2,364 reputation
11232
bio website berkeley.academia.edu/…
location Berkeley, CA
age 24
visits member for 3 years, 6 months
seen 4 mins ago

Currently a graduate student in mathematics at the University of California - Berkeley.

Previously obtained a MASt. in Applied Mathematics from the University of Cambridge (2013), and a B.S. in Mathematics and a B.A. in Physics from the University of Chicago (2012).


Nov
9
comment Derivation of the irreducible representations of SO(3)
This is exactly what I am suggesting. This has nothing to do with spin vs. orbital. If somebody has told you that spin in $SO(3)$ can change a sign after a rotation by $2\pi$, they were wrong. This can only happen for spin with $SU(2)$. The point is is that, in quantum mechanics, we don't really care about representations, but rather projective representations (basically representations up to phase). It turns out that projective representations of $SO(3)$ are in canonical one-to-one correspondence with actual representations of $SU(2)$.
Nov
9
comment Derivation of the irreducible representations of SO(3)
Suppose that the representation has spin $n/2$ with $n\in \mathbb{Z}^+$. Then, in particular, this representation will contain an element $v$ such that $S_zv=\frac{n}{2}v$. This is at the level of the Lie algebra. At the level of the Lie group, $R_z(\theta )=\exp (-i\theta S_z)$ is a rotation about the $z$-axis by an angle $\theta$. Hence, $R_z(\theta )v=\exp (-i\theta \frac{n}{2})$. For $\theta =2\pi$, we had better have that $R_z(2\pi )=R_z(0)$; however, using the formula above, $R_z(2\pi )=\exp (-i\pi)=-1$. Thus, this does not give us a representation of $SO(3)$ for half-integer spin.
Nov
9
comment Derivation of the irreducible representations of SO(3)
The spherical harmonics give you all the irreducible representations of $SO(3)$, but it does not give you all those of $SU(2)$. You'll note this does not make any direct reference to the Lie algebra of $SO(3)$
Feb
26
comment Unitary representations of the diffeomorphism group in curved spacetime
If this is the case, then I feel we should apply the Faddeev-Popov trick to the Einstein Hilbert action so as to find the corresponding BRST charge associated with this diffeomorphism symmetry, and perform the usual cohomological construction to find the space of states. Has this not been done before? What is the result?
Nov
19
comment Amplitudes in renormalized perturbation theory
@Neuneck Yes, but I thought the point is that the total result should be proportional to $\lambda ^2$, not just $\lambda$.
Nov
19
comment Amplitudes in renormalized perturbation theory
@Neuneck I don't understand. The Feynman rule corresponding to the $4$-point counterterm vertex is just $-\mathrm{i}\, \delta _\lambda$, not $-\mathrm{i}\, \delta _\lambda \lambda$ or anything like this, so this entire counterterm diagram should just be of order $\lambda$. What am I missing?
Nov
16
comment Amplitudes in renormalized perturbation theory
This was more or less what I was expecting, but when I went to check this myself, I didn't see how $\delta _\lambda$ was of order $\lambda ^2$. Indeed, according to Peskin and Schroeder's Eq. (10.17) on pg. 324, $\delta _\lambda =\lambda _0Z^2-\lambda$, which is just of order $\lambda$ . . . or am I missing something?
Oct
5
comment Is the Lorentz group compact (and if not, is U(1)?)
I also think it is mis-leading to only say that Lorentz boosts are parametrized by $v\in (-c,c)$. They are, but you mustn't forget that the group law is not simply addition in $\mathbb{R}$; instead, it is given by the usual velocity addition formula in special relativity. It turns out that this group law makes $(-c,c)$ into a Lie group which is isomorphic (as Lie groups) to $(\mathbb{R},+)$ via $\mathrm{arctanh}$.
Oct
5
comment Is the Lorentz group compact (and if not, is U(1)?)
In some sense periodic identification is absolutely crucial. There are precisely two connected $1$-dimensional Lie groups: $S^1$ and $\mathbb{R}$. These two are distinguished precisely by 'periodic identification', and furthermore, $S^1$ is compact while $\mathbb{R}$ is not. Even though $[0,c]$ or $[-c,c]$ might be compact, they are not Lie groups.
Sep
27
comment Renormalizability of the Polyakov Action
