3
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ 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 May 25 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 May 25 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 May 25 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 May 25 at 19:45
2
$\begingroup$

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.

| improve this answer | |
$\endgroup$
  • $\begingroup$ 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 May 25 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 May 25 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 May 25 at 18:06
  • 1
    $\begingroup$ 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 May 25 at 18:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.