How to prove that the query oracle is unitary?Why are oracles Hermitian by construction?How is the oracle in Grover's search algorithm implemented?Implementation of the oracle of Grover's algorithm on IBM Q using three qubitsHow would I implement the quantum oracle in Deutsch's algorithm?How is the Deutsch-Jozsa algorithm faster than classical for practical implementation?Balanced vs unbalanced superposition distinguisherWhy doesn't Deutsch-Jozsa Algorithm show that P ≠ BQP?Is it correct to say that we need controlled gates because unitary matrices are reversible?Equivalent unitary transformationsImplementing an oracleDeutsch–Jozsa algorithm: why is $f$ constant?
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How to prove that the query oracle is unitary?
Why are oracles Hermitian by construction?How is the oracle in Grover's search algorithm implemented?Implementation of the oracle of Grover's algorithm on IBM Q using three qubitsHow would I implement the quantum oracle in Deutsch's algorithm?How is the Deutsch-Jozsa algorithm faster than classical for practical implementation?Balanced vs unbalanced superposition distinguisherWhy doesn't Deutsch-Jozsa Algorithm show that P ≠ BQP?Is it correct to say that we need controlled gates because unitary matrices are reversible?Equivalent unitary transformationsImplementing an oracleDeutsch–Jozsa algorithm: why is $f$ constant?
$begingroup$
The query oracle: $O_x|irangle|brangle = |irangle|b oplus x_irangle$ used in algorithms like Deutsch Jozsa is unitary. How do I prove it is unitary?
algorithm quantum-gate unitarity deutsch-jozsa-algorithm
$endgroup$
add a comment |
$begingroup$
The query oracle: $O_x|irangle|brangle = |irangle|b oplus x_irangle$ used in algorithms like Deutsch Jozsa is unitary. How do I prove it is unitary?
algorithm quantum-gate unitarity deutsch-jozsa-algorithm
$endgroup$
$begingroup$
Related: Why are oracles Hermitian by construction?
$endgroup$
– Blue♦
Mar 26 at 12:25
add a comment |
$begingroup$
The query oracle: $O_x|irangle|brangle = |irangle|b oplus x_irangle$ used in algorithms like Deutsch Jozsa is unitary. How do I prove it is unitary?
algorithm quantum-gate unitarity deutsch-jozsa-algorithm
$endgroup$
The query oracle: $O_x|irangle|brangle = |irangle|b oplus x_irangle$ used in algorithms like Deutsch Jozsa is unitary. How do I prove it is unitary?
algorithm quantum-gate unitarity deutsch-jozsa-algorithm
algorithm quantum-gate unitarity deutsch-jozsa-algorithm
edited Mar 26 at 12:15
Blue♦
6,63141556
6,63141556
asked Mar 26 at 12:11
DivyatDivyat
283
283
$begingroup$
Related: Why are oracles Hermitian by construction?
$endgroup$
– Blue♦
Mar 26 at 12:25
add a comment |
$begingroup$
Related: Why are oracles Hermitian by construction?
$endgroup$
– Blue♦
Mar 26 at 12:25
$begingroup$
Related: Why are oracles Hermitian by construction?
$endgroup$
– Blue♦
Mar 26 at 12:25
$begingroup$
Related: Why are oracles Hermitian by construction?
$endgroup$
– Blue♦
Mar 26 at 12:25
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Notice that $mathcal O_x$ is a permutation matrix.
The matrix elements are
$$langle j, crvertmathcal O_xlvert i,brangle
=delta_ijlangle crvert boplus x_irangle
=delta_ijdelta_c,boplus x_i.$$
In other words, $mathcal O_x$ is diagonal with respect to the first register, and, for each block corresponding to a given $i$, connects all and only the indices $b,c$ such that $boplus c=x_i$ (remember that here $b,c,x_iinmathbb Z_2^otimes n$ are length-$n$ bit strings).
Also, notice that for a given $b$ there is no more than one $c$ such that $boplus c=x_i$ (more precisely, there isn't any such $c$ if $x_i=0$, and there is exactly one if $x_ineq 0$).
It follows that $mathcal O_x$ is a (real) permutation matrix, and such matrices are always unitarily diagonalizable with unit eigenvalues (and therefore unitary).
