Recent content by dyn

Both books in the OP are great. If you want simpler explanations than Boas then Martin & Shaw is the book for you. RHB is at a higher level and covers much more. If you get both books they will take you from pre-Boas to post-Boas

The trig formula 2sinAsinB = cos (A-B) - cos (A+B) applies for A=B and A≠B so when i use that formula in the integral why does it not apply for A=B ? I see your point that it gives cos(A-B) which is 1 when A=B so integrating 1 ends up giving the correct answer. But what i don't understand is why...

If i integrate cos (πx/L) ( n-k) , i get L/(π(n-k))sin (πx/L)(n-k) which is undefined at n=k

Hi If i calculate the definite integral between the limits of L and 0 of sin(nπx/L)sin(kπx/L) using the trig formula 2sinAsinB = cos (A-B) - cos (A+B) it is undefined when n=k because (n-k) appears in the denominator. If i calculate the same integral with n=k using the formula sin2(nπx/L) = (...

I also recommend that book ( and his other books ) but you need to keep a close eye on the typos. If you follow any derivations in the book , you should find the typos easy to spot

The derivation of the path integral in QM on P211-212 of "QFT for the gifted amateur" by Lancaster & Blundell

Yes . i know that but x can act on a position eigenstate and p can act on a momentum eigenstate. I just want to check that once the operators act on states to produce eigenvalues , that eigenvalues commute because they are just numbers ?

Hi I just wanted to check my understanding of something which has come up when first studying path integrals in QM. If x and px are operators then [ x , px ] = iħ but if x and px operate on states to produce eigenvalues then the eigenvalues x and px commute because they are just numbers. Is...

Thank you very much for your help

Hi I am using the textbook "Modern Particle Physics" by Thomson. Working from the K-G equation and comparing with the continuity equation he states that the probability density is given by ρ = i ( ψ*(∂ψ/∂t) - ψ(∂ψ*/∂t) ) He then states that the factor of i is included to ensure that the...

So if i have x2 = 9 and i take the square root of both sides i get x = 3 which is obviously not the full solution, so taking the square root of both sides does not give the full solution ?

Thanks everyone. If the inequality is turned into an equals sign and i have 9 = x2 and i take the square root of both sides do i have ±3 = ±x ?

Thanks. So . for a first step if i take the square root of both sides of 9 > x2 what do i get ?

Hi If i have the inequality , 9 > x2 then i know the answer is , -3 < x < +3 but my confusion lies in the following ; if i take the square root of both sides of the inequality i get , ± 3 > ±x Is that correct ? If so , it leads to the following solutions x < 3 , x > -3 , x < -3 , x >3 ...

I might be confusing myself here but if A is not Hermitian and A2 = AA and A3 = AAA then how do i write the expectation values of these 2 quantities ?