How to prove that the query oracle is unitary?
$begingroup$
The query oracle: $O_{x}|irangle|brangle = |irangle|b oplus x_{i}rangle$ used in algorithms like Deutsch Jozsa is unitary. How do I prove it is unitary?
algorithm quantum-gate unitarity deutsch-jozsa-algorithm
New contributor
$endgroup$
add a comment |
$begingroup$
The query oracle: $O_{x}|irangle|brangle = |irangle|b oplus x_{i}rangle$ used in algorithms like Deutsch Jozsa is unitary. How do I prove it is unitary?
algorithm quantum-gate unitarity deutsch-jozsa-algorithm
New contributor
$endgroup$
$begingroup$
Related: Why are oracles Hermitian by construction?
$endgroup$
– Blue♦
2 days ago
add a comment |
$begingroup$
The query oracle: $O_{x}|irangle|brangle = |irangle|b oplus x_{i}rangle$ used in algorithms like Deutsch Jozsa is unitary. How do I prove it is unitary?
algorithm quantum-gate unitarity deutsch-jozsa-algorithm
New contributor
$endgroup$
The query oracle: $O_{x}|irangle|brangle = |irangle|b oplus x_{i}rangle$ 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
New contributor
New contributor
edited 2 days ago
Blue♦
6,57541555
6,57541555
New contributor
asked 2 days ago
DivyatDivyat
283
283
New contributor
New contributor
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Related: Why are oracles Hermitian by construction?
$endgroup$
– Blue♦
2 days ago
add a comment |
$begingroup$
Related: Why are oracles Hermitian by construction?
$endgroup$
– Blue♦
2 days ago
$begingroup$
Related: Why are oracles Hermitian by construction?
$endgroup$
– Blue♦
2 days ago
$begingroup$
Related: Why are oracles Hermitian by construction?
$endgroup$
– Blue♦
2 days ago
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Notice that $mathcal O_x$ is a permutation matrix.
The matrix elements are
$$langle j, crvertmathcal O_xlvert i,brangle
=delta_{ij}langle crvert boplus x_irangle
=delta_{ij}delta_{c,boplus x_i}.$$
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
$$
|irangle(|0rangle+|1rangle)quadtext{and}quad|irangle(|0rangle-|1rangle)
$$
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,
$$
O_xO_x^star=mathbb{I},
$$
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
2 days ago
$begingroup$
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
2 days ago
$begingroup$
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
2 days ago
$begingroup$
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 $begin{pmatrix}costheta & 2sintheta \ sintheta/2 & -costhetaend{pmatrix}$, which is a square root of the identity, but not (unitarily) diagonalizable, and not unitary, even though its eigenvalues are $pm 1$.
$endgroup$
– glS
2 days ago
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
yesterday
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
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active
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$begingroup$
Notice that $mathcal O_x$ is a permutation matrix.
The matrix elements are
$$langle j, crvertmathcal O_xlvert i,brangle
=delta_{ij}langle crvert boplus x_irangle
=delta_{ij}delta_{c,boplus x_i}.$$
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
$$langle j, crvertmathcal O_xlvert i,brangle
=delta_{ij}langle crvert boplus x_irangle
=delta_{ij}delta_{c,boplus x_i}.$$
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
$$langle j, crvertmathcal O_xlvert i,brangle
=delta_{ij}langle crvert boplus x_irangle
=delta_{ij}delta_{c,boplus x_i}.$$
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
$$langle j, crvertmathcal O_xlvert i,brangle
=delta_{ij}langle crvert boplus x_irangle
=delta_{ij}delta_{c,boplus x_i}.$$
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 2 days ago
answered 2 days ago
glSglS
4,308739
4,308739
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
$$
|irangle(|0rangle+|1rangle)quadtext{and}quad|irangle(|0rangle-|1rangle)
$$
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,
$$
O_xO_x^star=mathbb{I},
$$
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
2 days ago
$begingroup$
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
2 days ago
$begingroup$
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
2 days ago
$begingroup$
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 $begin{pmatrix}costheta & 2sintheta \ sintheta/2 & -costhetaend{pmatrix}$, which is a square root of the identity, but not (unitarily) diagonalizable, and not unitary, even though its eigenvalues are $pm 1$.
$endgroup$
– glS
2 days ago
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
yesterday
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
$$
|irangle(|0rangle+|1rangle)quadtext{and}quad|irangle(|0rangle-|1rangle)
$$
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,
$$
O_xO_x^star=mathbb{I},
$$
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
2 days ago
$begingroup$
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
2 days ago
$begingroup$
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
2 days ago
$begingroup$
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 $begin{pmatrix}costheta & 2sintheta \ sintheta/2 & -costhetaend{pmatrix}$, which is a square root of the identity, but not (unitarily) diagonalizable, and not unitary, even though its eigenvalues are $pm 1$.
$endgroup$
– glS
2 days ago
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
yesterday
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
$$
|irangle(|0rangle+|1rangle)quadtext{and}quad|irangle(|0rangle-|1rangle)
$$
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,
$$
O_xO_x^star=mathbb{I},
$$
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
$$
|irangle(|0rangle+|1rangle)quadtext{and}quad|irangle(|0rangle-|1rangle)
$$
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,
$$
O_xO_x^star=mathbb{I},
$$
as required for a unitary.
edited yesterday
answered 2 days ago
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
2 days ago
$begingroup$
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
2 days ago
$begingroup$
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
2 days ago
$begingroup$
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 $begin{pmatrix}costheta & 2sintheta \ sintheta/2 & -costhetaend{pmatrix}$, which is a square root of the identity, but not (unitarily) diagonalizable, and not unitary, even though its eigenvalues are $pm 1$.
$endgroup$
– glS
2 days ago
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
yesterday
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
2 days ago
$begingroup$
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
2 days ago
$begingroup$
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
2 days ago
$begingroup$
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 $begin{pmatrix}costheta & 2sintheta \ sintheta/2 & -costhetaend{pmatrix}$, which is a square root of the identity, but not (unitarily) diagonalizable, and not unitary, even though its eigenvalues are $pm 1$.
$endgroup$
– glS
2 days ago
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
yesterday
$begingroup$
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
2 days ago
$begingroup$
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
2 days ago
$begingroup$
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
2 days ago
$begingroup$
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
2 days ago
$begingroup$
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
2 days ago
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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
2 days ago
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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 $begin{pmatrix}costheta & 2sintheta \ sintheta/2 & -costhetaend{pmatrix}$, 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
2 days ago
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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 $begin{pmatrix}costheta & 2sintheta \ sintheta/2 & -costhetaend{pmatrix}$, 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
2 days ago
1
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
yesterday
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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
yesterday
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Related: Why are oracles Hermitian by construction?
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– Blue♦
2 days ago