In this case, we have that the eigenvalues of $mathcal O_x$ are $pm1$, and the eigenvectors are, in the case $x_ineq 0$,
$$lvert irangleotimes(lvert branglepmlvert boplus x_irangle)$$
for all $i$ and $b$. If $x_i=0$ then $mathcal O_x$ is the identity, and therefore its spectrum is trivial$^dagger$.
This shows explicitly that $mathcal O_x$ is unitarily diagonalizable with unit eigenvalues, and therefore is unitary.
$^dagger$ I'm actually being a bit sloppy for the sake of simplicity here. This analysis holds for each different block of $mathcal O_x$ corresponding to a given $i$. More precisely, I should say that $mathcal O_x$ is block-diagonal as it doesn't connect spaces with $ineq j$ on the first register, and each block is either the identity in the subspace in which it acts if $x_i=0$, or a permutation matrix that connects different pairs of basis states if $x_ineq 0$.
$endgroup$
add a comment |
$begingroup$
Apply it twice:
$$
O_xO_x|irangle|brangle=O_x|irangle|boplus x_irangle=|irangle|boplus x_ioplus x_irangle=|irangle|brangle
$$
Hence, $O_x$ is its own inverse, and therefore reversible.
To prove unitarity, it makes more sense to prove that $O_x$ has eigenvectors
$$
|irangle(|0rangle+|1rangle)quadtextandquad|irangle(|0rangle-|1rangle)
$$
for all $i$ with eigenvalues $1$ and $(-1)^x_i$ respectively. These are all orthonormal, and span the full Hilbert space. Consequently, the eigenvalues of $O_x$ are all $pm 1$, and therefore $O_x^star=O_x$ (where $^star$ represents the Hermitian conjugate). Thus,
$$
O_xO_x^star=mathbbI,
$$
as required for a unitary.
$endgroup$
$begingroup$
Thanks, I now understand $O_x$ is its own inverse but how does that make it unitary? For a unitary matrix, one would need to show that its inverse is equal to its conjugate transpose, $MM^*=I$. I think we further need to show $O_x$ is Hermitian, then it will be done.
$endgroup$
– Divyat
Mar 26 at 13:56
$begingroup$
One more doubt, I think the answer still assumes that $O_x$ is normal matrix, as then with real eigenvalues we can claim it to be Hermitian. Please tell whether it is okay to assume oracle as normal or some justifications for it?
$endgroup$
– Divyat
Mar 26 at 15:52
$begingroup$
This explanation is not quite correct, because you assumed that $O_x$ is diagonalizable, but it is not follow from $O_x^2=I$. The thing is $O_x$ maps basis vectors to basis vectors. Since $O_x^2=I$ then $Q_x$ is a permutation of basis vectors.
$endgroup$
– Danylo Y
Mar 26 at 16:17
$begingroup$
I actually asked some time ago on math.SE whether one can deduce that $sqrt A$ is diagonalizable from the fact that $A$ is, see here. As far as I understood the answers, $A^2=I$ is not enough to imply that $A$ is unitarily diagonalizable, a counterexample being $beginpmatrixcostheta & 2sintheta \ sintheta/2 & -costhetaendpmatrix$, which is a square root of the identity, but not (unitarily) diagonalizable, and not unitary, even though its eigenvalues are $pm 1$.
$endgroup$
– glS
Mar 26 at 18:01
1
$begingroup$
Fair enough, I was being a bit glib, particularly initially with regards to the difference between reversible and unitary, because usually its the reversibility aspect people are interested in.
$endgroup$
– DaftWullie
Mar 27 at 8:44
add a comment |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Notice that $mathcal O_x$ is a permutation matrix.
The matrix elements are
$$langle j, crvertmathcal O_xlvert i,brangle
=delta_ijlangle crvert boplus x_irangle
=delta_ijdelta_c,boplus x_i.$$
In other words, $mathcal O_x$ is diagonal with respect to the first register, and, for each block corresponding to a given $i$, connects all and only the indices $b,c$ such that $boplus c=x_i$ (remember that here $b,c,x_iinmathbb Z_2^otimes n$ are length-$n$ bit strings).
