Mixing
Dirac Delta Function and Position [duplicate]
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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 $langle x’|xrangle = delta(x’-x)$?
quantum-mechanics hilbert-space probability dirac-delta-distributions normalization
edited Jan 11 at 7:07
Qmechanic♦
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asked Jan 11 at 3:05
Christina DanielChristina Daniel
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Jan 11 at 7:50
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.
add a comment |
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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 $langle x’|xrangle = delta(x’-x)$?
quantum-mechanics hilbert-space probability dirac-delta-distributions normalization
edited Jan 11 at 7:07
Qmechanic♦
104k121871192
asked Jan 11 at 3:05
Christina DanielChristina Daniel
906
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Jan 11 at 7:50
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
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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.
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– J. Murray
Jan 11 at 4:46
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Possible duplicates: physics.stackexchange.com/q/89958/2451 , physics.stackexchange.com/q/273423/2451 , physics.stackexchange.com/q/330416/2451 and links therein.
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– Qmechanic♦
Jan 11 at 7:09
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Related : Hermiticity of Momentum Operator (matrix) Represented in Position Basis
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– Frobenius
Jan 11 at 9:59
add a comment |
$begingroup$
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 $langle x’|xrangle = delta(x’-x)$?
quantum-mechanics hilbert-space probability dirac-delta-distributions normalization
edited Jan 11 at 7:07
Qmechanic♦
104k121871192
asked Jan 11 at 3:05
Christina DanielChristina Daniel
906
$endgroup$
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
quantum-mechanics hilbert-space probability dirac-delta-distributions normalization
edited Jan 11 at 7:07
Qmechanic♦
104k121871192
asked Jan 11 at 3:05
Christina DanielChristina Daniel
906
edited Jan 11 at 7:07
Qmechanic♦
104k121871192
asked Jan 11 at 3:05
Christina DanielChristina Daniel
906
edited Jan 11 at 7:07
Qmechanic♦
104k121871192
edited Jan 11 at 7:07
Qmechanic♦
104k121871192
edited Jan 11 at 7:07
Qmechanic♦
104k121871192
104k121871192
asked Jan 11 at 3:05
Christina DanielChristina Daniel
906
asked Jan 11 at 3:05
Christina DanielChristina Daniel
906
asked Jan 11 at 3:05
Christina DanielChristina Daniel
906
906
marked as duplicate by Qmechanic♦ quantum-mechanics
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Jan 11 at 7:50
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Jan 11 at 7:50
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
$begingroup$
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.
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– J. Murray
Jan 11 at 4:46
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Possible duplicates: physics.stackexchange.com/q/89958/2451 , physics.stackexchange.com/q/273423/2451 , physics.stackexchange.com/q/330416/2451 and links therein.
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– Qmechanic♦
Jan 11 at 7:09
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Related : Hermiticity of Momentum Operator (matrix) Represented in Position Basis
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– Frobenius
Jan 11 at 9:59
add a comment |
2
$begingroup$
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.
$endgroup$
– J. Murray
Jan 11 at 4:46
$begingroup$
Possible duplicates: physics.stackexchange.com/q/89958/2451 , physics.stackexchange.com/q/273423/2451 , physics.stackexchange.com/q/330416/2451 and links therein.
$endgroup$
– Qmechanic♦
Jan 11 at 7:09
$begingroup$
Related : Hermiticity of Momentum Operator (matrix) Represented in Position Basis
$endgroup$
– Frobenius
Jan 11 at 9:59
2
2
$begingroup$
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.
$endgroup$
– J. Murray
Jan 11 at 4:46
$begingroup$
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.
$endgroup$
– J. Murray
Jan 11 at 4:46
$begingroup$
Possible duplicates: physics.stackexchange.com/q/89958/2451 , physics.stackexchange.com/q/273423/2451 , physics.stackexchange.com/q/330416/2451 and links therein.
$endgroup$
– Qmechanic♦
Jan 11 at 7:09
$begingroup$
Possible duplicates: physics.stackexchange.com/q/89958/2451 , physics.stackexchange.com/q/273423/2451 , physics.stackexchange.com/q/330416/2451 and links therein.
$endgroup$
– Qmechanic♦
Jan 11 at 7:09
$begingroup$
Related : Hermiticity of Momentum Operator (matrix) Represented in Position Basis
$endgroup$
– Frobenius
Jan 11 at 9:59
$begingroup$
Related : Hermiticity of Momentum Operator (matrix) Represented in Position Basis
$endgroup$
– Frobenius
Jan 11 at 9:59
add a comment |
1 Answer
1
active
oldest
votes
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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.
answered Jan 11 at 3:20
Aaron StevensAaron Stevens
10.3k31742
$endgroup$
1
$begingroup$
Thank you. Do you happen to know why it is necessary to make the eigenfunction infinitely tall when working in the continuous spectrum?
$endgroup$
– Christina Daniel
Jan 11 at 3:23
1
$begingroup$
What is the "infinite spectrum"?
$endgroup$
– ggcg
Jan 11 at 4:00
$begingroup$
@ggcg Where do you see that being used?
