From the condition $[A,B]=A$, what can I say about $B$?
$begingroup$
I'm struggling in understanding the meaning of this condition that I found in an operator equation:
begin{equation}
[A,B]=A
end{equation}
where both $A$ and $B$ are hermitian operators. What can I say about the operator $B$?
abstract-algebra operator-theory quantum-mechanics
asked Feb 15 '18 at 18:50
GbpGbp
635
$endgroup$
|
show 6 more comments
$begingroup$
I'm struggling in understanding the meaning of this condition that I found in an operator equation:
begin{equation}
[A,B]=A
end{equation}
where both $A$ and $B$ are hermitian operators. What can I say about the operator $B$?
abstract-algebra operator-theory quantum-mechanics
asked Feb 15 '18 at 18:50
GbpGbp
635
$endgroup$
1
$begingroup$
I think that's for commutator i.e. $[A,B]=AB-BA$.
$endgroup$
– user296113
Feb 15 '18 at 18:53
1
$begingroup$
Then I suppose you can assume that $AB - BA = A$.
$endgroup$
– Joe Johnson 126
Feb 15 '18 at 18:59
1
$begingroup$
...and you'll be able, for example, to conclude that tr.$,A=0;$ ...
$endgroup$
– DonAntonio
Feb 15 '18 at 19:04
3
$begingroup$
If you are asking what the equation «$[A,B]=A$» means, then it is simply the same as «$AB-BA=A$». If you are asking something else, you should explain...
$endgroup$
– Mariano Suárez-Álvarez
Feb 15 '18 at 19:06
2
$begingroup$
Perhaps you meant $[A,B]=iA$ when you say that the operators are hermitian (as the commutator is antihermitian)
$endgroup$
– user8268
Feb 15 '18 at 19:55
|
show 6 more comments
$begingroup$
I'm struggling in understanding the meaning of this condition that I found in an operator equation:
begin{equation}
[A,B]=A
end{equation}
where both $A$ and $B$ are hermitian operators. What can I say about the operator $B$?
abstract-algebra operator-theory quantum-mechanics
asked Feb 15 '18 at 18:50
GbpGbp
635
$endgroup$
I'm struggling in understanding the meaning of this condition that I found in an operator equation:
begin{equation}
[A,B]=A
end{equation}
where both $A$ and $B$ are hermitian operators. What can I say about the operator $B$?
abstract-algebra operator-theory quantum-mechanics
abstract-algebra operator-theory quantum-mechanics
asked Feb 15 '18 at 18:50
GbpGbp
635
asked Feb 15 '18 at 18:50
GbpGbp
635
edited Feb 15 '18 at 19:41
Gbp
asked Feb 15 '18 at 18:50
GbpGbp
635
asked Feb 15 '18 at 18:50
GbpGbp
635
asked Feb 15 '18 at 18:50
GbpGbp
635
635
1
$begingroup$
I think that's for commutator i.e. $[A,B]=AB-BA$.
$endgroup$
– user296113
Feb 15 '18 at 18:53
1
$begingroup$
Then I suppose you can assume that $AB - BA = A$.
$endgroup$
– Joe Johnson 126
Feb 15 '18 at 18:59
1
$begingroup$
...and you'll be able, for example, to conclude that tr.$,A=0;$ ...
$endgroup$
– DonAntonio
Feb 15 '18 at 19:04
3
$begingroup$
If you are asking what the equation «$[A,B]=A$» means, then it is simply the same as «$AB-BA=A$». If you are asking something else, you should explain...
$endgroup$
– Mariano Suárez-Álvarez
Feb 15 '18 at 19:06
2
$begingroup$
Perhaps you meant $[A,B]=iA$ when you say that the operators are hermitian (as the commutator is antihermitian)
$endgroup$
– user8268
Feb 15 '18 at 19:55
|
show 6 more comments
1
$begingroup$
I think that's for commutator i.e. $[A,B]=AB-BA$.
$endgroup$
– user296113
Feb 15 '18 at 18:53
1
$begingroup$
Then I suppose you can assume that $AB - BA = A$.
$endgroup$
– Joe Johnson 126
Feb 15 '18 at 18:59
1
$begingroup$
...and you'll be able, for example, to conclude that tr.$,A=0;$ ...
$endgroup$
– DonAntonio
Feb 15 '18 at 19:04
3
$begingroup$
If you are asking what the equation «$[A,B]=A$» means, then it is simply the same as «$AB-BA=A$». If you are asking something else, you should explain...
$endgroup$
– Mariano Suárez-Álvarez
Feb 15 '18 at 19:06
2
$begingroup$
Perhaps you meant $[A,B]=iA$ when you say that the operators are hermitian (as the commutator is antihermitian)
$endgroup$
– user8268
Feb 15 '18 at 19:55
1
1
$begingroup$
I think that's for commutator i.e. $[A,B]=AB-BA$.