Also, notice that for a given $b$ there is no more than one $c$ such that $boplus c=x_i$ (more precisely, there isn't any such $c$ if $x_i=0$, and there is exactly one if $x_ineq 0$).
It follows that $mathcal O_x$ is a (real) permutation matrix, and such matrices are always unitarily diagonalizable with unit eigenvalues (and therefore unitary).
In this case, we have that the eigenvalues of $mathcal O_x$ are $pm1$, and the eigenvectors are, in the case $x_ineq 0$,
$$lvert irangleotimes(lvert branglepmlvert boplus x_irangle)$$
for all $i$ and $b$. If $x_i=0$ then $mathcal O_x$ is the identity, and therefore its spectrum is trivial$^dagger$.
This shows explicitly that $mathcal O_x$ is unitarily diagonalizable with unit eigenvalues, and therefore is unitary.
$^dagger$ I'm actually being a bit sloppy for the sake of simplicity here. This analysis holds for each different block of $mathcal O_x$ corresponding to a given $i$. More precisely, I should say that $mathcal O_x$ is block-diagonal as it doesn't connect spaces with $ineq j$ on the first register, and each block is either the identity in the subspace in which it acts if $x_i=0$, or a permutation matrix that connects different pairs of basis states if $x_ineq 0$.
$endgroup$
add a comment |
$begingroup$
Notice that $mathcal O_x$ is a permutation matrix.
The matrix elements are
$$langle j, crvertmathcal O_xlvert i,brangle
=delta_ijlangle crvert boplus x_irangle
=delta_ijdelta_c,boplus x_i.$$
In other words, $mathcal O_x$ is diagonal with respect to the first register, and, for each block corresponding to a given $i$, connects all and only the indices $b,c$ such that $boplus c=x_i$ (remember that here $b,c,x_iinmathbb Z_2^otimes n$ are length-$n$ bit strings).
Also, notice that for a given $b$ there is no more than one $c$ such that $boplus c=x_i$ (more precisely, there isn't any such $c$ if $x_i=0$, and there is exactly one if $x_ineq 0$).
It follows that $mathcal O_x$ is a (real) permutation matrix, and such matrices are always unitarily diagonalizable with unit eigenvalues (and therefore unitary).
In this case, we have that the eigenvalues of $mathcal O_x$ are $pm1$, and the eigenvectors are, in the case $x_ineq 0$,
$$lvert irangleotimes(lvert branglepmlvert boplus x_irangle)$$
for all $i$ and $b$. If $x_i=0$ then $mathcal O_x$ is the identity, and therefore its spectrum is trivial$^dagger$.
This shows explicitly that $mathcal O_x$ is unitarily diagonalizable with unit eigenvalues, and therefore is unitary.
$^dagger$ I'm actually being a bit sloppy for the sake of simplicity here. This analysis holds for each different block of $mathcal O_x$ corresponding to a given $i$. More precisely, I should say that $mathcal O_x$ is block-diagonal as it doesn't connect spaces with $ineq j$ on the first register, and each block is either the identity in the subspace in which it acts if $x_i=0$, or a permutation matrix that connects different pairs of basis states if $x_ineq 0$.
$endgroup$
add a comment |
$begingroup$
Notice that $mathcal O_x$ is a permutation matrix.
The matrix elements are
$$langle j, crvertmathcal O_xlvert i,brangle
=delta_ijlangle crvert boplus x_irangle
=delta_ijdelta_c,boplus x_i.$$
In other words, $mathcal O_x$ is diagonal with respect to the first register, and, for each block corresponding to a given $i$, connects all and only the indices $b,c$ such that $boplus c=x_i$ (remember that here $b,c,x_iinmathbb Z_2^otimes n$ are length-$n$ bit strings).
Also, notice that for a given $b$ there is no more than one $c$ such that $boplus c=x_i$ (more precisely, there isn't any such $c$ if $x_i=0$, and there is exactly one if $x_ineq 0$).
It follows that $mathcal O_x$ is a (real) permutation matrix, and such matrices are always unitarily diagonalizable with unit eigenvalues (and therefore unitary).
