Okay, so I need to solve this quantum mechanics problem about four identical spin-1/2 particles in a one-dimensional infinite potential well. The question asks for the energies of the ground state, first excited state, and second excited state. The choices are given as options A to D. Let me start by recalling what I know about particles in a 1D infinite potential well and their energy levels.

First, for a single particle in a 1D infinite potential well of length L, the energy levels are quantized. The energy levels are given by the formula E_n = (n² π² ħ²)/(2mL²), where n is the principal quantum number (n=1,2,3,...). The problem states that E = π² ħ²/(2mL²), so each energy level can be written as E_n = n² E. That simplifies things a bit because all energies are multiples of E.

Now, the system has four identical spin-1/2 particles. Since they are identical fermions (as spin-1/2 particles are fermions), their total wavefunction must be antisymmetric under the exchange of any two particles. The wavefunction is a product of spatial and spin parts. For fermions, the Pauli exclusion principle applies, meaning no two particles can have the same quantum numbers. However, since the particles are spin-1/2, each spatial state can hold up to two particles with opposite spins (up and down).

But wait, in a 1D box, the spatial wavefunctions are determined by the quantum number n. Each spatial state (n) can have two particles (spin up and spin down). So, for four particles, we need to fill the lowest available energy levels with two particles each until all four are accommodated.

Let me think step by step. The ground state would be the lowest possible energy configuration. Since each energy level can hold two particles (due to spin), the first two particles would occupy the n=1 state with opposite spins. Then the next two would go into the next available level, n=2, each with opposite spins. So, the ground state would have two particles in n=1 and two in n=2.

But wait, the question is about the energies of the ground state, first excited state, and second excited state. Wait, no. The question says "calculate the energies of the ground state, first excited state, and second excited state of the system". So, each of these states refers to a different configuration of the particles. Let me clarify.

In a system with multiple particles, the total energy is the sum of the individual energies. However, when particles are fermions, their arrangement (the spatial wavefunction and spin) affects the total energy. However, I think in this problem, since they are spin-1/2 and the particles are identical, the spatial wavefunction has to be antisymmetric if the spin wavefunction is symmetric, and vice versa. But since the particles are spin-1/2, their spin state can be either symmetric (triplet) or antisymmetric (singlet). The spatial wavefunction would then be antisymmetric or symmetric accordingly.

Wait, but for identical fermions, the total wavefunction (spatial * spin) must be antisymmetric. So if the spin part is symmetric (triplet state), the spatial part must be antisymmetric. Conversely, if the spin part is antisymmetric (singlet state), the spatial part must be symmetric.

But in this case, since the particles are spin-1/2, the spin state can be either symmetric (triplet, which has S=1) or antisymmetric (singlet, S=0). However, the spatial part's symmetry depends on the spin. But how does this affect the energy?

Wait, but in a 1D box, the spatial wavefunctions are determined by the quantum numbers. The spin part doesn't affect the spatial energy. So the total energy of the system is the sum of the individual energy levels. So each particle's energy is determined by its quantum number n, regardless of spin. So the spin affects the arrangement of the particles (whether they can occupy the same spatial state or not), but the energy is just the sum of each particle's individual E_n.

Therefore, the problem reduces to finding the lowest possible arrangement of four particles (spin-1/2) in the 1D box, considering the Pauli exclusion principle. Each spatial level can hold two particles (with opposite spins), so the first two particles go into n=1, and the next two into n=2. Then, the next two would go into n=3, etc.

Wait, but there are four particles. Let's count:

n=1: 2 particles (up and down)
n=2: 2 particles (up and down)
So all four are accommodated. Therefore, the ground state would have two particles in n=1 and two in n=2. The total energy would be 2*(1² E) + 2*(2² E) = 2E + 8E = 10E. So the ground state energy is 10E.

But the question is asking for the energies of the ground state, first excited state, and second excited state. So what are these states?

