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







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  • 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♦
    19 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♦
    19 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 20 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♦
    19 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♦
    19 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♦
19 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♦
19 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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この記事は検証可能な参考文献や出典が全く示されていないか、不十分です。 出典を追加して記事の信頼性向上にご協力ください。 （ 2019年1月 ） 「 サソリ 」のその他の用法については「サソリ (曖昧さ回避)」をご覧ください。 「 スコーピオン 」と「 スコルピオン 」はこの項目へ転送されています。その他の用法については「スコーピオン」をご覧ください。 サソリ目 生息年代: 430–0 Ma PreЄ Є O S D C P T J K Pg N シルル紀前期-現世 ダイオウサソリ（ Pandinus Imperator ） 分類 界 : 動物界 Animalia 門 : 節足動物門 Arthropoda 亜門 : 鋏角亜門 Chelicerata 綱 : クモガタ綱 Arachnida 目 : サソリ目 Scorpiones 学名 Scorpiones C. L. Koch, 1837 英名 Scorpion 上科 Pseudochactoidea キョクトウサソリ上科 Buthoidea Chaeriloidea Chactoidea Iuroidea Scorpionoidea サソリ （ サソリ類 、 蠍 、 蝎 、 scorpion ）は、鋏角亜門・クモガタ綱の サソリ目 （ Scorpiones ）に属する節足動物の総称である。体の前端に鋏型の触肢、後端に毒針を有する捕食者である。 1700種以上を含め、最古の化石記録は4億3千万年前のシルル紀まで遡る。有毒生物として広く知られるが、人間の命にかかわる毒を持つものはごく一部の種類に過ぎない。 目次 1 形態 1.1 前体 1.2 後体 1.2.1 中体 1.2.2 終体 2 雌雄 3 生態 3.1 天敵 3.2 生殖 4 分布 5 毒 6 蛍光現象 7 分類
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広島県道265号伴広島線

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一般県道 広島県道265号伴広島線 路線延長 km 起点 広島県広島市安佐南区伴東7丁目 終点 広島県広島市中区榎町 接続する 主な道路 （記法） 広島県道38号広島豊平線 広島県道71号広島湯来線 山陽自動車道五日市IC 国道183号（ 国道261号重複） ■テンプレート（■ノート ■使い方） ■PJ道路 広島県道265号伴広島線 （ひろしまけんどう265ごう ともひろしません）は広島県広島市安佐南区から広島県広島市中区を結ぶ一般県道である。 目次 1 路線データ 2 通過する自治体 3 見どころ 4 接続道路 4.1 本線 4.2 旧道 5 別名 6 関連項目 路線データ 本線 起点：広島市安佐南区伴東7丁目・大原下橋南詰交差点（広島県道38号広島豊平線（本線・旧道）交点） 終点：広島市中区榎町・十日市交差点（国道183号（国道261号重用）交点） 総延長： 旧道 起点：広島市安佐南区伴東4丁目・新細坂交差点（広島県道38号広島豊平線交点） 終点：広島市安佐南区伴中央4丁目・寺組橋(南)交差点（広島県道265号伴広島線（本線）交点） 総延長： 通過する自治体 広島県 広島市（安佐南区、佐伯区、西区、中区） 見どころ 広島高速交通広島新交通1号線（アストラムライン） 広島広域公園 広島広域公園陸上競技場（広島ビッグアーチ） 己斐峠 己斐橋（太田川放水路） 天満橋（天満川） 接続道路 本線 広島県道38号広島豊平線（本線・旧道）（広島市安佐南区伴東7丁目・大原下橋南詰交差点〔起点〕） 広島県道71号広島湯来線（本線）（広島市安佐南区伴中央2丁目・伴交番入口交差点 - 広島市安佐南区伴中央4丁目・寺組橋(南)交差点間で重用） 広島県道71号広島湯来線（旧道）（広島県道265号伴広島線（旧道）重用）（広島市安佐南区伴中央4丁目・寺組橋(南)交差点） 広島県道71号広島湯来線（広島市安佐南区伴中央4丁目・寺組橋(南)交差点 - 広島市佐伯区五日市町石内・五月が丘(南)交差点間で重用）
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Setup Asymptote in Texstudio

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0 Good time of the day! I am trying to compile the following file: documentclass[12pt]{article} usepackage[inline]{asymptote} begin{document} Here is a venn diagram produced with Asymptote, drawn to width 4cm: defA{A} defB{V{B}} begin{figure} begin{center} begin{asy} size(4cm,0); pen colour1=red; pen colour2=green; pair z0=(0,0); pair z1=(-1,0); pair z2=(1,0); real r=1.5; path c1=circle(z1,r); path c2=circle(z2,r); fill(c1,colour1); fill(c2,colour2); picture intersection=new picture; fill(intersection,c1,colour1+colour2); clip(intersection,c2); add(intersection); draw(c1); draw(c2); //draw("$A$",box,z1)
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