In this case, we have that the eigenvalues of $mathcal O_x$ are $pm1$, and the eigenvectors are, in the case $x_ineq 0$,
$$lvert irangleotimes(lvert branglepmlvert boplus x_irangle)$$
for all $i$ and $b$. If $x_i=0$ then $mathcal O_x$ is the identity, and therefore its spectrum is trivial$^dagger$.
This shows explicitly that $mathcal O_x$ is unitarily diagonalizable with unit eigenvalues, and therefore is unitary.
$^dagger$ I'm actually being a bit sloppy for the sake of simplicity here. This analysis holds for each different block of $mathcal O_x$ corresponding to a given $i$. More precisely, I should say that $mathcal O_x$ is block-diagonal as it doesn't connect spaces with $ineq j$ on the first register, and each block is either the identity in the subspace in which it acts if $x_i=0$, or a permutation matrix that connects different pairs of basis states if $x_ineq 0$.
$endgroup$
Notice that $mathcal O_x$ is a permutation matrix.
The matrix elements are
$$langle j, crvertmathcal O_xlvert i,brangle
=delta_ijlangle crvert boplus x_irangle
=delta_ijdelta_c,boplus x_i.$$
In other words, $mathcal O_x$ is diagonal with respect to the first register, and, for each block corresponding to a given $i$, connects all and only the indices $b,c$ such that $boplus c=x_i$ (remember that here $b,c,x_iinmathbb Z_2^otimes n$ are length-$n$ bit strings).
Also, notice that for a given $b$ there is no more than one $c$ such that $boplus c=x_i$ (more precisely, there isn't any such $c$ if $x_i=0$, and there is exactly one if $x_ineq 0$).
It follows that $mathcal O_x$ is a (real) permutation matrix, and such matrices are always unitarily diagonalizable with unit eigenvalues (and therefore unitary).
In this case, we have that the eigenvalues of $mathcal O_x$ are $pm1$, and the eigenvectors are, in the case $x_ineq 0$,
$$lvert irangleotimes(lvert branglepmlvert boplus x_irangle)$$
for all $i$ and $b$. If $x_i=0$ then $mathcal O_x$ is the identity, and therefore its spectrum is trivial$^dagger$.
This shows explicitly that $mathcal O_x$ is unitarily diagonalizable with unit eigenvalues, and therefore is unitary.
$^dagger$ I'm actually being a bit sloppy for the sake of simplicity here. This analysis holds for each different block of $mathcal O_x$ corresponding to a given $i$. More precisely, I should say that $mathcal O_x$ is block-diagonal as it doesn't connect spaces with $ineq j$ on the first register, and each block is either the identity in the subspace in which it acts if $x_i=0$, or a permutation matrix that connects different pairs of basis states if $x_ineq 0$.
edited Mar 26 at 18:05
answered Mar 26 at 17:49
glSglS
4,333740
4,333740
add a comment |
add a comment |
$begingroup$
Apply it twice:
$$
O_xO_x|irangle|brangle=O_x|irangle|boplus x_irangle=|irangle|boplus x_ioplus x_irangle=|irangle|brangle
$$
Hence, $O_x$ is its own inverse, and therefore reversible.
To prove unitarity, it makes more sense to prove that $O_x$ has eigenvectors
$$
|irangle(|0rangle+|1rangle)quadtextandquad|irangle(|0rangle-|1rangle)
$$
for all $i$ with eigenvalues $1$ and $(-1)^x_i$ respectively. These are all orthonormal, and span the full Hilbert space. Consequently, the eigenvalues of $O_x$ are all $pm 1$, and therefore $O_x^star=O_x$ (where $^star$ represents the Hermitian conjugate). Thus,
$$
O_xO_x^star=mathbbI,
$$
as required for a unitary.
$endgroup$
$begingroup$
Thanks, I now understand $O_x$ is its own inverse but how does that make it unitary? For a unitary matrix, one would need to show that its inverse is equal to its conjugate transpose, $MM^*=I$. I think we further need to show $O_x$ is Hermitian, then it will be done.
$endgroup$
– Divyat
Mar 26 at 13:56
$begingroup$
One more doubt, I think the answer still assumes that $O_x$ is normal matrix, as then with real eigenvalues we can claim it to be Hermitian. Please tell whether it is okay to assume oracle as normal or some justifications for it?
