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Why is the magnetic Schrödinger operator positive?
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In the book 'Schrödinger Operators' by Cycon et al. they prove that the magnetic Schrödinger operator (as well as the Pauli operator) have essential spectrum $sigma_ess = [0,infty)$ if $B$ has decay at infinity and the potential $V$ is $-Delta$-compact. Remember that the magnetic Schrödinger operator is given by
$$H_ms := (-inabla -A)^2 +V $$
where $A$ is the vector potential. In their proof they claim that since $H(A,0)= (-inabla - A)^2$ is positive the spectrum must be contained in $[0,infty)$. Recall that for a self-adjoint operator $T$ to be positive means that $langle Tx,xrangle geq 0$ for all $xin X$ where $X$ is Banach. Now why is $H(A,0)$ positive? Is it because it is the square of an operator, specifically $-inabla -A$? If so is it trivial to see this?
functional-analysis mathematical-physics spectral-theory quantum-mechanics
edited 29 mins ago
Zvi
656111
asked 4 hours ago
Jakob Elias
630316
add a comment |Â
up vote
2
down vote
favorite
In the book 'Schrödinger Operators' by Cycon et al. they prove that the magnetic Schrödinger operator (as well as the Pauli operator) have essential spectrum $sigma_ess = [0,infty)$ if $B$ has decay at infinity and the potential $V$ is $-Delta$-compact. Remember that the magnetic Schrödinger operator is given by
$$H_ms := (-inabla -A)^2 +V $$
where $A$ is the vector potential. In their proof they claim that since $H(A,0)= (-inabla - A)^2$ is positive the spectrum must be contained in $[0,infty)$. Recall that for a self-adjoint operator $T$ to be positive means that $langle Tx,xrangle geq 0$ for all $xin X$ where $X$ is Banach. Now why is $H(A,0)$ positive? Is it because it is the square of an operator, specifically $-inabla -A$? If so is it trivial to see this?
functional-analysis mathematical-physics spectral-theory quantum-mechanics
edited 29 mins ago
Zvi
656111
asked 4 hours ago
Jakob Elias
630316
Hint: prove $-inabla-A$ is Hermitian.
– J.G.
1 hour ago
@J.G. This is clear to me. I was wondering whether the square of a hermitian operator is always positive.
– Jakob Elias
1 hour ago
@J.G. But $-inabla-A$ is not Hermitian because it does not map a complex-valued function to a complex-valued function. It takes a complex-value function and returns a vector-valued function. A Hermitian operator must map from a Hilbert space into the same space, by definition.
– Zvi
1 hour ago
1
@Zvi That's only a minor concern; the square of this vector-valued operator is a sum of squares of "ordinary" Hermitian operators, by Pythagoras. When I call a vector Hermitian, I just mean its components are, since that causes the vector to also be self-adjoint (if its adjoint is defined in the obvious way).
– J.G.
1 hour ago
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
In the book 'Schrödinger Operators' by Cycon et al. they prove that the magnetic Schrödinger operator (as well as the Pauli operator) have essential spectrum $sigma_ess = [0,infty)$ if $B$ has decay at infinity and the potential $V$ is $-Delta$-compact. Remember that the magnetic Schrödinger operator is given by
$$H_ms := (-inabla -A)^2 +V $$
where $A$ is the vector potential. In their proof they claim that since $H(A,0)= (-inabla - A)^2$ is positive the spectrum must be contained in $[0,infty)$. Recall that for a self-adjoint operator $T$ to be positive means that $langle Tx,xrangle geq 0$ for all $xin X$ where $X$ is Banach. Now why is $H(A,0)$ positive? Is it because it is the square of an operator, specifically $-inabla -A$? If so is it trivial to see this?
