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Can I use any basis for the spectral theorem or does it have to be the eigenbasis?
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For a Hermitian operator $A$ on a Hilbert space $H$ with eigenvalues $a_i$ and eigenvectors $v_i$ forming a complete basis, one can write $A$ as:
$$ A = sum_i a_i v_i v_i^dagger, $$
where $v^dagger$ is the Hermitian conjugate of $v$.
Suppose there is another complete basis $w_i$. Can I also use the spectral decomposition with this $w$-basis, instead of the eigenbasis $v_i$?
$$ A = sum_i a_i w_i w_i^dagger $$
linear-algebra eigenvalues-eigenvectors
asked Aug 2 at 7:20
Stephan
735
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For a Hermitian operator $A$ on a Hilbert space $H$ with eigenvalues $a_i$ and eigenvectors $v_i$ forming a complete basis, one can write $A$ as:
$$ A = sum_i a_i v_i v_i^dagger, $$
where $v^dagger$ is the Hermitian conjugate of $v$.
Suppose there is another complete basis $w_i$. Can I also use the spectral decomposition with this $w$-basis, instead of the eigenbasis $v_i$?
$$ A = sum_i a_i w_i w_i^dagger $$
linear-algebra eigenvalues-eigenvectors
asked Aug 2 at 7:20
Stephan
735
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
For a Hermitian operator $A$ on a Hilbert space $H$ with eigenvalues $a_i$ and eigenvectors $v_i$ forming a complete basis, one can write $A$ as:
$$ A = sum_i a_i v_i v_i^dagger, $$
where $v^dagger$ is the Hermitian conjugate of $v$.
Suppose there is another complete basis $w_i$. Can I also use the spectral decomposition with this $w$-basis, instead of the eigenbasis $v_i$?
$$ A = sum_i a_i w_i w_i^dagger $$
linear-algebra eigenvalues-eigenvectors
asked Aug 2 at 7:20
Stephan
735
For a Hermitian operator $A$ on a Hilbert space $H$ with eigenvalues $a_i$ and eigenvectors $v_i$ forming a complete basis, one can write $A$ as:
$$ A = sum_i a_i v_i v_i^dagger, $$
where $v^dagger$ is the Hermitian conjugate of $v$.
Suppose there is another complete basis $w_i$. Can I also use the spectral decomposition with this $w$-basis, instead of the eigenbasis $v_i$?
$$ A = sum_i a_i w_i w_i^dagger $$
linear-algebra eigenvalues-eigenvectors
asked Aug 2 at 7:20
Stephan
735
asked Aug 2 at 7:20
Stephan
735
asked Aug 2 at 7:20
Stephan
735
asked Aug 2 at 7:20
Stephan
735
735
add a comment |Â
add a comment |Â
1 Answer
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No, you cannot. The point of an eigenbasis is that it diagonalises the operator; it works on each basis vector separately and independently of the others. That diagonalisation is exactly what you see in the sum.
Take, for instance, the two-dimensional case with $A$ represented by $left[beginsmallmatrix1&0\0&2endsmallmatrixright]$ in some basis $u_i$. Then that basis is an eigenbais, and you get
$$
A = 1cdot u_1u_1^dagger + 2cdot u_2u_2^dagger
$$
If you take a new basis, say $v_1 = u_1+u_2$ and $v_2 = u_1-u_2$, then in that basis we have that $A$ is represented by $left[beginsmallmatrix1.5&-0.5\-0.5&1.5endsmallmatrixright]$. Of course, the matrix is still symmetric / Hermitian, because change of basis cannot change that fact. But if you want to decompose it in something like the above sum it's not enough with $v_1v_1^dagger$ and $v_2v_2^dagger$. You need $v_1v_2^dagger$ and $v_2v_1^dagger$ terms as well, as there are now non-zero off-diagonal elements:
$$
A = 1.5cdot v_1v_1^dagger + 1.5cdot v_2v_2^dagger - 0.5cdot v_1v_2^dagger - 0.5cdot v_2v_1^dagger
$$
answered Aug 2 at 7:30
Arthur
98.2k793174
Makes sense, thank you for answering. My question arises from the solutions of an exam, where the author apparently used the spectral theorem (exam: phys.uic.edu/docs/default-source/…, solutions: phys.uic.edu/docs/default-source/…). In question 1a), the spectral theorem with a non-eigenbasis is used to show that the operator B has only positive eigenvalues. If I follow your reasoning, this solution does not make sense, does it?
– Stephan
Aug 2 at 7:43
@Stephan No, that's something else. Here it's not the case that they have a $B$ and they decompose it using the eigenbasis of $A$. Rather it's that they use the eigenbasis and spectral decomposition of $A$ to construct $B$ in the first place.
