Mixing
Eigenvalues of a sum of operators
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If we have a quantum operator that is composed of the sum of other operators, let's say, $A=B+C$, and we want to find it's eigenvalues, is it the same as finding the eigenvalues of each operator and then just adding them?
In other words, for orthonormal basis $(|psi_nrangle)_nin N$,
$A|psirangle=a|psirangle$
is the same as
$Apsirangle=(B+C)|psirangle=(b+c)|psirangle$
and thus $a=b+c$.
From what I've read in Sakurai, this seems to be the case for a finite basis, but wonder if we can extend this to an infinite basis.
quantum-mechanics homework-and-exercises operators
asked Aug 18 at 21:29
Charlie
1248
add a comment |Â
up vote
4
down vote
favorite
If we have a quantum operator that is composed of the sum of other operators, let's say, $A=B+C$, and we want to find it's eigenvalues, is it the same as finding the eigenvalues of each operator and then just adding them?
In other words, for orthonormal basis $(|psi_nrangle)_nin N$,
$A|psirangle=a|psirangle$
is the same as
$Apsirangle=(B+C)|psirangle=(b+c)|psirangle$
and thus $a=b+c$.
From what I've read in Sakurai, this seems to be the case for a finite basis, but wonder if we can extend this to an infinite basis.
quantum-mechanics homework-and-exercises operators
asked Aug 18 at 21:29
Charlie
1248
3
I think the operators need to have a common eigenbasis for this to be true.
– Aaron Stevens
Aug 18 at 21:34
add a comment |Â
up vote
4
down vote
favorite
up vote
4
down vote
favorite
If we have a quantum operator that is composed of the sum of other operators, let's say, $A=B+C$, and we want to find it's eigenvalues, is it the same as finding the eigenvalues of each operator and then just adding them?
In other words, for orthonormal basis $(|psi_nrangle)_nin N$,
$A|psirangle=a|psirangle$
is the same as
$Apsirangle=(B+C)|psirangle=(b+c)|psirangle$
and thus $a=b+c$.
From what I've read in Sakurai, this seems to be the case for a finite basis, but wonder if we can extend this to an infinite basis.
quantum-mechanics homework-and-exercises operators
asked Aug 18 at 21:29
Charlie
1248
If we have a quantum operator that is composed of the sum of other operators, let's say, $A=B+C$, and we want to find it's eigenvalues, is it the same as finding the eigenvalues of each operator and then just adding them?
In other words, for orthonormal basis $(|psi_nrangle)_nin N$,
$A|psirangle=a|psirangle$
is the same as
$Apsirangle=(B+C)|psirangle=(b+c)|psirangle$
and thus $a=b+c$.
From what I've read in Sakurai, this seems to be the case for a finite basis, but wonder if we can extend this to an infinite basis.
quantum-mechanics homework-and-exercises operators
asked Aug 18 at 21:29
Charlie
1248
asked Aug 18 at 21:29
Charlie
1248
asked Aug 18 at 21:29
Charlie
1248
asked Aug 18 at 21:29
Charlie
1248
1248
3
I think the operators need to have a common eigenbasis for this to be true.
– Aaron Stevens
Aug 18 at 21:34
add a comment |Â
3
I think the operators need to have a common eigenbasis for this to be true.
– Aaron Stevens
Aug 18 at 21:34
3
3
I think the operators need to have a common eigenbasis for this to be true.
– Aaron Stevens
Aug 18 at 21:34
I think the operators need to have a common eigenbasis for this to be true.
– Aaron Stevens
Aug 18 at 21:34
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
8
down vote
accepted
No.
In general, the eigenvalues and eigenvectors of $A$ and $B$ have extremely tenuous connections to the eigenvalues and eigenvectors of $A+B$. This relationship is so weak that you can phrase a huge fraction of the hard, Very Hard, and Completely Unsolved problems in quantum mechanics as some form of
given $A$ and $B$ with known eigenvalues and eigenfunctions, find the eigenvalues of $A+B$.
(As one such example, try $A=frac12mp^2$ and $B=V(x)$, or their generalizations to multiple electrons.)
IF $A$ and $B$ commute, then they share a common eigenbasis and on that eigenbasis the sum-of-eigenvalues property does hold. For cases where they don't commute, it doesn't.
If you want an explicit example, try, say,
$$
A = beginpmatrix 1 & 0 \ 0 & -1 endpmatrix, quad
B = beginpmatrix 0 & 1 \ 1 & 0 endpmatrix, quad textand quad
C = beginpmatrix 1 & 1 \ 1 & -1 endpmatrix,
$$
where $B$ has eigenvectors $(1,1)$ and $(1,-1)$ with eigenvalues $1$ and $-1$, respectively, and $C$ has eigenvectors $(1+sqrt 2,1)$ and $(1-sqrt 2,1)$ with eigenvalues $-sqrt2$ and $sqrt2$ resp.
answered Aug 18 at 21:34
Emilio Pisanty
74.9k18180371
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
8
down vote
accepted
No.
In general, the eigenvalues and eigenvectors of $A$ and $B$ have extremely tenuous connections to the eigenvalues and eigenvectors of $A+B$. This relationship is so weak that you can phrase a huge fraction of the hard, Very Hard, and Completely Unsolved problems in quantum mechanics as some form of
given $A$ and $B$ with known eigenvalues and eigenfunctions, find the eigenvalues of $A+B$.
(As one such example, try $A=frac12mp^2$ and $B=V(x)$, or their generalizations to multiple electrons.)
