domain of the adjoint of the momentum operator.Adjoint of multiplication operator on Sobolev spaceSpectral theorem for momentum operatorCan we talk about the adjoint of a linear operator defined on a distribution space?Restriction of self-adjoint operator self-adjoint?Adjoint of an operator on $L^2$Finding the Adjoint Operator Under Inner-Products: Behaviour of Provided Function Values.About vectors that have bounded support representation on the spectrum of a self-adjoint operatorIs the sum of unbounded, symmetric operators with same domain self-adjoint?Domain of the spectral resolution of a self-adjoint operatorSelf-Adjoint Integral Operator
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domain of the adjoint of the momentum operator.
Adjoint of multiplication operator on Sobolev spaceSpectral theorem for momentum operatorCan we talk about the adjoint of a linear operator defined on a distribution space?Restriction of self-adjoint operator self-adjoint?Adjoint of an operator on $L^2$Finding the Adjoint Operator Under Inner-Products: Behaviour of Provided Function Values.About vectors that have bounded support representation on the spectrum of a self-adjoint operatorIs the sum of unbounded, symmetric operators with same domain self-adjoint?Domain of the spectral resolution of a self-adjoint operatorSelf-Adjoint Integral Operator
$begingroup$
I was reading the book by Roman where he discusses the linear momentum operator on the half line: $pf= -if’$, this operator is densely defined and in the book he finds the adjoint by using the inner product on $L^2(0,infty)$, one wishes to find the class of $h$ such that: $$
int _0 ^infty h^*(-ig’) dx = int _0 ^infty psi^* g : dx
$$
for some $psi$ and all $g$ in the domain of $p$. So clearly this is easy to do if $h$ is differentiable and with square integrable derivative by integration by parts. The question is: could there be non differentiable $h$ that satisfy this? I have been trying to show that there aren’t any such functions but can’t get anywhere and the book assumes this.
functional-analysis hilbert-spaces adjoint-operators
asked Mar 28 at 15:12
Michele GalliMichele Galli
1759
$endgroup$
add a comment |
$begingroup$
I was reading the book by Roman where he discusses the linear momentum operator on the half line: $pf= -if’$, this operator is densely defined and in the book he finds the adjoint by using the inner product on $L^2(0,infty)$, one wishes to find the class of $h$ such that: $$
int _0 ^infty h^*(-ig’) dx = int _0 ^infty psi^* g : dx
$$
for some $psi$ and all $g$ in the domain of $p$. So clearly this is easy to do if $h$ is differentiable and with square integrable derivative by integration by parts. The question is: could there be non differentiable $h$ that satisfy this? I have been trying to show that there aren’t any such functions but can’t get anywhere and the book assumes this.
functional-analysis hilbert-spaces adjoint-operators
asked Mar 28 at 15:12
Michele GalliMichele Galli
1759
$endgroup$
$begingroup$
In the context of quantum mechanics, don't functions have to be continuous? And usually they're differentiable as well, right? Certainly wave functions are, or they couldn't be solutions of the Schrodinger equation.
$endgroup$
– Adrian Keister
Mar 28 at 15:50
$begingroup$
no strictly that’s not true, in that what is of interest here is the momentum operator on a certain space and finding its adjoint, so basically my question is what the domain of this adjoint is, momentum is not an observable on this space because it’s not self adjoint
$endgroup$
– Michele Galli
Mar 28 at 15:57
add a comment |
$begingroup$
I was reading the book by Roman where he discusses the linear momentum operator on the half line: $pf= -if’$, this operator is densely defined and in the book he finds the adjoint by using the inner product on $L^2(0,infty)$, one wishes to find the class of $h$ such that: $$
int _0 ^infty h^*(-ig’) dx = int _0 ^infty psi^* g : dx
$$
for some $psi$ and all $g$ in the domain of $p$. So clearly this is easy to do if $h$ is differentiable and with square integrable derivative by integration by parts. The question is: could there be non differentiable $h$ that satisfy this? I have been trying to show that there aren’t any such functions but can’t get anywhere and the book assumes this.
