets (topological) in Rigged Hilbert SpaceHow does one define the complex distribution $1/z$?The elliptic regularity theorem for differential operators with variable coefficientsDiscrete spectra of generators of compact Lie groupPath to quantum geometry for mathematicians?Quantum mechanics: First order perturbation to eigenstatesTechnical operator theory question on Albeverio's “Solvable Models in quantum mechanics”
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Application of Schwartz Kernel Theorem to Quantum Mechanics
Application of Fubini Theorem in Quantum MechanicsHeat operator formalism via spectral projections and Dirac measureBiorthogonality of vectorsUse of Nets (topological) in Rigged Hilbert SpaceHow does one define the complex distribution $1/z$?The elliptic regularity theorem for differential operators with variable coefficientsDiscrete spectra of generators of compact Lie groupPath to quantum geometry for mathematicians?Quantum mechanics: First order perturbation to eigenstatesTechnical operator theory question on Albeverio's “Solvable Models in quantum mechanics”
$begingroup$
I am currently reading Quantum Mechanics for Mathematicians and have a question about a statement made in the book:
Remark. By the Schwartz kernel theorem, the operator B can be represented by an integral operator with distributional kernel $K(q,q')$. Then the commutativity $BQ = QB$ implies that, in the distributional sense,
$$
(q-q')K(q,q') = 0,
$$
so that $K$ is "proportional" to the Dirac delta-function, i .e.,
$$
K(q,q') = f(q)delta(q-q').
$$
This argument is usually given in physics textbooks.
$B$ and $Q$ are both operators in the coordinate representation. $Q$ is the position operator and $B$ is bounded.
So, first of all, I can't find much about the kernel theorem online. I've been using this as a reference. According to this document, there is a relation between bilinear forms and distributions. However, I don't know how to view $B$ as a bilinear form and thus apply the theorem. Can someone elucidate how the kernel theorem is applicable to $B$? Secondly, what arguments/notation in physics is the author (of the QM book) referencing?
functional-analysis operator-theory mathematical-physics quantum-mechanics
edited 2 days ago
Andrews
1,2691421
asked Aug 18 '13 at 19:41
scottscott
312
$endgroup$
|
show 5 more comments
$begingroup$
I am currently reading Quantum Mechanics for Mathematicians and have a question about a statement made in the book:
Remark. By the Schwartz kernel theorem, the operator B can be represented by an integral operator with distributional kernel $K(q,q')$. Then the commutativity $BQ = QB$ implies that, in the distributional sense,
$$
(q-q')K(q,q') = 0,
$$
so that $K$ is "proportional" to the Dirac delta-function, i .e.,
$$
K(q,q') = f(q)delta(q-q').
$$
This argument is usually given in physics textbooks.
$B$ and $Q$ are both operators in the coordinate representation. $Q$ is the position operator and $B$ is bounded.
So, first of all, I can't find much about the kernel theorem online. I've been using this as a reference. According to this document, there is a relation between bilinear forms and distributions. However, I don't know how to view $B$ as a bilinear form and thus apply the theorem. Can someone elucidate how the kernel theorem is applicable to $B$? Secondly, what arguments/notation in physics is the author (of the QM book) referencing?
functional-analysis operator-theory mathematical-physics quantum-mechanics
edited 2 days ago
Andrews
1,2691421
asked Aug 18 '13 at 19:41
scottscott
312
$endgroup$
$begingroup$
What are Q and B here?
$endgroup$
– Cameron Williams
Aug 18 '13 at 19:50
$begingroup$
B and Q are both operators in the coordinate representation. Q is the position operator.
$endgroup$
– scott
Aug 18 '13 at 19:55
$begingroup$
The coordinate representation being the complex square integrable functions on the real line.
$endgroup$
– scott
Aug 18 '13 at 19:58
$begingroup$
What is B? It seems to have been referenced in something before the remark? Are there any restrictions on the operator B?
$endgroup$
– Cameron Williams
Aug 18 '13 at 20:03
$begingroup$
Ah, I forgot to say B is bounded. The preceding lemma says that if B is bounded and if it commutes with Q, then B is a function of Q. I'll make an edit, thanks.
$endgroup$
– scott
Aug 18 '13 at 20:05
|
show 5 more comments
$begingroup$
I am currently reading Quantum Mechanics for Mathematicians and have a question about a statement made in the book:
Remark. By the Schwartz kernel theorem, the operator B can be represented by an integral operator with distributional kernel $K(q,q')$. Then the commutativity $BQ = QB$ implies that, in the distributional sense,
$$
(q-q')K(q,q') = 0,
$$
so that $K$ is "proportional" to the Dirac delta-function, i .e.,
$$
K(q,q') = f(q)delta(q-q').
$$
This argument is usually given in physics textbooks.
