Recent content by Ashish Somwanshi

Is the answer 0.748?

I was not able to attempt since I don't know which formula or method can be used to solve the problem

My solution to above assignment goes like this: Since | ×±> = 1/ sqrt(2) |z+> ± 1/sqrt(2) |z-> |x+> = 1/sqrt (2) |z+> + 1/sqrt(2) |z-> |x-> = 1/sqrt(2) |z+> - 1/sqrt(2) |z-> So eigenvalue equations are: Sx |x+> = 1/sqrt(2) { |z+> + |z->} Sx |x-> = 1/sqrt(2) { |z+> - |z->} So we can...

This screenshot contains the original assignment statement and I need help to solve it. I have also attached my attempt below. I need to know if my matrices were correct and my method and algebra to solve the problem was correct...

Thanks extremely, @kuruman and @TSny , I've passed the course and got the correct answer of 0.217nm for penetration depth for which the probability density reduces to 1/2.71828 of its initial value at z=0 of the potential step.

Isn't this problem already solved at chegg? Can anyone can check it up on chegg and give a hint or method of action?

I am not getting the value of eta from method suggested by @kuruman

I got A^2=(1/e)A^2, now how I am supposed to solve for eta?

Psi^2(z)=A^2*e^(-2z/eta)

Just square the absolute value of wavefunction it gives the probability density. So how do I find the penetration depth where the probability density reduces to 1/eof its initial value at z=0. There must be some well knownformula? Is it related to transmission probability?

There was no derivation of this formula in my notes. I found this formula from the net. It seems now I have to provide my original notes for further inspection. As soon as I finish writing my complete notes. I will post it here.

Formula is independent of probability or probability amplitude and autograder says my answer of 0.434nm is incorrect. I have just substituted values given E and Vo and mass of electron and hket in the above formula for penetration depth in case of finite potential well. What has probability and...

I don't understand where I went wrong, the formula and calculations which I have attached are correct...please do help if anyone can spot the mistake.

Thanks guys, I've finally solved the question, to find eta value for second energy we had to find intersection of curve with cot(eta) function since it has odd parity and hence find the energy. I've got the correct result for second part of the question as well.

I got two values of eta Eta 1=1.466 from which ground state energy was 0.00325 which was correct according to the autograder. Eta 2 = 4.395 from which second energy was 0.02922, for some reason autograder rejected it. How is it possible for the same method one correct and another incorrect answer?