# Identical particles – bosons and fermions revisited

Shankar, R. (1994), Principles of Quantum Mechanics, Plenum Press. Chapter 10, Exercises 10.3.1 – 10.3.3.

Although we’ve looked at the quantum treatment of identical particles as done by Griffiths, it’s worth summarizing Shankar’s treatment of the topic as it provides a few more insights.

In classical physics, suppose we have two identical particles, where ‘identical’ here means that all their physical properties such as mass, size, shape, charge and so on are the same. Suppose we do an experiment in which these two particles collide and rebound in some way. Can we tell which particle ends up in which location? We’re not allowed to label the particles by writing on them, for example, since then they would no longer be identical. In classical physics, we can determine which particle is which by tracing their histories. For example, if we start with particle 1 at position ${\mathbf{r}_{1}}$ and particle 2 at position ${\mathbf{r}_{2}}$, then let them collide, and finally measure their locations at some time after the collision, we might find that one particle ends up at position ${\mathbf{r}_{3}}$ and the other at position ${\mathbf{r}_{4}}$. If we videoed the collision event, we would see the two particles follow well-defined paths before and after the collision, so by observing which particle followed the path that leads from ${\mathbf{r}_{1}}$ to the collision and then out again, we can tell whether it ends up at ${\mathbf{r}_{3}}$ or ${\mathbf{r}_{4}}$. That is, the identification of a particle depends on our ability to watch it as it travels through space.

In quantum mechanics, because of the uncertainty principle, a particle does not have a well-defined trajectory, since in order to define such a trajectory, we would need to specify its position and momentum precisely at each instant of time as it travels. In terms of our collision experiment, if we measured one particle to be at starting position ${\mathbf{r}_{1}}$ at time ${t=0}$ then we know nothing about its momentum, because we specified the position exactly. Thus we can’t tell what trajectory this particle will follow. If we measure the two particles at positions ${\mathbf{r}_{1}}$ and ${\mathbf{r}_{2}}$ at ${t=0}$, and then at ${\mathbf{r}_{3}}$ and ${\mathbf{r}_{4}}$ at some later time, we have no way of knowing which particle ends up at ${\mathbf{r}_{3}}$ and which at ${\mathbf{r}_{4}}$. In terms of the state vector, this means that the physics in the state vector must be the same if we exchange the two particles within the wave function. Since multiplying a state vector ${\psi}$ by some complex constant ${\alpha}$ leaves the physics unchanged, this means that we require

$\displaystyle \psi\left(a,b\right)=\alpha\psi\left(b,a\right) \ \ \ \ \ (1)$

where ${a}$ and ${b}$ represent the two particles.

For a two-particle system, the vector space is spanned by a direct product of the two one-particle vector spaces. Thus the two basis vectors in this vector space that can describe the two particle ${a}$ and ${b}$ are ${\left|ab\right\rangle }$ and ${\left|ba\right\rangle }$. If these two particles are identical, then ${\psi}$ must be some linear combination of these two vectors that satisfies 1. That is

 $\displaystyle \psi\left(b,a\right)$ $\displaystyle =$ $\displaystyle \beta\left|ab\right\rangle +\gamma\left|ba\right\rangle \ \ \ \ \ (2)$ $\displaystyle \psi\left(a,b\right)$ $\displaystyle =$ $\displaystyle \alpha\psi\left(b,a\right)\ \ \ \ \ (3)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \alpha\left(\beta\left|ab\right\rangle +\gamma\left|ba\right\rangle \right) \ \ \ \ \ (4)$

However, ${\psi\left(a,b\right)}$ is also just ${\psi\left(b,a\right)}$ with ${a}$ swapped with ${b}$, that is

$\displaystyle \psi\left(a,b\right)=\beta\left|ba\right\rangle +\gamma\left|ab\right\rangle \ \ \ \ \ (5)$

Since ${\left|ab\right\rangle }$ and ${\left|ba\right\rangle }$ are independent, we can equate their coefficients in the last two equations to get

 $\displaystyle \alpha\beta$ $\displaystyle =$ $\displaystyle \gamma\ \ \ \ \ (6)$ $\displaystyle \alpha\gamma$ $\displaystyle =$ $\displaystyle \beta \ \ \ \ \ (7)$