– Arthur
Aug 2 at 7:53
I see, but how can they argue that $|lambda_i|$ are the eigenvalues of $B$ if this is not a spectral decomposition of $B$?
– Stephan
Aug 2 at 8:03
1
@Stephan It is a spectral decomposition of $B$. That's exactly what $B$ is. It's defined as the matrix with that exact spectral decomposition.
– Arthur
Aug 2 at 8:04
1
Ah, I think now I get it: we define $B$ in such a way that it shares its eigenbasis with $A$. So as a corollary, $[A,B]=0$, right?
– Stephan
Aug 2 at 8:06
 |Â
show 1 more comment
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
No, you cannot. The point of an eigenbasis is that it diagonalises the operator; it works on each basis vector separately and independently of the others. That diagonalisation is exactly what you see in the sum.
Take, for instance, the two-dimensional case with $A$ represented by $left[beginsmallmatrix1&0\0&2endsmallmatrixright]$ in some basis $u_i$. Then that basis is an eigenbais, and you get
$$
A = 1cdot u_1u_1^dagger + 2cdot u_2u_2^dagger
$$
If you take a new basis, say $v_1 = u_1+u_2$ and $v_2 = u_1-u_2$, then in that basis we have that $A$ is represented by $left[beginsmallmatrix1.5&-0.5\-0.5&1.5endsmallmatrixright]$. Of course, the matrix is still symmetric / Hermitian, because change of basis cannot change that fact. But if you want to decompose it in something like the above sum it's not enough with $v_1v_1^dagger$ and $v_2v_2^dagger$. You need $v_1v_2^dagger$ and $v_2v_1^dagger$ terms as well, as there are now non-zero off-diagonal elements:
$$
A = 1.5cdot v_1v_1^dagger + 1.5cdot v_2v_2^dagger - 0.5cdot v_1v_2^dagger - 0.5cdot v_2v_1^dagger
$$
answered Aug 2 at 7:30
Arthur
98.2k793174
Makes sense, thank you for answering. My question arises from the solutions of an exam, where the author apparently used the spectral theorem (exam: phys.uic.edu/docs/default-source/…, solutions: phys.uic.edu/docs/default-source/…). In question 1a), the spectral theorem with a non-eigenbasis is used to show that the operator B has only positive eigenvalues. If I follow your reasoning, this solution does not make sense, does it?
– Stephan
Aug 2 at 7:43
@Stephan No, that's something else. Here it's not the case that they have a $B$ and they decompose it using the eigenbasis of $A$. Rather it's that they use the eigenbasis and spectral decomposition of $A$ to construct $B$ in the first place.
– Arthur
Aug 2 at 7:53
I see, but how can they argue that $|lambda_i|$ are the eigenvalues of $B$ if this is not a spectral decomposition of $B$?
– Stephan
Aug 2 at 8:03
1
@Stephan It is a spectral decomposition of $B$. That's exactly what $B$ is. It's defined as the matrix with that exact spectral decomposition.
– Arthur
Aug 2 at 8:04
1
Ah, I think now I get it: we define $B$ in such a way that it shares its eigenbasis with $A$. So as a corollary, $[A,B]=0$, right?
– Stephan
Aug 2 at 8:06
 |Â
show 1 more comment
up vote
2
down vote
accepted
No, you cannot. The point of an eigenbasis is that it diagonalises the operator; it works on each basis vector separately and independently of the others. That diagonalisation is exactly what you see in the sum.
Take, for instance, the two-dimensional case with $A$ represented by $left[beginsmallmatrix1&0\0&2endsmallmatrixright]$ in some basis $u_i$. Then that basis is an eigenbais, and you get
$$
A = 1cdot u_1u_1^dagger + 2cdot u_2u_2^dagger
$$
If you take a new basis, say $v_1 = u_1+u_2$ and $v_2 = u_1-u_2$, then in that basis we have that $A$ is represented by $left[beginsmallmatrix1.5&-0.5\-0.5&1.5endsmallmatrixright]$. Of course, the matrix is still symmetric / Hermitian, because change of basis cannot change that fact. But if you want to decompose it in something like the above sum it's not enough with $v_1v_1^dagger$ and $v_2v_2^dagger$. You need $v_1v_2^dagger$ and $v_2v_1^dagger$ terms as well, as there are now non-zero off-diagonal elements:
$$
A = 1.5cdot v_1v_1^dagger + 1.5cdot v_2v_2^dagger - 0.5cdot v_1v_2^dagger - 0.5cdot v_2v_1^dagger
$$
answered Aug 2 at 7:30
Arthur
98.2k793174
Makes sense, thank you for answering. My question arises from the solutions of an exam, where the author apparently used the spectral theorem (exam: phys.uic.edu/docs/default-source/…, solutions: phys.uic.edu/docs/default-source/…). In question 1a), the spectral theorem with a non-eigenbasis is used to show that the operator B has only positive eigenvalues. If I follow your reasoning, this solution does not make sense, does it?
