Occupation (or particle) number operator. Eigenvalues and eigenvectors. [closed]
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https://homepage.univie.ac.at/reinhold.bertlmann/pdfs/T2_Skript_Ch_5.pdf
Help me please. I made a screen (below) from the article above and highlighted what I did not understand.
Why is it true?
enter image description here
ordinary-differential-equations eigenvalues-eigenvectors quantum-mechanics
asked Dec 13 '18 at 12:20
Иван ПетровИван Петров
61
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closed as off-topic by Brahadeesh, amWhy, TMM, José Carlos Santos, Daniel Moskovich Dec 13 '18 at 15:49
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Brahadeesh, amWhy, TMM, José Carlos Santos, Daniel Moskovich
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$begingroup$
https://homepage.univie.ac.at/reinhold.bertlmann/pdfs/T2_Skript_Ch_5.pdf
Help me please. I made a screen (below) from the article above and highlighted what I did not understand.
Why is it true?
enter image description here
ordinary-differential-equations eigenvalues-eigenvectors quantum-mechanics
asked Dec 13 '18 at 12:20
Иван ПетровИван Петров
61
$endgroup$
closed as off-topic by Brahadeesh, amWhy, TMM, José Carlos Santos, Daniel Moskovich Dec 13 '18 at 15:49
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Brahadeesh, amWhy, TMM, José Carlos Santos, Daniel Moskovich
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
https://homepage.univie.ac.at/reinhold.bertlmann/pdfs/T2_Skript_Ch_5.pdf
Help me please. I made a screen (below) from the article above and highlighted what I did not understand.
Why is it true?
enter image description here
ordinary-differential-equations eigenvalues-eigenvectors quantum-mechanics
asked Dec 13 '18 at 12:20
Иван ПетровИван Петров
61
$endgroup$
https://homepage.univie.ac.at/reinhold.bertlmann/pdfs/T2_Skript_Ch_5.pdf
Help me please. I made a screen (below) from the article above and highlighted what I did not understand.
Why is it true?
enter image description here
ordinary-differential-equations eigenvalues-eigenvectors quantum-mechanics
ordinary-differential-equations eigenvalues-eigenvectors quantum-mechanics
asked Dec 13 '18 at 12:20
Иван ПетровИван Петров
61
asked Dec 13 '18 at 12:20
Иван ПетровИван Петров
61
asked Dec 13 '18 at 12:20
Иван ПетровИван Петров
61
asked Dec 13 '18 at 12:20
Иван ПетровИван Петров
61
asked Dec 13 '18 at 12:20
Иван ПетровИван Петров
61
61
closed as off-topic by Brahadeesh, amWhy, TMM, José Carlos Santos, Daniel Moskovich Dec 13 '18 at 15:49
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Brahadeesh, amWhy, TMM, José Carlos Santos, Daniel Moskovich
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Brahadeesh, amWhy, TMM, José Carlos Santos, Daniel Moskovich Dec 13 '18 at 15:49
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Brahadeesh, amWhy, TMM, José Carlos Santos, Daniel Moskovich
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The operator $N$ works like this
$$
N psi_{😺} = 😺 psi_{😺} tag{1}
$$
Which you can read as: If I apply $N$ to a state, I get the label of state times the state (more technically, $psi_{😺}$ is an eigenvector of $N$ with eigenvalue $😺$).
In your problem you have
$$
N(a^daggerpsi_nu) = (nu + 1)(a^dagger psi_nu) tag{2}
$$
If you compare this with (1) you will conclude that
$$
a^daggerpsi_nu sim psi_{nu + 1}
$$
answered Dec 13 '18 at 12:45
caveraccaverac
14.6k31130
$endgroup$
$begingroup$
I will ask in another way. Why can't one eigenvalue correspond to two linearly independent vectors? In this situation.
$endgroup$
– Иван Петров
Dec 13 '18 at 13:08
$begingroup$
@ИванПетров You mean something like $a^{dagger} psi_{nu} = (nu + 1)( psi^{(1)}_{nu+1} + psi^{(2)}_{nu+1} )$?
$endgroup$
– caverac
Dec 13 '18 at 13:14
$begingroup$
I mean that a^daggerpsi_nu and psi_{nu + 1} may be linearly independent. Why not?