functional-analysis hilbert-spaces adjoint-operators
asked Mar 28 at 15:12
Michele GalliMichele Galli
1759
$endgroup$
I was reading the book by Roman where he discusses the linear momentum operator on the half line: $pf= -if’$, this operator is densely defined and in the book he finds the adjoint by using the inner product on $L^2(0,infty)$, one wishes to find the class of $h$ such that: $$
int _0 ^infty h^*(-ig’) dx = int _0 ^infty psi^* g : dx
$$
for some $psi$ and all $g$ in the domain of $p$. So clearly this is easy to do if $h$ is differentiable and with square integrable derivative by integration by parts. The question is: could there be non differentiable $h$ that satisfy this? I have been trying to show that there aren’t any such functions but can’t get anywhere and the book assumes this.
functional-analysis hilbert-spaces adjoint-operators
functional-analysis hilbert-spaces adjoint-operators
asked Mar 28 at 15:12
Michele GalliMichele Galli
1759
asked Mar 28 at 15:12
Michele GalliMichele Galli
1759
asked Mar 28 at 15:12
Michele GalliMichele Galli
1759
asked Mar 28 at 15:12
Michele GalliMichele Galli
1759
asked Mar 28 at 15:12
Michele GalliMichele Galli
1759
1759
$begingroup$
In the context of quantum mechanics, don't functions have to be continuous? And usually they're differentiable as well, right? Certainly wave functions are, or they couldn't be solutions of the Schrodinger equation.
$endgroup$
– Adrian Keister
Mar 28 at 15:50
$begingroup$
no strictly that’s not true, in that what is of interest here is the momentum operator on a certain space and finding its adjoint, so basically my question is what the domain of this adjoint is, momentum is not an observable on this space because it’s not self adjoint
$endgroup$
– Michele Galli
Mar 28 at 15:57
add a comment |
$begingroup$
In the context of quantum mechanics, don't functions have to be continuous? And usually they're differentiable as well, right? Certainly wave functions are, or they couldn't be solutions of the Schrodinger equation.
$endgroup$
– Adrian Keister
Mar 28 at 15:50
$begingroup$
no strictly that’s not true, in that what is of interest here is the momentum operator on a certain space and finding its adjoint, so basically my question is what the domain of this adjoint is, momentum is not an observable on this space because it’s not self adjoint
$endgroup$
– Michele Galli
Mar 28 at 15:57
$begingroup$
In the context of quantum mechanics, don't functions have to be continuous? And usually they're differentiable as well, right? Certainly wave functions are, or they couldn't be solutions of the Schrodinger equation.
$endgroup$
– Adrian Keister
Mar 28 at 15:50
$begingroup$
In the context of quantum mechanics, don't functions have to be continuous? And usually they're differentiable as well, right? Certainly wave functions are, or they couldn't be solutions of the Schrodinger equation.
$endgroup$
– Adrian Keister
Mar 28 at 15:50
$begingroup$
no strictly that’s not true, in that what is of interest here is the momentum operator on a certain space and finding its adjoint, so basically my question is what the domain of this adjoint is, momentum is not an observable on this space because it’s not self adjoint
$endgroup$
– Michele Galli
Mar 28 at 15:57
$begingroup$
no strictly that’s not true, in that what is of interest here is the momentum operator on a certain space and finding its adjoint, so basically my question is what the domain of this adjoint is, momentum is not an observable on this space because it’s not self adjoint
$endgroup$
– Michele Galli
Mar 28 at 15:57
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Suppose your equation holds for all $ginmathcalD(p)$. Let $g_y$ be the function that is $1$ on $[0,y]$, is linear on $[y,y+epsilon]$ and is $0$ at $y+epsilon$, as well as for all $x > y+epsilon$. Then your adjoint equation would require
$$
int_0^inftyh^* (-ig_y')dx=int_0^inftypsi^* g_ydx \
-ifrac1epsilonint_y^y+epsilonh^*dx=int_0^inftypsi^*g_ydx
$$
Now, letting $epsilondownarrow 0$ gives
$$
-ih^*(y)=int_0^ypsi^*dx
$$
Hence, $h$ must be differentiable a.e., $h(0)=0$, and
$$
h'(y)=-ipsi(y).
