Dirac delta function like a “function†of an operator
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Let $mathcalSsubset mathcalH=L^2(mathbbR^n)subset mathcalS^*$ be the Schwartz space, the Hilbert space and the space of tempered distributions respectively. Consider an algebra $mathcalA$ of operators with continuous spectrum. We see delta function as the element of $mathcalS^*$, i.e. continuous (in kernal topology with seminorms $f mapsto |Af|, Ain mathcalA, fin mathcalH$) linear functional on $mathcalS$.
Question: how can we correctly define an action of $delta(A-vecx), Ain mathcalA, vecxin mathbbR^n$ on $fin mathcalH,?$
For example, here we have the algebra $mathcalA$ generated by operators $X colon f(vecx)mapsto vecx f(vecx)$ and $P colon f(vecx)mapsto partial_vecx f(vecx)$ with $[X,P]=iI$. Emilio Pisanty tell that functions of the position operator should work $⟨x|F(X)=F(vecx)⟨x|$ but I don't understand how it work with delta function.
Also we can find (here on p.13 formula (74)) the Fourier transform of operator $rho(vecx)=delta(X-vecx)$: $$rho(veck)=intlimits_mathbbR^n e^-iveckcdot vecxrho(vecx),dvecx=e^-iveckX.$$
I will be very grateful for any remarks and comments.
functional-analysis operator-theory hilbert-spaces quantum-mechanics
asked Aug 18 at 1:07
QuantumDK
64
 |Â
show 1 more comment
up vote
0
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Let $mathcalSsubset mathcalH=L^2(mathbbR^n)subset mathcalS^*$ be the Schwartz space, the Hilbert space and the space of tempered distributions respectively. Consider an algebra $mathcalA$ of operators with continuous spectrum. We see delta function as the element of $mathcalS^*$, i.e. continuous (in kernal topology with seminorms $f mapsto |Af|, Ain mathcalA, fin mathcalH$) linear functional on $mathcalS$.
Question: how can we correctly define an action of $delta(A-vecx), Ain mathcalA, vecxin mathbbR^n$ on $fin mathcalH,?$
For example, here we have the algebra $mathcalA$ generated by operators $X colon f(vecx)mapsto vecx f(vecx)$ and $P colon f(vecx)mapsto partial_vecx f(vecx)$ with $[X,P]=iI$. Emilio Pisanty tell that functions of the position operator should work $⟨x|F(X)=F(vecx)⟨x|$ but I don't understand how it work with delta function.
Also we can find (here on p.13 formula (74)) the Fourier transform of operator $rho(vecx)=delta(X-vecx)$: $$rho(veck)=intlimits_mathbbR^n e^-iveckcdot vecxrho(vecx),dvecx=e^-iveckX.$$
I will be very grateful for any remarks and comments.
functional-analysis operator-theory hilbert-spaces quantum-mechanics
asked Aug 18 at 1:07
QuantumDK
64
What is $A-vecx$? and then what is $delta(A-vecx)$?
– DisintegratingByParts
Aug 18 at 4:14
@DisintegratingByParts, $(A-vecx)f(vecx)=Af-vecxf$ and I don't understand what is $delta(A-vecx)$, I think that it's wrong form of something correct. By anology we have $int f(vecx)delta(A-vecx), dvecx=f(A),$ but it looks senseless. Is it correct: $delta(A-vecx)=frac12pi int e^i(A-vecx)cdot vecy,dvecy , ?$
– QuantumDK
Aug 18 at 13:45
1
I don't know how to make sense of an operator minus a vector.
– DisintegratingByParts
Aug 18 at 15:27
@DisintegratingByParts, and how to make sense of $delta(A), Ain mathcalA ,?$
– QuantumDK
Aug 20 at 17:26
Maybe we can write formal expansion and put to it an operator: $$delta(A-x)=frac12pisum_k=-infty^inftye^ik(A-x)=frac12pisum_k=-infty^inftye^ik(-x)sum_n=0^inftyfrac(ikA)^nn! , ?$$
– QuantumDK
Aug 21 at 1:39
 |Â
show 1 more comment
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let $mathcalSsubset mathcalH=L^2(mathbbR^n)subset mathcalS^*$ be the Schwartz space, the Hilbert space and the space of tempered distributions respectively. Consider an algebra $mathcalA$ of operators with continuous spectrum. We see delta function as the element of $mathcalS^*$, i.e. continuous (in kernal topology with seminorms $f mapsto |Af|, Ain mathcalA, fin mathcalH$) linear functional on $mathcalS$.
