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Quantum mechanics describes the microscopic properties of nature in a regime where classical mechanics no longer applies. It explains phenomena such as the wave-particle duality, quantization of energy, and the uncertainty principle and is generally used in single-body systems. Use the quantum-field-theory tag for the theory of many-body quantum-mechanical systems.

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$SU(2)$ vs. $SO(3)$ transformation, spinor rotation and measurement

Let us analyze a more general setup. Suppose you have two spins, one has spin-$S_1$ and the other has spin-$S_2$. No restriction on $S_1$ or $S_2$. We can then consider any state in the two-spin Hilbe …
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Is unitarity equivalent to linearity plus conservation of the norm?

If norm of any state is preserved, then the inner product is preserved as well. You have already shown that the real part of $\langle a|b\rangle$ is preserved. Doing the same for $|a\rangle + i|b\rang …
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2 votes

How do quantum numbers combine when two quantum systems combined?

Mathematically, quantum numbers refer to irreducible representations (reps) of the (unitary) symmetry group. For example, parity corresponds to one-dimensional reps of $Z_2$, the baryon number/3-compo …
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Large gauge transformation in $\mathrm{U}(1)$ flux threading argument

Eq (2) has nothing to do with translation invariance. It also has very little to do with the actual form of the Hamiltonian. Let us just consider a one-dimensional chain and let the site index runs fr …
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3 votes

Is the Dirac monopole quantization condition out by 1/2?

The Dirac quantization condition does not say the wavefunction can not change when the charged particle goes around the monopole (actually, it is meaningless to say a point-like particle going around …
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2 votes

How do I diagonalize this Hamiltonian?

Let us redefine $p\rightarrow \omega p, q\rightarrow \frac{q}{\omega}$, so the Hamiltonian becomes $$ H = \frac{\omega^2}{2}(p^2+q^2)+\gamma(pq+qp)=\omega^2\left[\frac{1}{2}(p^2+q^2)+\frac{\gamma}{\om …
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5 votes

Can the wavefunction be inferred from the expectation values of operators?

Intuitively, if you know the expectation values of all possible observables, that should be enough to fix the state of the system. This almost sounds tautological, since the state is just a mathematic …
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Time reversal symmery and spectrum statistics of generic Hamiltonians

To discuss level statistics, first we consider whether the ensemble of Hamiltonians has any anti-unitary symmetry. Your Hamiltonian is invariant under the anti-unitary symmetry $T=K$ where $K$ is the …
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Berry's phase for an electron in a two-level system

The problem is that when you use this parametrization of the "spin-up" state, the wavefunction is not single-valued in $\theta$. Namely, $|\theta+2\pi,\phi\rangle=-|\theta,\phi\rangle$. The usual form …
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1 vote

Spectral flow in IQHE

To answer 1, let us consider the Hamiltonian $H_\Phi$ on page 53, and ignore $V(r,\theta)$ for now. Then you can easily check that the wavefunction in the lowest Landau level takes the form $$ e^{im\p …
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Solving Periodic Time Dependent Hamiltonians

You are right that in general for a time-dependent Hamiltonian, one can not just exponentiate the time integral to get time evolution operator. However, this example is more of a special case for a pa …
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Hamiltonian invariant under time reversal symmetry

$O$ should be a unitary operator. The anti-unitary part of time-reversal already puts the $*$ in $H$, so $O$ is unitary. Then you can move $O$ inside the time derivative on the left-hand side and the …
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3 votes

Eigenstates harmonic oscillator with mass matrix

I assume you actually meant the Hamiltonian is $H=\mathbf{p}^T M \mathbf{p}+\mathbf{x}^T\mathbf{x}$, where $\mathbf{p}=-i\nabla$. I write both $\mathbf{p}$ and $\mathbf{x}$ as column vectors. Diagonal …
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Anti-unitary operator and hamiltonian

There are a couple of confusing (or even wrong?) points in the post. First, I assume $U^*$ means $U^\dagger$, the adjoint of $U$. A unitary symmetry means $UHU^\dagger=H$. An anti-unitary operator is …
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Why do the boundary conditions change the eigenvalues between unitarily equivalent Hamiltoni...

For $\theta$ not an integer mutiple of $2\pi$, the transformation $U$ can not be single-valued. In other words, it can not be a legitimate operator in the Hilbert space of periodic, square-integrable …
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