Inserting the second equation into the first, we get

 $\displaystyle \alpha^{2}\gamma$ $\displaystyle =$ $\displaystyle \gamma\ \ \ \ \ (8)$ $\displaystyle \alpha^{2}$ $\displaystyle =$ $\displaystyle 1\ \ \ \ \ (9)$ $\displaystyle \alpha$ $\displaystyle =$ $\displaystyle \pm1 \ \ \ \ \ (10)$

Thus the two possible state functions 1 are combinations of ${\left|ab\right\rangle }$ and ${\left|ba\right\rangle }$ such that

$\displaystyle \psi\left(a,b\right)=\pm\psi\left(b,a\right) \ \ \ \ \ (11)$

The plus sign gives the symmetric state, which can be written as

$\displaystyle \psi\left(ab,S\right)=\frac{1}{\sqrt{2}}\left(\left|ab\right\rangle +\left|ba\right\rangle \right) \ \ \ \ \ (12)$

and the minus sign gives the antisymmetric state

$\displaystyle \psi\left(ab,A\right)=\frac{1}{\sqrt{2}}\left(\left|ab\right\rangle -\left|ba\right\rangle \right) \ \ \ \ \ (13)$

The ${\frac{1}{\sqrt{2}}}$ factor normalizes the states so that

 $\displaystyle \left\langle \psi\left(ab,S\right)\left|\psi\left(ab,S\right)\right.\right\rangle$ $\displaystyle =$ $\displaystyle 1\ \ \ \ \ (14)$ $\displaystyle \left\langle \psi\left(ab,A\right)\left|\psi\left(ab,A\right)\right.\right\rangle$ $\displaystyle =$ $\displaystyle 1 \ \ \ \ \ (15)$

This follows because the basis vectors ${\left|ab\right\rangle }$ and ${\left|ba\right\rangle }$ are orthonormal vectors.

Particles with symmetric states are called bosons and particles with antisymmetric states are called fermions. The Pauli exclusion principle for fermions follows directly from 13, since if we set the state variables of the two particles to be the same, that is, ${a=b}$, then

$\displaystyle \psi\left(aa,A\right)=\frac{1}{\sqrt{2}}\left(\left|aa\right\rangle -\left|aa\right\rangle \right)=0 \ \ \ \ \ (16)$

The symmetry or antisymmetry rules apply to all the properties of the particle taken as an aggregate. That is, the labels ${a}$ and ${b}$ can refer to the particle’s location plus its other quantum numbers such as spin, charge, and so on. In order for two fermions to be excluded, the states of the two fermions must be identical in all their quantum numbers, so that two fermions with the same orbital location (as two electrons in the same orbital within an atom, for example) are allowed if their spins are different.

Example 1 Suppose we have 2 identical bosons that are measured to be in states ${\left|\phi\right\rangle }$ and ${\left|\chi\right\rangle }$ where ${\left\langle \phi\left|\chi\right.\right\rangle \ne0}$. What is their combined state vector? Since they are bosons, their state vector must be symmetric, so we must have

 $\displaystyle \psi\left(\phi,\chi\right)$ $\displaystyle =$ $\displaystyle A\left|\phi\chi\right\rangle +B\left|\chi\phi\right\rangle \ \ \ \ \ (17)$

Because ${\psi}$ must be symmetric, we must have ${A=B}$, so that ${\psi\left(\phi,\chi\right)=\psi\left(\chi,\phi\right)}$. The 2-particle states can be written as direct products, so we have

$\displaystyle \psi\left(\phi,\chi\right)=A\left(\left|\phi\right\rangle \otimes\left|\chi\right\rangle +\left|\chi\right\rangle \otimes\left|\phi\right\rangle \right) \ \ \ \ \ (18)$

To normalize, we have, assuming that ${\left|\phi\right\rangle }$ and ${\left|\chi\right\rangle }$ are normalized:

 $\displaystyle \left|\psi\right|^{2}$ $\displaystyle =$ $\displaystyle 1\ \ \ \ \ (19)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left|A\right|^{2}\left(\left\langle \phi\right|\otimes\left\langle \chi\right|+\left\langle \chi\right|\otimes\left\langle \phi\right|\right)\left(\left|\phi\right\rangle \otimes\left|\chi\right\rangle +\left|\chi\right\rangle \otimes\left|\phi\right\rangle \right)\ \ \ \ \ (20)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left|A\right|^{2}\left(1+1+\left|\left\langle \phi\left|\chi\right.\right\rangle \right|^{2}+\left|\left\langle \chi\left|\phi\right.\right\rangle \right|^{2}\right)\ \ \ \ \ (21)$ $\displaystyle A$ $\displaystyle =$ $\displaystyle \frac{\pm1}{\sqrt{2\left(1+\left|\left\langle \phi\left|\chi\right.\right\rangle \right|^{2}\right)}} \ \ \ \ \ (22)$

