Dirac Delta Function and Position [duplicate]












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  • Normalization of basis vectors with a continuous index?

    3 answers



  • Why is $langle x| x' rangle=delta(x-x')$? [duplicate]

    2 answers



  • How is the “normalisation” of non-normalizable states chosen?

    2 answers




How does one prove that the Dirac Delta distribution is the eigenfunction of the position operator $hat{x}$? In math, why does $langle x’|xrangle = delta(x’-x)$?










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  • 2





    What kind of proof are you looking for? Are you looking for plausible explanations as to the normalization condition you mention, or are you looking for a rigorous and mathematically consistent proof? The former is fine, and Aaron's answer does an excellent job of it, but if you're looking for the latter then unfortunately you're out of luck because strictly speaking the position operator (and indeed any operator with a purely continuous spectrum) does not actually have any eigenfunctions.

    – J. Murray
    yesterday











  • Possible duplicates: physics.stackexchange.com/q/89958/2451 , physics.stackexchange.com/q/273423/2451 , physics.stackexchange.com/q/330416/2451 and links therein.

    – Qmechanic
    yesterday











  • Related : Hermiticity of Momentum Operator (matrix) Represented in Position Basis

    – Frobenius
    yesterday
















5
















This question already has an answer here:




  • Normalization of basis vectors with a continuous index?

    3 answers



  • Why is $langle x| x' rangle=delta(x-x')$? [duplicate]

    2 answers



  • How is the “normalisation” of non-normalizable states chosen?

    2 answers




How does one prove that the Dirac Delta distribution is the eigenfunction of the position operator $hat{x}$? In math, why does $langle x’|xrangle = delta(x’-x)$?










share|cite|improve this question















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yesterday


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.











  • 2





    What kind of proof are you looking for? Are you looking for plausible explanations as to the normalization condition you mention, or are you looking for a rigorous and mathematically consistent proof? The former is fine, and Aaron's answer does an excellent job of it, but if you're looking for the latter then unfortunately you're out of luck because strictly speaking the position operator (and indeed any operator with a purely continuous spectrum) does not actually have any eigenfunctions.

    – J. Murray
    yesterday











  • Possible duplicates: physics.stackexchange.com/q/89958/2451 , physics.stackexchange.com/q/273423/2451 , physics.stackexchange.com/q/330416/2451 and links therein.

    – Qmechanic
    yesterday











  • Related : Hermiticity of Momentum Operator (matrix) Represented in Position Basis

    – Frobenius
    yesterday














5












5








5









This question already has an answer here:




  • Normalization of basis vectors with a continuous index?

    3 answers



  • Why is $langle x| x' rangle=delta(x-x')$? [duplicate]

    2 answers



  • How is the “normalisation” of non-normalizable states chosen?

    2 answers




How does one prove that the Dirac Delta distribution is the eigenfunction of the position operator $hat{x}$? In math, why does $langle x’|xrangle = delta(x’-x)$?










share|cite|improve this question

















This question already has an answer here:




  • Normalization of basis vectors with a continuous index?

    3 answers



  • Why is $langle x| x' rangle=delta(x-x')$? [duplicate]

    2 answers



  • How is the “normalisation” of non-normalizable states chosen?

    2 answers




How does one prove that the Dirac Delta distribution is the eigenfunction of the position operator $hat{x}$? In math, why does $langle x’|xrangle = delta(x’-x)$?





This question already has an answer here:




  • Normalization of basis vectors with a continuous index?

    3 answers



  • Why is $langle x| x' rangle=delta(x-x')$? [duplicate]

    2 answers



  • How is the “normalisation” of non-normalizable states chosen?

    2 answers








quantum-mechanics hilbert-space probability dirac-delta-distributions normalization






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edited yesterday









Qmechanic

102k121831163




102k121831163










asked yesterday









Christina DanielChristina Daniel

906




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marked as duplicate by Qmechanic quantum-mechanics
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This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.






