521 reputation
514
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location Bangalore, India
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visits member for 2 years, 5 months
seen 2 days ago

I am a Christian. Although I have an abiding interest in science and philosophy, I view things from a distinctly Christian vantage point, which harmonizes human well-being with what we know of the natural world.

Here's a nice quote I came across:

"The spectacle of the universe seems all the more grand and beautiful and worthy of its Author, when one considers that it is all derived from a small number of laws laid down most wisely." -Maupertuis, 1746


Mar
31
comment Find the median given a table of relative frequencies
I'm curious -- why do you think it should be zero? I don't see that.
Mar
31
asked Find the median given a table of relative frequencies
Mar
30
asked Finding a number given its remainder when divided by other numbers
Mar
11
asked Showing that $\operatorname{div}\Psi\nabla\Psi^*-\operatorname{div}\Psi^*\nabla\Psi=\operatorname{div}[\Psi\nabla\Psi^*-\Psi^*\nabla\Psi]$
Dec
7
comment Solving the time-independent Schrodinger equation for particle in a potential well
Ah, careless.. thanks.
Dec
7
asked Solving the time-independent Schrodinger equation for particle in a potential well
Nov
28
revised Manipulating derivatives after substitution: $\xi=\gamma x$
added 29 characters in body
Nov
28
answered Manipulating derivatives after substitution: $\xi=\gamma x$
Nov
28
revised Manipulating derivatives after substitution: $\xi=\gamma x$
Added further attempt at problem
Nov
28
accepted Showing that $\langle p\rangle=\int\limits_{-\infty}^{+\infty}p |a(p)|^2 dp$
Nov
28
asked Manipulating derivatives after substitution: $\xi=\gamma x$
Nov
22
comment Showing that $\langle p\rangle=\int\limits_{-\infty}^{+\infty}p |a(p)|^2 dp$
Thanks for your clarifications. That takes care of the $a(p')$. But what about the $p'$ becoming a $p$?
Nov
22
comment Showing that $\langle p\rangle=\int\limits_{-\infty}^{+\infty}p |a(p)|^2 dp$
Yes, but in the penultimate step, you have both $a(p)^*$ and $a(p)$ being integrated over $p$. This seems contrary to step 2, where $a(p)^*$ goes with $p$ and $a(p')$ with $p'$. So shouldn't your penultimate step read $\int_{-\infty}^{\infty} p' a(p)^*a(p') \, dp$?
Nov
22
comment Modulus of a complex function: $\Psi(x)=A_0 e^{-kx^2} e^{i\alpha x}$
It seems to work, and I'll make a note of it in my notes, but I just wanted to make sure that it's the most general way to find the modulus. Somehow, I seem to have missed the class where they extended the idea of a modulus to cover complex functions. It's like a secret that everybody is in on except me!
Nov
22
comment Modulus of a complex function: $\Psi(x)=A_0 e^{-kx^2} e^{i\alpha x}$
Can I do $|\Psi|=[\Psi^* \Psi]^{\frac{1}{2}}$?
Nov
22
asked Modulus of a complex function: $\Psi(x)=A_0 e^{-kx^2} e^{i\alpha x}$
Nov
22
comment Showing that $\langle p\rangle=\int\limits_{-\infty}^{+\infty}p |a(p)|^2 dp$
Also, I don't understand why you introduced the primed variables only to drop them in the last 2 expressions.
Nov
22
comment Showing that $\langle p\rangle=\int\limits_{-\infty}^{+\infty}p |a(p)|^2 dp$
I don't understand how you got the expression for $\Psi^*$. Can we simply take the conjugate within the integral sign? Does this mean that the integral of a conjugate is equal to the conjugate of the integral?
Nov
21
asked Showing that $\langle p\rangle=\int\limits_{-\infty}^{+\infty}p |a(p)|^2 dp$
Nov
21
accepted Unwanted $i$ floating around when trying to calculate $\langle p\rangle$