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Where does the complex conjugate term generally come from in a Hamiltonian?

The simplest justification is probably that the Hamiltonian has to be Hermitian. After all, its eigenvalues are interpreted as the possible energies of the system, and hence they need to be real. This ...
Níckolas Alves's user avatar
3 votes
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Does the Hamiltonian always commute with the Time Evolution Operator?

What can be said about the general case, in which $H$ depends on time explicitly? Specifically: Do $U(t, t_0)$ and $H$ still commute? No, not in general. In general, you can write $U$ as a time-...
hft's user avatar
  • 20k
2 votes

Exercise on self-adjointness of Hamiltonian

Thanks to @ZeroTheHero who brought me onto the right track, I was able to find the solution to the problem myself. The Hamiltonian with respect to the basis $\{ | \psi \rangle, |\phi \rangle, |\Gamma\...
Octavius's user avatar
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1 vote

Where does the complex conjugate term generally come from in a Hamiltonian?

This type of Hamiltonian is a second quantisation one. The most important part is defining your operators $a$ and $a^\dagger$. For example, if you have two interacting dipoles the interaction energy ...
Olivier Masset's user avatar
1 vote

Understanding equation for eigenvalues of a Hamiltonian

The expression for the determinant of a square matrix $M$ of size $N$ is $$\det M = \sum_{\sigma \in S_N} \mathrm{sgn}(\sigma)M_{1,\sigma(1)}\ldots M_{N,\sigma(N)}.$$ In this case, $M_{mn} = A_n\...
Vincent Thacker's user avatar

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