What happened to the 1/2 in the MAXCUT Problem Unitary for QAOA

Question

I recently started to begin with QuantumComputing following the Qiskit tutorial. I worked through the basics and now reached Chapter 4.1.3 (https://qiskit.org/textbook/ch-applications/qaoa.html).

I always try to comprehend the math behind the examples, but at this point I don't understand it. In fact, the step when they construct the problem unitary based on the hermitian.

The hermitian is $H_p=\frac{1}{2}(Z_0⊗Z_1⊗I_2⊗I_3)+\frac{1}{2}(I_0⊗Z_1⊗Z_2⊗I_3)+...$.

When constructing the unitary the factor 1/2 is just ignored which results in $U(H_p)=e^{-i\gamma H_Pi}=e^{-i\gamma Z_0Z_1}e^{-i\gamma Z_1Z_2}e^{-i \gamma Z_2Z_3}e^{-i \gamma Z_3Z_0}$.

What happened to the 1/2? It probably doesn't influence the outcome. But I can't understand exactly why. When considering a simple part of the unitary like $e^{-i\gamma Z_0Z_1}$, this can represented by a $CNOT$ followed by $R_z(2\gamma)$ and another $CNOT$. As I understand this, the argument is in fact important and influences the outcome. Maybe this is an obvious question for some, but as Im relatively new to the topic, any help would be appreciated.

Thanks in advance :-)

Answer 1

You are right! To be precise, in the equation $$U\left(H_P\right)=e^{-i \gamma H_P}=e^{-i \gamma Z_0 Z_1} e^{-i \gamma Z_1 Z_2} e^{-i \gamma Z_2 Z_3} e^{-i \gamma Z_0 Z_3}$$ the parameter $\gamma$ should be replaced by $\gamma/2$. The factor of $1/2$ is ignored because parameter $\gamma$ will be optimized, making it insignificant.

This is basically saying that finding the minimum of some function $f(x/2)$, $x\in \mathbb{R}$ is the same as finding the minimum of $f(x)$, $x\in \mathbb{R}$.