# 7C: Two Identical Particles in a Box


Let us return briefly to the particle in a box model and ask what happens if we put two identical particles in the box. If they were classical particles, they would carry an imaginary label'' that would allow us to tell the particles apart. In quantum mechanics, this is no longer possible because all we can predict is the probability that one of the particles is in a region $$dx_1$$ about the point $$x_1$$ and the other is in a region $$dx_2$$ about the point $$x_2$$, but we cannot tell which particle is where. This fact will affect both the allowed energies and the wavefunctions.

The classical energy inside the box is, in this case

$\frac{p_{1}^{2}}{2m}+\frac{p_{2}^{2}}{2m}=E$

where $$p_1$$ is the momentum of particle 1, and $$p_2$$ is the momentum of particle 2. Because the energy is a sum of independent energies for particles 1 and 2, the energies can also be separated

$E=\varepsilon_1 +\varepsilon_2$

The wavefunction will depend on the positions $$x_1$$ and $$x_2$$, $$\psi (x_1 ,x_2)$$, and therefore, we have four boundary conditions $$\psi (0,x_2)=0$$, $$\psi (x_1 ,0)=0$$, $$\psi (L, x_2 )=0$$ and $$\psi (x_1 ,L)=0$$. These give rise to the allowed values of $$\varepsilon_1$$ and $$\varepsilon_2$$ in the usual way:

$\varepsilon_{n_1}=\frac{\hbar^2 \pi^2}{2mL^2}n_{1}^{2}\;\;\;\; \varepsilon_{n_2}=\frac{\hbar^2 \pi^2}{2mL^2}n_{2}^{2}$

so that the allowed values of the total energy are

$E_{n_1 n_2}=\frac{\hbar^2 \pi^2}{2mL^2}(n_{1}^{2}+n_{2}^{2})$

Here each particle gets an independent integer for enumerating the energy levels $$n_1$$ and $$n_2$$.

Just as we did for a single particle in a two-dimensional box, we might expect the wavefunctions to be a simple product

$\Psi_{n_1 n_2}(x_1 ,x_2)=\psi_{n_1}(x_1)\psi_{n_2}(x_2)$

where $$\psi_n (x)$$ is just the usual wavefunction for the particle in a box

$\psi_n (x)=\sqrt{\frac{2}{L}}sin \left ( \frac{n\pi x}{L} \right )$

However, we immediately run into two problems. First, if the particles are truly identical, how do we know whether to assign the energy level $$n_1$$ to particle 1 or 2? We could do either, assigning $$n_2$$ to the other, and the total energy would not change. However, the simple product wavefunction above specifies a definite assignment of $$n_1$$ to particle 1 and $$n_2$$ to particle 2! The second problem is that the simple product form says that there are points at which $$\Psi_{n_1 n_2}(x,x) \neq 0$$, implying that there is a nonzero probability of finding the two particles at exactly the same point in space, i.e. sitting right on top of each other!

In order to handle the first problem, we simply include two possible product forms, i.e.

$\psi_{n_1}(x_1)\psi_{n_2}(x_2)\;\;\;\; \psi_{n_2}(x_1)\psi_{n_1}(x_2)$

and put them together to form $$\Psi_{n_1 n_2}(x_1 ,x_2)$$ , which we can to in two ways:

$\Psi_{n_1 n_2}(x_1 ,x_2)=\frac{1}{\sqrt{2}}[\psi_{n_1}(x_1)\psi_{n_2}(x_2)\pm \psi_{n_2}(x_1)\psi_{n_1}(x_2)]$

The $$1/\sqrt{2}$$ is needed for proper normalization. But which combination should be take? If we take the first one:

$\Psi_{n_1 n_2}(x_1 ,x_2)=\frac{1}{\sqrt{2}}[\psi_{n_1}(x_1)\psi_{n_2}(x_2)+ \psi_{n_2}(x_1)\psi_{n_1}(x_2)]​$

then we have not circumvented the second problem. It will still be possible for the two particle to sit at the same point in space. In fact, there are certain exotic particles in quantum mechanics that can do this (they are called bosons), and this possibility is what is responsible for curious phenomena such as superfluidity and superconductivity. However, the more mundane objects we care about, such as protons and electrons, cannot sit on top of each other. Thus, for these particles, we need to take the second form

$\Psi_{n_1 n_2}(x_1 ,x_2)=\frac{1}{\sqrt{2}}[\psi_{n_1}(x_1)\psi_{n_2}(x_2)- \psi_{n_2}(x_1)\psi_{n_1}(x_2)]​$

This form implies two things. Note first that

$\Psi_{n_1 n_2}(x,x)=0$

which means that for any $$n_1$$ and $$n_2$$, there is zero probability to find both particles in small regions near the same point. The other thing we see is that

$\Psi_{nn}(x_1 ,x_2)=0$

which implies that the two particles cannot be in the same energy level. That this possibility is excluded is known as the Pauli exclusion principle, and it plays a key role in both the electronic configurations of atoms and molecules.

Example $$\PageIndex{2}$$:

Two particles of mass $$m$$ are in a one-dimensional box of length $$L$$. What is the lowest total energy the two particle- system can have?

Solution

Since $$n_1 =1,2,3,...$$ and $$n_2 =1,2,3,...$$, if we choose $$n_1 =1$$, then, since $$n_2$$ cannot be equal to $$n_1$$, the next lowest value we can choose is $$n_2 =2$$. Thus, the lowest energy is

$E_{12}=\frac{\hbar^2 \pi^2}{2mL^2}(1^2 +2^2)=\frac{5\hbar^2 \pi^2}{2mL^2}$

Notice that the calculation of probabilities changes a little because of the increased complexity of the wavefunction. Let us consider the probability that a simultaneous measurement of the position of particle 1 and of particle yields values of $$x_1$$ and $$x_2$$, $$x_1 \epsilon [a,b]$$, $$x_2 \epsilon [c,d]$$. For electrons, this is given by

\begin{align*}P(x_1 \epsilon [a,b] \ and \ x_2\epsilon [c,d]) &= \int_{a}^{b}dx_1 \int_{c}^{d}dx_2 |\Psi_{n_1 n_2}(x_1 ,x_2)|^2 \\ &= \frac{1}{2}\int_{a}^{b}dx_1\int_{c}^{d}dx_2 [\psi_{n_1}^{2}(x_1)\psi_{n_2}^{2}(x_2)-2\psi_{n_1}(x_1)\psi_{n_2}(x_1)\psi_{n_2}(x_2)\psi_{n_1}(x_2)+\psi_{n_2}^{2}(x_1)\psi_{n_1}^{2}(x_2)]\\ &= \frac{1}{2} \left [ \int_{a}^{b}\psi_{n_1}^{2}(x_1)dx_1 \right ] \left [ \int_{c}^{d} \psi_{n_2}^{2}(x_2)dx_2 \right ]\\ &- \left [ \int_{a}^{b} \psi_{n_1} (x_1)\psi_{n_2} (x_1)dx_1 \right ] \left [ \int_{c}^{d}\psi_{n_2}(x_2)\psi_{n_1}(x_2)dx_2 \right ]\\ &+ \frac{1}{2} \left [ \int_{a}^{b}\psi_{n_2}^{2}(x_1)dx_1 \right ] \left [ \int_{c}^{d} \psi_{n_1}^{2} (x_2)dx_2 \right ]\end{align*}