Suppose, as an example, that we have six fermions below the Fermi level.
This means that we can make at most \( 6p-6h \) excitations. If we have an infinity of single particle states above the Fermi level, we will obviously have an infinity of say \( 2p-2h \) excitations. Each such way to configure the particles is called a
configuration. We will always have to truncate in the basis of single-particle states.
This gives us a finite number of possible Slater determinants. Our Hamiltonian matrix would then look like (where each block can have a large dimensionalities):
| | \( 0p-0h \) | \( 1p-1h \) | \( 2p-2h \) | \( 3p-3h \) | \( 4p-4h \) | \( 5p-5h \) | \( 6p-6h \) |
| \( 0p-0h \) | x | x | x | 0 | 0 | 0 | 0 |
| \( 1p-1h \) | x | x | x | x | 0 | 0 | 0 |
| \( 2p-2h \) | x | x | x | x | x | 0 | 0 |
| \( 3p-3h \) | 0 | x | x | x | x | x | 0 |
| \( 4p-4h \) | 0 | 0 | x | x | x | x | x |
| \( 5p-5h \) | 0 | 0 | 0 | x | x | x | x |
| \( 6p-6h \) | 0 | 0 | 0 | 0 | x | x | x |
with a two-body force. Why are there non-zero blocks of elements?