Recent content by emol1414

Homework Statement The initial wave function \Psi (x,0) of a free particle is a normalized gaussian with unitary probability. Let \sigma = \Delta x be the initial variance (average of the square deviations) with respect to the position; determine the variance \sigma (t) in a moment later...

Hmm, ok! So it's a different problem It makes sense now, I forgot the conditions... You're right, it is \int_{0}^{a}! Thank you

Kuruman... indeed A = \sqrt{\frac{2}{a}} is appropriate for \psi as a sin function. But when checking the result i got, i found this solution for the exact same problem: http://img714.imageshack.us/img714/6186/64732846.jpg The integral limit is set to 0,a/2... I'm not sure why, and if...

Hi, I'm stuck in this Griffiths' Introduction to QM problem (#2.8) Homework Statement A particle in the infinite square well has the initial wave function \Psi(x,0) = Ax(a-x) Normalize \Psi(x,0) Homework Equations \int_{0}^{a} |\Psi(x)|^2 dx = 1 The Attempt at a Solution...

Got it! Integration by parts only in one of the two terms left and then add to the other, so the factor 1/2 is gone... but... there's a m missing in the denominator, right? Thank you! =) One more thing... why is that \Psi goes to 0 when x \rightarrow \pm \infty? Is it a "single-case" fact...

Right... I'm not really sure why this is true (??), but doing so... we perform integration by parts 2 times and then \frac {d<x>}{dt} = -\frac{i \hbar}{2m} { \int_{-\infty}^{\infty} \Psi^*(\frac{\partial \Psi}{\partial x}) - \Psi (\frac{\partial \Psi^*}{\partial x} )dx} That's it?

So, we have, for the product rule, that \int_{a}^{b} f \frac{dg}{dx} dx = fg {|}^{b}_{a} - \int_{a}^{b} g \frac{df}{dx} dx And choosing f = x\Psi^* and g = \frac{\partial \Psi}{\partial x} \frac {d<x>}{dt} = \frac{i \hbar}{2m} {x \Psi^* \frac{\partial \Psi}{\partial x}...

*Fixed a typo in the equation

In QM x don't depend on t, right?

Homework Statement Prove that <p> = m \frac{d<x>}{dt} Homework Equations Schrödinger Equation: i\hbar \frac{\partial \Psi} {\partial x} = - \frac{\hbar^2}{2m} \frac{\partial^2 \Psi}{\partial x^2} + V{} \Psi Respective complex conjugate from equation above Expectation Position: <x> =...

Hm... even using L'Hopital (e.g, for a constante \rho) i would get a \infty root-mean-square radius anyway. It doesn't make sense, i think

Thank you, Mark! ^^ Now... with the correct latex code =P Could anyone give me a enlightenment here, how to work with this? I thought... if i have \infty both "sides" up and down... i could use L'Hopital. But I keep thinking this would be just 'too easy' Oo Idk why, but doesn't sound right...

I'm really trying to edit it and make the expression look nice, but i can't figure out how to do it. Anywayz... it's the first integral (from 0 to infinity) divided by the second integral (also 0 to infinity). And then the square root of this.

Homework Statement It's a physics problem, where i have to evaluate the root-mean-square radius defined by the expression below. (First for a constant \rho, then for a "(r)dependent" \rho).Homework Equations (\int{_0}{^\infty} \rho r^4 dr / \int{_0}{^\infty} \rho r^2 dr) ^(1/2)The Attempt at a...

This is a silly doubt i guess... Homework Statement When you know an atom's radius you can easily determine its volume by considering it's a sphere. But when you're dealing with solids, that is, a set of atoms... and then you have bands insted of orbitals... this differente...