functional-analysis mathematical-physics spectral-theory quantum-mechanics
edited 29 mins ago
Zvi
656111
asked 4 hours ago
Jakob Elias
630316
In the book 'Schrödinger Operators' by Cycon et al. they prove that the magnetic Schrödinger operator (as well as the Pauli operator) have essential spectrum $sigma_ess = [0,infty)$ if $B$ has decay at infinity and the potential $V$ is $-Delta$-compact. Remember that the magnetic Schrödinger operator is given by
$$H_ms := (-inabla -A)^2 +V $$
where $A$ is the vector potential. In their proof they claim that since $H(A,0)= (-inabla - A)^2$ is positive the spectrum must be contained in $[0,infty)$. Recall that for a self-adjoint operator $T$ to be positive means that $langle Tx,xrangle geq 0$ for all $xin X$ where $X$ is Banach. Now why is $H(A,0)$ positive? Is it because it is the square of an operator, specifically $-inabla -A$? If so is it trivial to see this?
functional-analysis mathematical-physics spectral-theory quantum-mechanics
functional-analysis mathematical-physics spectral-theory quantum-mechanics
edited 29 mins ago
Zvi
656111
asked 4 hours ago
Jakob Elias
630316
edited 29 mins ago
Zvi
656111
asked 4 hours ago
Jakob Elias
630316
edited 29 mins ago
Zvi
656111
edited 29 mins ago
Zvi
656111
edited 29 mins ago
Zvi
656111
656111
asked 4 hours ago
Jakob Elias
630316
asked 4 hours ago
Jakob Elias
630316
asked 4 hours ago
Jakob Elias
630316
630316
Hint: prove $-inabla-A$ is Hermitian.
– J.G.
1 hour ago
@J.G. This is clear to me. I was wondering whether the square of a hermitian operator is always positive.
– Jakob Elias
1 hour ago
@J.G. But $-inabla-A$ is not Hermitian because it does not map a complex-valued function to a complex-valued function. It takes a complex-value function and returns a vector-valued function. A Hermitian operator must map from a Hilbert space into the same space, by definition.
– Zvi
1 hour ago
1
@Zvi That's only a minor concern; the square of this vector-valued operator is a sum of squares of "ordinary" Hermitian operators, by Pythagoras. When I call a vector Hermitian, I just mean its components are, since that causes the vector to also be self-adjoint (if its adjoint is defined in the obvious way).
– J.G.
1 hour ago
add a comment |Â
Hint: prove $-inabla-A$ is Hermitian.
– J.G.
1 hour ago
@J.G. This is clear to me. I was wondering whether the square of a hermitian operator is always positive.
– Jakob Elias
1 hour ago
@J.G. But $-inabla-A$ is not Hermitian because it does not map a complex-valued function to a complex-valued function. It takes a complex-value function and returns a vector-valued function. A Hermitian operator must map from a Hilbert space into the same space, by definition.
– Zvi
1 hour ago
1
@Zvi That's only a minor concern; the square of this vector-valued operator is a sum of squares of "ordinary" Hermitian operators, by Pythagoras. When I call a vector Hermitian, I just mean its components are, since that causes the vector to also be self-adjoint (if its adjoint is defined in the obvious way).
– J.G.
1 hour ago
Hint: prove $-inabla-A$ is Hermitian.
– J.G.
1 hour ago
Hint: prove $-inabla-A$ is Hermitian.
– J.G.
1 hour ago
@J.G. This is clear to me. I was wondering whether the square of a hermitian operator is always positive.
– Jakob Elias
1 hour ago
@J.G. This is clear to me. I was wondering whether the square of a hermitian operator is always positive.
– Jakob Elias
1 hour ago
@J.G. But $-inabla-A$ is not Hermitian because it does not map a complex-valued function to a complex-valued function. It takes a complex-value function and returns a vector-valued function. A Hermitian operator must map from a Hilbert space into the same space, by definition.