$endgroup$
– user296113
Feb 15 '18 at 18:53
$begingroup$
I think that's for commutator i.e. $[A,B]=AB-BA$.
$endgroup$
– user296113
Feb 15 '18 at 18:53
1
1
$begingroup$
Then I suppose you can assume that $AB - BA = A$.
$endgroup$
– Joe Johnson 126
Feb 15 '18 at 18:59
$begingroup$
Then I suppose you can assume that $AB - BA = A$.
$endgroup$
– Joe Johnson 126
Feb 15 '18 at 18:59
1
1
$begingroup$
...and you'll be able, for example, to conclude that tr.$,A=0;$ ...
$endgroup$
– DonAntonio
Feb 15 '18 at 19:04
$begingroup$
...and you'll be able, for example, to conclude that tr.$,A=0;$ ...
$endgroup$
– DonAntonio
Feb 15 '18 at 19:04
3
3
$begingroup$
If you are asking what the equation «$[A,B]=A$» means, then it is simply the same as «$AB-BA=A$». If you are asking something else, you should explain...
$endgroup$
– Mariano Suárez-Álvarez
Feb 15 '18 at 19:06
$begingroup$
If you are asking what the equation «$[A,B]=A$» means, then it is simply the same as «$AB-BA=A$». If you are asking something else, you should explain...
$endgroup$
– Mariano Suárez-Álvarez
Feb 15 '18 at 19:06
2
2
$begingroup$
Perhaps you meant $[A,B]=iA$ when you say that the operators are hermitian (as the commutator is antihermitian)
$endgroup$
– user8268
Feb 15 '18 at 19:55
$begingroup$
Perhaps you meant $[A,B]=iA$ when you say that the operators are hermitian (as the commutator is antihermitian)
$endgroup$
– user8268
Feb 15 '18 at 19:55
|
show 6 more comments
1 Answer
1
active
oldest
votes
$begingroup$
If A and B are hermitian, then you need, instead,
$$
[A,B]= i A ,
$$
for consistency under taking the adjoint. So A is neither hermitian nor antihermitian, even though, conventionally, B is taken to be hermitian in applications. Let's stick with $[A,B]= A$.
These are the prototypes for the shift and counter operators, used routinely in quantum mechanics.
Specifically, for an eigenvector v of B with real eigenvalue λ,
$$
B v = lambda v ,
$$
utilize your simplest nonabelian Lie algebra you specified, to observe
$$
BAv = A(B-1)v,
$$
and hence
$$
B(Av)=(lambda-1)(Av).
$$
That is, A defines new eigenvectors with down-shifted eigenvalues of B, it "ladders" the state v down.
A simple realization of the algebra is $A=x$ and $B=-xpartial_x$. Consider what A does to the eigenvectors $x^{-lambda}$ of B.
Formally, this is the simplest nonabelian Lie algebra there is, with just one commutator, that of the affine group in one dimension . The upper-triangular 2x2 matrix representation thereof is given in the WP link provided.
You probably do not need to dwell on its delightful group structure. In QM, people use number operators for B and lowering operators a for A.
answered Dec 25 '18 at 16:54
Cosmas ZachosCosmas Zachos
1,802522
$endgroup$
add a comment |
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$begingroup$
If A and B are hermitian, then you need, instead,
$$
[A,B]= i A ,
$$
for consistency under taking the adjoint. So A is neither hermitian nor antihermitian, even though, conventionally, B is taken to be hermitian in applications. Let's stick with $[A,B]= A$.
These are the prototypes for the shift and counter operators, used routinely in quantum mechanics.
Specifically, for an eigenvector v of B with real eigenvalue λ,
$$
B v = lambda v ,
$$
utilize your simplest nonabelian Lie algebra you specified, to observe
$$
BAv = A(B-1)v,
$$
and hence
$$
B(Av)=(lambda-1)(Av).
$$
That is, A defines new eigenvectors with down-shifted eigenvalues of B, it "ladders" the state v down.
A simple realization of the algebra is $A=x$ and $B=-xpartial_x$. Consider what A does to the eigenvectors $x^{-lambda}$ of B.
Formally, this is the simplest nonabelian Lie algebra there is, with just one commutator, that of the affine group in one dimension . The upper-triangular 2x2 matrix representation thereof is given in the WP link provided.
You probably do not need to dwell on its delightful group structure. In QM, people use number operators for B and lowering operators a for A.
answered Dec 25 '18 at 16:54
Cosmas ZachosCosmas Zachos
1,802522
$endgroup$
add a comment |
$begingroup$
If A and B are hermitian, then you need, instead,
$$
[A,B]= i A ,
$$
for consistency under taking the adjoint. So A is neither hermitian nor antihermitian, even though, conventionally, B is taken to be hermitian in applications. Let's stick with $[A,B]= A$.
These are the prototypes for the shift and counter operators, used routinely in quantum mechanics.
