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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:

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


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

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?

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up vote 3 down vote accepted

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.

In other words, the Hamiltonians are just projectors to states which are orthogonal to those we want to be the lowest energy eigenstates, making the orthogonal states to have eigenvalue 1 and the lowest energy eigenstates to have eigenvalue 0.

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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? – Omar Shehab Jul 24 '13 at 0:48
I don't know what is the "standard" way, but in 2D it is a pretty straightforward way. In higher dimensions all other states will have eigenvalue 1 and the Hamiltonian will be degenerate. Whether this is a problem or not depends on what you want to do. – Bubble Jul 24 '13 at 1:06

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