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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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Microstates and Macrostates in Canonical Ensemble
The wikipedia emphasizes the difference, because in a microcanonical ensemble the total energy of the system is fixed. The "multiplicity" is the number of microstates with a given energy (or sometimes …
2
votes
Bloch state and mirror symmetry
What the mirror symmetry implies is that for each Bloch eigenstate $\Psi(x)=e^{ikx}u_k(x)$, its mirror image $\Psi(-x)=e^{-ikx}u_k(-x)$ is another eigenstate with the same energy. Then one can form sy …
3
votes
Accepted
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| …
1
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What is the importance of the biquadratic interaction in the AKLT model?
You don't need the biquadratic interaction to realize the Haldane phase. The Heisenberg spin-1 chain already does the job. Adding the biquadratic term allows one to have a parent Hamiltonian for the e …
1
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Accepted
'schrodinger' picture in measurement based topological quantum computation
For four Majorana zero modes, if the total topological charge is $1$ there are two states $|0\rangle_{12}0\rangle_{34}$ and $|1\rangle_{12}|1\rangle_{34}$ ($i\gamma_1\gamma_2\cdot i\gamma_3\gamma_4=1$ …
1
vote
Accepted
Angular momentum partial components of a $k$-dependent pairing potential
Just focusing on the $\cos^l\theta$ term is probably not going to get you anywhere, since $\cos^l\theta$, being a completely analytical function, is by no means singular (and following your argument y …
1
vote
Accepted
Angle operator $\hat{\phi}$ doesn't exist when doing quantum mechanics on the circle $S^1$?
The rigorous way is to use $e^{i\phi}=c+is$ and the conjugate angular momentum operators, as the Hilbert space is spanned by single-valued functions on the circle. But physicists often get away with u …
3
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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 …
2
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Accepted
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 …
2
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Static structure function for non-interacting Fermi gas
With Wick's theorem it would be very straightforward. But in this case it is not too difficult to directly work out the expectation value:
$
\displaystyle S_q=\langle \phi_0|n_qn_{-q}|\phi_0\rangle=\ …
1
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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 …
2
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Accepted
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 …
14
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Accepted
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 …
1
vote
Accepted
Commuting with time evolution operator implies commuting with Hamiltonian
Assuming $A$ itself is time-independent. If $[A, U_t]=0$ for all $t$, then it can be proven that $[A, H]=0$: We have
$H=i \partial_t U_t\cdot U_t^\dagger$
Because $[A, U_t]=0$, it follows that $[A, …
4
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Accepted
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.