# 7.1: Identical Particles

- Page ID
- 17221

## Quantum States of Two or More Particles

When we first discussed the notion of a "quantum state" of a particle, we said that it is a collection of all the information available about that particle, and that this information can be organized as a vector in Hilbert space. When two particles are considered in tandem, they now constitute a new system, for which we can define a new quantum state.

Alert

*Throughout our discussion of multi-particle quantum states, we will assume that the particles involved do not interact with each other (such as through the coulomb potential). The particles can be individually affected by some external potential, but the effects we will be discussing come only from the fact that they are combined into a single system.*

The information contained in this state is essentially the combination of the information related to each individual particle. We can express this abstractly by combining the individual kets into what is called a *direct product space*. For two particles, we would write:

\[ \left|\;\Psi\left(two\;particle\;system\right)\;\right> = \left|\;\Psi\left(particle\;A\right)\;\right> \otimes \left|\;\Psi\left(particle\;B\right)\;\right> \]

With the particles not interacting with each other, their individual states satisfy the Schrödinger equation for whatever potential they happen to be experiencing. Continuing further without making some assumptions is little more than discussing two states simultaneously. So here goes...

- The particles are indistinguishable from each other (e.g. all are electrons).
- The particles are all in bound states of the same potential, which means they have the same assortment of quantum numbers available.

So for example, we could have \(N\) particles in a one-dimensional infinite square well, or in a two-dimensional harmonic oscillator, or bound by a coulomb potential. As each particle has its own quantum numbers defining its state, the combined state naturally requires twice as many quantum numbers to define it.

## Stationary State Wave Functions

The question that arises is, how is the wave function of the two-particle system related to the wave functions of the individual particles? To answer this question, we first have to interpret what a wave function for a two-particle system represents. Let's stay in one dimension for now...

For a one particle system in one-dimension, we know that \(P\left(x\right)dx=\psi^*\left(x\right)\psi\left(x\right)dx\) is the probability of the particle being found between \(x\) and \(x+dx\). For the two particle system, we need the probability density that describes the positions of both particles particles at once, so we write the following probability density of simultaneously finding one of the particles between \(x_1\) and \(x_1+dx_1\), while the other particle is found between \(x_2\) and \(x_2+dx_2\):

\[ P\left(x_1,x_2\right)dx_1 dx_2 = \psi^*\left(x_1,x_2\right)\psi\left(x_1,x_2\right)dx_1 dx_2 \]

Alert

*It is important to emphasize that since the particles are indistinguishable, this wave function doesn't tell us which particles are at \(x_1\) or \(x_2\) – just that one of them is at each position.*

This is the wave function of a quantum state, so if it possesses a definite energy, it satisfies the stationary state Schrödinger equation:

\[ \widehat H \; \psi\left(x_1,x_2\right) = E_ \;\psi\left(x_1,x_2\right) \]

How do we write the hamiltonian here? First of all, each particle has its own kinetic energy, and they are identical particles, so they have the same mass, and that operator is:

\[ \widehat {KE} = \dfrac{\widehat p^2_1}{2m} +\dfrac{\widehat p^2_2}{2m}=-\dfrac{\hbar^2}{2m}\dfrac{\partial^2}{\partial x_1^2}-\dfrac{\hbar^2}{2m}\dfrac{\partial^2}{\partial x_2^2} \]

Alert

*One more reminder: We don't know which particle has momentum \(p_1\) and which one has **momentum \(p_2\), only that these are the momenta of the two particles.*

Next comes the potential energy function. This function will depend upon the positions of both particles, \(x_1\) and \(x_2\), but since the particles are non-interacting, the potential for each particle needs to be able to vary independently. This is only true when the potential energy function has the property that it can be split into a sum of two potential energies:

\[ V\left(x_1,x_2\right) = V_1\left(x_1\right) + V_2\left(x_2\right) \]

We are also assuming that the two particles are experiencing the *same* potential, which means that the potential energy functions \(V_1\) and \(V_2\) have the same functional form, which allows us to drop the subscripts, and write the full potential simply as:

\[ V\left(x_1,x_2\right) = V\left(x_1\right) + V\left(x_2\right) \]

The full stationary-state Schrödinger equation is therefore:

\[ -\dfrac{\hbar^2}{2m}\dfrac{\partial^2}{\partial x_1^2}\psi\left(x_1,x_2\right)-\dfrac{\hbar^2}{2m}\dfrac{\partial^2}{\partial x_2^2}\psi\left(x_1,x_2\right) + V\left(x_1\right)\psi\left(x_1,x_2\right) + V\left(x_2\right)\psi\left(x_1,x_2\right) = E\;\psi\left(x_1,x_2\right)\]

