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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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Second quantization and Hamiltonian diagonalization

Diagonalizing the Hamiltonian means you want to bring it into the form $H=\omega b^\dagger b$, and it is pretty obvious that $b$ should be a linear combination of $a$ and $a^\dagger$, and $b$ should s …
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10 votes
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How projective representations can lead to 't Hooft anomalies in quantum mechanics?

Shu-Heng Shao's argument is essentially the statement that gauging means projecting to $G$-invariant subspace in (0+1)d. In a way this is the definition of gauging in (0+1)d in the Hamiltonian formali …
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9 votes
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Time reversal symmetry of transverse field Ising model

Basically, the answer is yes: $H$ is TRI because it is real. Reality condition really means that the Hamiltonian obeys a certain anti-unitary symmetry. In this case, the time-reversal operation is sim …
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5 votes
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Commutator of the Hamiltonian and crystal momentum

The equation follows from the a general result called Hellman-Feynman theorem: suppose one has a Hamiltonian $H(\lambda)$ depending on a parameter $\lambda$ (e.g. your $q$), with eigenstates $|n(\lamb …
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5 votes

Iterative projection into ground state

For your question 2, I think the authors were saying nothing more profound or deeper than "it is too hard to calculate $e^{-\tau H}$ directly for a finite $\tau$" because of the non-commutativity betw …
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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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5 votes
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Relation between bosonization and conformal field theory

Bosonization means you map some problems (i.e. interacting fermions) to free (compact) scalar field theory, which is perhaps the simplest example of 1+1 CFT.
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4 votes

How do states in Hilbert Space act like irreducible representations?

Whenever you have a symmetry group $G$, it means that for each $g\in G$ there is an operator $U(g)$ (usually unitary) in the system corresponding to the action of $g$. "states behave like irrep of $G$ …
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4 votes
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which are the non-abelian anyons for universal quantum computation

All $\mathrm{SU}(2)_k$ with $k>2, k\neq 4$ are universal. For a proof see http://arxiv.org/abs/math/0103200.
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Can one define wavefunction for Bogoliubov quasiparticle excitation in a superconductor?

The localization of Majorana zero modes has a well-defined meaning: consider a Kitaev chain with two ends. Because of the zero modes, there are two nearly degenerate ground states, let us call them $| …
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3 votes
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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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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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3 votes
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Landau levels degeneracy in symmetric gauge

$r_\text{max}$ is the location where $|\psi|^2$ is maximized. Even after multiplying the wavefunction by $z^m$, $|\psi|^2$ is still symmetric under rotation around the origin (only a function of $|z| …
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3 votes
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Does time reversal symmetry hold for (Kitaev model) 1D spinless $p$-wave superconductor?

Obviously if one starts from spin-1/2 physical electrons and want to get effective spinless fermions, one has to break time-reversal symmetry. But let us imagine that we just have spinless electrons …
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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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