Thus the normalized state vector is (choosing the + sign):

 $\displaystyle \psi\left(\phi,\chi\right)$ $\displaystyle =$ $\displaystyle \frac{1}{\sqrt{2\left(1+\left|\left\langle \phi\left|\chi\right.\right\rangle \right|^{2}\right)}}\left(\left|\phi\chi\right\rangle +\left|\chi\phi\right\rangle \right) \ \ \ \ \ (23)$

Notice that this reduces to 12 if ${\left\langle \phi\left|\chi\right.\right\rangle =0}$.

For more than 2 particles, we need to form state vectors that are either totally symmetric or totally antisymmetric.

Example 2 Suppose we have 3 identical bosons, and they are measured to be in states 3, 3 and 4. Since two of them are in the same state, there are 3 possible combinations, which we can write as ${\left|334\right\rangle ,}$ ${\left|343\right\rangle }$ and ${\left|433\right\rangle }$. Assuming these states are orthonormal, the full normalized state vector is

$\displaystyle \psi\left(3,3,4\right)=\frac{1}{\sqrt{3}}\left(\left|334\right\rangle +\left|343\right\rangle +\left|433\right\rangle \right) \ \ \ \ \ (24)$

The ${\frac{1}{\sqrt{3}}}$ ensures that ${\left|\psi\left(3,3,4\right)\right|^{2}=1}$.

Incidentally, for ${N\ge3}$ particles, it turns out to be impossible to construct a linear combination of the basis states such that the overall state vector is symmetric with respect to the interchange of some pairs of particles and antisymmetric with respect to the interchange of other pairs. A general proof for all ${N}$ requires group theory, but for ${N=3}$ we can show this by brute force. There are ${3!=6}$ basis vectors

$\displaystyle \left|123\right\rangle ,\left|231\right\rangle ,\left|312\right\rangle ,\left|132\right\rangle ,\left|321\right\rangle ,\left|213\right\rangle \ \ \ \ \ (25)$

Suppose we require the compound state vector to be symmetric with respect to exchanging 1 and 2. We then must have

$\displaystyle \psi=A\left(\left|123\right\rangle +\left|213\right\rangle \right)+B\left(\left|231\right\rangle +\left|132\right\rangle \right)+C\left(\left|312\right\rangle +\left|321\right\rangle \right) \ \ \ \ \ (26)$

If we now try to make ${\psi}$ antisymmetric with respect to exchanging 2 and 3, we must have

$\displaystyle \psi=D\left(\left|123\right\rangle -\left|132\right\rangle \right)+E\left(\left|231\right\rangle -\left|321\right\rangle \right)+F\left(\left|312\right\rangle -\left|213\right\rangle \right) \ \ \ \ \ (27)$

Comparing the two, we see that

 $\displaystyle A$ $\displaystyle =$ $\displaystyle D=-F\ \ \ \ \ (28)$ $\displaystyle B$ $\displaystyle =$ $\displaystyle E=-D\ \ \ \ \ (29)$ $\displaystyle C$ $\displaystyle =$ $\displaystyle F=-E \ \ \ \ \ (30)$

Eliminating ${A,B}$, and ${C}$ we have, combining the 3 equations:

$\displaystyle D=-E=F \ \ \ \ \ (31)$

But from the first equation, we have ${D=-F}$, so ${F=-F=0}$. From the other equations, this implies that ${D=-F=0}$ and ${E=-F=0}$, and thus that ${A=B=C=0}$. So there is no non-trivial solution that allows both a symmetric and antisymmetric particle exchange within the same state vector.

Example 3 Suppose we have 3 particles and only 3 distinct states that each particle can have. If the particles are distinguishable (not identical) the total number of states is found by considering the possibilities. If all 3 particles are in different states, then there are ${3!=6}$ possible overall states. If two particles are in one state and one particle in another, there are ${\binom{3}{2}=3}$ ways of choosing the two states, for each of which there are 2 ways of partitioning these two states (that is, which state has 2 particles and which has the other one), and for each of those there are 3 possible configurations, so there are ${3\times2\times3=18}$ possible configurations. Finally, if all 3 particles are in the same state, there are 3 possibilities. Thus the total for distinguishable particles is ${6+18+3=27}$.

