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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.

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    $\begingroup$ Alexander, the equation (1) (or slight modification of it) is applicable also for one qubit case: for one qubit $H = \sum_{\alpha} h_{\alpha} \sigma_{\alpha} = h_i I + h_x \sigma_x + h_y \sigma_y + h_z \sigma_z$. $\endgroup$
    – Davit Khachatryan
    Commented May 25, 2020 at 18:28
  • $\begingroup$ And in the second (similarly for the third) sum of the equation (1) we don't have separate $h_{\alpha}$ and $h_{\beta}$, instead, there should be $h_{\alpha \beta}^{ij}$ that is not (necessarily) equal to $h_{\alpha}^i \cdot h_{\beta}^j$. $\endgroup$
    – Davit Khachatryan
    Commented May 25, 2020 at 18:41
  • $\begingroup$ @DavitKhachatryan I'm pretty sure I applied my erroneous thinking in the middle of transcribing it which further reinforced said thinking, lol $\endgroup$
    – Alexander Soare
    Commented May 25, 2020 at 19:22
  • $\begingroup$ Alexander, it is ok :). Sometimes the notations are not clear. BTW here is my Qiskit implementation/tutorial for one qubit VQE that might be interesting: github.com/DavitKhach/quantum-algorithms-tutorials/blob/master/… $\endgroup$
    – Davit Khachatryan
    Commented May 25, 2020 at 19:45

1 Answer 1

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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.

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  • $\begingroup$ I could be confused here, but I'm not actually referring to the tensor product (which would be for the 2-qubit case). I'm talking about the operator which is formed by the sequential application of $\sigma_z$ then $\sigma_y$, which is what I understand the paper means with equation (1) from my original question. $\endgroup$
    – Alexander Soare
    Commented May 25, 2020 at 17:57
  • $\begingroup$ @AlexanderSoare, I know the paper and how I understand sometimes/often people drop the $\otimes$ symbol, and in this particular case they also have dropped the $\otimes$ symbol. So in equation (1) they have tensor products. $\endgroup$
    – Davit Khachatryan
    Commented May 25, 2020 at 18:02
  • $\begingroup$ A quote from the paper: "Any Hamiltonian may be written as", it already means that the equation (1) is not written only for the one qubit case. $\endgroup$
    – Davit Khachatryan
    Commented May 25, 2020 at 18:06
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    $\begingroup$ Ah understood, I had the whole premise wrong. Thank you! I will accept your answer although I suppose it may make sense (if you have time) to add an edit fully clarifying this last part for anyone else who might have my issue. $\endgroup$
    – Alexander Soare
    Commented May 25, 2020 at 18:18

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