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Dec
26
answered Path integral with zero energy modes
Dec
23
comment Classical vs. Quantum use of the spin 4-vector
Sorry for the late response, I added an update explaining the ultra-relativistic case.
Dec
23
revised Classical vs. Quantum use of the spin 4-vector
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Dec
21
comment Classical vs. Quantum use of the spin 4-vector
The boost transformation of the spatial components of the Pauli Lubanski vector is the third equation copied from the reference article. For a massive particle, the spin is the value of the spatial component vector at its rest frame. Knowing the momentum and the Pauli-Lubanski vector, one can perform a boost to a frame where the particle is at rest and get its spin. In a frame where the particle is not at rest the spatial components of the Pauli Lubanski vector still satisfy the spin commutation relations since they are composed from the spin and an angular momentum, thus should be quantized.
Dec
20
comment Classical vs. Quantum use of the spin 4-vector
Basically, yes, but please notice that the spatial spin components in the rest frame satisfy the angular momentum commutation relations, and the x and z components cannot be measured simultinuously. Thus the numerical values of the Pauli- Lubanski vector should be understood as expectations.
Dec
19
revised Classical vs. Quantum use of the spin 4-vector
added 1 characters in body
Dec
19
answered Classical vs. Quantum use of the spin 4-vector
Dec
13
revised few fermions in a harmonic trap — position density matrix from diagrammatics
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Dec
13
answered few fermions in a harmonic trap — position density matrix from diagrammatics
Dec
11
comment Beyond WKB approximation for energies
@Emilio You may find these lecture note by B.C. Hall useful: math.cinvestav.mx/~NorteSur/Taller10/notas/Hall.pdf
Dec
10
answered Beyond WKB approximation for energies
Dec
6
awarded  Necromancer
Dec
3
answered Other Gross-Neveu like theories?
Nov
27
answered Hawking Radiation from the WKB Approximation
Nov
27
awarded  Caucus
Nov
22
answered Constructing Supersymmetric Lagrangians
Nov
20
comment Finding the energy levels of an electron in a plane perpendicular to a uniform magnetic field
If I may add, if you write $a = x+ip$ and $a^{\dagger} = x-ip$, you get the harmonic oscillator Hamiltonian in the usual representation, but the harmonic osillator coordinates $x$ and $p$ are not the original coordinates that you started with.
Nov
20
comment Finding the energy levels of an electron in a plane perpendicular to a uniform magnetic field
If you substitute the expressions of $a$ and $a^{\dagger}$ given in the answer in the Hamiltonian $H=\hbar\omega(a a^{\dagger}+\frac{1}{2})$ you get the Hamiltonian you started with expressed in terms of $X$ and $P$.
Nov
20
comment Finding the energy levels of an electron in a plane perpendicular to a uniform magnetic field
The Hamiltonian expressed in terms of the creation and annihilation operators has exactly the form of the harmonic oscillator Hamiltonian. Thus the energy levels are exactly equal to those of the harmonic oscillator, however with infinite degeneracy per level (The harmonic oscillator energy levels are nondegenerate). The advantage of using this method is that it allows an algebraic solution of the energy levels (i.e. without solving differential equations), please see the quantum harmonic oscillator Wikipedia page: en.wikipedia.org/wiki/Quantum_harmonic_oscillator
Nov
20
answered Finding the energy levels of an electron in a plane perpendicular to a uniform magnetic field