– Stephan
Aug 2 at 7:43
@Stephan No, that's something else. Here it's not the case that they have a $B$ and they decompose it using the eigenbasis of $A$. Rather it's that they use the eigenbasis and spectral decomposition of $A$ to construct $B$ in the first place.
– Arthur
Aug 2 at 7:53
I see, but how can they argue that $|lambda_i|$ are the eigenvalues of $B$ if this is not a spectral decomposition of $B$?
– Stephan
Aug 2 at 8:03
1
@Stephan It is a spectral decomposition of $B$. That's exactly what $B$ is. It's defined as the matrix with that exact spectral decomposition.
– Arthur
Aug 2 at 8:04
1
Ah, I think now I get it: we define $B$ in such a way that it shares its eigenbasis with $A$. So as a corollary, $[A,B]=0$, right?
– Stephan
Aug 2 at 8:06
 |Â
show 1 more comment
up vote
2
down vote
accepted
up vote
2
down vote
accepted
No, you cannot. The point of an eigenbasis is that it diagonalises the operator; it works on each basis vector separately and independently of the others. That diagonalisation is exactly what you see in the sum.
Take, for instance, the two-dimensional case with $A$ represented by $left[beginsmallmatrix1&0\0&2endsmallmatrixright]$ in some basis $u_i$. Then that basis is an eigenbais, and you get
$$
A = 1cdot u_1u_1^dagger + 2cdot u_2u_2^dagger
$$
If you take a new basis, say $v_1 = u_1+u_2$ and $v_2 = u_1-u_2$, then in that basis we have that $A$ is represented by $left[beginsmallmatrix1.5&-0.5\-0.5&1.5endsmallmatrixright]$. Of course, the matrix is still symmetric / Hermitian, because change of basis cannot change that fact. But if you want to decompose it in something like the above sum it's not enough with $v_1v_1^dagger$ and $v_2v_2^dagger$. You need $v_1v_2^dagger$ and $v_2v_1^dagger$ terms as well, as there are now non-zero off-diagonal elements:
$$
A = 1.5cdot v_1v_1^dagger + 1.5cdot v_2v_2^dagger - 0.5cdot v_1v_2^dagger - 0.5cdot v_2v_1^dagger
$$
answered Aug 2 at 7:30
Arthur
98.2k793174
No, you cannot. The point of an eigenbasis is that it diagonalises the operator; it works on each basis vector separately and independently of the others. That diagonalisation is exactly what you see in the sum.
Take, for instance, the two-dimensional case with $A$ represented by $left[beginsmallmatrix1&0\0&2endsmallmatrixright]$ in some basis $u_i$. Then that basis is an eigenbais, and you get
$$
A = 1cdot u_1u_1^dagger + 2cdot u_2u_2^dagger
$$
If you take a new basis, say $v_1 = u_1+u_2$ and $v_2 = u_1-u_2$, then in that basis we have that $A$ is represented by $left[beginsmallmatrix1.5&-0.5\-0.5&1.5endsmallmatrixright]$. Of course, the matrix is still symmetric / Hermitian, because change of basis cannot change that fact. But if you want to decompose it in something like the above sum it's not enough with $v_1v_1^dagger$ and $v_2v_2^dagger$. You need $v_1v_2^dagger$ and $v_2v_1^dagger$ terms as well, as there are now non-zero off-diagonal elements:
$$
A = 1.5cdot v_1v_1^dagger + 1.5cdot v_2v_2^dagger - 0.5cdot v_1v_2^dagger - 0.5cdot v_2v_1^dagger
$$
answered Aug 2 at 7:30
Arthur
98.2k793174
edited Aug 2 at 7:36
answered Aug 2 at 7:30
Arthur
98.2k793174
answered Aug 2 at 7:30
Arthur
98.2k793174
answered Aug 2 at 7:30
Arthur
98.2k793174
98.2k793174
Makes sense, thank you for answering. My question arises from the solutions of an exam, where the author apparently used the spectral theorem (exam: phys.uic.edu/docs/default-source/…, solutions: phys.uic.edu/docs/default-source/…). In question 1a), the spectral theorem with a non-eigenbasis is used to show that the operator B has only positive eigenvalues. If I follow your reasoning, this solution does not make sense, does it?
