3 typos corrected

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 $$f$$.