I was reading the paper Quantum computation, quantum state engineering, and quantum phase transitions driven by dissipation . The claim is that universal quantum computation can be achieved using the dissipative approach. That is, we let interactions with an environment drive quantum computation. For this, we need to find a Liouvillian in Lindblad form: \begin{equation} \mathcal{L}(\rho) = \sum_k L_k \rho L_k^{\dagger} - \frac{1}{2} \left \{ L_k^{\dagger}L_k, \rho \right \}. \end{equation} In the above equation, $\rho$ is a density matrix, $\{\cdot,\cdot \}$ is an anti-commutator, and $L_k$ are Lindblad operators representing couplings to an environment.

The quantum computation that should be realized through dissipation is $$|\psi_t\rangle = U_t U_{t-1} \cdots U_1 |0\rangle_1 \cdots |0\rangle_N.$$

To achieve this computation, the authors define two registers:

  1. The main register is initialized as $|0\rangle_1 \ldots |0\rangle_N$.
  2. An auxiliary time-register with states $\{|t\rangle\}_{t=0}^T$.

Given the definitions above, the authors choose the following Lindblad operators:

\begin{align} \tag{1} L_i &= |0\rangle_i\langle1| \otimes |0\rangle_t\langle 0|, \\ \tag{2} L_t &= U_t \otimes |t+1\rangle \langle t| + U^{\dagger}_t\otimes |t\rangle\langle t+1|, \end{align} where $i=1,\ldots,N$ and $t=0,\ldots, T$.

The simple challenge I'm facing:

I absolutely don't understand the sub-index notation on LHS and RHS of Eq (1) and Eq (2). To me, it seems they defined $L_i$ and then redefined it with $L_t$, which has a different meaning. I also don't understand the meaning of subindex in $|0\rangle_i$ and $|0\rangle_t$.

Moreover, since $i \in \{1, \ldots, N\}$ and $t \in \{0,\ldots, T\}$, we can easily have something like $k:=i=t$, then how would one understand what equation $L_k$ represents?

  • 1
    $\begingroup$ This reminds me of Feynman's clock Hamiltonian $H=\sum_{i=0}^{k-1}q_{i+1}^\dagger q_i A_{i+1}$ (see e.g. page 15 in "Quantum Mechanical Computers") where $q$ is the annihilation operator for the program counter register and $A_i$ are a sequence of operations to perform. $\endgroup$ Jan 6, 2023 at 20:44

1 Answer 1


The notation is perhaps a little lazy. There ar two distinct sets of Lindblad operators: $\{L_i\}$ and $\{L_t\}$ and you need both. One does not over-write the other. Yes, it's unfortunate that if you literally put indices in, both have an $L_1$, but you do need both.

As for the indices, $$ |0\rangle_i\langle 1|\otimes |0\rangle_t\langle 0| $$ simply means the operator that acts as $|0\rangle\langle 1|$ on qubit $i$ on the main register ($I$ on every other qubit) and $|0\rangle\langle 0|$ on the whole state of the time register.

Similarly, $$ U_t\otimes |t+1\rangle\langle t| $$ means apply unitary $U_t$ on the main register and, at the same time, change the state of the time register from $|t\rangle$ to $|t+1\rangle$.

  • $\begingroup$ Thanks! If $|0\rangle_i\langle 1|$ acts on $i$th qubit, then $|0\rangle_t \langle 0 |$ should also act on $t$th qubit of time register? $\endgroup$
    – MonteNero
    Jan 6, 2023 at 14:49
  • $\begingroup$ The time register is not explicitly set up as separate qubits. It just has a set of basis elements 0 to T. $\endgroup$
    – DaftWullie
    Jan 6, 2023 at 15:05
  • $\begingroup$ Should it be the tensor product of two outer products - i.e. $|0\rangle_i\langle 1|\otimes |0\rangle_t\langle 0|$ instead of $|0\rangle_i|1\rangle\otimes |0\rangle_t\langle 0|$? $\endgroup$ Jan 18, 2023 at 22:02
  • $\begingroup$ @MarkS Yes it should! $\endgroup$
    – DaftWullie
    Jan 19, 2023 at 9:21

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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