Wait, the ground state is the lowest possible energy configuration, which is the one I just described: two particles in n=1 and two in n=2. The first excited state would be the next lowest energy configuration. How do we get excited states?

An excited state would be when we have a different arrangement of the particles in the spatial states. Since the particles are fermions, the minimal energy is achieved when the lowest available spatial states are filled with the maximum number of particles. So to get the first excited state, we need to have the lowest possible energy configuration that's not the ground state.

The ground state is two particles in n=1 and two in n=2. To get the first excited state, perhaps we promote one of the particles from n=2 to a higher n=3 state. But since each n can hold two particles, we can't have more than two in any n. So if we move one particle from n=2 to n=3, but then we have three particles in n=1 and n=2 and one in n=3. Wait, but each n can hold two particles. So if we have two in n=1, one in n=2, and one in n=3. But that's only three particles. Wait, no, the total number must be four.

Wait, let me think again. The ground state is two in n=1 and two in n=2. If we want the first excited state, we need to have a configuration where the total energy is higher than the ground state. So the minimal change would be to have one particle in a higher state. Let's try to see.

If we promote one particle from n=2 to n=3, then the configuration becomes two in n=1, one in n=2, and one in n=3. Wait, but each n can hold two particles. So if we have two in n=1, one in n=2 (since we can't have two there anymore if we promote one), and one in n=3. Wait, but that's three particles. We need four. Wait, no, the total number is four. So if we promote one particle from n=2 to n=3, then we have two in n=1, one in n=2, and one in n=3. But that's only four particles. So the energy would be 2*(1² E) + 1*(2² E) + 1*(3² E) = 2E + 4E + 9E = 15E. So the first excited state would be 15E?

But wait, is there a lower energy configuration than that? Let's see. Alternatively, perhaps we can have two in n=1, two in n=2, but with a different spin arrangement. But no, the spin arrangement doesn't affect the spatial energies. So the energy is solely determined by the sum of the n² terms.

Alternatively, maybe the first excited state is when three particles are in n=1 and one in n=2. But wait, n=1 can hold two particles. So three particles in n=1 would violate the Pauli exclusion principle. Therefore, that's not allowed. So the next possibility is to have two in n=1, two in n=2. To get an excited state, we need to have a configuration where some particles are in higher n states.

Wait, perhaps the first excited state is when one particle is in n=1, one in n=2, and two in n=3. Let's check the total particles: 1+1+2=4. So the energy would be 1² E + 1² E + 2*3² E = E + E + 18E = 20E. But 20E is higher than 15E, so 15E is lower. So 15E would be the first excited state? But wait, another possibility: if we have two in n=1, one in n=2, and one in n=3. That's four particles. The energy is 2*1 + 1*4 + 1*9 = 2 + 4 + 9 = 15. So that's 15E. But is that the first excited state?

Alternatively, could we have two in n=1, two in n=2, but with a different spin configuration? Wait, but spin configuration doesn't affect the spatial energy. The energy is just the sum of the individual energies. So maybe the first excited state is when one particle is in n=1, two in n=2, and one in n=3. Wait, but that's 1+2+1=4. The energy would be 1 + 8 + 9 = 18. Wait, but that's 18E. But 18E is higher than 15E. So 15E is lower. So 15E would be the first excited state?

Wait, but wait. Let me check again. Let's think about how the particles fill up the states. For four particles, each n can hold two. So the ground state is n=1 (2 particles), n=2 (2 particles). The first excited state would be when one particle is promoted to n=3. So the configuration is n=1 (2), n=2 (1), n=3 (1). That's four particles. The energy is 2*1 + 1*4 + 1*9 = 2 + 4 + 9 = 15E. So the first excited state is 15E.

Then the second excited state would be the next higher configuration. Let's see. After 15E, the next possible configuration could be when another particle is promoted. For example, if two particles are in n=3. But wait, n=3 can hold two particles. So the configuration could be n=1 (2), n=2 (2), n=3 (2). Wait, but that's six particles. We only have four. So perhaps, the second excited state is when one particle is in n=3 and another in n=4. Let's check.

