Projection Method in Hubbard model This is a question from Altland and Simons book "Condensed Matter Field Theory".
In the second exercise on page 64, the book claims that if we define $\hat P_s, \hat P_d$ to be the operators that project onto the singly and doubly occupied subspaces respectively, then at half-filling the following equality holds
$\hat P_s \hat H_t \hat P_s = 0$ and $\hat P_d \hat H_t \hat P_d =0$, 
where $\hat H_t$ is the hopping term in the Hubbard model. I am having trouble to see why the second equality is true. In the doubly occupied subspace, it is not guaranteed that the hopping will produce only singly occupied states. Do I have to write out the projection operator in terms of creation and annihilation operators and directly calculate?
Thanks in advance.
 A: Note that this is a two-site problem, so the many-body Hilbert space is finite (and actually just four). Therefore one can simply express all the operators in terms of $4\times 4$ matrices. Let us order the many-body basis states as $|s_1\rangle=a_{1\uparrow}^\dagger a_{2\downarrow}^\dagger|\Omega\rangle$, $|s_2\rangle=a_{2\uparrow}^\dagger a_{1\downarrow}^\dagger|\Omega\rangle$, $|d_1\rangle=a_{1\uparrow}^\dagger a_{1\downarrow}^\dagger|\Omega\rangle$, $|d_2\rangle=a_{2\uparrow}^\dagger a_{2\downarrow}^\dagger|\Omega\rangle$. Then the hopping term $H_t=-t\sum_{\sigma=\uparrow,\downarrow}a_{1\sigma}^\dagger a_{2\sigma}+h.c.$ is represented as
$$H_t=\left[\begin{matrix}
 0 & 0 &-t&-t\\
 0 & 0 & t& t\\
-t & t & 0& 0\\
-t & t & 0& 0
\end{matrix}\right]$$
Obviously, hopping only connects singly  occupied subspaces to doubly occupied subspaces, and vice versa. The projection operators also have matrix representations,
$$P_s=\left[\begin{matrix}
 1 & 0 & 0 & 0 \\
 0 & 1 & 0 & 0 \\
 0 & 0 & 0 & 0 \\
 0 & 0 & 0 & 0
\end{matrix}\right], \quad 
P_d=\left[\begin{matrix}
 0 & 0 & 0 & 0 \\
 0 & 0 & 0 & 0 \\
 0 & 0 & 1 & 0 \\
 0 & 0 & 0 & 1
\end{matrix}\right]$$
Just by matrix multiplication, it can be verified that $P_s^2=P_d^2=1$, $P_sP_d=0$, and $H_t=P_sH_tP_d+P_dH_tP_s$, as well as $P_sH_tP_s=P_dH_tP_d=0$. These are useful results for the following deductions in the book.
