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The central term in the hamiltonian formalism. Can be interpreted as an energy input, or "true" energy.
0
votes
Accepted
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}
$$
and accordingly, the element of Hamiltonian …
5
votes
Canonical transformation for the "nearly diagonal" Hamiltonians
There are $n$ ortho-normal vectors, providing the bases for the orthogonal transformation $\mathbf{R}$ to the diagonalisation of the Hamiltonian.
$$
\mathbf{R} =
\begin{bmatrix}
A_0 & A_1 & A_2 & A_3& …
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 $U^T H U$. …
2
votes
Accepted
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 $E_2 = -\sqrt{2}$. …
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 …