1
$\begingroup$

The Adiabatic Quantum Computer is implemented by slowly increasing the parameter s from 0 to 1 in the intermediate Hamiltonian $[\hat{H}(s) = \hat{H}_{input} + (1-s)\hat{H}_{init} + s\hat{H}_{circuit}]$

How to show that this process evolves the state within the manifold spanned by $\{ |\alpha(0)\rangle|0_c\rangle,|\alpha(1)\rangle|1_c\rangle...|\alpha(L)\rangle|L_c\rangle$. In other words I meant to say how to show that $\hat{H}(s)$ does not map any state in this manifold to a state outside this manifold.

where $|\alpha(L)\rangle$ is is the data qubit state after applying the $l_{th}$ logic gate ,i.e $|\alpha(l)\rangle = \hat{U}_l...\hat{U}_1|0\rangle^{\otimes n}$

$\hat{H}_{input} = \sum_{i=1}^n|1_i\rangle\langle1_i|\otimes|0_c\rangle\langle0_c|$

$\hat{H}_{init} = \sum_{j=1}^L|j_c\rangle\langle j_c|$

$\hat{H}_{circuit} = \sum_{l=1}^L \hat{I} \otimes |(l-1)_c\rangle\langle(l-1)_c| - \hat{U}_l \otimes |l_c\rangle\langle(l-1)_c| - \hat{U}_l^\dagger \otimes |(l-1)_c\rangle\langle l_c|+ \hat{I} \otimes |l_c\rangle\langle l_c|$

$\endgroup$
6
+50
$\begingroup$

This is a very particular application of Adiabatic Quantum Computing so I think it's worth briefly mentioning some context.

Roughly speaking, one wants to show that given a quantum circuit defined as a sequence of unitary gates $U_1,U_2,\ldots,U_L$ it is possible to define a version of the quantum adiabatic algorithm that reproduces (a state with a large overlap with) the output state of the circuit: $\lvert \alpha(L) \rangle = U_L \cdots U_1 \lvert 0 \rangle^{\otimes n}$. Here we are assuming the circuit is initialized in the state $\lvert 0 \rangle^{\otimes n}$ at the beginning of the computation.

The Hamiltonian $H(s)$ that does this is defined over a Hilbert space $\mathcal{H} = \mathcal{H}_{comp} \otimes \mathcal{H}_{clock}$ that is the tensor product of a "computational" register (which is the same Hilbert space over which the quantum circuit acts) and an auxiliary "clock" register. The standard reference for these constructions is the article Adiabatic Quantum Computation is Equivalent to Standard Quantum Computation.

Now, in answering your question I will assume you've chosen your clock states $\lvert l_c \rangle$ to be orthonormal (as you should have). Following the notation of the reference above, define

$$ \lvert \gamma_l \rangle \equiv \lvert \alpha(l) \rangle \lvert l_c \rangle $$ Then your question $H(s)\lvert \gamma_l \rangle \in \mathcal{S}_0$ where \begin{equation} \mathcal{S}_0 \equiv \operatorname{span}\Big\{ \lvert \gamma_l \rangle \;\Big| \;l=0,1,\ldots,L \Big\} \end{equation} boils down to showing that for any choice of $0 \leq s \leq 1$ and $l=0,1,\ldots,L$ the state $ H(s) \lvert \gamma_l \rangle $ can be written as a linear combination of the states $\{\lvert \gamma_l \rangle \mid l=1,0\ldots,L\}$.