$endgroup$
– Иван Петров
Dec 13 '18 at 13:27
$begingroup$
@ИванПетров And have the same eigenvalue?
$endgroup$
– caverac
Dec 13 '18 at 13:36
$begingroup$
Yes, the same eigenvalue
$endgroup$
– Иван Петров
Dec 13 '18 at 13:49
|
show 3 more comments
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The operator $N$ works like this
$$
N psi_{😺} = 😺 psi_{😺} tag{1}
$$
Which you can read as: If I apply $N$ to a state, I get the label of state times the state (more technically, $psi_{😺}$ is an eigenvector of $N$ with eigenvalue $😺$).
In your problem you have
$$
N(a^daggerpsi_nu) = (nu + 1)(a^dagger psi_nu) tag{2}
$$
If you compare this with (1) you will conclude that
$$
a^daggerpsi_nu sim psi_{nu + 1}
$$
answered Dec 13 '18 at 12:45
caveraccaverac
14.6k31130
$endgroup$
$begingroup$
I will ask in another way. Why can't one eigenvalue correspond to two linearly independent vectors? In this situation.
$endgroup$
– Иван Петров
Dec 13 '18 at 13:08
$begingroup$
@ИванПетров You mean something like $a^{dagger} psi_{nu} = (nu + 1)( psi^{(1)}_{nu+1} + psi^{(2)}_{nu+1} )$?
$endgroup$
– caverac
Dec 13 '18 at 13:14
$begingroup$
I mean that a^daggerpsi_nu and psi_{nu + 1} may be linearly independent. Why not?
$endgroup$
– Иван Петров
Dec 13 '18 at 13:27
$begingroup$
@ИванПетров And have the same eigenvalue?
$endgroup$
– caverac
Dec 13 '18 at 13:36
$begingroup$
Yes, the same eigenvalue
$endgroup$
– Иван Петров
Dec 13 '18 at 13:49
|
show 3 more comments
$begingroup$
The operator $N$ works like this
$$
N psi_{😺} = 😺 psi_{😺} tag{1}
$$
Which you can read as: If I apply $N$ to a state, I get the label of state times the state (more technically, $psi_{😺}$ is an eigenvector of $N$ with eigenvalue $😺$).
In your problem you have
$$
N(a^daggerpsi_nu) = (nu + 1)(a^dagger psi_nu) tag{2}
$$
If you compare this with (1) you will conclude that
$$
a^daggerpsi_nu sim psi_{nu + 1}
$$
answered Dec 13 '18 at 12:45
caveraccaverac
14.6k31130
$endgroup$
$begingroup$
I will ask in another way. Why can't one eigenvalue correspond to two linearly independent vectors? In this situation.
$endgroup$
– Иван Петров
Dec 13 '18 at 13:08
$begingroup$
@ИванПетров You mean something like $a^{dagger} psi_{nu} = (nu + 1)( psi^{(1)}_{nu+1} + psi^{(2)}_{nu+1} )$?
$endgroup$
– caverac
Dec 13 '18 at 13:14
$begingroup$
I mean that a^daggerpsi_nu and psi_{nu + 1} may be linearly independent. Why not?
$endgroup$
– Иван Петров
Dec 13 '18 at 13:27
$begingroup$
@ИванПетров And have the same eigenvalue?
$endgroup$
– caverac
Dec 13 '18 at 13:36
$begingroup$
Yes, the same eigenvalue
$endgroup$
– Иван Петров
Dec 13 '18 at 13:49
|
show 3 more comments
$begingroup$
The operator $N$ works like this
$$
N psi_{😺} = 😺 psi_{😺} tag{1}
$$
Which you can read as: If I apply $N$ to a state, I get the label of state times the state (more technically, $psi_{😺}$ is an eigenvector of $N$ with eigenvalue $😺$).
In your problem you have
$$
N(a^daggerpsi_nu) = (nu + 1)(a^dagger psi_nu) tag{2}
$$
If you compare this with (1) you will conclude that
$$
a^daggerpsi_nu sim psi_{nu + 1}
$$
answered Dec 13 '18 at 12:45
caveraccaverac
14.6k31130
$endgroup$
The operator $N$ works like this
$$
N psi_{😺} = 😺 psi_{😺} tag{1}
$$
Which you can read as: If I apply $N$ to a state, I get the label of state times the state (more technically, $psi_{😺}$ is an eigenvector of $N$ with eigenvalue $😺$).
