Understanding Fermionic Particles and Quantum States

The question asks about a system containing two fermionic particles that can occupy any of three available quantum states. We need to find the ratio of the probability that the two particles are in the same state to the probability that they are in different states.

Fermionic particles are governed by Fermi-Dirac statistics, the most important consequence of which is the Pauli Exclusion Principle.

Pauli Exclusion Principle for Fermions

The Pauli Exclusion Principle states that no two identical fermions can occupy the same single-particle quantum state simultaneously. This means if we have two identical fermions, they must always be in different quantum states.

Probability of Particles in the Same State

We have two identical fermionic particles. There are three possible quantum states available for them to occupy. According to the Pauli Exclusion Principle, it is physically impossible for two identical fermions to be in the exact same quantum state (same state number like state 1 and state 1). Therefore, the number of possible configurations where the two particles are in the same state is zero.

Since there are no allowed configurations where the particles are in the same state, the probability of finding the two fermionic particles in the same state is 0.

Probability (Same State) = 0

Probability of Particles in Different States

Since identical fermions cannot be in the same state, they must occupy different states. We have three available quantum states: State 1, State 2, and State 3.

We need to find the number of ways to place the two identical fermions into two different states out of the three available states. This is equivalent to choosing 2 distinct states out of 3, which can be calculated using combinations.

The number of ways to choose 2 states from 3 is given by the combination formula:

\(\binom{n}{k} = \frac{n!}{k!(n-k)!}\)

Here, \(n = 3\) (total number of states) and \(k = 2\) (number of states to choose for the two particles).

Number of ways = \(\binom{3}{2} = \frac{3!}{2!(3-2)!} = \frac{3!}{2!1!} = \frac{3 \times 2 \times 1}{(2 \times 1) \times 1} = 3\)

The possible pairs of distinct states the two particles can occupy are:

• State 1 and State 2
• State 1 and State 3
• State 2 and State 3

These are the only physically allowed configurations for two identical fermions in three states. In all these allowed configurations, the particles are in different states.

Let's consider the total number of accessible quantum states for the two-fermion system. Based on the Pauli exclusion principle, only the configurations where the two particles occupy different states are accessible. Thus, the total number of accessible states is 3.

Since all 3 accessible states correspond to the particles being in different states, the probability of finding the particles in different states is 1 (assuming all accessible states are equally likely).

Probability (Different States) = 1

Calculating the Ratio

The question asks for the ratio of the probability that two particles are in the same state to that when the two particles are in different states.

Ratio = \(\frac{\text{Probability (Same State)}}{\text{Probability (Different States)}}\)

Ratio = \(\frac{0}{1} = 0\)

Therefore, the ratio of the probability that two fermionic particles are in the same state to that when they are in different states is 0.