Recent content by sensou

ahh i think i understand now. Thanks for the deeper insight into the question and helping me understand it better. I am new to this forum and i plan to stay and help others if i can since the help i received was so good.

so, i \hbar \int \sum_m \dot{c_m} (t) \psi_n^* \psi_m \,dx = i \hbar \sum_m \delta_{n,m} \dot{c_m(t)} = i \hbar \dot{c_m(t)} or \dot{c_n(t)} ? if it is the i \hbar \dot{c_n(t)} then my answer works out but i don't think the fact that it is c_n rather than c_m makes sense to me or is it...

i haven't used bra-ket notation before but i can understand what you wrote i will try to use latex code now hopefully it will turn out correctly what i wrote in the initial question was i \hbar \frac {d c_n(t)} {dt} = \sum_m H_n_m c_m(t) H_n_m = \int_{-\infty}^{+\infty} \psi_n^* \hat{H}...

I have been asked to show that the Schroding equation is equivalent to: i(hbar)d/dt(cn(t))=sum over m (Hnm*cm(t)) where Hnm=integral over all space of (complex conjugate of psin)*Hamiltonian operating on psim psi=sum over n (cn(t)*psin) But i don't know how to even start this question.

I am guessing you need to relate intermolecular distance to volume first, then to pressure.

I just realized that i haven't been able to get the <p^n> part right where p=-i(h-bar)d/dx i get <p^2>=(h-bar)^2 *a/2*sqrt(pi/a) <p^4>=(-3/4)*(h-bar)^4 *a^2*sqrt(pi/a) <p^6>=(-165/8)*(h-bar)^6 *a^4*sqrt(pi/a) <p^8>=(4605/16)*(h-bar)^8 *a^6*sqrt(pi/a) And this doesn't give me a nice...

Thanks for your quick replies. I am now able to get the answer :smile: I should have done it that way from the start but for some reason that infinite series thing led me on a bit. Thanks for all the help.

I need to calculate <x^n> and <p^n> for psi(x)=exp(-ax^2/2) for n even. For <x^n>: <x^n>=integral(exp(-ax^2)*x^n )dx from -inf to +inf then i use integration by parts to get an infinite series and i use a formula to find the finite sum of the series =[exp(-ax^2)*x^(n+1)/((n+1-2a*(n+1)^2)]...