# Intuition behind Hamiltonian

I am reading this paper by Das et al. which converts Deutsch's algorithm into an adiabatic quantum algorithm. I don't get the intuition behind the initial and final Hamiltonians.

If defines the initial state vector as:

$$|\Psi_0\rangle = \frac{1}{\sqrt{2}} (|0\rangle + |1\rangle)$$

and the final state vector as:

$$|\Psi_1\rangle = \alpha |0\rangle + \beta |1\rangle$$

where,

$$\alpha = \frac{1}{2} |(-1)^{f(0)} + (-1)^{f(1)}| \\ \beta = \frac{1}{2} |(-1)^{f(0)} - (-1)^{f(1)}| \\ \alpha + \beta = 1 \\ \alpha^2 = \alpha \\ \beta^2 = \beta \\ \alpha \beta = 0$$

The definitions of vectors make sense to me. The initial one is just an easy to make state and the final one encodes the condition for the result of algorithm.

Here is the confusing part i.e. definitions of the Hamiltonians:

$$H_0 = I - |\Psi_0\rangle \langle \Psi_0| \\ H_1 = I - |\Psi_1\rangle \langle \Psi_1| \\$$

Why does the author subtract the product of the vectors from identity matrix to create the Hamiltonian? What is the physical significance of this subtraction? Why from identity matrix?

Notice that we want the system to remain in the lowest eigenstate of the Hamiltonian. Both Hamiltonians are two dimensional and their eigenstates span a two dimensional Hilbert space. The Hamiltonians contain a projector to $|\Psi_0\rangle$ ($|\Psi_1\rangle$). Define $|\Psi_0^O\rangle$ ($|\Psi_1^O\rangle$) as being normalized and orthogonal to $|\Psi_0\rangle$ ($|\Psi_1\rangle$). The identity may be written then as $$I=|\Psi_0^O\rangle\langle\Psi_0^O|+|\Psi_0\rangle \langle\Psi_0|=|\Psi_1^O\rangle\langle\Psi_1^O|+|\Psi_1\rangle\langle\Psi_1|$$. Insert that into the definition of the Hamiltonian and you will see that $|\Psi_0\rangle$ ($|\Psi_1\rangle$) are eigenstate of the first (last) Hamiltonian with eigenvalue 0, and their orthogonal vectors are eigenstates with eigenvalue 1. They have managed to make the states they wanted to be the lowest eigenstates of their Hamiltonians.
• Thanks for the detailed answer. So, the standard way to create a Hamiltonian, $H$, which has a state, $|\Psi\rangle$, as the lowest eigenstate is as follows: $$H = I - |\Psi\rangle \langle \Psi |$$. Am I right? Commented Jul 24, 2013 at 0:48