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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.

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2 Answers 2

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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 $|\psi_N\rangle$.

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  • $\begingroup$ thank you very much! $\endgroup$
    – MonteNero
    May 13 at 1:41
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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 $$ \frac{1}{2}\left(1-\sqrt{1-s(s-1)\frac{N-1}{N}}\right). $$

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  • $\begingroup$ 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 $\endgroup$
    – MonteNero
    May 13 at 1:40

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