Recent content by Juqon

I was already figuring the DNA as one tube. In your article it says "the elastic cost of bending is totally negligible", but what is the limit for this negligence?

Persistence length: What are the beginning and end points in an polymer? Hello, can you tell me where the starting point and the ending point of the (parts of the) persistence length in a polymer are? I thought the persistence length was the greatest length that only just is not bent...

You are right, the integrals were wrong, thank you. This is my calculation with the right ones: E_{0}=\frac{\frac{\pi}{\alpha^{2}}\left\{ \frac{\hbar^{2}}{\mu}\alpha-e^{2}\right\} +\frac{\pi}{\alpha^{3}}\cdot\left(-\frac{\hbar^{2}}{2\mu}\alpha^{2}\right)}{\frac{\pi}{\alpha^{3}}}...

Mh, I didn't change the other integrals. E_{0}= \frac{<\psi|\hat{H}|\psi>}{<\psi|\psi>}=\frac{\int_{-\infty}^{\infty}d\vec{r}\psi^{*}\left\{ -\frac{\hbar^{2}}{2\mu}\left[\alpha^{2}-\frac{2}{r}\alpha\right]\psi-\frac{e^{2}}{r}\psi\right\} }{\int_{-\infty}^{\infty}d\vec{r}\psi^{*}\psi}...

This is my newest version. I think I showed in #13 what the result of the integral ist, but now that differentiations make it 0 I deleted the whole L term. E_{0}= \frac{<\psi|\hat{H}|\psi>}{<\psi|\psi>}=\frac{\int_{-\infty}^{\infty}dr\cdot r^{2}\cdot\psi^{*}\left\{...

Thanks, now here is my newest problem. [EDIT: I think I found it.]E_{0}=\frac{<\psi|\hat{H}|\psi>}{<\psi|\psi>}= \frac{\int_{-\infty}^{\infty}dr\cdot r^{2}\cdot\psi^{*}\left\{ -\frac{\hbar^{2}}{2\mu}\left[\alpha^{2}-\frac{2}{r}\alpha\right]\psi+\frac{1}{4\pi\epsilon_{0}}...

\langle \psi \vert \psi \rangle = \int_0^\infty \int_0^\pi \int_0^{2\pi} e^{-2\alpha r}r^2\sin \theta \,dr\,d\theta\,d\phi=\frac{1}{4}\alpha^{3} \cdot 2\pi \cdot\left[-cos\right]^\pi _0 = \frac{1}{4}\alpha^{3} \cdot 2\pi \cdot\left[-(-1) - (-(1))\right]=\frac{1}{4}\alpha^{3} \cdot 4\pi...

\langle \psi \vert \psi \rangle = \int_0^\infty \int_0^\pi \int_0^{2\pi} e^{-2\alpha r}r^2\sin \theta \,dr\,d\theta\,d\phi=\frac{1}{4}\alpha^{3} \cdot 2\pi² And with L^2 = -\hbar^2 \left(\frac{1}{\sin\theta}\frac{\partial}{\partial \theta}\left( \sin\theta \frac{\partial}{\partial...

\langle \psi \vert \psi \rangle = \int_{0}^\infty \int_{0}^\pi \int_{0}^{2\pi} r²\cdot sin² (\theta) \psi^*(r, \theta,\phi) \psi(r,\theta,\phi)\ ,dr\,d \theta\, d\phi

I'd say that is in spherical coordinates: \langle \psi \vert \psi \rangle = \int_{-\infty}^\infty \int_{-\infty}^\infty \int_{-\infty}^\infty r²\cdot sin² (\theta) \psi^*(r, \theta,\phi)\psi(r,\theta,\phi)\ ,dr\,d \theta\,d\phi. I understand of course that \vec{r}=(r cos \theta cos \phi, r...

Wikipedia says L^2 = -\hbar^2 \left(\frac{1}{\sin\theta}\frac{\partial}{\partial \theta}\left( \sin\theta \frac{\partial}{\partial \theta}\right) + \frac{1}{\sin^2\theta}\frac{\partial^2}{\partial \phi^2}\right). But that would be more complicated, because I don't see it go away by a derivation...

Ok, I will correct the sign, sorry. The example was very helpful, thanks. Did I use the integral measure correctly? And is the result really that complicated? E_{0}=\frac{<\psi|\hat{H}|\psi>}{<\psi|\psi>}= \frac{\int_{-\infty}^{\infty} dr\cdot r^{2}\cdot\psi^{*}\left\{...

Homework Statement Calculate the zero-point energy of the hydrogen atom using the Ritz variaton principle and the approach \psi_{\alpha}. Hint: The stationary Schroedinger equation in spherical coordinates is ... Homework Equations \left\{...

In quantum mechanics, but it was not directly a task, just a way to find the Gaussian packet and read the expectation value for x and also the Delta_X², and since it was in the book I wanted to do it myself. But it seems there are easier ways to do this.

[7] =\frac{A^{2}}{4\pi\sqrt{d^{2}+\frac{\hbar^{2}t^{2}}{4m^{2}}}}\cdot\exp\left\{ \frac{\left(-x^{2}d^{2}+2\frac{\hbar t}{m}d^{2}k_{0}x-\frac{\hbar^{2}t^{2}}{m^{2}}d^2k_{0}^{2}\right)}{2d^{4}+\frac{\hbar^{2}t^{2}}{2m}}\right\} [8] \frac{A^{2}}{4\pi\sqrt{d^{2} +\frac{\hbar^{2}t^{2}}{4m^{2}}}}...