Question: how can we correctly define an action of $delta(A-vecx), Ain mathcalA, vecxin mathbbR^n$ on $fin mathcalH,?$
For example, here we have the algebra $mathcalA$ generated by operators $X colon f(vecx)mapsto vecx f(vecx)$ and $P colon f(vecx)mapsto partial_vecx f(vecx)$ with $[X,P]=iI$. Emilio Pisanty tell that functions of the position operator should work $⟨x|F(X)=F(vecx)⟨x|$ but I don't understand how it work with delta function.
Also we can find (here on p.13 formula (74)) the Fourier transform of operator $rho(vecx)=delta(X-vecx)$: $$rho(veck)=intlimits_mathbbR^n e^-iveckcdot vecxrho(vecx),dvecx=e^-iveckX.$$
I will be very grateful for any remarks and comments.
functional-analysis operator-theory hilbert-spaces quantum-mechanics
asked Aug 18 at 1:07
QuantumDK
64
Let $mathcalSsubset mathcalH=L^2(mathbbR^n)subset mathcalS^*$ be the Schwartz space, the Hilbert space and the space of tempered distributions respectively. Consider an algebra $mathcalA$ of operators with continuous spectrum. We see delta function as the element of $mathcalS^*$, i.e. continuous (in kernal topology with seminorms $f mapsto |Af|, Ain mathcalA, fin mathcalH$) linear functional on $mathcalS$.
Question: how can we correctly define an action of $delta(A-vecx), Ain mathcalA, vecxin mathbbR^n$ on $fin mathcalH,?$
For example, here we have the algebra $mathcalA$ generated by operators $X colon f(vecx)mapsto vecx f(vecx)$ and $P colon f(vecx)mapsto partial_vecx f(vecx)$ with $[X,P]=iI$. Emilio Pisanty tell that functions of the position operator should work $⟨x|F(X)=F(vecx)⟨x|$ but I don't understand how it work with delta function.
Also we can find (here on p.13 formula (74)) the Fourier transform of operator $rho(vecx)=delta(X-vecx)$: $$rho(veck)=intlimits_mathbbR^n e^-iveckcdot vecxrho(vecx),dvecx=e^-iveckX.$$
I will be very grateful for any remarks and comments.
functional-analysis operator-theory hilbert-spaces quantum-mechanics
asked Aug 18 at 1:07
QuantumDK
64
asked Aug 18 at 1:07
QuantumDK
64
asked Aug 18 at 1:07
QuantumDK
64
asked Aug 18 at 1:07
QuantumDK
64
64
What is $A-vecx$? and then what is $delta(A-vecx)$?
– DisintegratingByParts
Aug 18 at 4:14
@DisintegratingByParts, $(A-vecx)f(vecx)=Af-vecxf$ and I don't understand what is $delta(A-vecx)$, I think that it's wrong form of something correct. By anology we have $int f(vecx)delta(A-vecx), dvecx=f(A),$ but it looks senseless. Is it correct: $delta(A-vecx)=frac12pi int e^i(A-vecx)cdot vecy,dvecy , ?$
– QuantumDK
Aug 18 at 13:45
1
I don't know how to make sense of an operator minus a vector.
– DisintegratingByParts
Aug 18 at 15:27
@DisintegratingByParts, and how to make sense of $delta(A), Ain mathcalA ,?$
– QuantumDK
Aug 20 at 17:26
Maybe we can write formal expansion and put to it an operator: $$delta(A-x)=frac12pisum_k=-infty^inftye^ik(A-x)=frac12pisum_k=-infty^inftye^ik(-x)sum_n=0^inftyfrac(ikA)^nn! , ?$$
– QuantumDK
Aug 21 at 1:39
 |Â
show 1 more comment
What is $A-vecx$? and then what is $delta(A-vecx)$?