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yesterday


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.










  • 2





    What kind of proof are you looking for? Are you looking for plausible explanations as to the normalization condition you mention, or are you looking for a rigorous and mathematically consistent proof? The former is fine, and Aaron's answer does an excellent job of it, but if you're looking for the latter then unfortunately you're out of luck because strictly speaking the position operator (and indeed any operator with a purely continuous spectrum) does not actually have any eigenfunctions.

    – J. Murray
    yesterday











  • Possible duplicates: physics.stackexchange.com/q/89958/2451 , physics.stackexchange.com/q/273423/2451 , physics.stackexchange.com/q/330416/2451 and links therein.

    – Qmechanic
    yesterday











  • Related : Hermiticity of Momentum Operator (matrix) Represented in Position Basis

    – Frobenius
    yesterday














  • 2





    What kind of proof are you looking for? Are you looking for plausible explanations as to the normalization condition you mention, or are you looking for a rigorous and mathematically consistent proof? The former is fine, and Aaron's answer does an excellent job of it, but if you're looking for the latter then unfortunately you're out of luck because strictly speaking the position operator (and indeed any operator with a purely continuous spectrum) does not actually have any eigenfunctions.

    – J. Murray
    yesterday











  • Possible duplicates: physics.stackexchange.com/q/89958/2451 , physics.stackexchange.com/q/273423/2451 , physics.stackexchange.com/q/330416/2451 and links therein.

    – Qmechanic
    yesterday











  • Related : Hermiticity of Momentum Operator (matrix) Represented in Position Basis

    – Frobenius
    yesterday








2




2





What kind of proof are you looking for? Are you looking for plausible explanations as to the normalization condition you mention, or are you looking for a rigorous and mathematically consistent proof? The former is fine, and Aaron's answer does an excellent job of it, but if you're looking for the latter then unfortunately you're out of luck because strictly speaking the position operator (and indeed any operator with a purely continuous spectrum) does not actually have any eigenfunctions.

– J. Murray
yesterday





What kind of proof are you looking for? Are you looking for plausible explanations as to the normalization condition you mention, or are you looking for a rigorous and mathematically consistent proof? The former is fine, and Aaron's answer does an excellent job of it, but if you're looking for the latter then unfortunately you're out of luck because strictly speaking the position operator (and indeed any operator with a purely continuous spectrum) does not actually have any eigenfunctions.

– J. Murray
yesterday













Possible duplicates: physics.stackexchange.com/q/89958/2451 , physics.stackexchange.com/q/273423/2451 , physics.stackexchange.com/q/330416/2451 and links therein.

– Qmechanic
yesterday





Possible duplicates: physics.stackexchange.com/q/89958/2451 , physics.stackexchange.com/q/273423/2451 , physics.stackexchange.com/q/330416/2451 and links therein.

– Qmechanic
yesterday













Related : Hermiticity of Momentum Operator (matrix) Represented in Position Basis

– Frobenius
yesterday





Related : Hermiticity of Momentum Operator (matrix) Represented in Position Basis

– Frobenius
yesterday










1 Answer
1






active

oldest

votes


















6














This is true not just for position eigenvectors. This is true for all eigenvectors in their own (orthonormal) eigenbasis.



For an operator with discrete eigenvalues $n$ with eigenvectors $|nrangle$,
$$langle n'|nrangle=delta_{n,n'}$$
where $delta_{n,n'}$ is the Kronecker Delta that is $1$ when $n=n'$ and $0$ otherwise.



For an operator with a continuous spectrum of eigenvalues $p$ with eigenvectors $|prangle$,
$$langle p'|prangle=delta(p'-p)$$
where the $delta$ this time is the Dirac delta function.



This is all just because when expressing a vector in a basis where that vector is a basis vector, it only has a component along that one vector (itself).





This isn't part of the main question, but I think it relates to the general idea of what the above expressions really mean and why they are used in the way they are used.