$endgroup$
– Aaron Stevens
Jan 11 at 4:11
$begingroup$
@ChristinaDaniel I have edited my question.
$endgroup$
– Aaron Stevens
Jan 11 at 4:12
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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.
answered Jan 11 at 3:20
Aaron StevensAaron Stevens
10.3k31742
$endgroup$
1
$begingroup$
Thank you. Do you happen to know why it is necessary to make the eigenfunction infinitely tall when working in the continuous spectrum?
$endgroup$
– Christina Daniel
Jan 11 at 3:23
1
$begingroup$
What is the "infinite spectrum"?
$endgroup$
– ggcg
Jan 11 at 4:00
$begingroup$
@ggcg Where do you see that being used?
$endgroup$
– Aaron Stevens
Jan 11 at 4:11
$begingroup$
@ChristinaDaniel I have edited my question.
$endgroup$
– Aaron Stevens
Jan 11 at 4:12
add a comment |
$begingroup$
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.
answered Jan 11 at 3:20
Aaron StevensAaron Stevens
10.3k31742
$endgroup$
1
$begingroup$
Thank you. Do you happen to know why it is necessary to make the eigenfunction infinitely tall when working in the continuous spectrum?
$endgroup$
– Christina Daniel
Jan 11 at 3:23
1
$begingroup$
What is the "infinite spectrum"?
$endgroup$
– ggcg
Jan 11 at 4:00
$begingroup$
@ggcg Where do you see that being used?
$endgroup$
– Aaron Stevens
Jan 11 at 4:11
$begingroup$
@ChristinaDaniel I have edited my question.
$endgroup$
– Aaron Stevens
Jan 11 at 4:12
add a comment |
$begingroup$
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.
answered Jan 11 at 3:20
Aaron StevensAaron Stevens
10.3k31742
$endgroup$
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.
answered Jan 11 at 3:20
Aaron StevensAaron Stevens
10.3k31742
edited Jan 11 at 4:15
answered Jan 11 at 3:20
Aaron StevensAaron Stevens
10.3k31742
answered Jan 11 at 3:20
Aaron StevensAaron Stevens
10.3k31742
answered Jan 11 at 3:20
Aaron StevensAaron Stevens
10.3k31742
10.3k31742
1
$begingroup$
Thank you. Do you happen to know why it is necessary to make the eigenfunction infinitely tall when working in the continuous spectrum?
$endgroup$
– Christina Daniel
Jan 11 at 3:23
1
$begingroup$
What is the "infinite spectrum"?
$endgroup$
– ggcg
Jan 11 at 4:00
$begingroup$
@ggcg Where do you see that being used?
$endgroup$
– Aaron Stevens
Jan 11 at 4:11
$begingroup$
@ChristinaDaniel I have edited my question.
$endgroup$
– Aaron Stevens
Jan 11 at 4:12
add a comment |
1
$begingroup$
Thank you. Do you happen to know why it is necessary to make the eigenfunction infinitely tall when working in the continuous spectrum?
$endgroup$
– Christina Daniel
Jan 11 at 3:23
1
$begingroup$
What is the "infinite spectrum"?
$endgroup$
– ggcg
Jan 11 at 4:00
$begingroup$
@ggcg Where do you see that being used?
$endgroup$
– Aaron Stevens
Jan 11 at 4:11
$begingroup$
@ChristinaDaniel I have edited my question.
$endgroup$
– Aaron Stevens
Jan 11 at 4:12
1
1
$begingroup$
Thank you. Do you happen to know why it is necessary to make the eigenfunction infinitely tall when working in the continuous spectrum?
$endgroup$
– Christina Daniel
Jan 11 at 3:23
$begingroup$
Thank you. Do you happen to know why it is necessary to make the eigenfunction infinitely tall when working in the continuous spectrum?
$endgroup$
– Christina Daniel
Jan 11 at 3:23
1
1
$begingroup$
What is the "infinite spectrum"?
$endgroup$
– ggcg
Jan 11 at 4:00
$begingroup$
What is the "infinite spectrum"?
$endgroup$
– ggcg
Jan 11 at 4:00
$begingroup$
@ggcg Where do you see that being used?
$endgroup$
– Aaron Stevens
Jan 11 at 4:11
$begingroup$
@ggcg Where do you see that being used?
$endgroup$
– Aaron Stevens
Jan 11 at 4:11
$begingroup$
@ChristinaDaniel I have edited my question.
$endgroup$
– Aaron Stevens
Jan 11 at 4:12
$begingroup$
@ChristinaDaniel I have edited my question.
$endgroup$
– Aaron Stevens
Jan 11 at 4:12
add a comment |
2
$begingroup$
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.
$endgroup$
– J. Murray
Jan 11 at 4:46
$begingroup$
Possible duplicates: physics.stackexchange.com/q/89958/2451 , physics.stackexchange.com/q/273423/2451 , physics.stackexchange.com/q/330416/2451 and links therein.
$endgroup$
– Qmechanic♦
Jan 11 at 7:09
$begingroup$
Related : Hermiticity of Momentum Operator (matrix) Represented in Position Basis
$endgroup$
– Frobenius
Jan 11 at 9:59