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The Hamiltonian operator is defined on the graph $G$ as $H_{A}(t) = \exp(itA)$ where $A$ is the adjacency matrix of the graph $G$. It is said that this operator is a transition matrix and represents the continuous-time quantum walk on the graph $G$.

I am not familiar with the concepts in quantum theory and I am curious to know:

  1. the meaning and idea behind this operator (as a transition matrix, for example when we have $H_{A}(t) z$ where $z$ is a vector) in an understandable and clear way.

  2. What is the impact of adjacency matrix in those transitions (I mean what happen on the vertices)?

I appreciate any help.

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In Quantum Mechanics, we describe the state of the system as a vector in some vector space. For example, one might have a 2-state system with a state vector: $$\psi = \left( \begin{matrix} x\\y\end{matrix} \right)$$

The time derivative of a vector is defined by the multiplication of a matrix (which is called the Hamiltonian) with the vector. This is written as:

$$ \dot{\psi} = i H \psi $$

This is the famous Schrodinger's equation. So over time, we would have:

$$ \psi(t) = \exp(i H t) \psi(t=0) $$

This is a continuous process. For example for a Hamiltoninan:

$$ H = g \left( \begin{matrix}{0 \ 1}\\{1\ 0} \end{matrix} \right)$$

The solution would be: $$\psi(t) = \exp(i H t) \left( \begin{matrix} 1\\0\end{matrix} \right)= \cos(g t) \left( \begin{matrix} 1\\0\end{matrix} \right) + i \sin (g t) \left( \begin{matrix} 0\\1\end{matrix} \right)$$

So over time, there is a continuous transition between the two states, defined by the "adjacency matrix" H. Notice that there are some $i$'s wondering around. In physics their meaning is a bit tricky, but you should only think of the "weight" each state has by the end, ignoring the complex phase. The total weight will always be conserved.

The adjacency matrix describe exactly which transition will occur. A state vector describe a "weight of being" in each vertex, and using the adjacency as the Hamiltonian, means that states will evolve to their adjacent vertecies. You will get a similar result for more complex matrices.

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  • $\begingroup$ thanks. But as I said, I am not familiar with this area and the notations you used is quite unclear for me. Would you please write your answer in more clear way to me? $\endgroup$
    – Amin
    Commented Feb 6, 2019 at 14:57
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    $\begingroup$ The vector $\vec z$ mentioned in your question is usually represented with a notation called "bra-ket" notation invented by Paul Dirac as $|z\rangle$, which is an element (ket) of a Hilbert space, and its dual $\langle z|$ (bra). $|0\rangle$, $|1\rangle$ denote the standard basis elements (sometimes with different interpretations in physics), and $|x\rangle\langle y|$ denotes the tensor product $\vec x \otimes \vec y$. $\endgroup$
    – GodotMisogi
    Commented Feb 6, 2019 at 15:02
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    $\begingroup$ I tried again, I hope this is better. $\endgroup$
    – Yotam Vaknin
    Commented Feb 6, 2019 at 15:09
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    $\begingroup$ g is some constant so $g t$ would have units of angles (and it is the transition rate, in the physical system). The last equality requires some boring calculations (which every QM course forces you to do), I might missed a sign somewhere. You can just exponentiate the matrix yourself (as a series) using Mathematica or something. $\endgroup$
    – Yotam Vaknin
    Commented Feb 6, 2019 at 15:31
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    $\begingroup$ The last line shows the transition from (1,0) to a combination of (1,0) and (0,1). For example for $t g = \pi/2$ gives a transition from (1,0) to (0,1). $\endgroup$
    – Yotam Vaknin
    Commented Feb 6, 2019 at 15:32

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