$endgroup$
– Divyat
Mar 26 at 15:52
$begingroup$
This explanation is not quite correct, because you assumed that $O_x$ is diagonalizable, but it is not follow from $O_x^2=I$. The thing is $O_x$ maps basis vectors to basis vectors. Since $O_x^2=I$ then $Q_x$ is a permutation of basis vectors.
$endgroup$
– Danylo Y
Mar 26 at 16:17
$begingroup$
I actually asked some time ago on math.SE whether one can deduce that $sqrt A$ is diagonalizable from the fact that $A$ is, see here. As far as I understood the answers, $A^2=I$ is not enough to imply that $A$ is unitarily diagonalizable, a counterexample being $beginpmatrixcostheta & 2sintheta \ sintheta/2 & -costhetaendpmatrix$, which is a square root of the identity, but not (unitarily) diagonalizable, and not unitary, even though its eigenvalues are $pm 1$.
$endgroup$
– glS
Mar 26 at 18:01
1
$begingroup$
Fair enough, I was being a bit glib, particularly initially with regards to the difference between reversible and unitary, because usually its the reversibility aspect people are interested in.
$endgroup$
– DaftWullie
Mar 27 at 8:44
add a comment |
$begingroup$
Apply it twice:
$$
O_xO_x|irangle|brangle=O_x|irangle|boplus x_irangle=|irangle|boplus x_ioplus x_irangle=|irangle|brangle
$$
Hence, $O_x$ is its own inverse, and therefore reversible.
To prove unitarity, it makes more sense to prove that $O_x$ has eigenvectors
$$
|irangle(|0rangle+|1rangle)quadtextandquad|irangle(|0rangle-|1rangle)
$$
for all $i$ with eigenvalues $1$ and $(-1)^x_i$ respectively. These are all orthonormal, and span the full Hilbert space. Consequently, the eigenvalues of $O_x$ are all $pm 1$, and therefore $O_x^star=O_x$ (where $^star$ represents the Hermitian conjugate). Thus,
$$
O_xO_x^star=mathbbI,
$$
as required for a unitary.
$endgroup$
$begingroup$
Thanks, I now understand $O_x$ is its own inverse but how does that make it unitary? For a unitary matrix, one would need to show that its inverse is equal to its conjugate transpose, $MM^*=I$. I think we further need to show $O_x$ is Hermitian, then it will be done.
$endgroup$
– Divyat
Mar 26 at 13:56
$begingroup$
One more doubt, I think the answer still assumes that $O_x$ is normal matrix, as then with real eigenvalues we can claim it to be Hermitian. Please tell whether it is okay to assume oracle as normal or some justifications for it?
$endgroup$
– Divyat
Mar 26 at 15:52
$begingroup$
This explanation is not quite correct, because you assumed that $O_x$ is diagonalizable, but it is not follow from $O_x^2=I$. The thing is $O_x$ maps basis vectors to basis vectors. Since $O_x^2=I$ then $Q_x$ is a permutation of basis vectors.
$endgroup$
– Danylo Y
Mar 26 at 16:17
$begingroup$
I actually asked some time ago on math.SE whether one can deduce that $sqrt A$ is diagonalizable from the fact that $A$ is, see here. As far as I understood the answers, $A^2=I$ is not enough to imply that $A$ is unitarily diagonalizable, a counterexample being $beginpmatrixcostheta & 2sintheta \ sintheta/2 & -costhetaendpmatrix$, which is a square root of the identity, but not (unitarily) diagonalizable, and not unitary, even though its eigenvalues are $pm 1$.
$endgroup$
– glS
Mar 26 at 18:01
1
$begingroup$
Fair enough, I was being a bit glib, particularly initially with regards to the difference between reversible and unitary, because usually its the reversibility aspect people are interested in.
$endgroup$
– DaftWullie
Mar 27 at 8:44
add a comment |
$begingroup$
Apply it twice:
$$
O_xO_x|irangle|brangle=O_x|irangle|boplus x_irangle=|irangle|boplus x_ioplus x_irangle=|irangle|brangle
$$
Hence, $O_x$ is its own inverse, and therefore reversible.