If the particles are bosons, then if all 3 are in different states, there is only 1 symmetric combination of the 6 basis states. If two particles are in one state and one particle in another, there are ${3\times2=6}$ ways of partitioning the states, each of which contributes only one symmetric overall state. Finally, if all 3 particles are in the same state, there are 3 possibilities. Thus the total for bosons is ${1+6+3=10}$.

For fermions, all three particles must be in different states, so there is only 1 possibility.

# Exchange force: infinite square well

Required math: calculus

Required physics: Schrödinger equation

Reference: Griffiths, David J. (2005), Introduction to Quantum Mechanics, 2nd Edition; Pearson Education – Section 5.1.2 & Problem 5.6.

We’ve seen that distinguishable particles and identical particles must be treated differently in quantum mechanics, resulting in different combinations of the single-particle wave functions in 2-particle systems. It’s useful to work out what this means for some of the observables in a 2-particle system.

We can begin by looking at possibly the simplest case: the average positions of the two particles. If the particles are distinguishable, then the wave function is ${\psi(x_{a},x_{b})=\psi_{1}\left(x_{a}\right)\psi_{2}\left(x_{b}\right)}$ and

 $\displaystyle \left\langle x_{a}\right\rangle$ $\displaystyle =$ $\displaystyle \left\langle \psi\left|x_{a}\right|\psi\right\rangle \ \ \ \ \ (1)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left\langle \psi_{1a}\left|x_{a}\right|\psi_{1a}\right\rangle \left\langle \left.\psi_{2b}\right|\psi_{2b}\right\rangle \ \ \ \ \ (2)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left\langle \psi_{1a}\left|x_{a}\right|\psi_{1a}\right\rangle \ \ \ \ \ (3)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left\langle x\right\rangle _{1} \ \ \ \ \ (4)$

where the notation ${\left|\psi_{1a}\right\rangle \equiv\psi_{1}\left(x_{a}\right)}$ and so on.

That is, ${\left\langle x\right\rangle }$ is the mean value of ${x}$ in state ${\psi_{1}}$. We can drop the suffix ${a}$ here, since ${x_{a}}$ is just a dummy name for the integration variable in ${\left\langle \psi_{1a}\left|x_{a}\right|\psi_{1a}\right\rangle }$.

For identical particles,

$\displaystyle \psi_{\pm}\left(\mathbf{r}_{a},\mathbf{r}_{b}\right)=\frac{1}{\sqrt{2}}\left[\psi_{1}\left(\mathbf{r}_{a}\right)\psi_{2}\left(\mathbf{r}_{b}\right)\pm\psi_{2}\left(\mathbf{r}_{a}\right)\psi_{1}\left(\mathbf{r}_{b}\right)\right] \ \ \ \ \ (5)$

This time, working out ${\left\langle x_{a}\right\rangle }$ is a bit messier but not too bad if we use the orthogonality of the two states.

 $\displaystyle 2\left\langle x_{a}\right\rangle$ $\displaystyle =$ $\displaystyle \left\langle \psi_{1a}\left|x_{a}\right|\psi_{1a}\right\rangle \left\langle \left.\psi_{2b}\right|\psi_{2b}\right\rangle +\left\langle \psi_{2a}\left|x_{a}\right|\psi_{2a}\right\rangle \left\langle \left.\psi_{1b}\right|\psi_{1b}\right\rangle \ \ \ \ \ (6)$ $\displaystyle$ $\displaystyle$ $\displaystyle \pm\left\langle \psi_{1a}\left|x_{a}\right|\psi_{2a}\right\rangle \left\langle \left.\psi_{2b}\right|\psi_{1b}\right\rangle \pm\left\langle \psi_{2a}\left|x_{a}\right|\psi_{1a}\right\rangle \left\langle \left.\psi_{1b}\right|\psi_{2b}\right\rangle \ \ \ \ \ (7)$ $\displaystyle \left\langle x_{a}\right\rangle$ $\displaystyle =$ $\displaystyle \frac{1}{2}\left(\left\langle x\right\rangle _{1}+\left\langle x\right\rangle _{2}\right) \ \ \ \ \ (8)$

Thus the mean position of particle ${a}$ is the average of its positions in the two states, which isn’t all that surprising. We’d get the same result for particle ${b}$ of course, since the two particles are identical. This result is true for both bosons and fermions, since the plus/minus terms work out to zero due to the orthogonality of the states ${\psi_{1}}$ and ${\psi_{2}}$.