– Zvi
1 hour ago
@J.G. But $-inabla-A$ is not Hermitian because it does not map a complex-valued function to a complex-valued function. It takes a complex-value function and returns a vector-valued function. A Hermitian operator must map from a Hilbert space into the same space, by definition.
– Zvi
1 hour ago
1
1
@Zvi That's only a minor concern; the square of this vector-valued operator is a sum of squares of "ordinary" Hermitian operators, by Pythagoras. When I call a vector Hermitian, I just mean its components are, since that causes the vector to also be self-adjoint (if its adjoint is defined in the obvious way).
– J.G.
1 hour ago
@Zvi That's only a minor concern; the square of this vector-valued operator is a sum of squares of "ordinary" Hermitian operators, by Pythagoras. When I call a vector Hermitian, I just mean its components are, since that causes the vector to also be self-adjoint (if its adjoint is defined in the obvious way).
– J.G.
1 hour ago
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
3
down vote
accepted
If $X$ is a vector with Hermitian components, $$langlepsi |Xcdot X|psirangle=sum_ilanglepsi |X_i^2| psirangle=sum_ilanglepsi |X_i^TX_i| psirangle=sum_iVert X_i|psirangleVert^2ge 0.$$
answered 1 hour ago
J.G.
16.2k11829
Well that solves it.
– Jakob Elias
1 hour ago
@JakobElias This solution needs to be somehow fixed in my opinion. Read my comment under the question. This map $X=-inabla-A$ is not Hermitian.
– Zvi
1 hour ago
@Zvi Sorry for replying to your comment underneath the OP instead of this question; I'm having internet problems ATM that are making it hard to keep up with the page's updates.
– J.G.
1 hour ago
I see how your argument works now. But, if you don't mind, I will add your comment in the answer. You can remove it if you don't like it.
– Zvi
1 hour ago
1
@Zvi Following a community rejection of your suggested edit, I've modified the algebra a little to incorporate the comment's point.
– J.G.
55 mins ago
add a comment |Â
up vote
2
down vote
Let $mathcalH$ denote the Hilbert space of wave functions $psi:mathbbR^3tomathbbC$. Recall that
$$langle u|vrangle =iiint_mathbbR^3baru(x,y,z) v(x,y,z) dx dy dz$$
for all $u,vin mathcalH$. Write $h$ for the operator $(-inabla -A)^2$. Observe that
beginalignlangle hu|vrangle &=iiint_mathbbR^3overlinehu(x,y,z) v(x,y,z) dx dy dz
\&=iiint_mathbbR^3(inabla -A)^2baru(x,y,z) v(x,y,z) dx dy dz
\&=iiint_mathbbR^3(inabla -A)cdot overlinePhi u(x,y,z) v(x,y,z) dx dy dz,
endalign
where $Phi=-inabla-A$. That is,
beginalignlangle hu|vrangle &=iiiint_mathbbR^3nablacdot overlinePhi u(x,y,z) v(x,y,z) dx dy dz-iiint_mathbbR^3 overlinePhi u(x,y,z)cdot Av(x,y,z) dx dy dz
\&=-iiiint_mathbbR^3overlinePhi u(x,y,z)cdot nabla v(x,y,z) dx dy dz
-iiint_mathbbR^3overlinePhi u(x,y,z)cdot Av(x,y,z) dx dy dz,endalign
where we apply integration by parts in higher dimension, assuming that $u(x,y,z)$ and $v(x,y,z)$ vanish quickly when $(x,y,z)$ is large. That is,
beginalignlangle hu|vrangle &=iiint_mathbbR^3overlinePhi u(x,y,z)cdot(-inabla-A)v(x,y,z) dx dy dz
\&=iiint_mathbbR^3overlinePhi u(x,y,z)cdotPhi v(x,y,z) dx dy dz.
endalign
In particular,
beginalignlangle hu|urangle &=int_mathbbR^3overlinePhi u(x,y,z)cdotPhi u(x,y,z) dx dy dz\&=int_mathbbR^3bigVertPhi u(x,y,z)bigVert^2 dx dy dzgeq 0endalign
answered 1 hour ago
Zvi
656111
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
If $X$ is a vector with Hermitian components, $$langlepsi |Xcdot X|psirangle=sum_ilanglepsi |X_i^2| psirangle=sum_ilanglepsi |X_i^TX_i| psirangle=sum_iVert X_i|psirangleVert^2ge 0.$$
answered 1 hour ago
J.G.
