Recent content by Elekko

On the book "Introduction to Solid State Physics" by Kittel, on page 98 he derived the roots for optical and acoustical branches for the equation: M_1 M_2 \omega^4-2C(M_1+M_2)\omega^2+2C^2(1-cos(Ka))=0 where the roots are: \omega^2=2C(\frac{1}{M_1}+\frac{1}{M_2}) and...

Geometric distribution

Homework Statement Let X_n \in Ge(\lambda/(n+\lambda)) \lambda>0. (geometric distribution) Show that \frac{X_n}{n} converges in distribution to Exp(\frac{1}{\lambda}) Homework Equations I was wondering if some kind of law is required to use here, but I don't know what Does anyone know how this...

Homework Statement Given that, in free space the probability density for a wave function (free particle) is \mid \Psi(x,t)\mid^2=P(x,t)=\frac{\sigma_0}{\mid \alpha \mid^2\sqrt{\pi}}exp(-(\frac{\sigma_0}{\mid \alpha \mid})^4\frac{(x-x_0-p_0t/m)^2}{\sigma_0^2}) What is need to be done is to...

That's a perfect explanation. Thank you very much!

Need to mention that of course f = 8 MHz

Can this be used as the uncertainty in energy in the energy-time uncertainty principle? (Im not very sure about this, science I don't have a solution for this)

Homework Statement The smallest uncertainty in the frequency of emitted light when excited atoms return to ground state for molecules is estimated to be 8 MHz. Use this information to estimate the lifetime of the excited states. I would like to know if I'm thinking correct. Homework...

Thanks! I got it now! In case of quantum mechanics we use the waveFUNCTION \psi in order to prove the commutator. I appreciate for all your help guys!

I think I got it now guys. Thank you! Tiny-tim, nevermind. It seems that I don't understand why we even do have to apply \psi on the commutator to begin with. Its just simply xp-px where x = x, p = -ihd/dx

Wow you're quick in replying. So let me see if I have got it right: the expectation value \overline{Q} = \int \Psi^\ast (x,t) \hat{Q}\Psi(x,t) dx has nothing to do with to derive the commutator. The commutator of momentum and position is simply i\hbar by using the observables in the...

So here, it means that when we derive the commutator for momentum and position to i\hbar \psi the \psi is always applied and it can be seen as and identity operator as tiny-tim mentioned?

Hi all, I appreciate for all your reply! So this means that in the case of momentum and position, the "identity" operator is: [itex]\Psi[itex] ?

Hi! In my textbook the explanation of the expectation value in general covering any observable Q is: \overline{Q} = \int \Psi^\ast (x,t) \hat{Q}\Psi(x,t) dx Then they define the commutator as: [\hat{A},\hat{B}] = \hat{A}\hat{B}-\hat{B}\hat{A} Now for position and momentum...