Do you happen to know why it is necessary to make the eigenfunction infinitely tall when working in the continuous spectrum?




This is a poor picture of the Dirac delta function. Lets first look at the Kronecker delta. If we have this in a sum
$$sum_{n'} delta_{n,n'}c_{n'}$$
We know this sum will just evaluate to $c_n$. The Kronecker delta "kills off" all other $c_{n'}$ terms.



Now let's move to the continuous version of this example with the Dirac delta function:
$$int delta(p-p')c(p')text d p'=c(p)$$
Now the integral is a "continuous sum" of the terms $c(p')delta(p-p')text d p'$. Notice how here we have $text d p'$ in each term. Therefore, if we want to pick out just $c(p)$ from this sum, we need $delta(p-p')$ to be equal to $1/text d p'$ when $p'=p$ and $0$ otherwise. Since $text d p'$ is an infinitesimal amount, $1/text d p'$ is an infinite amount. This is why you usually see that the Dirac delta function as an infinite spike, but saying this somewhat obscures the above analogy with the discrete case.






share|cite|improve this answer





















  • 1





    Thank you. Do you happen to know why it is necessary to make the eigenfunction infinitely tall when working in the continuous spectrum?

    – Christina Daniel
    yesterday






  • 1





    What is the "infinite spectrum"?

    – ggcg
    yesterday











  • @ggcg Where do you see that being used?

    – Aaron Stevens
    yesterday













  • @ChristinaDaniel I have edited my question.

    – Aaron Stevens
    yesterday


















1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









6














This is true not just for position eigenvectors. This is true for all eigenvectors in their own (orthonormal) eigenbasis.



For an operator with discrete eigenvalues $n$ with eigenvectors $|nrangle$,
$$langle n'|nrangle=delta_{n,n'}$$
where $delta_{n,n'}$ is the Kronecker Delta that is $1$ when $n=n'$ and $0$ otherwise.



For an operator with a continuous spectrum of eigenvalues $p$ with eigenvectors $|prangle$,
$$langle p'|prangle=delta(p'-p)$$
where the $delta$ this time is the Dirac delta function.



This is all just because when expressing a vector in a basis where that vector is a basis vector, it only has a component along that one vector (itself).





This isn't part of the main question, but I think it relates to the general idea of what the above expressions really mean and why they are used in the way they are used.




Do you happen to know why it is necessary to make the eigenfunction infinitely tall when working in the continuous spectrum?




This is a poor picture of the Dirac delta function. Lets first look at the Kronecker delta. If we have this in a sum
$$sum_{n'} delta_{n,n'}c_{n'}$$
We know this sum will just evaluate to $c_n$. The Kronecker delta "kills off" all other $c_{n'}$ terms.



Now let's move to the continuous version of this example with the Dirac delta function:
$$int delta(p-p')c(p')text d p'=c(p)$$
Now the integral is a "continuous sum" of the terms $c(p')delta(p-p')text d p'$. Notice how here we have $text d p'$ in each term. Therefore, if we want to pick out just $c(p)$ from this sum, we need $delta(p-p')$ to be equal to $1/text d p'$ when $p'=p$ and $0$ otherwise. Since $text d p'$ is an infinitesimal amount, $1/text d p'$ is an infinite amount. This is why you usually see that the Dirac delta function as an infinite spike, but saying this somewhat obscures the above analogy with the discrete case.






share|cite|improve this answer





















  • 1





    Thank you. Do you happen to know why it is necessary to make the eigenfunction infinitely tall when working in the continuous spectrum?

    – Christina Daniel
    yesterday






  • 1





    What is the "infinite spectrum"?

    – ggcg
    yesterday











  • @ggcg Where do you see that being used?

    – Aaron Stevens
    yesterday













  • @ChristinaDaniel I have edited my question.

