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# Section 13.2: Two Particles in a One-dimensional Infinite Well

Please wait for the animation to completely load.

What if we allowed *two identical particle*s into a single infinite well? If they are non-interacting, the particles would move independently of each other. (They would not even collide; they would pass directly through each other!) For one particle we find that for an infinite well from 0 < *x* < *L*, ψ_{n}(*x*) = (2/*L*)^{1/2} sin(*n*π*x*/*L*) and *E*_{n }= *n*^{2}π^{2}*ħ*^{2}/(2*mL*^{2}). Now add another particle. The Hamiltonian has changed with the addition of terms that involve the second particle. The total wave function must now be a function of both positions *x*_{1} and *x*_{2}:

[−(*ħ*^{2}/2*m*)(∂^{2}/∂*x*_{1}^{2}) − (*ħ*^{2}/2*m*)(∂^{2}/∂*x*_{2}^{2})] Ψ(*x*_{1},*x*_{2}) = *E*Ψ(*x*_{1},*x*_{2}) .

Since each particle (one at *x*_{1} and one at *x*_{2}) is independent of each other, solving this problem proceeds like the two-dimensional case in Section 13.1 and we find that Ψ_{n1 n2}(*x*_{1},*x*_{2}) = (2/*L*)sin(*n*_{1}π*x*_{1}/*L*)sin(*n*_{2}π*x*_{2}/*L*) and

*E*_{n1 n2 } = *n*_{1}^{2}π^{2}*ħ*^{2}/(8*mL*^{2}) + *n*_{2}^{2}π^{2}*ħ*^{2}/(8*mL*^{2}) = π^{2}*ħ*^{2}/(8*mL*^{2}) [*n*_{1}^{2} + *n*_{2}^{2}] ,

which yields the same energy spectrum as in Section 13.1. This degeneracy, however, is due to the *exchang*e of particle one (*n*_{1}) at position *x*_{1} with particle two (*n*_{2}) at position *x*_{2}. This is called an exchange degeneracy. Any time we have two solutions that give the same energy, we can consider a linear combination of these solutions as another solution (since this linear solution will have the same energy as its composite states). For the case of two particles in a one-dimensional well, we combine the ψ_{12}(x_{1},x_{2}) with the ψ_{12}(x_{2},x_{1}) solution as

Ψ_{S}(*x*_{1},*x*_{2}) = 2^{−1/2} [Ψ_{12}(*x*_{1},*x*_{2}) + Ψ_{12 }(*x*_{2},*x*_{1})], (13.5)

and

Ψ_{A}(*x*_{1},*x*_{2}) = 2^{−1/2} [Ψ_{12}(*x*_{1},*x*_{2}) −Ψ_{12 }(*x*_{2},*x*_{1})], (13.6)

as a symmetric and an antisymmetric solution, respectively. Ψ_{S}(x_{1},x_{2}) is symmetric under interchange because Ψ_{S}(*x*_{1},*x*_{2}) = Ψ_{S}(*x*_{2},*x*_{1}). Likewise, Ψ_{A}(*x*_{1},*x*_{2}) is antisymmetric under interchange because Ψ_{A}(*x*_{1},*x*_{2}) = −Ψ_{A}(*x*_{2},*x*_{1}). We can take the square |ψ|^{2} of each wave function and we find that these expressions are invariant to the exchange of the particles

|Ψ_{S}(*x*_{1},*x*_{2})|^{2} = |Ψ_{S}(*x*_{2},*x*_{1})|^{2} (13.7)

and

|Ψ_{A}(*x*_{1},*x*_{2})|^{2} = |Ψ_{A}(*x*_{2},*x*_{1})|^{2}, (13.8)

respectively, where |Ψ_{S}(*x*_{1},*x*_{2})|^{2 }≠ |Ψ_{A}(*x*_{1},*x*_{2})|^{2}. Shown in the animation are the probability densities for two identical particles in an infinite well with *L* = 2. The *x* axis corresponds to the position of particle 1 and the *y* axis corresponds to the position of particle 2. The probability densities depicted are for one of the particles in the ground state and the other particle in the first excited state. Animation 1 shows the symmetric solution, while Animation 2 shows the antisymmetric solution. Note that the diagonal line across the animation (*x* = *y*) corresponds to *x*_{1 }= *x*_{2}, which is where both particles are at the same position. For the antisymmetric case, one never finds both particles at the same position (probability density is zero along this line). For the symmetric case, one is likely to find both particles at the same position (probability density is non-zero, and is actually the largest at certain points along this line).

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