16.2k11829
Well that solves it.
– Jakob Elias
1 hour ago
@JakobElias This solution needs to be somehow fixed in my opinion. Read my comment under the question. This map $X=-inabla-A$ is not Hermitian.
– Zvi
1 hour ago
@Zvi Sorry for replying to your comment underneath the OP instead of this question; I'm having internet problems ATM that are making it hard to keep up with the page's updates.
– J.G.
1 hour ago
I see how your argument works now. But, if you don't mind, I will add your comment in the answer. You can remove it if you don't like it.
– Zvi
1 hour ago
1
@Zvi Following a community rejection of your suggested edit, I've modified the algebra a little to incorporate the comment's point.
– J.G.
55 mins ago
add a comment |Â
up vote
3
down vote
accepted
If $X$ is a vector with Hermitian components, $$langlepsi |Xcdot X|psirangle=sum_ilanglepsi |X_i^2| psirangle=sum_ilanglepsi |X_i^TX_i| psirangle=sum_iVert X_i|psirangleVert^2ge 0.$$
answered 1 hour ago
J.G.
16.2k11829
Well that solves it.
– Jakob Elias
1 hour ago
@JakobElias This solution needs to be somehow fixed in my opinion. Read my comment under the question. This map $X=-inabla-A$ is not Hermitian.
– Zvi
1 hour ago
@Zvi Sorry for replying to your comment underneath the OP instead of this question; I'm having internet problems ATM that are making it hard to keep up with the page's updates.
– J.G.
1 hour ago
I see how your argument works now. But, if you don't mind, I will add your comment in the answer. You can remove it if you don't like it.
– Zvi
1 hour ago
1
@Zvi Following a community rejection of your suggested edit, I've modified the algebra a little to incorporate the comment's point.
– J.G.
55 mins ago
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
If $X$ is a vector with Hermitian components, $$langlepsi |Xcdot X|psirangle=sum_ilanglepsi |X_i^2| psirangle=sum_ilanglepsi |X_i^TX_i| psirangle=sum_iVert X_i|psirangleVert^2ge 0.$$
answered 1 hour ago
J.G.
16.2k11829
If $X$ is a vector with Hermitian components, $$langlepsi |Xcdot X|psirangle=sum_ilanglepsi |X_i^2| psirangle=sum_ilanglepsi |X_i^TX_i| psirangle=sum_iVert X_i|psirangleVert^2ge 0.$$
answered 1 hour ago
J.G.
16.2k11829
edited 56 mins ago
answered 1 hour ago
J.G.
16.2k11829
answered 1 hour ago
J.G.
16.2k11829
answered 1 hour ago
J.G.
16.2k11829
16.2k11829
Well that solves it.
– Jakob Elias
1 hour ago
@JakobElias This solution needs to be somehow fixed in my opinion. Read my comment under the question. This map $X=-inabla-A$ is not Hermitian.
– Zvi
1 hour ago
@Zvi Sorry for replying to your comment underneath the OP instead of this question; I'm having internet problems ATM that are making it hard to keep up with the page's updates.
– J.G.
1 hour ago
I see how your argument works now. But, if you don't mind, I will add your comment in the answer. You can remove it if you don't like it.
– Zvi
1 hour ago
1
@Zvi Following a community rejection of your suggested edit, I've modified the algebra a little to incorporate the comment's point.
– J.G.
55 mins ago
add a comment |Â
Well that solves it.