    – Aaron Stevens
    yesterday
















6














This is true not just for position eigenvectors. This is true for all eigenvectors in their own (orthonormal) eigenbasis.



For an operator with discrete eigenvalues $n$ with eigenvectors $|nrangle$,
$$langle n'|nrangle=delta_{n,n'}$$
where $delta_{n,n'}$ is the Kronecker Delta that is $1$ when $n=n'$ and $0$ otherwise.



For an operator with a continuous spectrum of eigenvalues $p$ with eigenvectors $|prangle$,
$$langle p'|prangle=delta(p'-p)$$
where the $delta$ this time is the Dirac delta function.



This is all just because when expressing a vector in a basis where that vector is a basis vector, it only has a component along that one vector (itself).





This isn't part of the main question, but I think it relates to the general idea of what the above expressions really mean and why they are used in the way they are used.




Do you happen to know why it is necessary to make the eigenfunction infinitely tall when working in the continuous spectrum?




This is a poor picture of the Dirac delta function. Lets first look at the Kronecker delta. If we have this in a sum
$$sum_{n'} delta_{n,n'}c_{n'}$$
We know this sum will just evaluate to $c_n$. The Kronecker delta "kills off" all other $c_{n'}$ terms.



Now let's move to the continuous version of this example with the Dirac delta function:
$$int delta(p-p')c(p')text d p'=c(p)$$
Now the integral is a "continuous sum" of the terms $c(p')delta(p-p')text d p'$. Notice how here we have $text d p'$ in each term. Therefore, if we want to pick out just $c(p)$ from this sum, we need $delta(p-p')$ to be equal to $1/text d p'$ when $p'=p$ and $0$ otherwise. Since $text d p'$ is an infinitesimal amount, $1/text d p'$ is an infinite amount. This is why you usually see that the Dirac delta function as an infinite spike, but saying this somewhat obscures the above analogy with the discrete case.






share|cite|improve this answer





















  • 1





    Thank you. Do you happen to know why it is necessary to make the eigenfunction infinitely tall when working in the continuous spectrum?

    – Christina Daniel
    yesterday






  • 1





    What is the "infinite spectrum"?

    – ggcg
    yesterday











  • @ggcg Where do you see that being used?

    – Aaron Stevens
    yesterday













  • @ChristinaDaniel I have edited my question.

    – Aaron Stevens
    yesterday














6












6








6







This is true not just for position eigenvectors. This is true for all eigenvectors in their own (orthonormal) eigenbasis.



For an operator with discrete eigenvalues $n$ with eigenvectors $|nrangle$,
$$langle n'|nrangle=delta_{n,n'}$$
where $delta_{n,n'}$ is the Kronecker Delta that is $1$ when $n=n'$ and $0$ otherwise.



For an operator with a continuous spectrum of eigenvalues $p$ with eigenvectors $|prangle$,
$$langle p'|prangle=delta(p'-p)$$
where the $delta$ this time is the Dirac delta function.



This is all just because when expressing a vector in a basis where that vector is a basis vector, it only has a component along that one vector (itself).





This isn't part of the main question, but I think it relates to the general idea of what the above expressions really mean and why they are used in the way they are used.




Do you happen to know why it is necessary to make the eigenfunction infinitely tall when working in the continuous spectrum?




This is a poor picture of the Dirac delta function. Lets first look at the Kronecker delta. If we have this in a sum
$$sum_{n'} delta_{n,n'}c_{n'}$$
We know this sum will just evaluate to $c_n$. The Kronecker delta "kills off" all other $c_{n'}$ terms.



