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The central term in the hamiltonian formalism. Can be interpreted as an energy input, or "true" energy.

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Time-independent constant-Hamiltonian Schrodinger equation solution when at time 0 there's a...

For example, I have 2d lattice $2 \times 2$ $$ \begin{matrix} (1,1) & (1, 2) \\ (2, 1) & (2,2) \end{matrix} $$ and the Hamiltonian matrix element has form $\langle(i,j)|H|(n,m)\rangle … vec{V} = \begin{bmatrix} (1,1) \\ (1, 2) \\ (2, 1) \\ (2,2) \end{bmatrix} = \begin{bmatrix} v_1 \\ v_2 \\ v_3 \\ v_4 \end{bmatrix} $$ and accordingly, the element of Hamiltonian
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5 votes

Canonical transformation for the "nearly diagonal" Hamiltonians

There are $n$ ortho-normal vectors, providing the bases for the orthogonal transformation $\mathbf{R}$ to the diagonalisation of the Hamiltonian. $$ \mathbf{R} = \begin{bmatrix} A_0 & A_1 & A_2 & A_3& …
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1 vote

Two-site fermion system

Act Hamiltonian on the states $\vert 00,00\rangle $ and $\vert 11,11\rangle $. … The Hamiltonian will become diagonal by $U^T H U$. …
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2 votes
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Time evolution operator

$\renewcommand{\ket}[1]{\left \lvert #1 \right\rangle}$ $\renewcommand{\bra}[1]{\left \langle #1 \right\rvert}$ To my best guess, the problem started with a Hamiltonian from some bases $\ket{1}$ and $\ … $$\tag{1} H = \left(\ket{1} \bra{1} - \ket{2}\bra{2} -i \ket{1}\bra{2} + i \ket{2} \bra{1}\right) \, . $$ This Hamiltonian has two eigen values $E_1 = \sqrt{2}$ and $E_2 = -\sqrt{2}$. …
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3 votes

Hamiltonian of two coupled oscillators

\tag{3} $$ Thus, we can diagonal the hamiltonian using the eigen vectors of matrix $\mathbf B$. They are mutually orthogonal. … Diagonalization of Hamiltonian: Write Hamiltoniam in terms of $\vec y$, defined as $$ \vec y(t) \equiv \begin{pmatrix} \sqrt{m_1} x_1(t) \\ \sqrt{m_2} x_2(t) \end{pmatrix}. $$ The hamiltonian becomes …
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