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...
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...
Homework Statement
Given that we the following elliptic problem on a rectangular region:
\nabla^2 T=0, \ (x,y)\in \Omega
T(0,y)=300, \ T(4,y)=600, \ 0 \leq y \leq 2
\frac{\partial T}{\partial y}(x,0)=0, \frac{\partial T}{\partial y}(x,2) = 0, \ 0\leq x \leq 4
We want to solve this problem...
I'm wondering how this actually can be solved since i think there is one piece missing.
Homework Statement
Determin the diffusion length L on a P+N diode with Nd = 10^16 cm^-3 in the N region.
Homework Equations
Known is that Diffusion length can be found from L=sqrt(Dt)
The...
Homework Statement
I'm studying about gravitational lensing and got stuck on an article:
http://arxiv.org/pdf/0708.2684v1.pdf
where the equation 2.1 is abit strange to me. The authour uses the reference
http://arxiv.org/pdf/astro-ph/9703103v1.pdf
for deriving the equation.
According to me...
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...
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?