Ground state and 1st excited state energy of 3 Fermions

[catpotato]

Homework Statement

So in my problem, there's a given of 3 non interacting fermions in a harmonic well potential. I already got the wavefunction but i have problems in obtaining the ground state energy and its 1st excited state energy for 3 fermions (assuming they are non interacting and identical)

Homework Equations

E_n = E₁ + E₂ + E₃
E_n = (n + 1/2) ħω

The Attempt at a Solution

Since they are fermions, i tried applying pauli's exclusion principle so my E_n would be equal to: (n₁ + 1/2) ħω + (n₂ + 1/2) ħω + (n₃ + 1/2) ħω
where n₁ = 1, n₂ = 2, n₃ = 3 for the ground state.

I am not sure if this is right for fermions so I'm kinda stuck and still skeptic on how to solve for the 1st excited state. Any help would be much appreciated.

 

[Charles Link]

I'm no expert in this type of problem, but with opposite spins, shouldn't it be possible to have two particles in the lowest energy state?

 

[catpotato]

I'm no expert in this type of problem, but with opposite spins, shouldn't it be possible to have two particles in the lowest energy state?

i think so too, if there were two particles in the lowest energy then n₁ and n₂ will both be equal to 1? and n₃ will be equal to 2?

 

[Charles Link]

i think so too, if there were two particles in the lowest energy then n₁ and n₂ will both be equal to 1? and n₃ will be equal to 2?

I think the lowest state has $n=0$. I think I have it right, but I am hardly an expert in this area. $\\$ Additional comment: For the wave function, I think it is necessary to use a Slater determinant type waveform, but I would like to hear from someone with considerable Q.M. expertise. $\\$@bhobba Might you have an input here?

 

[DrClaude]

Homework Equations

E_n = E₁ + E₂ + E₃
E_n = (n + 1/2) ħω

You have to be careful with the notation. Does the index n label the particle or the state of the h.o.?

I think the lowest state has $n=0$. I think I have it right, but I am hardly an expert in this area.

Yes, that is correct.

The Attempt at a Solution

Since they are fermions, i tried applying pauli's exclusion principle so my E_n would be equal to: (n₁ + 1/2) ħω + (n₂ + 1/2) ħω + (n₃ + 1/2) ħω
where n₁ = 1, n₂ = 2, n₃ = 3 for the ground state.

I am not sure if this is right for fermions so I'm kinda stuck and still skeptic on how to solve for the 1st excited state. Any help would be much appreciated.

You have to find the combinations of n that give you the lowest energy and then the next lowest energy without violating the Pauli exclusion principle. There might be more than one possible solution for the first excited state.

By the way, you haven't specified the spin of the fermions. It is very important here.

 

[catpotato]

You have to be careful with the notation. Does the index n label the particle or the state of the h.o.?Yes, that is correct.You have to find the combinations of n that give you the lowest energy and then the next lowest energy without violating the Pauli exclusion principle. There might be more than one possible solution for the first excited state.

By the way, you haven't specified the spin of the fermions. It is very important here.

oh i forgot, the E_n there is supposed to be the total energy of the particles and the spin wasn't specified in our homework.

 

[DrClaude]

oh i forgot, the E_n there is supposed to be the total energy of the particles and the spin wasn't specified in our homework.

I'm still not sure what the index n represents. If spin was not specified, then the solution is not unique.