$B$ and $Q$ are both operators in the coordinate representation. $Q$ is the position operator and $B$ is bounded.
So, first of all, I can't find much about the kernel theorem online. I've been using this as a reference. According to this document, there is a relation between bilinear forms and distributions. However, I don't know how to view $B$ as a bilinear form and thus apply the theorem. Can someone elucidate how the kernel theorem is applicable to $B$? Secondly, what arguments/notation in physics is the author (of the QM book) referencing?
functional-analysis operator-theory mathematical-physics quantum-mechanics
edited 2 days ago
Andrews
1,2691421
asked Aug 18 '13 at 19:41
scottscott
312
$endgroup$
I am currently reading Quantum Mechanics for Mathematicians and have a question about a statement made in the book:
Remark. By the Schwartz kernel theorem, the operator B can be represented by an integral operator with distributional kernel $K(q,q')$. Then the commutativity $BQ = QB$ implies that, in the distributional sense,
$$
(q-q')K(q,q') = 0,
$$
so that $K$ is "proportional" to the Dirac delta-function, i .e.,
$$
K(q,q') = f(q)delta(q-q').
$$
This argument is usually given in physics textbooks.
$B$ and $Q$ are both operators in the coordinate representation. $Q$ is the position operator and $B$ is bounded.
So, first of all, I can't find much about the kernel theorem online. I've been using this as a reference. According to this document, there is a relation between bilinear forms and distributions. However, I don't know how to view $B$ as a bilinear form and thus apply the theorem. Can someone elucidate how the kernel theorem is applicable to $B$? Secondly, what arguments/notation in physics is the author (of the QM book) referencing?
functional-analysis operator-theory mathematical-physics quantum-mechanics
functional-analysis operator-theory mathematical-physics quantum-mechanics
edited 2 days ago
Andrews
1,2691421
asked Aug 18 '13 at 19:41
scottscott
312
edited 2 days ago
Andrews
1,2691421
asked Aug 18 '13 at 19:41
scottscott
312
edited 2 days ago
Andrews
1,2691421
edited 2 days ago
Andrews
1,2691421
edited 2 days ago
Andrews
1,2691421
1,2691421
asked Aug 18 '13 at 19:41
scottscott
312
asked Aug 18 '13 at 19:41
scottscott
312
asked Aug 18 '13 at 19:41
scottscott
312
312
$begingroup$
What are Q and B here?
$endgroup$
– Cameron Williams
Aug 18 '13 at 19:50
$begingroup$
B and Q are both operators in the coordinate representation. Q is the position operator.
$endgroup$
– scott
Aug 18 '13 at 19:55
$begingroup$
The coordinate representation being the complex square integrable functions on the real line.
$endgroup$
– scott
Aug 18 '13 at 19:58
$begingroup$
What is B? It seems to have been referenced in something before the remark? Are there any restrictions on the operator B?
$endgroup$
– Cameron Williams
Aug 18 '13 at 20:03
$begingroup$
Ah, I forgot to say B is bounded. The preceding lemma says that if B is bounded and if it commutes with Q, then B is a function of Q. I'll make an edit, thanks.
$endgroup$
– scott
Aug 18 '13 at 20:05
|
show 5 more comments
$begingroup$
What are Q and B here?
$endgroup$
– Cameron Williams
Aug 18 '13 at 19:50
$begingroup$
B and Q are both operators in the coordinate representation. Q is the position operator.
$endgroup$
– scott
Aug 18 '13 at 19:55
$begingroup$
The coordinate representation being the complex square integrable functions on the real line.
$endgroup$
– scott
Aug 18 '13 at 19:58
$begingroup$
What is B? It seems to have been referenced in something before the remark? Are there any restrictions on the operator B?
$endgroup$
– Cameron Williams
Aug 18 '13 at 20:03
$begingroup$
Ah, I forgot to say B is bounded. The preceding lemma says that if B is bounded and if it commutes with Q, then B is a function of Q. I'll make an edit, thanks.
$endgroup$
– scott
Aug 18 '13 at 20:05
$begingroup$
What are Q and B here?
$endgroup$
– Cameron Williams
Aug 18 '13 at 19:50
$begingroup$
What are Q and B here?
$endgroup$
– Cameron Williams
Aug 18 '13 at 19:50
$begingroup$
B and Q are both operators in the coordinate representation. Q is the position operator.
$endgroup$
– scott
Aug 18 '13 at 19:55
$begingroup$
B and Q are both operators in the coordinate representation. Q is the position operator.
$endgroup$
– scott
Aug 18 '13 at 19:55
$begingroup$
The coordinate representation being the complex square integrable functions on the real line.