Since $H(s)$ is a sum of terms proportional to $H_{input}$, $H_{init}$ and $H_{circuit}$, it is sufficient to show that the states \begin{equation} H_{input}\lvert \gamma_l \rangle, \qquad H_{init}\lvert \gamma_l \rangle, \qquad \text{and} \qquad H_{circuit}\lvert \gamma_l \rangle \end{equation} are all linear combinations of the $\lvert \gamma_l \rangle$s. This can be done by inspection:

$$ H_{input}\lvert \gamma_l \rangle = \sum_{i=1}^n \lvert 1_i \rangle \langle 1_i | \alpha(l) \rangle \otimes \lvert 0_c \rangle \langle 0_c | l_c \rangle = \mathbf{0} $$ because if $l \neq 0$ then $\langle 0_c | l_c \rangle=0$ while for $l=0$ we have $\lvert \alpha(0) \rangle = \lvert 0 \rangle^{\otimes n}$ so all the single-qubit projectors $\lvert 1_i \rangle \langle 1_i \rvert$ will annihilate the state $\lvert \alpha(0) \rangle$. Then \begin{equation} H_{init} \lvert \gamma_l \rangle = \lvert \alpha(l) \rangle \otimes \sum_{j=1}^L \lvert j_c \rangle \langle j_c | l_c \rangle = \lvert \alpha(l) \rangle \lvert l_c \rangle \end{equation} since here the sum is over all clock states (which are orthonormal) and $\lvert l \rangle$ is one of them. Finally for $H_{circuit}$ you can split the terms \begin{align} \sum_{k=1}^L \lvert \alpha(l) \rangle \otimes \lvert (k-1)_c \rangle \langle (k-1)_c | l_c \rangle &= \begin{cases} 0 & \text{if $l=L$}\\ \lvert \alpha(l) \rangle \lvert l_c \rangle & \text{otherwise}\\ \end{cases}\\ \sum_{k=1}^L U_l \lvert \alpha(l) \rangle \otimes \lvert k_c \rangle \langle (k-1)_c | l_c \rangle &= \begin{cases} 0 & \text{if $l=L$}\\ \lvert \alpha(l+1) \rangle \lvert (l+1)_c \rangle & \text{otherwise} \end{cases}\\ \sum_{k=1}^L U^{\dagger}_l \lvert \alpha(l) \rangle \otimes \lvert (k-1)_c \rangle \langle k_c | l_c \rangle &= \begin{cases} 0 & \text{if $l=0$}\\ \lvert \alpha(l-1) \rangle \lvert (l-1)_c \rangle & \text{otherwise} \end{cases}\\ \sum_{k=1}^L \lvert \alpha(l) \rangle \otimes \lvert k_c \rangle \langle k_c | l_c \rangle &= \begin{cases} 0 & \text{if $l=0$}\\ \lvert \alpha(l) \rangle \lvert l_c \rangle & \text{otherwise}\\ \end{cases} \end{align} so that you have \begin{equation} H_{circuit} \lvert \gamma_l \rangle = \begin{cases} \lvert \alpha(0) \rangle \lvert 0_c \rangle - \lvert \alpha(1) \rangle \lvert 1_c \rangle& \text{if $l=0$}\\ 2\lvert \alpha(l) \rangle \lvert l_c \rangle - \lvert \alpha(l+1) \rangle \lvert (l+1)_c \rangle - \lvert \alpha(l-1) \rangle \lvert (l-1)_c \rangle & \text{if $0 < l < L$}\\ \lvert \alpha(L) \rangle \lvert L_c \rangle - \lvert \alpha(L-1) \rangle \lvert (L-1)_c \rangle & \text{if $l=L$} \end{cases} \end{equation} so in all cases $H_{circuit} \lvert \gamma_l \rangle$ is a linear combination of the $\lvert \gamma_l \rangle$ states. Putting everything together observe that \begin{equation} H(s)\lvert \gamma_l \rangle = (1-s)\lvert \gamma_l \rangle + sH_{circuit} \lvert \gamma_l \rangle \end{equation} so $H(s)\lvert \gamma_l \rangle \in \mathcal{S}_0$ and this completes the proof.

$\endgroup$
2
  • $\begingroup$ +1 and welcome to our community! We hope to see much more of you, and thank you for your contribution! Not a bad first answer!! $\endgroup$ Mar 26 at 20:59
  • $\begingroup$ @user1271772 thanks! I will try my best :) $\endgroup$ Mar 27 at 17:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.