– DisintegratingByParts
Aug 18 at 4:14
@DisintegratingByParts, $(A-vecx)f(vecx)=Af-vecxf$ and I don't understand what is $delta(A-vecx)$, I think that it's wrong form of something correct. By anology we have $int f(vecx)delta(A-vecx), dvecx=f(A),$ but it looks senseless. Is it correct: $delta(A-vecx)=frac12pi int e^i(A-vecx)cdot vecy,dvecy , ?$
– QuantumDK
Aug 18 at 13:45
1
I don't know how to make sense of an operator minus a vector.
– DisintegratingByParts
Aug 18 at 15:27
@DisintegratingByParts, and how to make sense of $delta(A), Ain mathcalA ,?$
– QuantumDK
Aug 20 at 17:26
Maybe we can write formal expansion and put to it an operator: $$delta(A-x)=frac12pisum_k=-infty^inftye^ik(A-x)=frac12pisum_k=-infty^inftye^ik(-x)sum_n=0^inftyfrac(ikA)^nn! , ?$$
– QuantumDK
Aug 21 at 1:39
What is $A-vecx$? and then what is $delta(A-vecx)$?
– DisintegratingByParts
Aug 18 at 4:14
What is $A-vecx$? and then what is $delta(A-vecx)$?
– DisintegratingByParts
Aug 18 at 4:14
@DisintegratingByParts, $(A-vecx)f(vecx)=Af-vecxf$ and I don't understand what is $delta(A-vecx)$, I think that it's wrong form of something correct. By anology we have $int f(vecx)delta(A-vecx), dvecx=f(A),$ but it looks senseless. Is it correct: $delta(A-vecx)=frac12pi int e^i(A-vecx)cdot vecy,dvecy , ?$
– QuantumDK
Aug 18 at 13:45
@DisintegratingByParts, $(A-vecx)f(vecx)=Af-vecxf$ and I don't understand what is $delta(A-vecx)$, I think that it's wrong form of something correct. By anology we have $int f(vecx)delta(A-vecx), dvecx=f(A),$ but it looks senseless. Is it correct: $delta(A-vecx)=frac12pi int e^i(A-vecx)cdot vecy,dvecy , ?$
– QuantumDK
Aug 18 at 13:45
1
1
I don't know how to make sense of an operator minus a vector.
– DisintegratingByParts
Aug 18 at 15:27
I don't know how to make sense of an operator minus a vector.
– DisintegratingByParts
Aug 18 at 15:27
@DisintegratingByParts, and how to make sense of $delta(A), Ain mathcalA ,?$
– QuantumDK
Aug 20 at 17:26
@DisintegratingByParts, and how to make sense of $delta(A), Ain mathcalA ,?$
– QuantumDK
Aug 20 at 17:26
Maybe we can write formal expansion and put to it an operator: $$delta(A-x)=frac12pisum_k=-infty^inftye^ik(A-x)=frac12pisum_k=-infty^inftye^ik(-x)sum_n=0^inftyfrac(ikA)^nn! , ?$$
– QuantumDK
Aug 21 at 1:39
Maybe we can write formal expansion and put to it an operator: $$delta(A-x)=frac12pisum_k=-infty^inftye^ik(A-x)=frac12pisum_k=-infty^inftye^ik(-x)sum_n=0^inftyfrac(ikA)^nn! , ?$$
– QuantumDK
Aug 21 at 1:39
 |Â
show 1 more comment
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What is $A-vecx$? and then what is $delta(A-vecx)$?
– DisintegratingByParts
Aug 18 at 4:14
@DisintegratingByParts, $(A-vecx)f(vecx)=Af-vecxf$ and I don't understand what is $delta(A-vecx)$, I think that it's wrong form of something correct. By anology we have $int f(vecx)delta(A-vecx), dvecx=f(A),$ but it looks senseless. Is it correct: $delta(A-vecx)=frac12pi int e^i(A-vecx)cdot vecy,dvecy , ?$
– QuantumDK
Aug 18 at 13:45
1
I don't know how to make sense of an operator minus a vector.
– DisintegratingByParts
Aug 18 at 15:27
@DisintegratingByParts, and how to make sense of $delta(A), Ain mathcalA ,?$
– QuantumDK
Aug 20 at 17:26
Maybe we can write formal expansion and put to it an operator: $$delta(A-x)=frac12pisum_k=-infty^inftye^ik(A-x)=frac12pisum_k=-infty^inftye^ik(-x)sum_n=0^inftyfrac(ikA)^nn! , ?$$
– QuantumDK
Aug 21 at 1:39