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How can I classify all two-qubit Hamiltonians $H$ such that $e^{-iHt}|00\rangle$ is unentangled for all times $t?$

This question is warm-up for several related questions of mine: 816017, 797030, but I also find it interesting in its own right.

It is easy to find several (overlapping) classes of Hamiltonians that leave $|00\rangle$ unentangled.

These include

  1. Completely separable time evolution $\{O_1 \otimes I + I \otimes O_2\}$
  2. Right spin is fixed $\{O_1 \otimes I + O_2 \otimes \sigma^z\}$
  3. Left spin is fixed $\{I \otimes O_1 + \sigma^z \otimes O_2 \}$
  4. Hamiltonians of the form $P_{00} O_{12} P_{00} + (1-P_{00}) O_{12} (1-P_{00})$, which include, for example, $\sigma^x \otimes \sigma^x + \sigma^y \otimes \sigma^y$.

Here $O_1$ and $O_2$ are arbitrary single-qubit operators, and $O_{12}$ is an arbitrary two-qubit operator. $P_{00} = |00\rangle \langle 00|$ is the projector onto $|00\rangle$; 4 is the class of $H$ with $|00\rangle$ as an eigenstate.

It is rather tempting that the union of these classes captures all Hamiltonians that leave $|00\rangle$ unentangled. However, I am unsure how to prove this.

To check for separability, we can use that the purity of the reduced density matrix is $1$. For the case of a pure state $a|00\rangle + b|01\rangle + c|10\rangle + d|11\rangle$ of two qubits, there is an even simpler constraint that $ad=bc$.

Thus separability of $e^{-iHt}|00\rangle$ means that

$$\langle 00| e^{-iHt}|00\rangle\langle 11| e^{-iHt}|00\rangle = \langle 10| e^{-iHt}|00\rangle\langle 01| e^{-iHt}|00\rangle$$

To get this true for all $t$, we can Taylor expand and equate coefficients. That is, for all $n$,

$$\sum_{k=0}^n \binom{n}{k} \langle 00| H^k |00\rangle\langle 11| H^{n-k}|00\rangle = \sum_{k=0}^n \binom{n}{k} \langle 10| H^k |00\rangle\langle 01| H^{n-k}|00\rangle$$

For the first few $n$, these constraints are, after rearranging a little

$$\langle 11| H|00\rangle = 0$$ $$\langle 11|H^2|00\rangle = 2 \langle 01| H |00\rangle \langle 10| H |00\rangle $$ $$\langle 11|H^3|00\rangle = 3 \langle 01| H^2 |00\rangle \langle 10| H |00\rangle + 3 \langle 01| H |00\rangle \langle 10| H^2 |00\rangle - 6\langle 00|H|00\rangle\langle 10|H|00\rangle\langle 01|H|00\rangle$$

However, I do not see a nice way of using these constraints to rule out every other Hamiltonian besides the ones in sets 1 to 4. In particular, the constraints become nonlinear in the entries of $H$, and solving elements of $H$ via the constraints starts to give rather cumbersome expressions already by $n=5$.

user196574
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1 Answers1

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Following ideas from this paper, the most general non-entangling operation on $\vert 00 \rangle$ is a product of two unitaries $$ U = (W_1 \otimes W_2 ) (1 \oplus V), $$ where $W_1$ acts on qubit 1 and $W_2$ on qubit 2, $W_1, W_2 \in \text{SU(2)}$. $1 \oplus V$ is a shorthand for the unitary that leaves $\vert 00 \rangle$ invariant (I absorbed a possible global phase $e^{i\phi}$ into $W_1$) and applies a unitary $V\in \text{U(3)}$ on the subspace $\text{span}\{\vert 01\rangle, \vert 10\rangle, \vert 11\rangle\}$. The generators of $W_{1/2}$ are exactly the Paulis acting on only one party (i.e. your class 1) and the generators of $V$ are the 8 Gell-Mann matrices plus the 3-dimensional identity operator. In order to not have redundant generators, you might want to divide out the subgroup $\text{SU(2)} \subset \text{U(3)}$, which only acts on qubit 2, as well as the three generators

$${\rm diag}(1,0,0) \qquad \begin{bmatrix} 0 & 0 & 1 \\0 & 0 & 0 \\1 & 0 & 0 \\ \end{bmatrix} \qquad \begin{bmatrix} 0 & 0 & -i \\0 & 0 & 0 \\ i & 0 & 0 \\ \end{bmatrix}$$

as these are effective single-qubit operations. Let's call $\sigma_a^{(i)}, a = 1,2,3$, the Paulis on qubit $i$ and $\lambda_\mu, \mu=1, ..., 3$, the generators of $\text{U(3)} / (\text{SU(2)} \times \text{SU}(2))$. Then, the most general time-dependent Hamiltonian leaving $\vert 00 \rangle$ unentangled could be written as $$ H(t) = {\rm rect}(t) \sum_{i=1,2 \\ a = 1,2,3} c_{a}^{i} \sigma_a^{(i)} + {\rm rect}(t+1) \sum_{\mu = 1}^3 c_\mu \lambda_\mu, $$ with real coefficients $c_a^i$ and $c_\mu$ and the rectangular function, which just toggles the gates from the product formula above. If you insist on having a time-independent Hamiltonian, you can always formally define the logarithm of the product formula, which will be nasty, but well-defined: $$ H = \frac{i}{2} \log \left[ \mathcal{T}\exp \left( -i \int_{-1}^1 dt H(t) \right) \right]. $$ If you add a layer of single-qubit gates in the beginning, you are back to the most generic $\text{SU(4)}$ gate.

Refik Mansuroglu
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