Wait, no. Let's think step by step. The ground state is 10E. First excited state is 15E. Then the second excited state would be the next possible configuration. Let's see. After 15E, what's the next possible configuration? Let's see. The next possible would be when one particle is in n=3 and another in n=4. But that would require two particles in n=3 and one in n=4, but that's three particles. Wait, no. Let's think again.

Alternatively, perhaps the second excited state is when two particles are in n=3. Wait, but n=3 can hold two particles. So if we have two in n=3, that's four particles. But that's the same as the ground state but with two particles in n=3 instead of n=2. Wait, no. Wait, n=3 can hold two particles. So if we have two in n=3, that's two particles. Then, we need two more particles. But n=1 can hold two. So the configuration would be n=1 (2), n=3 (2). The energy would be 2*1 + 2*9 = 2 + 18 = 20E. But that's higher than 15E. So maybe that's not the second excited state.

Wait, but another configuration could be n=1 (2), n=2 (1), n=3 (1), n=4 (0). Wait, but that's the same as the first excited state. Hmm. Alternatively, maybe the second excited state is when two particles are in n=2 and two in n=3. Let's see. The energy would be 2*(2²) + 2*(3²) = 8 + 18 = 26E. That's higher than 15E. So 26E would be higher than 15E. But that's a higher energy than the first excited state. So maybe the order of excited states is:

Ground state: 10E

First excited state: 15E

Second excited state: 20E?

Wait, but how? Let me think. If the first excited state is 15E (two in n=1, one in n=2, one in n=3), then the second excited state would be when another particle is promoted. For example, moving one particle from n=3 to n=4. So the configuration would be two in n=1, one in n=2, one in n=4. The energy would be 2*1 + 1*4 + 1*16 = 2 + 4 + 16 = 22E. But that's higher than 20E. Alternatively, maybe there's a lower configuration.

Wait, perhaps the second excited state is when two particles are in n=3. Wait, but that would require two particles in n=3, but n=3 can hold two. So if we have two in n=3, that's four particles (since n=1 and n=2 are empty). The energy would be 2*9 = 18E. But 18E is higher than 15E. So 18E would be the second excited state? But 18E is higher than 15E, so that's possible.

Wait, but let's list all possible configurations in order of increasing energy. The ground state is 10E. The first excited state is when a single particle is promoted from n=2 to n=3, resulting in 15E. The next possible promotion could be moving another particle from n=2 to n=3, but that would require two in n=3, which would make the configuration two in n=1, two in n=3. The energy would be 2*1 + 2*9 = 2 + 18 = 20E. Alternatively, moving a particle from n=3 to n=4, but that's higher.

Alternatively, perhaps the second excited state is when three particles are in n=1 and one in n=2. But that's not possible because n=1 can hold two particles. So three in n=1 is not allowed. So the next possible is two in n=1, one in n=2, and one in n=3 (15E), then two in n=1, one in n=2, and two in n=3 (20E). Wait, but two in n=3 would require n=3 to have two particles, but earlier in the first excited state, only one was in n=3. So if we have two in n=3, that's a different configuration. Let's check.

Wait, but in the first excited state, we have two in n=1, one in n=2, and one in n=3. The total is four particles. The second excited state would be when another particle is in a higher state. For example, moving one particle from n=3 to n=4. So configuration: two in n=1, one in n=2, and two in n=4. The energy would be 2*1 + 1*4 + 2*16 = 2 + 4 + 32 = 38E. That's even higher. Alternatively, moving two particles from n=2 to n=3. So two in n=1, two in n=3. The energy is 2*1 + 2*9 = 2 + 18 = 20E. So 20E would be the second excited state.

But wait, is there a configuration between 15E and 20E? For example, what if we have one particle in n=1, two in n=2, and one in n=3. The total is 1+2+1=4. The energy would be 1 + 8 + 9 = 18E. So 18E is between 15E and 20E. So is that a possible configuration?