@user10001 Thanks much. Do you want to write up what you said in a comment as an answer? At this point, I could write it up myself, but I feel as if it would make more sense to give you the credit for it.
Sep
23
comment Renormalizability of the Polyakov Action
@user10001 This works (in fact, with just a bit more detail, I would accept this as an answer), but there is at least one thing that worries me about this. Presumably by $1$ in $G=1+aX+(aX)^2+\cdots$, you mean $\eta _{\mu \nu}$. If this is the case, you find that the kinetic term is given by $-\frac{1}{2}\partial _\alpha Y^\mu \partial ^\alpha Y^\nu \eta _{\mu \nu}$, which means that the field $Y^0$ has the wrong sign for its kinetic term.
Sep
23
comment Quantization of Nambu–Goto action in multiples of Planck's constant?
@LubošMotl When you say ". . . the (theory defined by) the Nambu-Goto action may be quantized and and what one gets is known as string theory . . .", do you mean, in particular, that not only is the is the Nambu-Goto action equivalent to the Polyakov action at the classical level, but it is in fact equivalent to the Polyakov action even at the quantum level? Is this unique to the string ($p=1$), or is this result true for the corresponding actions for $p$-branes as well? In particular, is it true for point-particles ($p=0$)?
Sep
12
comment Derivation of the Polyakov Action
. . . Of course, I guess you might just say that the simplest one that works is the way to go (in this case, an action with just one extra field). Still, I have to admit, I'm not completely satisfied with this answer.
Sep
12
comment Derivation of the Polyakov Action
. . . And hell, if we're going to add in yet another field, we could probably find get another symmetry along with it.
Sep
12
comment Derivation of the Polyakov Action
I like this, but this method doesn't really convince me that putting in another auxiliary field is the way to go. Let's say in QFT we want a theory with a complex scalar field with a $U(1)$ symmetry. Of course, we could always introduce other fields into the theory, but that's not what one usually does unless you wanted the extra fields to begin with. And even if you decide that introducing a new field is the way to go, why stop at one? Surely we could add two new fields that respected all the symmetries we wanted . . .
Sep
3
comment Divergence of One and Two Graviton Exchanges
One last question, and then I think I got it. How do we determine the kinetic term for $h_{\mu \nu}$? (I ask because this will allow me to determine the appropriate dimensions of a spin $2$ field, and hence the appropriate dimensions of the coupling constant.)
Sep
2
comment Divergence of One and Two Graviton Exchanges
Are you assuming that the particle is a scalar, so that the interaction is of the form $\partial ^\mu \phi \partial ^\nu \phi g_{\mu \nu}$? This is the only way I could see you getting momenta coming into what appears to be your vertex factors. But if my dimensional analysis is correct, $[\partial ^\mu \phi \partial ^\nu \phi g_{\mu \nu}]=L^{-4}$, so that the corresponding coupling constant would be dimension-less, and in particular, could not be proportional to $1/M_P$. What's going on?
Sep
1
comment Vacuum Expectation Value and the Minima of the Potential
. . . In practice, you do this by writing the Lagrangian in terms of $\phi :=\phi _0-v$, where $v$ is some minimum of the potential and $\phi _0$ is the original field. My question could then be equivalently phrased as "Why does this guarantee that $\langle 0|\phi (x)|0\rangle =0$?".
Sep
1
comment Vacuum Expectation Value and the Minima of the Potential
@MichaelBrown Indeed, I was under the impression that you had to have this as a re-normalization condition in order to apply LSZ (that is, a hypothesis require for the LSZ Reduction Formula to hold was that $\langle 0|\phi (x)|0\rangle =0$). In fact, I thought this was the entire idea behind the symmetry breaking: you must re-write your Lagrangian in terms of the re-normalized field (with vanishing VEV), and if the bare field had a non-vanishing VEV, this will 'break' the symmetry . . .
Sep
1
comment Vacuum Expectation Value and the Minima of the Potential
So then what is the proof of this leading-order approximation?