Recent content by mrjohns

Ok excellent, thanks for the help, will post if I have any issues.

Thanks. Does this mean I should really do it the first way you've said (rather than subtracting) as this would satisfy the lecturer better - or does it not really matter. And if I do it that way, does that mean the f(z)=1/z case won't cause any issues anywhere?

A complex analysis question. Homework Statement Verify the Cauchy theorem by calculating the contour integrals. Where ω is the appropriately orientated boundary of the annulus/donut defined by 1/3 ≤ IzI ≤ 2 for the following analytic functions: i. f(z)=z^2 ii. f(z)=1/z Homework...

I know that for a constant: S(kN,KV,KU)=kS(N,V,U) But I'm not sure how that restricts the power to a third.

I don't have an attempt because I'm completely stumped. The units on the left hand side are J/K, and on the right they are J^(1/3) m - which don't match. I can't see anything in the postulates that helps either: P1 - There exist equilibrium states characterised completely by U, V, N. P2 -...

Homework Statement Consider relationship for a thermodynamic system: S=A[UVN]^d , where A is a constant and d a real number. I need to explain why d=1/3 is the only allowed value consistent with the postulates of thermodynamics. The Attempt at a Solution I'm having a hard time...

Cheers, thanks so much for your help.

Ah, so: Prob:(ψ1) = 1/4 [2+e^-i(ω1-ω2)t+e^+i(ω1-ω2)t] Prob:(ψ1) = 1/2 {1+cos[(ω1-ω2)t]} Prob:(ψ2) = 1/4 [2-(e^-i(ω1-ω2)t+e^+i(ω1-ω2)t)] Prob:(ψ2) = 1/2 {1-cos[(ω1-ω2)t]} Now I can see that looks a lot better - the probabilities for each are bound between 0 and 1, and Prob:(ψ1)=1 at t=0. When...

Whoops, so I should have typed: Prob:(ψ1) = 1/4 [e^-iω1t + e^-iω2t][e^+iω1t + e^+iω2t] Prob:(ψ1) = 1/4 [1+e^-i(ω1-ω2)t+e^-i(ω2-ω1)t+1] Prob:(ψ1) = 1/4 [2+e^-i(ω1-ω2)t+e^-i(ω2-ω1)t] And similarly: Prob:(ψ2) = 1/4 [2-e^-i(ω1-ω2)t-e^-i(ω2-ω1)t] Thus Prob:(ψ1) + Prob:(ψ2) + Prob:(ψ3) = 1/4...

So does the methodology here look right? <ψ1|ψ(t)> = {1/√2 [< 2 1 -1 | - < 2 1 1 |]}{1/√2 [e-iω1t | 2 1 -1 > - e-iω2t | 2 1 1 > ] } <ψ1|ψ(t)> = 1/2 [< 2 1 -1 | e-iω1t | 2 1 -1 > + < 2 1 1 | e-iω2t | 2 1 1 >] <ψ1|ψ(t)> = 1/2 [e^-iω1t + e^-iω2t] by orthonormality And similarly: <ψ2|ψ(t)> = 1/2...

Using matrix multiplication and putting bras to kets i get: Prob amp for |ψ1> = 1/2 [e^-iω1t + e^-iω2t] => Probability is one when e^-iω1t = e^-iω2t = 1 Prob amp for |ψ2> = 1/2 [e^-iω1t - e^-iω2t] => Probability is one when e^-iω1t = -e^-iω2t = 1 Prob amp for |ψ3> = 0 for all time Now...

Thanks for the reply. If I say the energies are E=E(n)-E(b)*mh I get by time evolution: |psi(t)> = 1/sqr(2) [ e^-i(E(n)+E(b))t | 2 1 -1 > - e^-i(E(n)-E(b))t | 2 1 1 > ] Which by Eulers formula: =1/sqrt(2) { [ cos[(-E(n)-E(b))t] + isin[(-E(n)-E(b))t] ] | 2 1 -1 > - [ cos[(-E(n)+E(b))t] +...

I'm studying the hydrogen atom and have this question. Apparently it can be solved without perturbation theory, however I'm having trouble justifying it. Homework Statement 2. The attempt at a solution Avoiding perturbation theory I simply get: E = E(n) - constant*(mh) where m...