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Oct
18
comment Time evolution operator of a periodic Hamiltonian
Which static Hamiltonian do you mean? You can find $H_{eff}$ by solving the Dyson series for the evolution operator (or a Magnus expansion), and you can truncate the series at some order to get an approximation. But let's say that you manage to find $H_{eff}$ exactly. Then if you view the system in stroboscopic time (periods of $T$), the system looks exactly like it involves under the static Hamiltonian $H_{eff}$, that's not an approximation.
Oct
17
comment Time evolution operator of a periodic Hamiltonian
The answer is yes. $U_2 = U_1 U_1$. You can prove it. Furthermore writing $U_1$ as an exponential of an effective Hamiltonian is not an approximation, any unitary can be written like as an exponential of $i$ times a Hermitian operator. So the equation defines the effective Hamiltonian.
Oct
1
comment What is the definition of a quantum integrable model?
This definition is useless because in any system there will always be $n$ independent operators that commute with $H$ and among themselves - the projectors onto the eigenstates. That's just a restatement that the Hamiltonian can be diagonalized. So clearly integrability is not the same as diagonalizability, and some notion of locality or whatever is required of your operators $K_i$.
Sep
24
awarded  Autobiographer
Aug
1
awarded  Nice Question
Jul
2
awarded  Curious
Jun
24
comment Decoupling of Holomorphic and Anti-holomorphic parts in 2D CFT
quantum hall droplet is a chiral CFT. From this example note that the decoupling of holomorphic and antiholomorphic part is only true for CFTs on the $\mathbb{C}$- plane. For a CFT on a plane with a boundary, or a defect, for examples, the Virasoro algebra does not decouple and you cannot solve for the holomorphic and antiholomoprhic parts separately.
Jun
24
comment Decoupling of Holomorphic and Anti-holomorphic parts in 2D CFT
say, $l_n$, and then simply copy the result for the other set. But we must always remember that the full theory is comprised of the tensor products of both sets of generators, thus, we need to tensor the holomoprhic and antiholomoprhic parts together. The physical significance of these two parts has to do with time-reversal. In a time reversal symmetric system the left and right moving parts must pair up together to cancel out, to give no net chirality. But there are chiral (time reversal broken) CFTs which have only movers in one direction. for e.g., the theory on the boundary of a fractional
Jun
24
comment Decoupling of Holomorphic and Anti-holomorphic parts in 2D CFT
Why are you talking about the Lorentz group only? Your question is about CFT, so, you want to look at the conformal group (which subsumes the Lorentz group). Now to study the conformal group we look at its Lie algebra. On the 2D plane this happens to be the Witt algebra (quantum version is the Virasoro algebra). In both cases the generators of the Lie algebra fall under two sets, $l_n$ and $\bar{l}_n$, and they obey the same commutation relations within each set, and they commute between sets: $[l_n, \bar{l}_m] = 0$. So, that means we only need to use representation theory on one set,
Jun
3
awarded  Popular Question
May
12
awarded  Nice Question
Mar
5
comment Valley meaning explanation for foreigner
@BrandonEnright No, that's not right. Valley in this case refers to Dirac points as noted below. Why valley? Because in the most commonly analyzed example of graphene, at zero doping, all the states in the bottom Dirac cone are filled, which leaves the top Dirac cone empty. This empty Dirac cone looks like a valley, like the letter 'V', and the interesting physics comes from exciting electrons into V.
Mar
2
awarded  Nice Question
Feb
16
asked Eigenvalue problem for differential equations in QM
Feb
7
comment Finding the ground state of the toric code Hamiltonian
see socrates.berkeley.edu/~jemoore/Physics_250_files/…
Feb
2
comment Confusion regarding field operators
I fail to understand the point you're trying to make about the Green's function. It is just a matter of convention. Would it help if in the field theory context I wrote $\phi = \phi_+ + \phi_-$ (see P&S)? Obviously $\phi, \pi$ will obey different commutation relations from $\phi_+, \phi_-$, but they are related to one another. Then $\phi_+, \phi_-$ in the field theory context will be equivalent to $\phi, \phi^\dagger$ in the many-body context.
Feb
2
comment What does a $SU(2)$ doublet really mean?
@SanathDevalapurkar The isospin transformation asked in this case is not a gauge transformation, i.e. it does not vary from point to point, but is rather just a global (internal) transformation. Weak isospin on the other hand is a gauge transformation
Feb
1
comment Is the spin 1/2 rotation matrix taken to be counterclockwise?
@Hunter No, what K-boy means is that you take $e^{i \theta S_z}$ (2x2) and act on it to each 2x2 component of the 3x1 column vector $\vec{S}$. You end up with a new column vector which is 3x1 with different 2x2 components. It is fine.
Jan
30
comment A point between two charges has an electric potential of zero, but a charge placed at this point will gain kinetic energy. Why?
@ByronS The two charges (or any collection of charges) will create a potential field that permeates all space. A test charge put into this system will acquire a potential energy, due to this potential field, of strength $q V$, as measured from the potential energy at infinity, $0$.
Jan
30
revised Interpretation of the 1D transverve field Ising model vacuum state in a spin-language
added 141 characters in body