# Solving Hamiltonian eigenvalue problem

I would like to solve an eigenvalue problem of a Hamiltonian. I was able to find the lowest eigenvalue by converting the Hamiltonian into a matrix and applying linear algebra eigenvalue techniques. But this method is extremely cumbersome and does not generalize to arbitrary-sized Hamiltonians. I was hoping somebody could point to a more general approach. Here is the definition of the problem:

Let $$\vert \psi_N \rangle$$ denote the uniform superposition, $$\vert \psi_N \rangle = \frac{1}{\sqrt{N}}\sum^{N-1}_{i=0}\lvert i \rangle.$$ Then $$\vert \psi_N \rangle$$ is the ground state of the Hamiltonian $$H_0 = I - \lvert \psi_N \rangle \langle \psi_N \lvert$$ with the lowest eigenvalue $$0$$. Let $$\vert m \rangle = \vert 1 0...0 \rangle$$. Then it is the ground state of the Hamiltonian $$H_m = I - \vert m \rangle \langle m \vert$$.

For $$s \in [0,1]$$ define the Hamiltonian $$H(s) = (1-s)H_0 + s H_m.$$

What would be the general approach to solving the following eigenvalue problem for an arbitrary $$N$$ \begin{align} H(s) \lvert E_k, s \rangle = E_k(s) \lvert E_k, s\rangle \end{align} where $$E_k(s)$$ is the $$k$$th eigenvalue at time $$s$$.

I was able to solve the problem for $$N = 4$$ by converting the Hamiltonian into a matrix and then using computer algebra I got $$E_0(s) = \displaystyle \frac{1}{2} - \frac{\sqrt{3 s^{2} - 3 s + 1}}{2}.$$ The problem with this approach is that it is not general and requires conversion to matrices and then solving the eigenvalue problem. I suspect that it is possible to get the answers in terms of $$N$$ and $$s$$ without fixing the size $$N$$ and expressing the Hamiltonian as a matrix.

You can solve this by referring to this question.

To estimate the eigenvalues of $$H\left( s \right) =\left( 1-s \right) H_0+sH_m=I-\left( 1-s \right) |\psi _N\rangle \langle \psi _N|-s|m\rangle \langle m|$$, we can only calculate the eigenvalues of $$\left( 1-s \right) |\psi _N\rangle \langle \psi _N|+s|m\rangle \langle m|$$. Then, with the method of the link, this equals to calculate the eigenvalues of $$\left( \begin{matrix} 1-s& \frac{\sqrt{\left( 1-s \right) s}}{\sqrt{N}}\\ \frac{\sqrt{\left( 1-s \right) s}}{\sqrt{N}}& s\\ \end{matrix} \right) .$$ Solving this we get the eigenvalues should be $$\lambda =\frac{1\pm \sqrt{1-4\left[ s-s^2-\frac{\left( 1-s \right) s}{N} \right]}}{2}.$$ By replacing $$N=4$$, we get your special case.

Above only gives two eigenvalues, other eigenvalues of $$H\left( s \right)$$ are all $$1$$ with eigenvectors orthogonal to the space spanned by $$|m\rangle$$ and $$|\psi_N\rangle$$.

• thank you very much! May 13 at 1:41

Have you ever seen a derivation of Grover's search? The approach that you want is very similar.

Start by defining two states, perhaps $$|a\rangle=|\psi_N\rangle, \qquad |b\rangle=|m\rangle-|a\rangle\langle a|m\rangle,$$ where I've only given $$|b\rangle$$ up to normalisation. The point is that these two states should be orthonormal and span the space spanned by $$|\psi_N\rangle$$ and $$|m\rangle$$.

Any state $$|\phi\rangle$$ that is not in this span automatically satisfies $$H|\phi\rangle=(1-s)|\phi\rangle+s|\phi\rangle=|\phi\rangle$$ and is hence a $$+1$$ eigenstate.

For any state in the span, you can think about a linear combination $$\alpha|a\rangle+\beta|b\rangle$$ and how $$H$$ acts on this. The outcome is always a state in the same span. Hence, we can talk about this as a two-dimensional subspace and just write out a $$2\times 2$$ matrix. It looks something like $$H_\text{sub}=\begin{bmatrix} s\frac{N-1}{N} & -s\frac{\sqrt{N-1}}{N} \\ -s\frac{\sqrt{N-1}}{N} & 1-s\frac{N-1}{N} \end{bmatrix}.$$ So, you should be able to evaluate the two eigenvalues of this matrix: $$\lambda^2-\lambda-s(s-1)\frac{N-1}{N}=0$$ and thus $$\lambda=\frac{1}{2}\left(1\pm\sqrt{1-s(s-1)\frac{N-1}{N}}\right).$$ The ground state energy is thus $$\frac{1}{2}\left(1-\sqrt{1-s(s-1)\frac{N-1}{N}}\right).$$

• Thank you very much for the explanation. Your answer is very close to the correct answer. The correct answer is given by the user narip May 13 at 1:40