What is a bit more interesting is the mean square separation of the two particles, that is ${\left\langle \left(x_{a}-x_{b}\right)^{2}\right\rangle }$. This can be worked out using the same procedure as above, and is done by Griffiths in his section 5.1.2, although his notation is a bit different from mine. (I’ve used a numerical suffix on the wave function, since this is the usual notation used for stationary states. Thus a letter suffix indicates which particle and a number suffix indicates which stationary state.) The results are, in my notation, first for distinguishable particles:

$\displaystyle \left\langle \left(x_{a}-x_{b}\right)^{2}\right\rangle =\left\langle x^{2}\right\rangle _{1}+\left\langle x^{2}\right\rangle _{2}-2\left\langle x\right\rangle _{1}\left\langle x\right\rangle _{2} \ \ \ \ \ (9)$

For identical particles, we get

$\displaystyle \left\langle \left(x_{a}-x_{b}\right)^{2}\right\rangle _{\pm}=\left\langle x^{2}\right\rangle _{1}+\left\langle x^{2}\right\rangle _{2}-2\left\langle x\right\rangle _{1}\left\langle x\right\rangle _{2}\mp2\left|\left\langle x\right\rangle _{12}\right|^{2} \ \ \ \ \ (10)$

where the plus (minus) sign on the left and minus (plus) on the right refer to bosons (fermions), and

$\displaystyle \left\langle x\right\rangle _{12}\equiv\left\langle \psi_{1}\left|x_{\;}\right|\psi_{2}\right\rangle \ \ \ \ \ (11)$

In general, then, since the term ${2\left|\left\langle x\right\rangle _{12}\right|^{2}}$ is always positive, bosons tend to be closer together than distinguishable particles while fermions are further apart. This is a sort of pseudo-force which is an entirely quantum mechanical effect of the symmetries of the wave functions. Although it’s not really a force in the sense that electromagnetism and gravity are, it’s known as the exchange force.

As an example, consider 2 particles in the infinite square well. The wave functions for a single particle are

$\displaystyle \psi\left(x\right)=\sqrt{\frac{2}{a}}\sin\frac{n\pi x}{a} \ \ \ \ \ (12)$

where ${a}$ is the width of the well. If the total wave function is a combination of states ${l}$ and ${n}$, then if the particles are distinguishable

 $\displaystyle \left\langle \left(x_{a}-x_{b}\right)^{2}\right\rangle$ $\displaystyle =$ $\displaystyle \left\langle x^{2}\right\rangle _{1}+\left\langle x^{2}\right\rangle _{2}-2\left\langle x\right\rangle _{1}\left\langle x\right\rangle _{2}\ \ \ \ \ (13)$ $\displaystyle$ $\displaystyle =$ $\displaystyle a^{2}\left(\frac{1}{3}-\frac{1}{2l^{2}\pi^{2}}\right)+a^{2}\left(\frac{1}{3}-\frac{1}{2n^{2}\pi^{2}}\right)-2\left(\frac{a}{2}\right)\left(\frac{a}{2}\right)\ \ \ \ \ (14)$ $\displaystyle$ $\displaystyle =$ $\displaystyle a^{2}\left(\frac{1}{6}-\frac{l^{2}+n^{2}}{2(\pi ln)^{2}}\right) \ \ \ \ \ (15)$

In line 2, we used the results of our earlier calculations for the infinite square well.

If the particles are identical, then

 $\displaystyle \left\langle x\right\rangle _{ln}$ $\displaystyle =$ $\displaystyle \left\langle \psi_{l}\left|x_{\;}\right|\psi_{n}\right\rangle \ \ \ \ \ (16)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{2}{a}\int_{0}^{a}\sin\left(\frac{l\pi x}{a}\right)\sin\left(\frac{n\pi x}{a}\right)xdx\ \ \ \ \ (17)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left(-1+\left(-1\right)^{n+l}\right)\frac{4anl}{\left[\pi\left(n^{2}-l^{2}\right)\right]^{2}} \ \ \ \ \ (18)$

This term is zero if ${n+l}$ is even, so there is a difference in the separation only when ${n+l}$ is odd. In general, we have