– Jakob Elias
1 hour ago
@JakobElias This solution needs to be somehow fixed in my opinion. Read my comment under the question. This map $X=-inabla-A$ is not Hermitian.
– Zvi
1 hour ago
@Zvi Sorry for replying to your comment underneath the OP instead of this question; I'm having internet problems ATM that are making it hard to keep up with the page's updates.
– J.G.
1 hour ago
I see how your argument works now. But, if you don't mind, I will add your comment in the answer. You can remove it if you don't like it.
– Zvi
1 hour ago
1
@Zvi Following a community rejection of your suggested edit, I've modified the algebra a little to incorporate the comment's point.
– J.G.
55 mins ago
Well that solves it.
– Jakob Elias
1 hour ago
Well that solves it.
– Jakob Elias
1 hour ago
@JakobElias This solution needs to be somehow fixed in my opinion. Read my comment under the question. This map $X=-inabla-A$ is not Hermitian.
– Zvi
1 hour ago
@JakobElias This solution needs to be somehow fixed in my opinion. Read my comment under the question. This map $X=-inabla-A$ is not Hermitian.
– Zvi
1 hour ago
@Zvi Sorry for replying to your comment underneath the OP instead of this question; I'm having internet problems ATM that are making it hard to keep up with the page's updates.
– J.G.
1 hour ago
@Zvi Sorry for replying to your comment underneath the OP instead of this question; I'm having internet problems ATM that are making it hard to keep up with the page's updates.
– J.G.
1 hour ago
I see how your argument works now. But, if you don't mind, I will add your comment in the answer. You can remove it if you don't like it.
– Zvi
1 hour ago
I see how your argument works now. But, if you don't mind, I will add your comment in the answer. You can remove it if you don't like it.
– Zvi
1 hour ago
1
1
@Zvi Following a community rejection of your suggested edit, I've modified the algebra a little to incorporate the comment's point.
– J.G.
55 mins ago
@Zvi Following a community rejection of your suggested edit, I've modified the algebra a little to incorporate the comment's point.
– J.G.
55 mins ago
add a comment |Â
up vote
2
down vote
Let $mathcalH$ denote the Hilbert space of wave functions $psi:mathbbR^3tomathbbC$. Recall that
$$langle u|vrangle =iiint_mathbbR^3baru(x,y,z) v(x,y,z) dx dy dz$$
for all $u,vin mathcalH$. Write $h$ for the operator $(-inabla -A)^2$. Observe that
beginalignlangle hu|vrangle &=iiint_mathbbR^3overlinehu(x,y,z) v(x,y,z) dx dy dz
\&=iiint_mathbbR^3(inabla -A)^2baru(x,y,z) v(x,y,z) dx dy dz
\&=iiint_mathbbR^3(inabla -A)cdot overlinePhi u(x,y,z) v(x,y,z) dx dy dz,
endalign
where $Phi=-inabla-A$. That is,
beginalignlangle hu|vrangle &=iiiint_mathbbR^3nablacdot overlinePhi u(x,y,z) v(x,y,z) dx dy dz-iiint_mathbbR^3 overlinePhi u(x,y,z)cdot Av(x,y,z) dx dy dz
\&=-iiiint_mathbbR^3overlinePhi u(x,y,z)cdot nabla v(x,y,z) dx dy dz
-iiint_mathbbR^3overlinePhi u(x,y,z)cdot Av(x,y,z) dx dy dz,endalign
where we apply integration by parts in higher dimension, assuming that $u(x,y,z)$ and $v(x,y,z)$ vanish quickly when $(x,y,z)$ is large. That is,
beginalignlangle hu|vrangle &=iiint_mathbbR^3overlinePhi u(x,y,z)cdot(-inabla-A)v(x,y,z) dx dy dz
\&=iiint_mathbbR^3overlinePhi u(x,y,z)cdotPhi v(x,y,z) dx dy dz.
