Recent content by dipanshum

Okay so, yes Orodruin you are right, and I was wrong. The final normalised function would be ψ=1/√3 ψ1 + √(2/3)ψ 2. Tell me if I'm right this time.

no only probability of ψ is given, but that is sufficient data. and what you suggested already helped me and the answer is right. what happened is after normalization one will get α2+β2=1 then calculate and put values of α, β in the equation ψ=αψ1+βψ2

The probability of ψ1 is given in the question.

Ok, I worked out the sum. Please tell me if I'm right. The normalized wavefunction is: ψ = 1/3 ψ1 + 2√2/3 ψ2

Should I treat ψ1 as ψ and ψ 2 as ψ*?

You first need to prove [x , pn] = niħpn-1. In order to do that take arbitrary values of n=1, 2, 3 and check. You'll see for n=1 [x , p] = iħ result holds (as it is given and also we know) for n=2 [x , p] = p[x , p] +[x , p]p = piħ +ipħ = 2ipħ = 2iħp2-1 again result holds...