Confusion regarding second quantised notation Suppose Hamiltonian of the system in 2nd quantised notation is
$H= t\sum_{x=1}^{N}d_{x}^{\dagger}d_{x}$.
Does this mean that eigenstates of the Hamiltonian is $N$-fold degenerate with energy $t$?
 A: Yes, indeed, there are $N$ identical states, each having energy $t$.
Remark:
Note that usually $t$ is used for a hopping integral, as in
$$
H= \epsilon_0\sum_{x=1}^Nd_x^\dagger d_x + t\sum_{x=1}^N\left(d_x^\dagger d_{x+1} + h.c.\right)
$$
Remark
In view of the comments and the other answer, it is necessary to note that what we mean here depends on the context. Specifically:

*

*one may speak about a single-particle Hamiltonian in second quantized notation (or simply the fact that the system is composed of $N$ identical orbitals), which si how I understood the OP

*one may also speak about many-body states, as suggested by the use of the second quantization.

Note however that second quantization is widely used to deal with one-particle problems, so one cannot reasonably claim that only one interpretation is correct.
A: The degeneracy of the state with energy $nt$ is the number of ways you can make the integer $n$ out of $N$ numbers eg if $N=3$ and then we can get $4t$ as
$$
4+0+0\\
0+4+0\\
\vdots
$$
etc, and
$$
3+1+1\\
1+3+1\\
\vdots
$$
or
$$
2+0+2\\
0+2+2\\
\vdots
$$
I leave it to you to compute the degeracy of the level $E=nt$!
