From the condition $[A,B]=A$, what can I say about $B$?
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 at 18:50
Gbp
585
|
show 6 more comments
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 at 18:50
Gbp
585
1
I think that's for commutator i.e. $[A,B]=AB-BA$.
– user296113
Feb 15 at 18:53
1
Then I suppose you can assume that $AB - BA = A$.
– Joe Johnson 126
Feb 15 at 18:59
1
...and you'll be able, for example, to conclude that tr.$,A=0;$ ...
– DonAntonio
Feb 15 at 19:04
3
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...
– Mariano Suárez-Álvarez
Feb 15 at 19:06
2
Perhaps you meant $[A,B]=iA$ when you say that the operators are hermitian (as the commutator is antihermitian)
– user8268
Feb 15 at 19:55
|
show 6 more comments
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 at 18:50
Gbp
585
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 at 18:50
Gbp
585
asked Feb 15 at 18:50
Gbp
585
edited Feb 15 at 19:41
asked Feb 15 at 18:50
Gbp
585
asked Feb 15 at 18:50
Gbp
585
asked Feb 15 at 18:50
Gbp
585
585
1
I think that's for commutator i.e. $[A,B]=AB-BA$.
– user296113
Feb 15 at 18:53
1
Then I suppose you can assume that $AB - BA = A$.
– Joe Johnson 126
Feb 15 at 18:59
1
...and you'll be able, for example, to conclude that tr.$,A=0;$ ...
– DonAntonio
Feb 15 at 19:04
3
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...
– Mariano Suárez-Álvarez
Feb 15 at 19:06
2
Perhaps you meant $[A,B]=iA$ when you say that the operators are hermitian (as the commutator is antihermitian)
– user8268
Feb 15 at 19:55
|
show 6 more comments
1
I think that's for commutator i.e. $[A,B]=AB-BA$.
– user296113
Feb 15 at 18:53
1
Then I suppose you can assume that $AB - BA = A$.
– Joe Johnson 126
Feb 15 at 18:59
1
...and you'll be able, for example, to conclude that tr.$,A=0;$ ...
– DonAntonio
Feb 15 at 19:04
3
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...
– Mariano Suárez-Álvarez
Feb 15 at 19:06
2
Perhaps you meant $[A,B]=iA$ when you say that the operators are hermitian (as the commutator is antihermitian)
– user8268
Feb 15 at 19:55
1
1
I think that's for commutator i.e. $[A,B]=AB-BA$.
– user296113
Feb 15 at 18:53
I think that's for commutator i.e. $[A,B]=AB-BA$.
– user296113
Feb 15 at 18:53
1
1
Then I suppose you can assume that $AB - BA = A$.
– Joe Johnson 126
Feb 15 at 18:59
Then I suppose you can assume that $AB - BA = A$.
– Joe Johnson 126
Feb 15 at 18:59
1
1
...and you'll be able, for example, to conclude that tr.$,A=0;$ ...
– DonAntonio
Feb 15 at 19:04
...and you'll be able, for example, to conclude that tr.$,A=0;$ ...
– DonAntonio
Feb 15 at 19:04
3
3
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...
– Mariano Suárez-Álvarez
Feb 15 at 19:06
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...
– Mariano Suárez-Álvarez
Feb 15 at 19:06
2
2
Perhaps you meant $[A,B]=iA$ when you say that the operators are hermitian (as the commutator is antihermitian)
– user8268
Feb 15 at 19:55
Perhaps you meant $[A,B]=iA$ when you say that the operators are hermitian (as the commutator is antihermitian)
– user8268
Feb 15 at 19:55
|
show 6 more comments
1 Answer
1
active
oldest
votes
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 2 days ago
Cosmas Zachos
1,498520
add a comment |
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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 2 days ago
Cosmas Zachos
1,498520
add a comment |
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 2 days ago
Cosmas Zachos
1,498520
add a comment |
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 2 days ago
Cosmas Zachos
1,498520
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 2 days ago
Cosmas Zachos
1,498520
edited yesterday
answered 2 days ago
Cosmas Zachos
1,498520
answered 2 days ago
Cosmas Zachos
1,498520
answered 2 days ago
Cosmas Zachos
1,498520
1,498520
add a comment |
add a comment |
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1
I think that's for commutator i.e. $[A,B]=AB-BA$.
– user296113
Feb 15 at 18:53
1
Then I suppose you can assume that $AB - BA = A$.
– Joe Johnson 126
Feb 15 at 18:59
1
...and you'll be able, for example, to conclude that tr.$,A=0;$ ...
– DonAntonio
Feb 15 at 19:04
3
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...
– Mariano Suárez-Álvarez
Feb 15 at 19:06
2
Perhaps you meant $[A,B]=iA$ when you say that the operators are hermitian (as the commutator is antihermitian)
– user8268
Feb 15 at 19:55