509 reputation
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location Bangalore, India
age
visits member for 2 years, 3 months
seen Oct 14 at 15:55

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
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 $\langle p\rangle$
Nov
21
asked Unwanted $i$ floating around when trying to calculate $\langle p\rangle$
Nov
21
revised Which is the easier way to do integration by parts when there is an exponential term?
clarified answer by qualifying it
Nov
21
revised Which is the easier way to do integration by parts when there is an exponential term?
added I_0