IF $A$ and $B$ commute, then they share a common eigenbasis and on that eigenbasis the sum-of-eigenvalues property does hold. For cases where they don't commute, it doesn't.
If you want an explicit example, try, say,
$$
A = beginpmatrix 1 & 0 \ 0 & -1 endpmatrix, quad
B = beginpmatrix 0 & 1 \ 1 & 0 endpmatrix, quad textand quad
C = beginpmatrix 1 & 1 \ 1 & -1 endpmatrix,
$$
where $B$ has eigenvectors $(1,1)$ and $(1,-1)$ with eigenvalues $1$ and $-1$, respectively, and $C$ has eigenvectors $(1+sqrt 2,1)$ and $(1-sqrt 2,1)$ with eigenvalues $-sqrt2$ and $sqrt2$ resp.
answered Aug 18 at 21:34
Emilio Pisanty
74.9k18180371
add a comment |Â
up vote
8
down vote
accepted
No.
In general, the eigenvalues and eigenvectors of $A$ and $B$ have extremely tenuous connections to the eigenvalues and eigenvectors of $A+B$. This relationship is so weak that you can phrase a huge fraction of the hard, Very Hard, and Completely Unsolved problems in quantum mechanics as some form of
given $A$ and $B$ with known eigenvalues and eigenfunctions, find the eigenvalues of $A+B$.
(As one such example, try $A=frac12mp^2$ and $B=V(x)$, or their generalizations to multiple electrons.)
IF $A$ and $B$ commute, then they share a common eigenbasis and on that eigenbasis the sum-of-eigenvalues property does hold. For cases where they don't commute, it doesn't.
If you want an explicit example, try, say,
$$
A = beginpmatrix 1 & 0 \ 0 & -1 endpmatrix, quad
B = beginpmatrix 0 & 1 \ 1 & 0 endpmatrix, quad textand quad
C = beginpmatrix 1 & 1 \ 1 & -1 endpmatrix,
$$
where $B$ has eigenvectors $(1,1)$ and $(1,-1)$ with eigenvalues $1$ and $-1$, respectively, and $C$ has eigenvectors $(1+sqrt 2,1)$ and $(1-sqrt 2,1)$ with eigenvalues $-sqrt2$ and $sqrt2$ resp.
answered Aug 18 at 21:34
Emilio Pisanty
74.9k18180371
add a comment |Â
up vote
8
down vote
accepted
up vote
8
down vote
accepted
No.
In general, the eigenvalues and eigenvectors of $A$ and $B$ have extremely tenuous connections to the eigenvalues and eigenvectors of $A+B$. This relationship is so weak that you can phrase a huge fraction of the hard, Very Hard, and Completely Unsolved problems in quantum mechanics as some form of
given $A$ and $B$ with known eigenvalues and eigenfunctions, find the eigenvalues of $A+B$.
(As one such example, try $A=frac12mp^2$ and $B=V(x)$, or their generalizations to multiple electrons.)
IF $A$ and $B$ commute, then they share a common eigenbasis and on that eigenbasis the sum-of-eigenvalues property does hold. For cases where they don't commute, it doesn't.
If you want an explicit example, try, say,
$$
A = beginpmatrix 1 & 0 \ 0 & -1 endpmatrix, quad
B = beginpmatrix 0 & 1 \ 1 & 0 endpmatrix, quad textand quad
C = beginpmatrix 1 & 1 \ 1 & -1 endpmatrix,
$$
where $B$ has eigenvectors $(1,1)$ and $(1,-1)$ with eigenvalues $1$ and $-1$, respectively, and $C$ has eigenvectors $(1+sqrt 2,1)$ and $(1-sqrt 2,1)$ with eigenvalues $-sqrt2$ and $sqrt2$ resp.
answered Aug 18 at 21:34
Emilio Pisanty
74.9k18180371
No.
In general, the eigenvalues and eigenvectors of $A$ and $B$ have extremely tenuous connections to the eigenvalues and eigenvectors of $A+B$. This relationship is so weak that you can phrase a huge fraction of the hard, Very Hard, and Completely Unsolved problems in quantum mechanics as some form of
given $A$ and $B$ with known eigenvalues and eigenfunctions, find the eigenvalues of $A+B$.
(As one such example, try $A=frac12mp^2$ and $B=V(x)$, or their generalizations to multiple electrons.)
IF $A$ and $B$ commute, then they share a common eigenbasis and on that eigenbasis the sum-of-eigenvalues property does hold. For cases where they don't commute, it doesn't.
If you want an explicit example, try, say,
$$
A = beginpmatrix 1 & 0 \ 0 & -1 endpmatrix, quad
B = beginpmatrix 0 & 1 \ 1 & 0 endpmatrix, quad textand quad
C = beginpmatrix 1 & 1 \ 1 & -1 endpmatrix,
$$
where $B$ has eigenvectors $(1,1)$ and $(1,-1)$ with eigenvalues $1$ and $-1$, respectively, and $C$ has eigenvectors $(1+sqrt 2,1)$ and $(1-sqrt 2,1)$ with eigenvalues $-sqrt2$ and $sqrt2$ resp.
answered Aug 18 at 21:34
Emilio Pisanty
74.9k18180371
edited Aug 18 at 21:44
answered Aug 18 at 21:34
Emilio Pisanty
74.9k18180371
answered Aug 18 at 21:34
Emilio Pisanty
74.9k18180371
answered Aug 18 at 21:34
Emilio Pisanty
74.9k18180371
74.9k18180371
add a comment |Â
add a comment |Â
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3
I think the operators need to have a common eigenbasis for this to be true.
– Aaron Stevens
Aug 18 at 21:34