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bio website bored
location your fireplace
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visits member for 1 year, 11 months
seen Dec 26 '13 at 0:47

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Dec
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comment Gap exponents and homogeneous functions
@Ignore my last comment, I think I understand. So $g_f(x)$ is not strictly speaking a homogeneous function, but in the large and small limits of $x$ behaves approximately like a homogeneous function...?
Dec
9
comment Gap exponents and homogeneous functions
If you have a homogeneous function $g_f(x)$ of order $k>0$ then its limit should be $0$. If it's of order $0$ then it is independent of $x$. (I think I am missing something.) So here is my confusion. If we assume that it's homogeneous of order $0$, then even in the limit that $x$ gets large it shouldn't change.
Dec
9
comment Gap exponents and homogeneous functions
this is circular reasoning, no? We show the limit is fixed by fixing the limit...
Dec
9
comment Gap exponents and homogeneous functions
Thanks. Adam, do you mind elaborating specifically on how you get that $\lim {x \to 0} g_f(x)=c$ Why doesn't $c=0$? Edit to comment: looking at your last line I think is what I needed. How do you show that $b=0$?
Dec
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asked Gap exponents and homogeneous functions
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Oct
31
comment Quantum Entanglement - What's the big deal?
@joshphysics No biggie.
Oct
31
comment Quantum Entanglement - What's the big deal?
@joshphysics So Josh.. I'm going to nitpick a little... The paradoxical thing, at least to me, is that when you measure an electron in NY, the other electron in LA will always be flipped the other way, even if you measure it before enough time has passed for the "information" to reach it. It's not just like marbles where nobody knows which marble is which. It's like you have two marbles, each flickering black and white, and when you measure one to be black, the other, while still flickering, will always turn out to be white -- even if you measure it "instantly" afterwards.
Oct
26
comment Intuitively, how can the work done on an object be equal to zero?
There's no reason to make the font this big.
Oct
18
accepted A “Hermitian” operator with imaginary eigenvalues
Oct
18
comment A “Hermitian” operator with imaginary eigenvalues
@EmilioPisanty furthermore, what is the erroneous step in the following calculation? $ \langle\psi_{\lambda}(x)|{\bf H}\psi_{\lambda}(x)\rangle= \langle{\bf H}^{\dagger}\psi_{\lambda}(x)|\psi_{\lambda}(x)\rangle=\langle{\bf H}\psi_{\lambda}(x)|\psi_{\lambda}(x)\rangle=\langle\psi_{\lambda}(x)|{\bf H}\psi_{\lambda}(x)\rangle^{*}.$ All inner products are presumably evaluated in the same Hilbert space.