In your problem you have
$$
N(a^daggerpsi_nu) = (nu + 1)(a^dagger psi_nu) tag{2}
$$
If you compare this with (1) you will conclude that
$$
a^daggerpsi_nu sim psi_{nu + 1}
$$
answered Dec 13 '18 at 12:45
caveraccaverac
14.6k31130
answered Dec 13 '18 at 12:45
caveraccaverac
14.6k31130
answered Dec 13 '18 at 12:45
caveraccaverac
14.6k31130
answered Dec 13 '18 at 12:45
caveraccaverac
14.6k31130
14.6k31130
$begingroup$
I will ask in another way. Why can't one eigenvalue correspond to two linearly independent vectors? In this situation.
$endgroup$
– Иван Петров
Dec 13 '18 at 13:08
$begingroup$
@ИванПетров You mean something like $a^{dagger} psi_{nu} = (nu + 1)( psi^{(1)}_{nu+1} + psi^{(2)}_{nu+1} )$?
$endgroup$
– caverac
Dec 13 '18 at 13:14
$begingroup$
I mean that a^daggerpsi_nu and psi_{nu + 1} may be linearly independent. Why not?
$endgroup$
– Иван Петров
Dec 13 '18 at 13:27
$begingroup$
@ИванПетров And have the same eigenvalue?
$endgroup$
– caverac
Dec 13 '18 at 13:36
$begingroup$
Yes, the same eigenvalue
$endgroup$
– Иван Петров
Dec 13 '18 at 13:49
|
show 3 more comments
$begingroup$
I will ask in another way. Why can't one eigenvalue correspond to two linearly independent vectors? In this situation.
$endgroup$
– Иван Петров
Dec 13 '18 at 13:08
$begingroup$
@ИванПетров You mean something like $a^{dagger} psi_{nu} = (nu + 1)( psi^{(1)}_{nu+1} + psi^{(2)}_{nu+1} )$?
$endgroup$
– caverac
Dec 13 '18 at 13:14
$begingroup$
I mean that a^daggerpsi_nu and psi_{nu + 1} may be linearly independent. Why not?
$endgroup$
– Иван Петров
Dec 13 '18 at 13:27
$begingroup$
@ИванПетров And have the same eigenvalue?
$endgroup$
– caverac
Dec 13 '18 at 13:36
$begingroup$
Yes, the same eigenvalue
$endgroup$
– Иван Петров
Dec 13 '18 at 13:49
$begingroup$
I will ask in another way. Why can't one eigenvalue correspond to two linearly independent vectors? In this situation.
$endgroup$
– Иван Петров
Dec 13 '18 at 13:08
$begingroup$
I will ask in another way. Why can't one eigenvalue correspond to two linearly independent vectors? In this situation.
$endgroup$
– Иван Петров
Dec 13 '18 at 13:08
$begingroup$
@ИванПетров You mean something like $a^{dagger} psi_{nu} = (nu + 1)( psi^{(1)}_{nu+1} + psi^{(2)}_{nu+1} )$?
$endgroup$
– caverac
Dec 13 '18 at 13:14
$begingroup$
@ИванПетров You mean something like $a^{dagger} psi_{nu} = (nu + 1)( psi^{(1)}_{nu+1} + psi^{(2)}_{nu+1} )$?
$endgroup$
– caverac
Dec 13 '18 at 13:14
$begingroup$
I mean that a^daggerpsi_nu and psi_{nu + 1} may be linearly independent. Why not?
$endgroup$
– Иван Петров
Dec 13 '18 at 13:27
$begingroup$
I mean that a^daggerpsi_nu and psi_{nu + 1} may be linearly independent. Why not?
$endgroup$
– Иван Петров
Dec 13 '18 at 13:27
$begingroup$
@ИванПетров And have the same eigenvalue?
$endgroup$
– caverac
Dec 13 '18 at 13:36
$begingroup$
@ИванПетров And have the same eigenvalue?
$endgroup$
– caverac
Dec 13 '18 at 13:36
$begingroup$
Yes, the same eigenvalue
$endgroup$
– Иван Петров
Dec 13 '18 at 13:49
$begingroup$
Yes, the same eigenvalue
$endgroup$
– Иван Петров
Dec 13 '18 at 13:49
|
show 3 more comments
