2,070 reputation
1320
bio website
location
age
visits member for 2 years, 3 months
seen 2 days ago

Undergrad


2d
awarded  Popular Question
Apr
17
comment Is time reversal symmetry broken in (conventional) superconductors?
I'm asking because let's say I'm given a random spin Hamiltonian. Or perhaps random fermionic Hamiltonian. How can I tell if it's time reversal invariant or not? I see many expositions like "because H is invariant under complex conjugation so it is T-invariant"... that doesn't mean anything to me.. How does one define complex conjugation in a basis independent fashion? (For example, the Pauli MATRIX $\sigma^y$ is not-time reversal invariant because of the special basis I've chosen for it, but what's special about this basis? I would like to think of it as obeying the Pauli algebra only)
Apr
17
comment Is time reversal symmetry broken in (conventional) superconductors?
I don't quite understand why time reversal symmetry is effected as $c_{k \uparrow} \to c_{-k \downarrow}$ and $c_{k \downarrow} \to -c_{-k \uparrow}$... I know this is because of a lack of my conceptual understanding, but how does one define the action of time reversal symmetry on spin variables, fermionic variables, Majorana modes from first principles?
Apr
16
revised First and second order phase transitions
The previous edit is wrong. Terms like m^3, m^5... are allowed because they explicitly break the symmetry. That is how you model a FIRST order phase transition. sigh.. To maintain positive-definiteness the largest power must be even.
Mar
24
comment Double semion model on a square lattice
Hi @No.9999, I'm interested in the correct form of the double semion Hamiltonian on the square lattice. May I know what it is? Thanks.
Jan
7
awarded  Popular Question
Dec
31
awarded  Yearling
Nov
28
revised The spin and weight of a primary field in CFT
edited body
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.