Specifically, for an eigenvector v of B with real eigenvalue λ,
$$
B v = lambda v ,
$$
utilize your simplest nonabelian Lie algebra you specified, to observe
$$
BAv = A(B-1)v,
$$
and hence
$$
B(Av)=(lambda-1)(Av).
$$
That is, A defines new eigenvectors with down-shifted eigenvalues of B, it "ladders" the state v down.
A simple realization of the algebra is $A=x$ and $B=-xpartial_x$. Consider what A does to the eigenvectors $x^{-lambda}$ of B.
Formally, this is the simplest nonabelian Lie algebra there is, with just one commutator, that of the affine group in one dimension . The upper-triangular 2x2 matrix representation thereof is given in the WP link provided.
You probably do not need to dwell on its delightful group structure. In QM, people use number operators for B and lowering operators a for A.
answered Dec 25 '18 at 16:54
Cosmas ZachosCosmas Zachos
1,802522
$endgroup$
add a comment |
$begingroup$
If A and B are hermitian, then you need, instead,
$$
[A,B]= i A ,
$$
for consistency under taking the adjoint. So A is neither hermitian nor antihermitian, even though, conventionally, B is taken to be hermitian in applications. Let's stick with $[A,B]= A$.
These are the prototypes for the shift and counter operators, used routinely in quantum mechanics.
Specifically, for an eigenvector v of B with real eigenvalue λ,
$$
B v = lambda v ,
$$
utilize your simplest nonabelian Lie algebra you specified, to observe
$$
BAv = A(B-1)v,
$$
and hence
$$
B(Av)=(lambda-1)(Av).
$$
That is, A defines new eigenvectors with down-shifted eigenvalues of B, it "ladders" the state v down.
A simple realization of the algebra is $A=x$ and $B=-xpartial_x$. Consider what A does to the eigenvectors $x^{-lambda}$ of B.
Formally, this is the simplest nonabelian Lie algebra there is, with just one commutator, that of the affine group in one dimension . The upper-triangular 2x2 matrix representation thereof is given in the WP link provided.
You probably do not need to dwell on its delightful group structure. In QM, people use number operators for B and lowering operators a for A.
answered Dec 25 '18 at 16:54
Cosmas ZachosCosmas Zachos
1,802522
$endgroup$
If A and B are hermitian, then you need, instead,
$$
[A,B]= i A ,
$$
for consistency under taking the adjoint. So A is neither hermitian nor antihermitian, even though, conventionally, B is taken to be hermitian in applications. Let's stick with $[A,B]= A$.
These are the prototypes for the shift and counter operators, used routinely in quantum mechanics.
Specifically, for an eigenvector v of B with real eigenvalue λ,
$$
B v = lambda v ,
$$
utilize your simplest nonabelian Lie algebra you specified, to observe
$$
BAv = A(B-1)v,
$$
and hence
$$
B(Av)=(lambda-1)(Av).
$$
That is, A defines new eigenvectors with down-shifted eigenvalues of B, it "ladders" the state v down.
A simple realization of the algebra is $A=x$ and $B=-xpartial_x$. Consider what A does to the eigenvectors $x^{-lambda}$ of B.
Formally, this is the simplest nonabelian Lie algebra there is, with just one commutator, that of the affine group in one dimension . The upper-triangular 2x2 matrix representation thereof is given in the WP link provided.
You probably do not need to dwell on its delightful group structure. In QM, people use number operators for B and lowering operators a for A.
answered Dec 25 '18 at 16:54
Cosmas ZachosCosmas Zachos
1,802522
edited Dec 26 '18 at 16:32
answered Dec 25 '18 at 16:54
Cosmas ZachosCosmas Zachos
1,802522
answered Dec 25 '18 at 16:54
Cosmas ZachosCosmas Zachos
1,802522
answered Dec 25 '18 at 16:54
Cosmas ZachosCosmas Zachos
1,802522
1,802522
add a comment |
add a comment |
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1
$begingroup$
I think that's for commutator i.e. $[A,B]=AB-BA$.
$endgroup$
– user296113
Feb 15 '18 at 18:53
1
$begingroup$
Then I suppose you can assume that $AB - BA = A$.
$endgroup$
– Joe Johnson 126
Feb 15 '18 at 18:59
1
$begingroup$
...and you'll be able, for example, to conclude that tr.$,A=0;$ ...
$endgroup$
– DonAntonio
Feb 15 '18 at 19:04
3
$begingroup$
If you are asking what the equation «$[A,B]=A$» means, then it is simply the same as «$AB-BA=A$». If you are asking something else, you should explain...
$endgroup$
– Mariano Suárez-Álvarez
Feb 15 '18 at 19:06
2
$begingroup$
Perhaps you meant $[A,B]=iA$ when you say that the operators are hermitian (as the commutator is antihermitian)
$endgroup$
– user8268
Feb 15 '18 at 19:55