Do I need to consider the spin of electrons when they are in infinite potential well

This is the problem I have And this is one of my books tell me what should I do And my question is: Do I need to consider spin in this case?(that is, I don't think the book is right...)

I found an example of 1-dimensional infinite potential well in griffith's introduction of quantum mechanics, and in that case the author didn't consider the spin.

And as far as I have learned, I think that Pauli exclusion principle state that fermions like electron can't occupy the same state, that is they can't occupy the same quantum number simultaneously. So in my perspection, no need for the consideration of spins since they don't have the same quantum number both in 3-d and 1-d infinite potential well.

While in comparison ,the spin is a quantum number in a hydrogen case we must consider.

Am I right? Or both concepts are right?

• It all depends on the Hamiltonian you are considering. If it doesn't include a term with spin, then your spin states are decoupled. The only thing that you would need to consider is if you are calculating the total energy of many particles. In that case you can only have two per energy state. – Greg Petersen Feb 7 '16 at 16:14
• @Greg Do you mean that I should take spin into consideration in this case? I'm so confusing because if the spin does matter here, maybe the problem should tell me how many electrons there are and what the spin they are, since the spin is an intrinsic angular momentum, right? – cindy Feb 8 '16 at 1:28
• No, you just need to solve the non interacting problem. Normally it would be explicitly stated. – Greg Petersen Feb 8 '16 at 4:47
• But the one who just answered made me more confusing.....so since the problem doesn't mention anything about spin I should ignore that haha – cindy Feb 8 '16 at 12:40
• Well spin is there, however there is no interaction. Thus the Hamiltonian is already diagonal in spin space. So instead of solving the Hamiltonian for spin up and down, you are just solving it for one and can use it for both cases as they are the same. The comment is talking about the stacking in your second picture. I actually don't think the second picture is related to the first problem. – Greg Petersen Feb 8 '16 at 15:55