Wait, but can we have one particle in n=1, two in n=2, and one in n=3? Let me check the Pauli principle. Each spatial state can hold two particles (spin up and down). So n=1 can have two, n=2 can have two, n=3 can have two. So yes, having one in n=1, two in n=2, and one in n=3 is allowed. The total energy would be 1*1 + 2*4 + 1*9 = 1 + 8 + 9 = 18E. So 18E is a possible energy. So that would be the second excited state. Then the third excited state would be 20E. Wait, but then the order would be:

Ground state: 10E

First excited state: 15E

Second excited state: 18E

Third excited state: 20E

But the problem only asks for the first three states: ground, first excited, second excited. So in the problem statement, the options are asking for the energies of the ground state, first excited, second excited. Let's see the options again.

Options:

A. 4E, 10E, 50E

B. 30E, 39E, 50E

C. 4E, 10E, 15E

D. 10E, 15E, 18E

Wait, option D is 10E,15E,18E. So according to my previous reasoning, the ground state is 10E (two in n=1, two in n=2), first excited 15E (two in n=1, one in n=2, one in n=3), second excited 18E (one in n=1, two in n=2, one in n=3). So option D matches this. But wait, let me check again.

Wait, another possibility: if the first excited state is when three particles are in n=1 and one in n=2. But that's not allowed because n=1 can only hold two. So the next possible is two in n=1, one in n=2, one in n=3. Then the second excited state would be two in n=1, two in n=3. Let's compute the energies.

First excited state (15E):

Two in n=1: 2*1=2E

One in n=2: 1*4=4E

One in n=3: 1*9=9E

Total: 15E.

Second excited state (18E):

Two in n=1: 2E

Two in n=3: 2*9=18E

Total: 20E? Wait, no. Wait, if we have two in n=3, then the configuration is two in n=1, two in n=3. So 2E + 2*9E = 2 + 18 = 20E. But earlier, the configuration with one in n=1, two in n=2, one in n=3 gives 18E. So which one is lower?

18E is lower than 20E. So the correct order would be:

Ground state: 10E

First excited: 15E

Second excited: 18E

Third excited: 20E

So the energies would be 10E,15E,18E, which matches option D. But let me check the options again.

Option D is 10E,15E,18E. But the problem states "the energies of the ground state, first excited state, and second excited state". So according to this, the correct answer would be D.

But wait, let me check if there's a configuration that results in 10E,15E,20E. Because if the second excited state is 20E, then option D would have 18E, which might not be correct.

Alternatively, perhaps the second excited state is when two particles are in n=3, which would be 20E. Let me check the possible configurations.

Wait, let's list all possible configurations and their energies:

1. All four particles in n=1: Not allowed, since n=1 can only hold two.

2. Two in n=1, two in n=2: Ground state (10E).

3. Two in n=1, one in n=2, one in n=3: 15E.

4. One in n=1, two in n=2, one in n=3: 1*1 + 2*4 + 1*9 = 18E.

5. Two in n=1, two in n=3: 2*1 + 2*9 = 20E.

6. One in n=1, two in n=3, one in n=4: 1 + 2*9 + 1*16 = 1 + 18 +16=35E.

7. Two in n=2, two in n=3: 2*4 + 2*9=8 +18=26E.

So the order of energies from lowest to highest would be:

10E (ground), 15E, 18E, 20E, 26E, 35E, etc.

Therefore, the energies are 10E,15E,18E. So the correct answer is option D: 10E,15E,18E.

But wait, the problem says "four identical spin-1/2 particles", so each can have spin up or down. However, the spin state affects the total wavefunction. But the energy is determined by the spatial part. So the spin state (singlet or triplet) doesn't affect the spatial energy. Therefore, the energies are determined purely by the spatial wavefunctions, which are determined by the quantum numbers n. Therefore, the calculation as above holds.

Therefore, the correct answer should be D: 10E,15E,18E.
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