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I'm trying to understand the DeutchDeutsch-Josza algorithm from an adiabatic perspective as presented in Adiabatic quantum computing A: Review of modern physics, vol 90, (2018) pp 015002-1 (arxivarXiv version).

When explaining the unitary interpolation technique, the authors begin with:

The initial Hamiltonian is chosen such that its ground state is the uniform superposition state [$|\phi\rangle = |+\rangle^{\otimes n}$], i.e., $H(0) = w\sum^{n}_{i = 1}|-\rangle_i \langle-|$, where $w$ is the energy scale.

  • How can I calculate the ground state of $H(0)$?

Also, I have seen, but don't remember exactly where, an observation that $H(0)$ is introduced in a such a way that a penalty is provided for any state having a contribution of $|-\rangle$. What does it mean?

Then the paper gogoes on and states that:

An adiabatic implementation requires a final Hamiltonian $H(1)$ such that its ground state is $|\Psi(1)\rangle = U|\Psi(0)\rangle$, and that this can be accomplished via a unitary transformation of $H(0)$, i.e. $H(1) = UH(0)U^\dagger$.

where $U$ is a diagonal matrix such that:

$$U = diag[(-1)^{f(0)}, \dots,(-1)^{f(2^n-1)}] $$

At this point, I don't see why bother with this; since to arrive at the answer using adiabatic quantum computation I need to construct a unitary in such a way that I will already have the answer if $f$ is balanced or constant. Am I overlooking anything?


Trying to workout an example, setting $n = 1$ and making $f(x) = 1$ (constant 1).

$$H(0) = w|-\rangle\langle-| = w\pmatrix{1 & -1 \\ -1 & 1}$$

I will set $w = 1$ to get it out of the way. Then,

$$U = \pmatrix{-1 & 0 \\ 0 & -1} $$ $$H(1) = UH(0)U^\dagger = \pmatrix{-1 & 0 \\ 0 & -1}\pmatrix{1 & -1 \\ -1 & 1}\pmatrix{-1 & 0 \\ 0 & -1}$$ $$H(1) = \pmatrix{1 & -1 \\ -1 & 1}$$

Meaning that the ground state of $H(1)$ is $|+\rangle$.

If now I do the function $f(x) = x$, then $$U = \pmatrix{1 & 0 \\ 0 & -1} $$

and

$$H(1) = \pmatrix{1 & 1 \\ 1 & 1}$$

which, I suppose, has ground state $|-\rangle$. And with this we can differentiate between a constant and a balanced function $f$.

I'm trying to understand the Deutch-Josza algorithm from adiabatic perspective as presented in Adiabatic quantum computing A: Review of modern physics, vol 90, (2018) pp 015002-1 (arxiv version).

When explaining the unitary interpolation technique, the authors begin with:

The initial Hamiltonian is chosen such that its ground state is the uniform superposition state [$|\phi\rangle = |+\rangle^{\otimes n}$], i.e., $H(0) = w\sum^{n}_{i = 1}|-\rangle_i \langle-|$, where $w$ is the energy scale.

  • How can I calculate the ground state of $H(0)$?

Also, I have seen, but don't remember exactly where, an observation that $H(0)$ is introduced in a such a way that a penalty is provided for any state having a contribution of $|-\rangle$. What does it mean?

Then the paper go on and states that:

An adiabatic implementation requires a final Hamiltonian $H(1)$ such that its ground state is $|\Psi(1)\rangle = U|\Psi(0)\rangle$, and that this can be accomplished via a unitary transformation of $H(0)$, i.e. $H(1) = UH(0)U^\dagger$.

where $U$ is a diagonal matrix such that:

$$U = diag[(-1)^{f(0)}, \dots,(-1)^{f(2^n-1)}] $$

At this point, I don't see why bother with this; since to arrive at the answer using adiabatic quantum computation I need to construct a unitary in such a way that I will already have the answer if $f$ is balanced or constant. Am I overlooking anything?


Trying to workout an example, setting $n = 1$ and making $f(x) = 1$ (constant 1).

$$H(0) = w|-\rangle\langle-| = w\pmatrix{1 & -1 \\ -1 & 1}$$

I will set $w = 1$ to get it out of the way. Then,

$$U = \pmatrix{-1 & 0 \\ 0 & -1} $$ $$H(1) = UH(0)U^\dagger = \pmatrix{-1 & 0 \\ 0 & -1}\pmatrix{1 & -1 \\ -1 & 1}\pmatrix{-1 & 0 \\ 0 & -1}$$ $$H(1) = \pmatrix{1 & -1 \\ -1 & 1}$$

Meaning that the ground state of $H(1)$ is $|+\rangle$.

If now I do the function $f(x) = x$, then $$U = \pmatrix{1 & 0 \\ 0 & -1} $$

and

$$H(1) = \pmatrix{1 & 1 \\ 1 & 1}$$

which, I suppose, has ground state $|-\rangle$. And with this we can differentiate between a constant and a balanced function $f$.

