Recent content by QuarksAbove

r = 1.21x10-4 /year If I have N C-14 atoms, then 0.7 of those N atoms decay each minute.. 0.7 decay/minute = 367920 atom decays per year. I'm not sure how it's related to the decay rate. Sorry, I haven't done a problem with two rates before and it's really confusing.

Homework Statement Suppose you have a 1.4g sample of old charcoal. It produces 0.7 beta decays per minute. How old is the charcoal. Given: 1g of carbon current day has 6.36x1010 atoms of 14C Homework Equations N = Noe-rt N = number of atoms in the sample (current-day) No =...

Oh right, I forgot. After normalization, |\psi> = 1/\sqrt{2}(|\uparrow\downarrow> - |\downarrow\uparrow>) +1/\sqrt{2}(|\uparrow\downarrow> + |\downarrow\uparrow>) leaving the probability to be in the triplet state (S=1) to be 1/2. Thanks for your help! I think I got it!

Hello =) so the singlet state is |\psi_{singlet}>= 1/(√2)(|\uparrow\downarrow> - |\downarrow\uparrow>) and the triplet state is |\psi_{triplet}>= 1/(√2)(|\uparrow\downarrow> + |\downarrow\uparrow>) so, |\psi> = 1/2(|\psi_{singlet}> + |\psi_{triplet}>) in order to get S=1...

Homework Statement Two particles with spin 1/2 and are in the spin state: |\psi> = |\uparrow_{1}>|\downarrow_{2}> where |\uparrow_{1}> is a state where particle 1 has spin up along the z-axis and |\downarrow_{2}> is a state where particle 2 is spin down along the z-axis. If...

Ah! yes! I completely forgot about that. H = (a^{\dagger}a + 1/2)\hbar \omega I know what to do now. I compute the commutator using Z, and the U's will go away because they commute with H. Thank you! in the end I got that Z(t) = Z because dz(t)/dt = 0 which means that z(t) =...

Homework Statement A particle of mass m is in a harmonic oscillator potential with spring constant k. An observable quantity is given in the Schrödinger picture by the operator: Z = a^{\dagger}a a^{\dagger} a a) Determine the equation of motion of the operator in the Heisenberg...

Homework Statement Calculate <i(\hat{a} - \hat{a^{t}})> Homework Equations |\psi > = e^{-\alpha ^{2}/2} \sum \frac{(\alpha e^{i\phi })^n}{\sqrt{n!}} |n> \hat{a}|n> = \sqrt{n}|n-1> I derived: \hat{a}|\psi> = (\alpha e^{i\phi})^{-1}|\psi> The Attempt at a Solution...