Problem with the mathematical formulation of “qubitization”
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In this research paper, the authors introduce a new algorithm to perform Hamiltonian simulation.
The beginning of their abstract is
Given a Hermitian operator $hat{H} = langle Gvert hat{U} vert Grangle$ that is the projection of an oracle $hat{U}$ by state $vert Grangle$
created with oracle $hat{G}$, the problem of Hamiltonian simulation is approximating the time evolution operator $e^{-ihat{H}t}$ at time $t$ with error $epsilon$.
In the article:
$hat{G}$ and $hat{U}$ are called "oracles".
$hat{H}$ is an Hermitian operator in $mathbb{C}^{2^n} times mathbb{C}^{2^n}$.
$vert G rangle in mathbb{C}^d$ (legend of Table 1).
My question is the following: what means $hat{H} = langle Gvert hat{U} vert Grangle$? More precisely, I do not understand what $langle Gvert hat{U} vert Grangle$ represents when $hat{U}$ is an oracle and $vert G rangle$ a quantum state.
mathematics hamiltonian-simulation notation
$endgroup$
add a comment |
$begingroup$
In this research paper, the authors introduce a new algorithm to perform Hamiltonian simulation.
The beginning of their abstract is
Given a Hermitian operator $hat{H} = langle Gvert hat{U} vert Grangle$ that is the projection of an oracle $hat{U}$ by state $vert Grangle$
created with oracle $hat{G}$, the problem of Hamiltonian simulation is approximating the time evolution operator $e^{-ihat{H}t}$ at time $t$ with error $epsilon$.
In the article:
$hat{G}$ and $hat{U}$ are called "oracles".
$hat{H}$ is an Hermitian operator in $mathbb{C}^{2^n} times mathbb{C}^{2^n}$.
$vert G rangle in mathbb{C}^d$ (legend of Table 1).
My question is the following: what means $hat{H} = langle Gvert hat{U} vert Grangle$? More precisely, I do not understand what $langle Gvert hat{U} vert Grangle$ represents when $hat{U}$ is an oracle and $vert G rangle$ a quantum state.
mathematics hamiltonian-simulation notation
$endgroup$
add a comment |
$begingroup$
In this research paper, the authors introduce a new algorithm to perform Hamiltonian simulation.
The beginning of their abstract is
Given a Hermitian operator $hat{H} = langle Gvert hat{U} vert Grangle$ that is the projection of an oracle $hat{U}$ by state $vert Grangle$
created with oracle $hat{G}$, the problem of Hamiltonian simulation is approximating the time evolution operator $e^{-ihat{H}t}$ at time $t$ with error $epsilon$.
In the article:
$hat{G}$ and $hat{U}$ are called "oracles".
$hat{H}$ is an Hermitian operator in $mathbb{C}^{2^n} times mathbb{C}^{2^n}$.
$vert G rangle in mathbb{C}^d$ (legend of Table 1).
My question is the following: what means $hat{H} = langle Gvert hat{U} vert Grangle$? More precisely, I do not understand what $langle Gvert hat{U} vert Grangle$ represents when $hat{U}$ is an oracle and $vert G rangle$ a quantum state.
mathematics hamiltonian-simulation notation
$endgroup$
In this research paper, the authors introduce a new algorithm to perform Hamiltonian simulation.
The beginning of their abstract is
Given a Hermitian operator $hat{H} = langle Gvert hat{U} vert Grangle$ that is the projection of an oracle $hat{U}$ by state $vert Grangle$
created with oracle $hat{G}$, the problem of Hamiltonian simulation is approximating the time evolution operator $e^{-ihat{H}t}$ at time $t$ with error $epsilon$.
In the article:
$hat{G}$ and $hat{U}$ are called "oracles".
$hat{H}$ is an Hermitian operator in $mathbb{C}^{2^n} times mathbb{C}^{2^n}$.
$vert G rangle in mathbb{C}^d$ (legend of Table 1).
My question is the following: what means $hat{H} = langle Gvert hat{U} vert Grangle$? More precisely, I do not understand what $langle Gvert hat{U} vert Grangle$ represents when $hat{U}$ is an oracle and $vert G rangle$ a quantum state.
mathematics hamiltonian-simulation notation
mathematics hamiltonian-simulation notation
edited Dec 11 '18 at 13:33
Nelimee
asked Dec 11 '18 at 12:58
NelimeeNelimee
1,570326
1,570326
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1 Answer
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$begingroup$
You want to start by being careful with the sizes of the operators. $hat U$ acts on $q$ qubits, and $hat H$ acts on $n<q$ qubits. I believe that $|Grangle$ is a state of $q-n$ qubits. So, what we really need to talk about is two distinct sets of qubits. Let me call them sets $A$ and $B$. $A$ contains $n$ qubits, and $B$ contains $q-n$ qubits. I'll use subscripts to denote which qubits the different operators and states act upon:
$$
hat H_A=(langle G|_Botimesmathbb{I}_A)hat U_{AB}(|Grangle_Botimesmathbb{I}_A)
$$
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$begingroup$
@Nelimee I'm not sure if this is sufficient to resolve your confusion? Or is there something more that you're asking?
