# time-dependent Hartree-Fock for two-component bosons

How does the ansatz for the time-dependent Hartree-Fock wavefunction look like in second quantization if we have two-component boson system and in one case the Hamiltonian commutes with number of particles in each component $[\hat{H}, \hat{N}_a] = [\hat{H}, \hat{N}_b] = 0$, but in the second case it doesn't? For example, the Hamiltonian may contain the following term: $$\hat{\Psi}_a^{\dagger}\hat{\Psi}_a^{\dagger}\hat{\Psi}_b\hat{\Psi}_b + \text{h.c.}$$

For the case when number of particle of each flavor are conserved. Assuming ${\Psi_{a/b}^{\dagger}(p/q)}$ denote set of creation operators for creating bosons of flavors a/b to create in state p/q. Dimensionality of basis for each boson flavor single particle space assumed to be $M_{a/b}$. Assuming there are $N_{a/b}$ number of bosons of flavor a/b in the system under consideration. Time dependent Hartree-Fock ansatz wavefunction (best possible "Slater permanent") in second quantized form can be expressed as :
$$|\Psi(t)\rangle=\prod_{k_a=1}^{N_a}\prod_{k_b=1}^{N_b}\sum_{m_{k_a}=1}^{M_a}\sum_{n_{k_b}=1}^{M_b}U_{m,k_a}(t)V_{n,k_b}(t)\Psi_{a}^{\dagger}(m_{k_a})\Psi_{b}^{\dagger}(n_{k_b})|\text{vac}\rangle$$ where $|\text{vac}\rangle$ is the reference vaccuum state.
The equation of motion for the matrix elements $U_{m,k_a}(t)$ and $V_{n,k_b}(t)$ can be derived by applying the Dirac-Frenkel variational principle, namely : $$\mathop{\text{Extremize}}_{{{U_{m,k_a}(t)},{V_{n,k_b}(t)}}}\langle\Psi(t)|i\hbar\frac{d}{dt}-H_\text{system}^{}|\Psi(t)\rangle.$$ To simplify the above expectation value, you have to use Wick's theorem before extremization.
For the case when bosons of different flavor are not conserved, it is redundant to describe the bosons of two kinds and the ansatz : $$|\Psi(t)\rangle=\prod_{k=1}^{N_\text{tot}}\sum_{m_{k}=1}^{M_\text{tot}}U_{m,k}(t)\Psi_{}^{\dagger}(m_{k})|\text{vac}\rangle$$ should be the answer.