endalign
In particular,
beginalignlangle hu|urangle &=int_mathbbR^3overlinePhi u(x,y,z)cdotPhi u(x,y,z) dx dy dz\&=int_mathbbR^3bigVertPhi u(x,y,z)bigVert^2 dx dy dzgeq 0endalign
answered 1 hour ago
Zvi
656111
add a comment |Â
up vote
2
down vote
Let $mathcalH$ denote the Hilbert space of wave functions $psi:mathbbR^3tomathbbC$. Recall that
$$langle u|vrangle =iiint_mathbbR^3baru(x,y,z) v(x,y,z) dx dy dz$$
for all $u,vin mathcalH$. Write $h$ for the operator $(-inabla -A)^2$. Observe that
beginalignlangle hu|vrangle &=iiint_mathbbR^3overlinehu(x,y,z) v(x,y,z) dx dy dz
\&=iiint_mathbbR^3(inabla -A)^2baru(x,y,z) v(x,y,z) dx dy dz
\&=iiint_mathbbR^3(inabla -A)cdot overlinePhi u(x,y,z) v(x,y,z) dx dy dz,
endalign
where $Phi=-inabla-A$. That is,
beginalignlangle hu|vrangle &=iiiint_mathbbR^3nablacdot overlinePhi u(x,y,z) v(x,y,z) dx dy dz-iiint_mathbbR^3 overlinePhi u(x,y,z)cdot Av(x,y,z) dx dy dz
\&=-iiiint_mathbbR^3overlinePhi u(x,y,z)cdot nabla v(x,y,z) dx dy dz
-iiint_mathbbR^3overlinePhi u(x,y,z)cdot Av(x,y,z) dx dy dz,endalign
where we apply integration by parts in higher dimension, assuming that $u(x,y,z)$ and $v(x,y,z)$ vanish quickly when $(x,y,z)$ is large. That is,
beginalignlangle hu|vrangle &=iiint_mathbbR^3overlinePhi u(x,y,z)cdot(-inabla-A)v(x,y,z) dx dy dz
\&=iiint_mathbbR^3overlinePhi u(x,y,z)cdotPhi v(x,y,z) dx dy dz.
endalign
In particular,
beginalignlangle hu|urangle &=int_mathbbR^3overlinePhi u(x,y,z)cdotPhi u(x,y,z) dx dy dz\&=int_mathbbR^3bigVertPhi u(x,y,z)bigVert^2 dx dy dzgeq 0endalign
answered 1 hour ago
Zvi
656111
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Let $mathcalH$ denote the Hilbert space of wave functions $psi:mathbbR^3tomathbbC$. Recall that
$$langle u|vrangle =iiint_mathbbR^3baru(x,y,z) v(x,y,z) dx dy dz$$
for all $u,vin mathcalH$. Write $h$ for the operator $(-inabla -A)^2$. Observe that
beginalignlangle hu|vrangle &=iiint_mathbbR^3overlinehu(x,y,z) v(x,y,z) dx dy dz
\&=iiint_mathbbR^3(inabla -A)^2baru(x,y,z) v(x,y,z) dx dy dz
\&=iiint_mathbbR^3(inabla -A)cdot overlinePhi u(x,y,z) v(x,y,z) dx dy dz,
endalign
where $Phi=-inabla-A$. That is,
beginalignlangle hu|vrangle &=iiiint_mathbbR^3nablacdot overlinePhi u(x,y,z) v(x,y,z) dx dy dz-iiint_mathbbR^3 overlinePhi u(x,y,z)cdot Av(x,y,z) dx dy dz
\&=-iiiint_mathbbR^3overlinePhi u(x,y,z)cdot nabla v(x,y,z) dx dy dz
-iiint_mathbbR^3overlinePhi u(x,y,z)cdot Av(x,y,z) dx dy dz,endalign
where we apply integration by parts in higher dimension, assuming that $u(x,y,z)$ and $v(x,y,z)$ vanish quickly when $(x,y,z)$ is large. That is,
beginalignlangle hu|vrangle &=iiint_mathbbR^3overlinePhi u(x,y,z)cdot(-inabla-A)v(x,y,z) dx dy dz
\&=iiint_mathbbR^3overlinePhi u(x,y,z)cdotPhi v(x,y,z) dx dy dz.