I'm trying to understand the Deutsch-Josza algorithm from an adiabatic perspective as presented in Adiabatic quantum computing A: Review of modern physics, vol 90, (2018) pp 015002-1 (arXiv version).

When explaining the unitary interpolation technique, the authors begin with:

The initial Hamiltonian is chosen such that its ground state is the uniform superposition state [$|\phi\rangle = |+\rangle^{\otimes n}$], i.e., $H(0) = w\sum^{n}_{i = 1}|-\rangle_i \langle-|$, where $w$ is the energy scale.

  • How can I calculate the ground state of $H(0)$?

Also, I have seen, but don't remember exactly where, an observation that $H(0)$ is introduced in a such a way that a penalty is provided for any state having a contribution of $|-\rangle$. What does it mean?

Then the paper goes on and states that:

An adiabatic implementation requires a final Hamiltonian $H(1)$ such that its ground state is $|\Psi(1)\rangle = U|\Psi(0)\rangle$, and that this can be accomplished via a unitary transformation of $H(0)$, i.e. $H(1) = UH(0)U^\dagger$.

where $U$ is a diagonal matrix such that:

$$U = diag[(-1)^{f(0)}, \dots,(-1)^{f(2^n-1)}] $$

At this point, I don't see why bother with this; since to arrive at the answer using adiabatic quantum computation I need to construct a unitary in such a way that I will already have the answer if $f$ is balanced or constant. Am I overlooking anything?


Trying to workout an example, setting $n = 1$ and making $f(x) = 1$ (constant 1).

$$H(0) = w|-\rangle\langle-| = w\pmatrix{1 & -1 \\ -1 & 1}$$

I will set $w = 1$ to get it out of the way. Then,

$$U = \pmatrix{-1 & 0 \\ 0 & -1} $$ $$H(1) = UH(0)U^\dagger = \pmatrix{-1 & 0 \\ 0 & -1}\pmatrix{1 & -1 \\ -1 & 1}\pmatrix{-1 & 0 \\ 0 & -1}$$ $$H(1) = \pmatrix{1 & -1 \\ -1 & 1}$$

Meaning that the ground state of $H(1)$ is $|+\rangle$.

If now I do the function $f(x) = x$, then $$U = \pmatrix{1 & 0 \\ 0 & -1} $$

and

$$H(1) = \pmatrix{1 & 1 \\ 1 & 1}$$

which, I suppose, has ground state $|-\rangle$. And with this we can differentiate between a constant and a balanced function $f$.

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How to understand Deutsch-Jozsa algorithm from an adiabatic perspective?

I'm trying to understand the Deutch-Josza algorithm from adiabatic perspective as presented in Adiabatic quantum computing A: Review of modern physics, vol 90, (2018) pp 015002-1 (arxiv version).

When explaining the unitary interpolation technique, the authors begin with:

The initial Hamiltonian is chosen such that its ground state is the uniform superposition state [$|\phi\rangle = |+\rangle^{\otimes n}$], i.e., $H(0) = w\sum^{n}_{i = 1}|-\rangle_i \langle-|$, where $w$ is the energy scale.

  • How can I calculate the ground state of $H(0)$?

Also, I have seen, but don't remember exactly where, an observation that $H(0)$ is introduced in a such a way that a penalty is provided for any state having a contribution of $|-\rangle$. What does it mean?

Then the paper go on and states that:

An adiabatic implementation requires a final Hamiltonian $H(1)$ such that its ground state is $|\Psi(1)\rangle = U|\Psi(0)\rangle$, and that this can be accomplished via a unitary transformation of $H(0)$, i.e. $H(1) = UH(0)U^\dagger$.

where $U$ is a diagonal matrix such that:

$$U = diag[(-1)^{f(0)}, \dots,(-1)^{f(2^n-1)}] $$

At this point, I don't see why bother with this; since to arrive at the answer using adiabatic quantum computation I need to construct a unitary in such a way that I will already have the answer if $f$ is balanced or constant. Am I overlooking anything?


Trying to workout an example, setting $n = 1$ and making $f(x) = 1$ (constant 1).

$$H(0) = w|-\rangle\langle-| = w\pmatrix{1 & -1 \\ -1 & 1}$$

I will set $w = 1$ to get it out of the way. Then,

$$U = \pmatrix{-1 & 0 \\ 0 & -1} $$ $$H(1) = UH(0)U^\dagger = \pmatrix{-1 & 0 \\ 0 & -1}\pmatrix{1 & -1 \\ -1 & 1}\pmatrix{-1 & 0 \\ 0 & -1}$$ $$H(1) = \pmatrix{1 & -1 \\ -1 & 1}$$

Meaning that the ground state of $H(1)$ is $|+\rangle$.

If now I do the function $f(x) = x$, then $$U = \pmatrix{1 & 0 \\ 0 & -1} $$

and

$$H(1) = \pmatrix{1 & 1 \\ 1 & 1}$$

which, I suppose, has ground state $|-\rangle$. And with this we can differentiate between a constant and a balanced function $f$.