Dirac Delta Function and Position












5














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






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







share|cite|improve this question




















  • 1




    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
    2 hours ago










  • Possible duplicates: physics.stackexchange.com/q/89958/2451 , physics.stackexchange.com/q/273423/2451 , physics.stackexchange.com/q/330416/2451 and links therein.
    – Qmechanic♦
    17 mins ago
















5














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






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







share|cite|improve this question




















  • 1




    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
    2 hours ago










  • Possible duplicates: physics.stackexchange.com/q/89958/2451 , physics.stackexchange.com/q/273423/2451 , physics.stackexchange.com/q/330416/2451 and links therein.
    – Qmechanic♦
    17 mins ago














5












5








5







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






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







share|cite|improve this question















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






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




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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 19 mins ago









Qmechanic♦

102k121831161




102k121831161










asked 4 hours ago









Christina DanielChristina Daniel

886




886








  • 1




    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
    2 hours ago










  • Possible duplicates: physics.stackexchange.com/q/89958/2451 , physics.stackexchange.com/q/273423/2451 , physics.stackexchange.com/q/330416/2451 and links therein.
    – Qmechanic♦
    17 mins ago














  • 1




    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
    2 hours ago










  • Possible duplicates: physics.stackexchange.com/q/89958/2451 , physics.stackexchange.com/q/273423/2451 , physics.stackexchange.com/q/330416/2451 and links therein.
    – Qmechanic♦
    17 mins ago








1




1




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
2 hours ago




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
2 hours ago












Possible duplicates: physics.stackexchange.com/q/89958/2451 , physics.stackexchange.com/q/273423/2451 , physics.stackexchange.com/q/330416/2451 and links therein.
– Qmechanic♦
17 mins ago




Possible duplicates: physics.stackexchange.com/q/89958/2451 , physics.stackexchange.com/q/273423/2451 , physics.stackexchange.com/q/330416/2451 and links therein.
– Qmechanic♦
17 mins ago










1 Answer
1






active

oldest

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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
    4 hours ago






  • 1




    What is the "infinite spectrum"?
    – ggcg
    3 hours ago










  • @ggcg Where do you see that being used?
    – Aaron Stevens
    3 hours ago












  • @ChristinaDaniel I have edited my question.
    – Aaron Stevens
    3 hours ago











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1 Answer
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active

oldest

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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
    4 hours ago






  • 1




    What is the "infinite spectrum"?
    – ggcg
    3 hours ago










  • @ggcg Where do you see that being used?
    – Aaron Stevens
    3 hours ago












  • @ChristinaDaniel I have edited my question.
    – Aaron Stevens
    3 hours ago
















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
    4 hours ago






  • 1




    What is the "infinite spectrum"?
    – ggcg
    3 hours ago










  • @ggcg Where do you see that being used?
    – Aaron Stevens
    3 hours ago












  • @ChristinaDaniel I have edited my question.
    – Aaron Stevens
    3 hours ago














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 3 hours ago

























answered 4 hours ago









Aaron StevensAaron Stevens

9,60631741




9,60631741








  • 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
    4 hours ago






  • 1




    What is the "infinite spectrum"?
    – ggcg
    3 hours ago










  • @ggcg Where do you see that being used?
    – Aaron Stevens
    3 hours ago












  • @ChristinaDaniel I have edited my question.
    – Aaron Stevens
    3 hours ago














  • 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
    4 hours ago






  • 1




    What is the "infinite spectrum"?
    – ggcg
    3 hours ago










  • @ggcg Where do you see that being used?
    – Aaron Stevens
    3 hours ago












  • @ChristinaDaniel I have edited my question.
    – Aaron Stevens
    3 hours ago








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
4 hours ago




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
4 hours ago




1




1




What is the "infinite spectrum"?
– ggcg
3 hours ago




What is the "infinite spectrum"?
– ggcg
3 hours ago












@ggcg Where do you see that being used?
– Aaron Stevens
3 hours ago






@ggcg Where do you see that being used?
– Aaron Stevens
3 hours ago














@ChristinaDaniel I have edited my question.
– Aaron Stevens
3 hours ago




@ChristinaDaniel I have edited my question.
– Aaron Stevens
3 hours ago


















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