Now let's move to the continuous version of this example with the Dirac delta function:
$$int delta(p-p')c(p')text d p'=c(p)$$
Now the integral is a "continuous sum" of the terms $c(p')delta(p-p')text d p'$. Notice how here we have $text d p'$ in each term. Therefore, if we want to pick out just $c(p)$ from this sum, we need $delta(p-p')$ to be equal to $1/text d p'$ when $p'=p$ and $0$ otherwise. Since $text d p'$ is an infinitesimal amount, $1/text d p'$ is an infinite amount. This is why you usually see that the Dirac delta function as an infinite spike, but saying this somewhat obscures the above analogy with the discrete case.






share|cite|improve this answer















This is true not just for position eigenvectors. This is true for all eigenvectors in their own (orthonormal) eigenbasis.



For an operator with discrete eigenvalues $n$ with eigenvectors $|nrangle$,
$$langle n'|nrangle=delta_{n,n'}$$
where $delta_{n,n'}$ is the Kronecker Delta that is $1$ when $n=n'$ and $0$ otherwise.



For an operator with a continuous spectrum of eigenvalues $p$ with eigenvectors $|prangle$,
$$langle p'|prangle=delta(p'-p)$$
where the $delta$ this time is the Dirac delta function.



This is all just because when expressing a vector in a basis where that vector is a basis vector, it only has a component along that one vector (itself).





This isn't part of the main question, but I think it relates to the general idea of what the above expressions really mean and why they are used in the way they are used.




Do you happen to know why it is necessary to make the eigenfunction infinitely tall when working in the continuous spectrum?




This is a poor picture of the Dirac delta function. Lets first look at the Kronecker delta. If we have this in a sum
$$sum_{n'} delta_{n,n'}c_{n'}$$
We know this sum will just evaluate to $c_n$. The Kronecker delta "kills off" all other $c_{n'}$ terms.



Now let's move to the continuous version of this example with the Dirac delta function:
$$int delta(p-p')c(p')text d p'=c(p)$$
Now the integral is a "continuous sum" of the terms $c(p')delta(p-p')text d p'$. Notice how here we have $text d p'$ in each term. Therefore, if we want to pick out just $c(p)$ from this sum, we need $delta(p-p')$ to be equal to $1/text d p'$ when $p'=p$ and $0$ otherwise. Since $text d p'$ is an infinitesimal amount, $1/text d p'$ is an infinite amount. This is why you usually see that the Dirac delta function as an infinite spike, but saying this somewhat obscures the above analogy with the discrete case.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited yesterday

























answered yesterday









Aaron StevensAaron Stevens

9,61931741




9,61931741








  • 1





    Thank you. Do you happen to know why it is necessary to make the eigenfunction infinitely tall when working in the continuous spectrum?

    – Christina Daniel
    yesterday






  • 1





    What is the "infinite spectrum"?

    – ggcg
    yesterday











  • @ggcg Where do you see that being used?

    – Aaron Stevens
    yesterday













  • @ChristinaDaniel I have edited my question.

    – Aaron Stevens
    yesterday














  • 1





    Thank you. Do you happen to know why it is necessary to make the eigenfunction infinitely tall when working in the continuous spectrum?

    – Christina Daniel
    yesterday






  • 1





    What is the "infinite spectrum"?

    – ggcg
    yesterday











  • @ggcg Where do you see that being used?

    – Aaron Stevens
    yesterday













  • @ChristinaDaniel I have edited my question.

    – Aaron Stevens
    yesterday








1




1





Thank you. Do you happen to know why it is necessary to make the eigenfunction infinitely tall when working in the continuous spectrum?

– Christina Daniel
yesterday





Thank you. Do you happen to know why it is necessary to make the eigenfunction infinitely tall when working in the continuous spectrum?

– Christina Daniel
yesterday




1




1





What is the "infinite spectrum"?

– ggcg
yesterday





What is the "infinite spectrum"?

– ggcg
yesterday













@ggcg Where do you see that being used?

– Aaron Stevens
yesterday







@ggcg Where do you see that being used?

– Aaron Stevens
yesterday















@ChristinaDaniel I have edited my question.

– Aaron Stevens
yesterday





@ChristinaDaniel I have edited my question.

– Aaron Stevens
yesterday



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