– Stephan
Aug 2 at 7:43
@Stephan No, that's something else. Here it's not the case that they have a $B$ and they decompose it using the eigenbasis of $A$. Rather it's that they use the eigenbasis and spectral decomposition of $A$ to construct $B$ in the first place.
– Arthur
Aug 2 at 7:53
I see, but how can they argue that $|lambda_i|$ are the eigenvalues of $B$ if this is not a spectral decomposition of $B$?
– Stephan
Aug 2 at 8:03
1
@Stephan It is a spectral decomposition of $B$. That's exactly what $B$ is. It's defined as the matrix with that exact spectral decomposition.
– Arthur
Aug 2 at 8:04
1
Ah, I think now I get it: we define $B$ in such a way that it shares its eigenbasis with $A$. So as a corollary, $[A,B]=0$, right?
– Stephan
Aug 2 at 8:06
 |Â
show 1 more comment
Makes sense, thank you for answering. My question arises from the solutions of an exam, where the author apparently used the spectral theorem (exam: phys.uic.edu/docs/default-source/…, solutions: phys.uic.edu/docs/default-source/…). In question 1a), the spectral theorem with a non-eigenbasis is used to show that the operator B has only positive eigenvalues. If I follow your reasoning, this solution does not make sense, does it?
– Stephan
Aug 2 at 7:43
@Stephan No, that's something else. Here it's not the case that they have a $B$ and they decompose it using the eigenbasis of $A$. Rather it's that they use the eigenbasis and spectral decomposition of $A$ to construct $B$ in the first place.
– Arthur
Aug 2 at 7:53
I see, but how can they argue that $|lambda_i|$ are the eigenvalues of $B$ if this is not a spectral decomposition of $B$?
– Stephan
Aug 2 at 8:03
1
@Stephan It is a spectral decomposition of $B$. That's exactly what $B$ is. It's defined as the matrix with that exact spectral decomposition.
– Arthur
Aug 2 at 8:04
1
Ah, I think now I get it: we define $B$ in such a way that it shares its eigenbasis with $A$. So as a corollary, $[A,B]=0$, right?
– Stephan
Aug 2 at 8:06
Makes sense, thank you for answering. My question arises from the solutions of an exam, where the author apparently used the spectral theorem (exam: phys.uic.edu/docs/default-source/…, solutions: phys.uic.edu/docs/default-source/…). In question 1a), the spectral theorem with a non-eigenbasis is used to show that the operator B has only positive eigenvalues. If I follow your reasoning, this solution does not make sense, does it?
– Stephan
Aug 2 at 7:43
Makes sense, thank you for answering. My question arises from the solutions of an exam, where the author apparently used the spectral theorem (exam: phys.uic.edu/docs/default-source/…, solutions: phys.uic.edu/docs/default-source/…). In question 1a), the spectral theorem with a non-eigenbasis is used to show that the operator B has only positive eigenvalues. If I follow your reasoning, this solution does not make sense, does it?
– Stephan
Aug 2 at 7:43
@Stephan No, that's something else. Here it's not the case that they have a $B$ and they decompose it using the eigenbasis of $A$. Rather it's that they use the eigenbasis and spectral decomposition of $A$ to construct $B$ in the first place.
– Arthur
Aug 2 at 7:53
@Stephan No, that's something else. Here it's not the case that they have a $B$ and they decompose it using the eigenbasis of $A$. Rather it's that they use the eigenbasis and spectral decomposition of $A$ to construct $B$ in the first place.
– Arthur
Aug 2 at 7:53
I see, but how can they argue that $|lambda_i|$ are the eigenvalues of $B$ if this is not a spectral decomposition of $B$?
– Stephan
Aug 2 at 8:03
I see, but how can they argue that $|lambda_i|$ are the eigenvalues of $B$ if this is not a spectral decomposition of $B$?
– Stephan
Aug 2 at 8:03
1
1
@Stephan It is a spectral decomposition of $B$. That's exactly what $B$ is. It's defined as the matrix with that exact spectral decomposition.
– Arthur
Aug 2 at 8:04
@Stephan It is a spectral decomposition of $B$. That's exactly what $B$ is. It's defined as the matrix with that exact spectral decomposition.
– Arthur
Aug 2 at 8:04
1
1
Ah, I think now I get it: we define $B$ in such a way that it shares its eigenbasis with $A$. So as a corollary, $[A,B]=0$, right?
– Stephan
Aug 2 at 8:06
Ah, I think now I get it: we define $B$ in such a way that it shares its eigenbasis with $A$. So as a corollary, $[A,B]=0$, right?
– Stephan
Aug 2 at 8:06
 |Â
show 1 more comment
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