Recent content by vj3336

I think the above expression evaluates to: \left.\left.\left.\left.\overset{\wedge }{1}|n\right\rangle +\frac{f^{(1)}(0)}{1!}\overset{\wedge }{n_k}{}^1|n\right\rangle +\frac{f^{(2)}(0)}{2!}\overset{\wedge }{n_k}{}^2|n\right\rangle +\text{...}+\frac{f^{(m)}(0)}{m!}\overset{\wedge...

sorry , I still don't know how to do it, can you give me a more explicit hint ?

The first excited state is when one of the oscillators is in the E1 state, the second excited state is when one of the oscillator is in the E2 state or one is in E1 state and some other is also in E1 state ... etc Isn't the answer to that question in how many ways I can distribute En energy...

ok, I think that I can find energy eigenstates in this way: \left.\left.H|n\rangle =E_n\right|n\right\rangle \left.\left.\left.\left.\left(\sum _k (\hbar -\mu )a_k{}^+a_k-\frac{1}{2}\hbar \right)\right|n\right\rangle =E_n\right|n\right\rangle \left.\left...

Homework Statement Compute the partition function Z = Tr(Exp(-βH)) and then the average number of particles in a quantum state <nα > for an assembly of identical simple harmonic oscillators. The Hamiltonian is: H = \sum _{k}[(nk+1/2)\hbar - \mu nk] with nk=ak+ak. Do the calculations once...

thanks, if I take x to be a vector with components (x y) then, x†M(a)x = a(xx† + yy†) + (1/4)i(xy† + x†y) this then means a(xx† + yy†)≥(1/4)i(xy† + x†y) but term on the left hand side is real and term on the right hand side may be complex, so how can I conclude when is x†M(a)x≥0 ? also is my...

Homework Statement Given a matrix M(a) = (a -(1/4)i ; (1/4)i a) (semicolon separates rows) a) Determine a so that M(a) is a density matrix. b) Show that the system is in a mixed state. c) Purify M(a) The Attempt at a Solution a) from conditions for a density matrices...

Thanks ! Your hints where very helpful pointing me in the right direction.

The general step is then : Var(1/n(X1+...+X2)) = 1/n*σ^2 + 2/n^2 *(1/2 * (n-1)n ) ρσ^2 = 1/n*σ^2 + (n-1)/n *ρσ^2 = ρσ^2 + 1/n*σ^2 + -1/n*ρσ^2 = ρσ^2 + 1/n*(1-ρ)σ^2

yes, I can do it , so: 1/2σ^2 + 1/2*ρ*σ^2 = 1/2*(2ρσ^2-ρσ^2) + 1/2*σ^2 = ρσ^2 - 1/2*ρσ^2 + 1/2*σ^2 = ρσ^2 + 1/2*(σ^2-ρσ^2) = ρσ^2 + 1/2*(1-ρ)σ ^2 So the next step is to try it with n instead of 2 ?, I'll try that now

but maybe I can rearrange this to your form ...I'll try it

The best thing I could do is : Var(1/2(X+Y))= Var(1/2*X + 1/2*Y)= 1/4σx^2 + 1/4σy^2 + 2*1/4*ρ Sqrt(σx^2)*Sqrt(σy^2) =1/4σ^2 + 1/4σ^2 +1/2*ρ*σ^2

right, so when X and Y are correlated Var(X+Y) = Var(X) + Var(Y) + 2ρ*Sqrt(Var(X))*Sqrt(Var(Y)) where ρ is corelattion coefficient between X and Y

I think I could, I would start with this : Var(X+Y) = Var(X) + Var(Y) = E[(X-μx)^2] - E[(Y-μy)^2] where μx is the mean of RV X, and μy the mean of RV Y

Homework Statement Show that for identically distributed, but not necessarily independent random variables with positive pairwise correlation ρ, the variance of their average is ρσ^2 + (1-ρ)σ^2/B. ρ - pairwise corellation σ^2 - variance of each variable B - number of samples...