The Exercise

A system of $n$ qubits has a state space of dimension $2^n$. Its operators are $2^n\times 2^n$ complex matrices, and its observables are the Hermitian ones, with $H^\dagger = H$. The exercise asks how many real numbers it takes to specify such a matrix when its trace is zero, and the answer is $4^n-1$.^✦

Hermitian Matrices and the Trace

Fix a size $d$. The Hermitian $d\times d$ complex matrices are closed under sums and real multiples, and form a real vector space $\mathrm{Herm}(d)$. The real trace $\tau_d$ on $\mathrm{Herm}(d)$ is the real linear functional that sends a Hermitian matrix $H$ to the real part of its trace: $$\tau_d(H) = \operatorname{Re}\operatorname{tr}H = \operatorname{Re}\sum_{i=1}^{d} H_{ii}.$$(1)^✦

The space $\mathcal{T}_d$ of traceless Hermitian $d\times d$ matrices is the real subspace of $\mathrm{Herm}(d)$ on which the real trace vanishes, the kernel of $\tau_d$: $$\mathcal{T}_d = \ker\tau_d = \{\, H\in\mathrm{Herm}(d) : \operatorname{Re}\operatorname{tr}H = 0 \,\}.$$(2)^✦

The Claim

For $n$ qubits the size is $d = 2^n$.

What Is Checked

The claim is stated over the space $\mathcal{T}_{2^n}$ of (2), cut out of $\mathrm{Herm}(2^n)$ by the real trace (1), with its dimension taken over the real numbers. About these constructions the library proves the dimension count (3), for each number of qubits.