If we renamed \(x_1\) and \(x_2\) to "\(x\)" and "\(y\)," this would look like exactly like an equation for a single-particle in two dimensions. We can therefore solve it the same way as we did when we moved to multiple dimensions - with our old friend, separation of variables. We write the wave function as a product of two partial wave functions, for which there are then two separate ordinary differential equations – one in each variable. Each partial wave function contributes its own quantum number, and the two quantum numbers can be varied independently. That is, we get:

## Exchange Symmetry

We have a bit of a problem with Equations 7.1.8. The position \(x_1\) is linked with the quantum number \(n_1\), and \(x_2\) is linked with \(n_2\). Why is this a problem? Well, the full wave function's magnitude-squared is supposed to be the probability density that one particle will be found at position \(x_1\), and the other at \(x_2\). But this magnitude-squared is:

If we throw in a factor of \(dx_1dx_2\), we get:

\[ \left|\psi\left(x_1,x_2\right)\right|^2 dx_1dx_2 = \left|\psi_{n_1}\left(x_1\right)\right|^2dx_1\;\left|\psi_{n_2}\left(x_2\right)\right|^2dx_2 \]

The left side of this we interpret above in Equation 7.1.2. The right side is a product of two probabilities: The probability that a particle *with quantum number* \(n_1\) will be found between \(x_1\) and \(x_1+dx_1\) and that a particle with *quantum number* \(n_2\) will be found between \(x_2\) and \(x_2+dx_2\). This product of probabilities produces the "and" that we need for both conditions to hold (i.e. finding a particle at both places), but there is an extra condition that doesn't belong – the restriction that the particle with quantum number \(n_1\) is specifically the one found at position \(x_1\). These particles are indistinguishable, which means that we can label them but these labels better not tell us anything about the state. In this case, if we label the particle we find at \(x_1\) as "particle #1," that's fine, but we cannot say, "Hey, that's the particle with quantum number \(n_1\)!" That would mean that the label distinguishes the particles, which cannot happen.

Let's put it another way. With the particles indistinguishable, the probabilities measured by the quantum state should not change if we swap the positions of the two particles. That is, if we swap the variables \(x_1\) and \(x_2\) in the two-particle wave function, the probability density should not change. It should be clear that this is not the case for Equation 7.1.9, but in case it is not, consider the diagram below, which depicts two partial wave functions for a one-dimensional two-particle system, and two positions along the \(x\)-axis.

**Figure 7.1.1 – Partial Wave Functions of a Two Particle State**

If we plug in the values \(\psi_{n_1}\left(x_1\right)\) and \(\psi_{n_2}\left(x_2\right)\) into Equation 7.1.9, we get some non-zero probability. But now suppose we swap the positions of the two identical particles – they are indistinguishable, so it should not change the probability density. But \(\psi_{n_2}\left(x_1\right)\) is *zero*, which means that the probabilities don't match when the particles switch positions.

So how do we fix the construction of the two-particle wave function from the partial wave functions? We do this by not linking the each quantum number with a particular position. In the original solution with separation of variables, we can acknowledge that although there are two quantum numbers, we have to leave it undetermined which one goes with each wave function. It's possible to show that there are two ways to construct the two-particle wave function from the partial wave functions, that both satisfies the stationary-state Schrödinger equation and yields the same probability when the positions are swapped. They are these two states:

\[ \psi_S\left(x_1,x_2\right) = \dfrac{1}{\sqrt 2}\left[\psi_{n_1}\left(x_1\right)\psi_{n_2}\left(x_2\right) + \psi_{n_1}\left(x_2\right)\psi_{n_2}\left(x_1\right)\right] \]

The subscripts "\(S\)" and "\(A\)" stand for "symmetric" and "antisymmetric," respectively, for obvious reasons. These two cases provide the two possibilities for *exchange symmetry* for quantum particles. The constant in front of these is there to normalize the two-particle wave function assuming that the partial wave functions are already normalized.

It turns out that the type of exchange symmetry that identical particles obey is determined by the kind of particles they are. Quanta of every possible variety satisfy one of these two exchange symmetries. Those that obey positive exchange symmetry (i.e. their two-particle wave function is unchanged by a swap of particles) are called *bosons*, while particles that obey negative exchange symmetry (i.e. their two-particle wave function flips its sign due to a swap of particles) are called *fermions.*