$$
answered Mar 29 at 2:58
DisintegratingByPartsDisintegratingByParts
60.3k42681
$endgroup$
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Suppose your equation holds for all $ginmathcalD(p)$. Let $g_y$ be the function that is $1$ on $[0,y]$, is linear on $[y,y+epsilon]$ and is $0$ at $y+epsilon$, as well as for all $x > y+epsilon$. Then your adjoint equation would require
$$
int_0^inftyh^* (-ig_y')dx=int_0^inftypsi^* g_ydx \
-ifrac1epsilonint_y^y+epsilonh^*dx=int_0^inftypsi^*g_ydx
$$
Now, letting $epsilondownarrow 0$ gives
$$
-ih^*(y)=int_0^ypsi^*dx
$$
Hence, $h$ must be differentiable a.e., $h(0)=0$, and
$$
h'(y)=-ipsi(y).
$$
answered Mar 29 at 2:58
DisintegratingByPartsDisintegratingByParts
60.3k42681
$endgroup$
add a comment |
$begingroup$
Suppose your equation holds for all $ginmathcalD(p)$. Let $g_y$ be the function that is $1$ on $[0,y]$, is linear on $[y,y+epsilon]$ and is $0$ at $y+epsilon$, as well as for all $x > y+epsilon$. Then your adjoint equation would require
$$
int_0^inftyh^* (-ig_y')dx=int_0^inftypsi^* g_ydx \
-ifrac1epsilonint_y^y+epsilonh^*dx=int_0^inftypsi^*g_ydx
$$
Now, letting $epsilondownarrow 0$ gives
$$
-ih^*(y)=int_0^ypsi^*dx
$$
Hence, $h$ must be differentiable a.e., $h(0)=0$, and
$$
h'(y)=-ipsi(y).
$$
answered Mar 29 at 2:58
DisintegratingByPartsDisintegratingByParts
60.3k42681
$endgroup$
add a comment |
$begingroup$
Suppose your equation holds for all $ginmathcalD(p)$. Let $g_y$ be the function that is $1$ on $[0,y]$, is linear on $[y,y+epsilon]$ and is $0$ at $y+epsilon$, as well as for all $x > y+epsilon$. Then your adjoint equation would require
$$
int_0^inftyh^* (-ig_y')dx=int_0^inftypsi^* g_ydx \
-ifrac1epsilonint_y^y+epsilonh^*dx=int_0^inftypsi^*g_ydx
$$
Now, letting $epsilondownarrow 0$ gives
$$
-ih^*(y)=int_0^ypsi^*dx
$$
Hence, $h$ must be differentiable a.e., $h(0)=0$, and
$$
h'(y)=-ipsi(y).
$$
answered Mar 29 at 2:58
DisintegratingByPartsDisintegratingByParts
60.3k42681
$endgroup$
Suppose your equation holds for all $ginmathcalD(p)$. Let $g_y$ be the function that is $1$ on $[0,y]$, is linear on $[y,y+epsilon]$ and is $0$ at $y+epsilon$, as well as for all $x > y+epsilon$. Then your adjoint equation would require
$$
int_0^inftyh^* (-ig_y')dx=int_0^inftypsi^* g_ydx \
-ifrac1epsilonint_y^y+epsilonh^*dx=int_0^inftypsi^*g_ydx
$$
Now, letting $epsilondownarrow 0$ gives
$$
-ih^*(y)=int_0^ypsi^*dx
$$
Hence, $h$ must be differentiable a.e., $h(0)=0$, and
$$
h'(y)=-ipsi(y).
$$
answered Mar 29 at 2:58
DisintegratingByPartsDisintegratingByParts
60.3k42681
answered Mar 29 at 2:58
DisintegratingByPartsDisintegratingByParts
60.3k42681
answered Mar 29 at 2:58
DisintegratingByPartsDisintegratingByParts
60.3k42681
answered Mar 29 at 2:58
DisintegratingByPartsDisintegratingByParts
60.3k42681
60.3k42681
add a comment |
add a comment |
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$begingroup$
In the context of quantum mechanics, don't functions have to be continuous? And usually they're differentiable as well, right? Certainly wave functions are, or they couldn't be solutions of the Schrodinger equation.
$endgroup$
– Adrian Keister
Mar 28 at 15:50
$begingroup$
no strictly that’s not true, in that what is of interest here is the momentum operator on a certain space and finding its adjoint, so basically my question is what the domain of this adjoint is, momentum is not an observable on this space because it’s not self adjoint
$endgroup$
– Michele Galli
Mar 28 at 15:57