Particles with Spin in a Box - Energy Levels

[erok81]

Homework Statement

This is a two part problem. I think I am pretty close, but I'm quite understanding it. Which sucks because this is probably an extremely easy question.

I have five particles in a box with S spin. I am to find the ground state of the five particles if they have spin 0,½,1,3/2.

Homework Equations

My energy in box (infinite potential well) is given by:

$$E=\frac{n^2 \pi^2 \hbar^2}{2mL^2}$$

The Attempt at a Solution

For S=0,1 all particles are Bosons and therefore can all fit in the same level. Therefore my ground state energy is given by:

$$E= \frac{5 \cdot 1^2 \pi^2 \hbar^2}{2mL^2}~=~ \frac{5 \pi^2 \hbar^2}{2mL^2}$$

For my S=½ I can fit 2n² where the factor of two arises because the allowed states are doubles since I can fit ± 1/2 spins per level. Since I have five particles, I will occupy up to n=2. Therefore my ground state energy is given by:

$$E= \frac{2(n_{1}^{2}+n_{2}^{2}) \pi^2 \hbar^2}{2mL^2}~=~ \frac{5 n^2 \pi^2 \hbar^2}{mL^2}$$

And finally for my S=3/2 I can fit a factor of 4 per level (±3/2,±1/2). Using the same justification as in S=½ I arrive at the ground energy of:

$$E= \frac{10 n^2 \pi^2 \hbar^2}{2mL^2}$$

Since the first level will contain four particles and n=2 will have one.

First. Is this correct so far?

The next part is where I am having the most trouble.

I have to find the excited state for all spins in the first half.

I know in order to do this I only need to move one particle to the next energy level. So for S=1,0 I think I have it. They are all in the ground state and only one needs to move up - four stay in the ground state and one moves up. Therefore I have.

$$E=\frac{4 \cdot 1^2 \pi^2 \hbar^2}{2mL^2}+\frac{2^2 \pi^2 \hbar^2}{2mL^2}~=~ \frac{4 n^2 \pi^2 \hbar^2}{mL^2}$$

Is this looking okay so far?

That's about as far as I am comfortable with. The next spin I tried was S=½.

Here I have two particles in n=1 and three particles in n=2. Similar to the above case I only need to move one particle up. So I would have two in n=1, two in n=2, and one in n=3 to raise it to the first excited state. I know that much, but when I try to write it out, I am stuck. I think it's that factor of two that throws me off. I don't know if I should include it in each level.

So how does the work look so far? Am I even close?

 

[vela]

Homework Statement

This is a two part problem. I think I am pretty close, but I'm quite understanding it. Which sucks because this is probably an extremely easy question.

I have five particles in a box with S spin. I am to find the ground state of the five particles if they have spin 0,½,1,3/2.

Homework Equations

My energy in box (infinite potential well) is given by:

$$E=\frac{n^2 \pi^2 \hbar^2}{2mL^2}$$

The Attempt at a Solution

For S=0,1 all particles are Bosons and therefore can all fit in the same level. Therefore my ground state energy is given by:

$$E= \frac{5 \cdot 1^2 \pi^2 \hbar^2}{2mL^2}~=~ \frac{5 \pi^2 \hbar^2}{2mL^2}$$

For my S=½ I can fit 2n² where the factor of two arises because the allowed states are doubles since I can fit ± 1/2 spins per level. Since I have five particles, I will occupy up to n=2. Therefore my ground state energy is given by:

$$E= \frac{2(n_{1}^{2}+n_{2}^{2}) \pi^2 \hbar^2}{2mL^2}~=~ \frac{5 n^2 \pi^2 \hbar^2}{mL^2}$$

And finally for my S=3/2 I can fit a factor of 4 per level (±3/2,±1/2). Using the same justification as in S=½ I arrive at the ground energy of:

$$E= \frac{10 n^2 \pi^2 \hbar^2}{2mL^2}$$

Since the first level will contain four particles and n=2 will have one.