$endgroup$
– DaftWullie
Dec 11 '18 at 13:24
$begingroup$
I am still trying to understand your answer but the sizes of the operators were definitely one of the points I missed! About your answer, what does $vert G rangle_B otimes mathbb{I}_A$ represent? A tensor product between a quantum state (a vector) and an operator (a matrix)?
$endgroup$
– Nelimee
Dec 11 '18 at 13:39
$begingroup$
Yes, exactly. Where, of course, you should think of a vector as a matrix where one of the dimensions is just 1.
$endgroup$
– DaftWullie
Dec 11 '18 at 13:45
$begingroup$
Ok that solved my problem! Thanks for the quick clarification :)
$endgroup$
– Nelimee
Dec 11 '18 at 13:54
add a comment |
Your Answer
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1 Answer
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1 Answer
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active
oldest
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$begingroup$
You want to start by being careful with the sizes of the operators. $hat U$ acts on $q$ qubits, and $hat H$ acts on $n<q$ qubits. I believe that $|Grangle$ is a state of $q-n$ qubits. So, what we really need to talk about is two distinct sets of qubits. Let me call them sets $A$ and $B$. $A$ contains $n$ qubits, and $B$ contains $q-n$ qubits. I'll use subscripts to denote which qubits the different operators and states act upon:
$$
hat H_A=(langle G|_Botimesmathbb{I}_A)hat U_{AB}(|Grangle_Botimesmathbb{I}_A)
$$
$endgroup$
$begingroup$
@Nelimee I'm not sure if this is sufficient to resolve your confusion? Or is there something more that you're asking?
$endgroup$
– DaftWullie
Dec 11 '18 at 13:24
$begingroup$
I am still trying to understand your answer but the sizes of the operators were definitely one of the points I missed! About your answer, what does $vert G rangle_B otimes mathbb{I}_A$ represent? A tensor product between a quantum state (a vector) and an operator (a matrix)?
$endgroup$
– Nelimee
Dec 11 '18 at 13:39
$begingroup$
Yes, exactly. Where, of course, you should think of a vector as a matrix where one of the dimensions is just 1.
$endgroup$
– DaftWullie
Dec 11 '18 at 13:45
$begingroup$
Ok that solved my problem! Thanks for the quick clarification :)
$endgroup$
– Nelimee
Dec 11 '18 at 13:54
add a comment |
$begingroup$
You want to start by being careful with the sizes of the operators. $hat U$ acts on $q$ qubits, and $hat H$ acts on $n<q$ qubits. I believe that $|Grangle$ is a state of $q-n$ qubits. So, what we really need to talk about is two distinct sets of qubits. Let me call them sets $A$ and $B$. $A$ contains $n$ qubits, and $B$ contains $q-n$ qubits. I'll use subscripts to denote which qubits the different operators and states act upon:
$$
hat H_A=(langle G|_Botimesmathbb{I}_A)hat U_{AB}(|Grangle_Botimesmathbb{I}_A)
$$
$endgroup$
$begingroup$
@Nelimee I'm not sure if this is sufficient to resolve your confusion? Or is there something more that you're asking?
$endgroup$
– DaftWullie
Dec 11 '18 at 13:24
$begingroup$
I am still trying to understand your answer but the sizes of the operators were definitely one of the points I missed! About your answer, what does $vert G rangle_B otimes mathbb{I}_A$ represent? A tensor product between a quantum state (a vector) and an operator (a matrix)?
$endgroup$
– Nelimee
Dec 11 '18 at 13:39
$begingroup$
Yes, exactly. Where, of course, you should think of a vector as a matrix where one of the dimensions is just 1.