To prove unitarity, it makes more sense to prove that $O_x$ has eigenvectors
$$
|irangle(|0rangle+|1rangle)quadtextandquad|irangle(|0rangle-|1rangle)
$$
for all $i$ with eigenvalues $1$ and $(-1)^x_i$ respectively. These are all orthonormal, and span the full Hilbert space. Consequently, the eigenvalues of $O_x$ are all $pm 1$, and therefore $O_x^star=O_x$ (where $^star$ represents the Hermitian conjugate). Thus,
$$
O_xO_x^star=mathbbI,
$$
as required for a unitary.
$endgroup$
Apply it twice:
$$
O_xO_x|irangle|brangle=O_x|irangle|boplus x_irangle=|irangle|boplus x_ioplus x_irangle=|irangle|brangle
$$
Hence, $O_x$ is its own inverse, and therefore reversible.
To prove unitarity, it makes more sense to prove that $O_x$ has eigenvectors
$$
|irangle(|0rangle+|1rangle)quadtextandquad|irangle(|0rangle-|1rangle)
$$
for all $i$ with eigenvalues $1$ and $(-1)^x_i$ respectively. These are all orthonormal, and span the full Hilbert space. Consequently, the eigenvalues of $O_x$ are all $pm 1$, and therefore $O_x^star=O_x$ (where $^star$ represents the Hermitian conjugate). Thus,
$$
O_xO_x^star=mathbbI,
$$
as required for a unitary.
edited Mar 27 at 8:51
answered Mar 26 at 12:18
DaftWullieDaftWullie
15.2k1542
15.2k1542
$begingroup$
Thanks, I now understand $O_x$ is its own inverse but how does that make it unitary? For a unitary matrix, one would need to show that its inverse is equal to its conjugate transpose, $MM^*=I$. I think we further need to show $O_x$ is Hermitian, then it will be done.
$endgroup$
– Divyat
Mar 26 at 13:56
$begingroup$
One more doubt, I think the answer still assumes that $O_x$ is normal matrix, as then with real eigenvalues we can claim it to be Hermitian. Please tell whether it is okay to assume oracle as normal or some justifications for it?
$endgroup$
– Divyat
Mar 26 at 15:52
$begingroup$
This explanation is not quite correct, because you assumed that $O_x$ is diagonalizable, but it is not follow from $O_x^2=I$. The thing is $O_x$ maps basis vectors to basis vectors. Since $O_x^2=I$ then $Q_x$ is a permutation of basis vectors.
$endgroup$
– Danylo Y
Mar 26 at 16:17
$begingroup$
I actually asked some time ago on math.SE whether one can deduce that $sqrt A$ is diagonalizable from the fact that $A$ is, see here. As far as I understood the answers, $A^2=I$ is not enough to imply that $A$ is unitarily diagonalizable, a counterexample being $beginpmatrixcostheta & 2sintheta \ sintheta/2 & -costhetaendpmatrix$, which is a square root of the identity, but not (unitarily) diagonalizable, and not unitary, even though its eigenvalues are $pm 1$.
$endgroup$
– glS
Mar 26 at 18:01
1
$begingroup$
Fair enough, I was being a bit glib, particularly initially with regards to the difference between reversible and unitary, because usually its the reversibility aspect people are interested in.
$endgroup$
– DaftWullie
Mar 27 at 8:44
add a comment |
$begingroup$
Thanks, I now understand $O_x$ is its own inverse but how does that make it unitary? For a unitary matrix, one would need to show that its inverse is equal to its conjugate transpose, $MM^*=I$. I think we further need to show $O_x$ is Hermitian, then it will be done.
$endgroup$
– Divyat
Mar 26 at 13:56
$begingroup$
One more doubt, I think the answer still assumes that $O_x$ is normal matrix, as then with real eigenvalues we can claim it to be Hermitian. Please tell whether it is okay to assume oracle as normal or some justifications for it?
$endgroup$
– Divyat
Mar 26 at 15:52
$begingroup$
This explanation is not quite correct, because you assumed that $O_x$ is diagonalizable, but it is not follow from $O_x^2=I$. The thing is $O_x$ maps basis vectors to basis vectors. Since $O_x^2=I$ then $Q_x$ is a permutation of basis vectors.
