VQE: Can I build a non-hermitian Hamiltonian with just Pauli matrices?

Question

From the VQE paper they claim that a Hamiltonian can be expressed as a polynomial series of pauli operators (equation 1).

While coding up VQE from scratch I made a function which would allow me to specify coefficients up to 2nd order to build the corresponding Hamiltonian (for 1 qubit).

But I noticed that $\sigma_y\sigma_z$ is in fact not hermitian, and so it doesn't give me purely real energy eigenvalues.

So is it not true the other way around? Can I not specify an arbitrary polynomial series of Pauli operators such that the result is a Hamiltonian for a closed system?

EDIT

See the accepted answer. I actually misunderstood the equation in the paper, not realising that the higher order terms were actually tensor products and only applicable to more-than-single-qubit systems.

Answer 1

The Hamiltonian of the closed system is by definition a Hermitian operator. A quote from M. Nielsen and I. Chuang textbook page 82:

Postulate 2': The time evolution of the state of a closed quantum system is described by the Schrödinger equation,

$$i \hbar \frac{d |\psi\rangle}{dt} = H |\psi\rangle$$

In this equation, $\hbar$ is a physical constant known as Planck’s constant whose value must be experimentally determined. The exact value is not important to us. In practice, it is common to absorb the factor $\hbar$ into $H$, effectively setting $\hbar$ = 1. $H$ is a fixed Hermitian operator known as the Hamiltonian of the closed system.

The operator $\sigma_y \otimes \sigma_z$ is Hermitian, because $(\sigma_y \otimes \sigma_z)^\dagger = \sigma_y^{\dagger} \otimes \sigma_z^{\dagger} = \sigma_y \otimes \sigma_z$

For decomposing the Hamiltonian matrix into the sum of Pauli terms look in this thread.

About non-Hermitian Hamiltonians (they are not conventional Hamiltonians) can be found in this answer.