$endgroup$
– scott
Aug 18 '13 at 19:58
$begingroup$
The coordinate representation being the complex square integrable functions on the real line.
$endgroup$
– scott
Aug 18 '13 at 19:58
$begingroup$
What is B? It seems to have been referenced in something before the remark? Are there any restrictions on the operator B?
$endgroup$
– Cameron Williams
Aug 18 '13 at 20:03
$begingroup$
What is B? It seems to have been referenced in something before the remark? Are there any restrictions on the operator B?
$endgroup$
– Cameron Williams
Aug 18 '13 at 20:03
$begingroup$
Ah, I forgot to say B is bounded. The preceding lemma says that if B is bounded and if it commutes with Q, then B is a function of Q. I'll make an edit, thanks.
$endgroup$
– scott
Aug 18 '13 at 20:05
$begingroup$
Ah, I forgot to say B is bounded. The preceding lemma says that if B is bounded and if it commutes with Q, then B is a function of Q. I'll make an edit, thanks.
$endgroup$
– scott
Aug 18 '13 at 20:05
|
show 5 more comments
1 Answer
1
active
oldest
votes
$begingroup$
If $T$ is a distribution such that $$forall g in C^infty_c, langle T, x g rangle = 0$$ then pick $phi in C^infty_c$ constant $=1$ on $[-1,1]$, let $$g = frac1x (f-f(0)phi)quadin C^infty_c$$ then $$ langle T,f rangle = langle T,f(0)phi rangle + langle T,xgrangle =f(0) langle T, phi rangle$$
answered 2 days ago
reunsreuns
21.2k21351
$endgroup$
add a comment |
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1 Answer
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oldest
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
If $T$ is a distribution such that $$forall g in C^infty_c, langle T, x g rangle = 0$$ then pick $phi in C^infty_c$ constant $=1$ on $[-1,1]$, let $$g = frac1x (f-f(0)phi)quadin C^infty_c$$ then $$ langle T,f rangle = langle T,f(0)phi rangle + langle T,xgrangle =f(0) langle T, phi rangle$$
answered 2 days ago
reunsreuns
21.2k21351
$endgroup$
add a comment |
$begingroup$
If $T$ is a distribution such that $$forall g in C^infty_c, langle T, x g rangle = 0$$ then pick $phi in C^infty_c$ constant $=1$ on $[-1,1]$, let $$g = frac1x (f-f(0)phi)quadin C^infty_c$$ then $$ langle T,f rangle = langle T,f(0)phi rangle + langle T,xgrangle =f(0) langle T, phi rangle$$
answered 2 days ago
reunsreuns
21.2k21351
$endgroup$
add a comment |
$begingroup$
If $T$ is a distribution such that $$forall g in C^infty_c, langle T, x g rangle = 0$$ then pick $phi in C^infty_c$ constant $=1$ on $[-1,1]$, let $$g = frac1x (f-f(0)phi)quadin C^infty_c$$ then $$ langle T,f rangle = langle T,f(0)phi rangle + langle T,xgrangle =f(0) langle T, phi rangle$$
answered 2 days ago
reunsreuns
21.2k21351
$endgroup$
If $T$ is a distribution such that $$forall g in C^infty_c, langle T, x g rangle = 0$$ then pick $phi in C^infty_c$ constant $=1$ on $[-1,1]$, let $$g = frac1x (f-f(0)phi)quadin C^infty_c$$ then $$ langle T,f rangle = langle T,f(0)phi rangle + langle T,xgrangle =f(0) langle T, phi rangle$$
answered 2 days ago
reunsreuns
21.2k21351
answered 2 days ago
reunsreuns
21.2k21351
answered 2 days ago
reunsreuns
21.2k21351
answered 2 days ago
reunsreuns
21.2k21351
21.2k21351
add a comment |
add a comment |
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1ko KUqsZtUa J66MX,kDA
$begingroup$
What are Q and B here?
$endgroup$
– Cameron Williams
Aug 18 '13 at 19:50
$begingroup$
B and Q are both operators in the coordinate representation. Q is the position operator.
$endgroup$
– scott
Aug 18 '13 at 19:55
$begingroup$
The coordinate representation being the complex square integrable functions on the real line.
$endgroup$
– scott
Aug 18 '13 at 19:58
$begingroup$
What is B? It seems to have been referenced in something before the remark? Are there any restrictions on the operator B?
$endgroup$
– Cameron Williams
Aug 18 '13 at 20:03
$begingroup$
Ah, I forgot to say B is bounded. The preceding lemma says that if B is bounded and if it commutes with Q, then B is a function of Q. I'll make an edit, thanks.
$endgroup$
– scott
Aug 18 '13 at 20:05