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– Danylo Y
Mar 26 at 16:17
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I actually asked some time ago on math.SE whether one can deduce that $sqrt A$ is diagonalizable from the fact that $A$ is, see here. As far as I understood the answers, $A^2=I$ is not enough to imply that $A$ is unitarily diagonalizable, a counterexample being $beginpmatrixcostheta & 2sintheta \ sintheta/2 & -costhetaendpmatrix$, which is a square root of the identity, but not (unitarily) diagonalizable, and not unitary, even though its eigenvalues are $pm 1$.
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– glS
Mar 26 at 18:01
1
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Fair enough, I was being a bit glib, particularly initially with regards to the difference between reversible and unitary, because usually its the reversibility aspect people are interested in.
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– DaftWullie
Mar 27 at 8:44
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Thanks, I now understand $O_x$ is its own inverse but how does that make it unitary? For a unitary matrix, one would need to show that its inverse is equal to its conjugate transpose, $MM^*=I$. I think we further need to show $O_x$ is Hermitian, then it will be done.
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– Divyat
Mar 26 at 13:56
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Thanks, I now understand $O_x$ is its own inverse but how does that make it unitary? For a unitary matrix, one would need to show that its inverse is equal to its conjugate transpose, $MM^*=I$. I think we further need to show $O_x$ is Hermitian, then it will be done.
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– Divyat
Mar 26 at 13:56
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One more doubt, I think the answer still assumes that $O_x$ is normal matrix, as then with real eigenvalues we can claim it to be Hermitian. Please tell whether it is okay to assume oracle as normal or some justifications for it?
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– Divyat
Mar 26 at 15:52
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One more doubt, I think the answer still assumes that $O_x$ is normal matrix, as then with real eigenvalues we can claim it to be Hermitian. Please tell whether it is okay to assume oracle as normal or some justifications for it?
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– Divyat
Mar 26 at 15:52
$begingroup$
This explanation is not quite correct, because you assumed that $O_x$ is diagonalizable, but it is not follow from $O_x^2=I$. The thing is $O_x$ maps basis vectors to basis vectors. Since $O_x^2=I$ then $Q_x$ is a permutation of basis vectors.
$endgroup$
– Danylo Y
Mar 26 at 16:17
$begingroup$
This explanation is not quite correct, because you assumed that $O_x$ is diagonalizable, but it is not follow from $O_x^2=I$. The thing is $O_x$ maps basis vectors to basis vectors. Since $O_x^2=I$ then $Q_x$ is a permutation of basis vectors.
$endgroup$
– Danylo Y
Mar 26 at 16:17
$begingroup$
I actually asked some time ago on math.SE whether one can deduce that $sqrt A$ is diagonalizable from the fact that $A$ is, see here. As far as I understood the answers, $A^2=I$ is not enough to imply that $A$ is unitarily diagonalizable, a counterexample being $beginpmatrixcostheta & 2sintheta \ sintheta/2 & -costhetaendpmatrix$, which is a square root of the identity, but not (unitarily) diagonalizable, and not unitary, even though its eigenvalues are $pm 1$.
$endgroup$
– glS
Mar 26 at 18:01
$begingroup$
I actually asked some time ago on math.SE whether one can deduce that $sqrt A$ is diagonalizable from the fact that $A$ is, see here. As far as I understood the answers, $A^2=I$ is not enough to imply that $A$ is unitarily diagonalizable, a counterexample being $beginpmatrixcostheta & 2sintheta \ sintheta/2 & -costhetaendpmatrix$, which is a square root of the identity, but not (unitarily) diagonalizable, and not unitary, even though its eigenvalues are $pm 1$.
$endgroup$
– glS
Mar 26 at 18:01
1
1
$begingroup$
Fair enough, I was being a bit glib, particularly initially with regards to the difference between reversible and unitary, because usually its the reversibility aspect people are interested in.
$endgroup$
– DaftWullie
Mar 27 at 8:44
$begingroup$
Fair enough, I was being a bit glib, particularly initially with regards to the difference between reversible and unitary, because usually its the reversibility aspect people are interested in.
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– DaftWullie
Mar 27 at 8:44
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Related: Why are oracles Hermitian by construction?
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– Blue♦
Mar 26 at 12:25