endalign
In particular,
beginalignlangle hu|urangle &=int_mathbbR^3overlinePhi u(x,y,z)cdotPhi u(x,y,z) dx dy dz\&=int_mathbbR^3bigVertPhi u(x,y,z)bigVert^2 dx dy dzgeq 0endalign
answered 1 hour ago
Zvi
656111
Let $mathcalH$ denote the Hilbert space of wave functions $psi:mathbbR^3tomathbbC$. Recall that
$$langle u|vrangle =iiint_mathbbR^3baru(x,y,z) v(x,y,z) dx dy dz$$
for all $u,vin mathcalH$. Write $h$ for the operator $(-inabla -A)^2$. Observe that
beginalignlangle hu|vrangle &=iiint_mathbbR^3overlinehu(x,y,z) v(x,y,z) dx dy dz
\&=iiint_mathbbR^3(inabla -A)^2baru(x,y,z) v(x,y,z) dx dy dz
\&=iiint_mathbbR^3(inabla -A)cdot overlinePhi u(x,y,z) v(x,y,z) dx dy dz,
endalign
where $Phi=-inabla-A$. That is,
beginalignlangle hu|vrangle &=iiiint_mathbbR^3nablacdot overlinePhi u(x,y,z) v(x,y,z) dx dy dz-iiint_mathbbR^3 overlinePhi u(x,y,z)cdot Av(x,y,z) dx dy dz
\&=-iiiint_mathbbR^3overlinePhi u(x,y,z)cdot nabla v(x,y,z) dx dy dz
-iiint_mathbbR^3overlinePhi u(x,y,z)cdot Av(x,y,z) dx dy dz,endalign
where we apply integration by parts in higher dimension, assuming that $u(x,y,z)$ and $v(x,y,z)$ vanish quickly when $(x,y,z)$ is large. That is,
beginalignlangle hu|vrangle &=iiint_mathbbR^3overlinePhi u(x,y,z)cdot(-inabla-A)v(x,y,z) dx dy dz
\&=iiint_mathbbR^3overlinePhi u(x,y,z)cdotPhi v(x,y,z) dx dy dz.
endalign
In particular,
beginalignlangle hu|urangle &=int_mathbbR^3overlinePhi u(x,y,z)cdotPhi u(x,y,z) dx dy dz\&=int_mathbbR^3bigVertPhi u(x,y,z)bigVert^2 dx dy dzgeq 0endalign
answered 1 hour ago
Zvi
656111
edited 55 mins ago
answered 1 hour ago
Zvi
656111
answered 1 hour ago
Zvi
656111
answered 1 hour ago
Zvi
656111
656111
add a comment |Â
add a comment |Â
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Hint: prove $-inabla-A$ is Hermitian.
– J.G.
1 hour ago
@J.G. This is clear to me. I was wondering whether the square of a hermitian operator is always positive.
– Jakob Elias
1 hour ago
@J.G. But $-inabla-A$ is not Hermitian because it does not map a complex-valued function to a complex-valued function. It takes a complex-value function and returns a vector-valued function. A Hermitian operator must map from a Hilbert space into the same space, by definition.
– Zvi
1 hour ago
1
@Zvi That's only a minor concern; the square of this vector-valued operator is a sum of squares of "ordinary" Hermitian operators, by Pythagoras. When I call a vector Hermitian, I just mean its components are, since that causes the vector to also be self-adjoint (if its adjoint is defined in the obvious way).
– J.G.
1 hour ago