$endgroup$
– DaftWullie
Dec 11 '18 at 13:45
$begingroup$
Ok that solved my problem! Thanks for the quick clarification :)
$endgroup$
– Nelimee
Dec 11 '18 at 13:54
add a comment |
$begingroup$
You want to start by being careful with the sizes of the operators. $hat U$ acts on $q$ qubits, and $hat H$ acts on $n<q$ qubits. I believe that $|Grangle$ is a state of $q-n$ qubits. So, what we really need to talk about is two distinct sets of qubits. Let me call them sets $A$ and $B$. $A$ contains $n$ qubits, and $B$ contains $q-n$ qubits. I'll use subscripts to denote which qubits the different operators and states act upon:
$$
hat H_A=(langle G|_Botimesmathbb{I}_A)hat U_{AB}(|Grangle_Botimesmathbb{I}_A)
$$
$endgroup$
You want to start by being careful with the sizes of the operators. $hat U$ acts on $q$ qubits, and $hat H$ acts on $n<q$ qubits. I believe that $|Grangle$ is a state of $q-n$ qubits. So, what we really need to talk about is two distinct sets of qubits. Let me call them sets $A$ and $B$. $A$ contains $n$ qubits, and $B$ contains $q-n$ qubits. I'll use subscripts to denote which qubits the different operators and states act upon:
$$
hat H_A=(langle G|_Botimesmathbb{I}_A)hat U_{AB}(|Grangle_Botimesmathbb{I}_A)
$$
answered Dec 11 '18 at 13:23
DaftWullieDaftWullie
13.6k1539
13.6k1539
$begingroup$
@Nelimee I'm not sure if this is sufficient to resolve your confusion? Or is there something more that you're asking?
$endgroup$
– DaftWullie
Dec 11 '18 at 13:24
$begingroup$
I am still trying to understand your answer but the sizes of the operators were definitely one of the points I missed! About your answer, what does $vert G rangle_B otimes mathbb{I}_A$ represent? A tensor product between a quantum state (a vector) and an operator (a matrix)?
$endgroup$
– Nelimee
Dec 11 '18 at 13:39
$begingroup$
Yes, exactly. Where, of course, you should think of a vector as a matrix where one of the dimensions is just 1.
$endgroup$
– DaftWullie
Dec 11 '18 at 13:45
$begingroup$
Ok that solved my problem! Thanks for the quick clarification :)
$endgroup$
– Nelimee
Dec 11 '18 at 13:54
add a comment |
$begingroup$
@Nelimee I'm not sure if this is sufficient to resolve your confusion? Or is there something more that you're asking?
$endgroup$
– DaftWullie
Dec 11 '18 at 13:24
$begingroup$
I am still trying to understand your answer but the sizes of the operators were definitely one of the points I missed! About your answer, what does $vert G rangle_B otimes mathbb{I}_A$ represent? A tensor product between a quantum state (a vector) and an operator (a matrix)?
$endgroup$
– Nelimee
Dec 11 '18 at 13:39
$begingroup$
Yes, exactly. Where, of course, you should think of a vector as a matrix where one of the dimensions is just 1.
$endgroup$
– DaftWullie
Dec 11 '18 at 13:45
$begingroup$
Ok that solved my problem! Thanks for the quick clarification :)
$endgroup$
– Nelimee
Dec 11 '18 at 13:54
$begingroup$
@Nelimee I'm not sure if this is sufficient to resolve your confusion? Or is there something more that you're asking?
$endgroup$
– DaftWullie
Dec 11 '18 at 13:24
$begingroup$
@Nelimee I'm not sure if this is sufficient to resolve your confusion? Or is there something more that you're asking?
$endgroup$
– DaftWullie
Dec 11 '18 at 13:24
$begingroup$
I am still trying to understand your answer but the sizes of the operators were definitely one of the points I missed! About your answer, what does $vert G rangle_B otimes mathbb{I}_A$ represent? A tensor product between a quantum state (a vector) and an operator (a matrix)?
$endgroup$
– Nelimee
Dec 11 '18 at 13:39
$begingroup$
I am still trying to understand your answer but the sizes of the operators were definitely one of the points I missed! About your answer, what does $vert G rangle_B otimes mathbb{I}_A$ represent? A tensor product between a quantum state (a vector) and an operator (a matrix)?
$endgroup$
– Nelimee
Dec 11 '18 at 13:39
$begingroup$
Yes, exactly. Where, of course, you should think of a vector as a matrix where one of the dimensions is just 1.
$endgroup$
– DaftWullie
Dec 11 '18 at 13:45
$begingroup$
Yes, exactly. Where, of course, you should think of a vector as a matrix where one of the dimensions is just 1.
$endgroup$
– DaftWullie
Dec 11 '18 at 13:45
$begingroup$
Ok that solved my problem! Thanks for the quick clarification :)
$endgroup$
– Nelimee
Dec 11 '18 at 13:54
$begingroup$
Ok that solved my problem! Thanks for the quick clarification :)
$endgroup$
– Nelimee
Dec 11 '18 at 13:54
add a comment |
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