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Constant of Motion in Quantum Mechanics for explicit time-dependent Operators

Take any time-independent operator $X$ which commutes with the Hamiltonian and multiply it by an arbitrary real-valued function of time : $$Y(t) = f(t)X$$ Then $Y(t)$ commutes with the Hamiltonian, ...
SolubleFish's user avatar
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Constant of Motion in Quantum Mechanics for explicit time-dependent Operators

If I understand you correctly, you want to compute $$\frac{d}{dt}\underbrace{\left(X(t)-\frac{P(t)}{m} t\right)}_{X^\prime(t)}$$ using the Heisenberg equation of motion $$\frac{d X^\prime(t)}{dt} =\...
Hyperon's user avatar
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How to deal with explicit time dependence in the Heisenberg picture?

The lecturer then went on to use Heisenberg's equation on $\hat{\mathbf{\Pi}}$: $$\frac{\mathrm{d}\hat{\mathbf{\Pi}}}{\mathrm{d}t} = \frac{1}{i\hslash}\left[\hat{\mathbf{\Pi}}, \frac{1}{2m}\hat{\...
hft's user avatar
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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
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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
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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
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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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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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How to tell whether Hamiltonian has rotational invariance? Conservation of angular momentum?

In the classical framework, the exchange interaction term S$_{i}$ $\cdot$ S$_{j}$ satisfies the rotation invariance of the x,y,z-direction, but the uniaxial anisotropy term ${(S_{i}^{z})}^2$ breaks ...
Xin's user avatar
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