First. Is this correct so far?

No, you didn't get the fermions right. There's only one spatial state for each n for the particle in the box, not n² states. You're thinking of the hydrogen atom.

The next part is where I am having the most trouble.

I have to find the excited state for all spins in the first half.

I know in order to do this I only need to move one particle to the next energy level. So for S=1,0 I think I have it. They are all in the ground state and only one needs to move up - four stay in the ground state and one moves up. Therefore I have.

$$E=\frac{4 \cdot 1^2 \pi^2 \hbar^2}{2mL^2}+\frac{2^2 \pi^2 \hbar^2}{2mL^2}~=~ \frac{4 n^2 \pi^2 \hbar^2}{mL^2}$$

Is this looking okay so far?

That's about as far as I am comfortable with. The next spin I tried was S=½.

Here I have two particles in n=1 and three particles in n=2. Similar to the above case I only need to move one particle up. So I would have two in n=1, two in n=2, and one in n=3 to raise it to the first excited state. I know that much, but when I try to write it out, I am stuck. I think it's that factor of two that throws me off. I don't know if I should include it in each level.

So how does the work look so far? Am I even close?

 

[erok81]

Ok...that's what I originally thought, but got off track looking around on Wikipedia.

So disregard my fermions above. New energies given by:

spin-1/2 I can have two per energy level. Therefore I need three energy levels to get all five.

$$E=\frac{(1^2+2^2+3^2) \pi^2 \hbar^2}{2mL^2}~=~ \frac{15 \pi^2 \hbar^2}{2mL^2}$$

Then for my spin-3/2 I can have four per level. Therefore I only need two levels for all five particles.

$$E=\frac{(1^2+2^2) \pi^2 \hbar^2}{2mL^2}~=~\frac{5 \pi^2 \hbar^2}{2mL^2}$$

Does that look better?

 

[vela]

Yes, that's right.

 

[erok81]

For the next part, to get to the next energy (spin-1/2) I will need to move two particles, right? Since I can't only move the one currently in n=3, since that will leave n=3 empty.

At the ground state I have two in n=1, two in n=2, and one in n=3.

The first excited state I would need two in n=1, one in n=2, one in n=3, and one in n=4?

 

[vela]

No, you should only have to move one. The only restriction is you can't end up with 3 particles in the same state.

 

[erok81]

Ooooh so it doesn't necessarily have to start a new energy level (like I did with n=4) just that one particle needs to move up an energy level. Of course keeping to the exclusion rule.

So in that case I'd only need to move one from n=2 to n=3.

Then I'd have two in n=1, one in n=2, and two in n=3?

 

[vela]

Yes. You didn't explicitly state it, but I assume you were looking for the configuration with the next lowest energy level after the ground state. If you're just looking for any excited state, you could bump any of the particles up as long as you don't end up with 3 with the same n.

 

[erok81]

Yeah, sorry. I am looking for the first excited state. In this case spin-1/2

I tried to write it out, but I'm not sure how to show this with my energy equation.

My ground state energy for spin-1/2 is

$$<br /> E=\frac{(1^2+2^2+3^2) \pi^2 \hbar^2}{2mL^2}~=~ \frac{15 \pi^2 \hbar^2}{2mL^2}<br />$$

I am still only occupying the same energy levels as the ground state (n=1,2,3). Except now I have one less particle in n=2 and one more in n=3.

So unless my ground state is wrong, I don't see how it will change.

 

[vela]

Sorry, I was mistaken in post 4. The energies you listed in post 3 aren't correct. You want to sum the energies of each particle, so if you have two particles in the n=1 state, the total energy would be 2E₁, for instance.

 

[erok81]

Ah, okay. That makes sense now. I can also see how to bump up the energies. I'll write these out and post again.

So for the spin-1/2 would I have:

$$E=\frac{(2(1^2)+1(2^2)+2(3^2)) \pi^2 \hbar^2}{2mL^2}~=~ \frac{12 \pi^2 \hbar^2}{mL^2}$$