$\displaystyle \left\langle \left(x_{a}-x_{b}\right)^{2}\right\rangle _{\pm}=a^{2}\left(\frac{1}{6}-\frac{l^{2}+n^{2}}{2(\pi ln)^{2}}\right)\mp2\left[\left(-1+\left(-1\right)^{n+l}\right)\frac{4anl}{\left[\pi\left(n^{2}-l^{2}\right)\right]^{2}}\right]^{2} \ \ \ \ \ (19)$

# Infinite square well: 2 particle systems

Required math: calculus

Required physics: Schrödinger equation

Reference: Griffiths, David J. (2005), Introduction to Quantum Mechanics, 2nd Edition; Pearson Education – Problem 5.5.

As an example of the differences between distinguishable particles and identical particles, consider 2 particles in the infinite square well. The wave functions for a single particle are

$\displaystyle \psi\left(x\right)=\sqrt{\frac{2}{a}}\sin\frac{n\pi x}{a} \ \ \ \ \ (1)$

where ${a}$ is the width of the well. The energy (eigenvalue) of this state is

$\displaystyle E_{n}=\frac{n^{2}\pi^{2}\hbar^{2}}{2ma^{2}}\equiv n^{2}K \ \ \ \ \ (2)$

If we have two noninteracting particles, the hamiltonian consists of terms for each particle:

$\displaystyle H\Psi=-\frac{\hbar^{2}}{2m}\frac{\partial^{2}\Psi}{\partial x_{1}^{2}}-\frac{\hbar^{2}}{2m}\frac{\partial^{2}\Psi}{\partial x_{2}^{2}} \ \ \ \ \ (3)$

for ${0.

If the particles are distinguishable, then there is no need to arrange the wave functions so that we can’t tell which particle is in which state. That is, we can form a simple product:

$\displaystyle \psi_{d}\left(x_{1},x_{2}\right)=\psi_{n_{1}}\left(x_{1}\right)\psi_{n_{2}}\left(x_{2}\right) \ \ \ \ \ (4)$

If we apply the hamiltonian to this function, the total energy is just the sum of the two individual energies:

$\displaystyle H\psi_{d}\left(x_{1},x_{2}\right)=E_{n_{1}}+E_{n_{2}} \ \ \ \ \ (5)$

The ground state has ${n_{1}=n_{2}=1}$, and thus an energy of ${2K}$. The first excited state has either ${n_{1}=1;\; n_{2}=2}$ or ${n_{1}=2;\; n_{2}=1}$. Since the particles are not identical, these are distinct states, so the first excited state has a degeneracy of 2, and an energy of ${5K}$. After that, the next state has ${n_{1}=n_{2}=2}$ (degeneracy 1, energy ${8K}$), then ${n_{1}=1;\; n_{2}=3}$ or ${n_{1}=3;\; n_{2}=1}$ (degeneracy 2, energy ${10K}$).

If the particles are bosons, then they are identical and the wave function is a symmetric sum, so we get

$\displaystyle \psi_{b}\left(x_{1},x_{2}\right)=\frac{1}{\sqrt{2}}\left[\psi_{n_{1}}\left(x_{1}\right)\psi_{n_{2}}\left(x_{2}\right)+\psi_{n_{1}}\left(x_{2}\right)\psi_{n_{2}}\left(x_{1}\right)\right] \ \ \ \ \ (6)$

The ground state has ${n_{1}=n_{2}=1}$, and thus an energy of ${2K}$ as before. The first excited state has ${n_{1}=1;\; n_{2}=2}$. This time, however, setting ${n_{1}=2;\; n_{2}=1}$ does not give us a separate state, as you can see by plugging in the numbers into ${\psi_{b}}$. Thus the ${5K}$ energy state has degeneracy 1, not 2.

We then get ${n_{1}=n_{2}=2}$ (degeneracy 1, energy ${8K}$), then ${n_{1}=1;\; n_{2}=3}$ (degeneracy 1, energy ${10K}$).

For fermions, the wave function is an antisymmetric sum:

$\displaystyle \psi_{f}\left(x_{1},x_{2}\right)=\frac{1}{\sqrt{2}}\left[\psi_{n_{1}}\left(x_{1}\right)\psi_{n_{2}}\left(x_{2}\right)-\psi_{n_{1}}\left(x_{2}\right)\psi_{n_{2}}\left(x_{1}\right)\right] \ \ \ \ \ (7)$

This time, whenever ${n_{1}=n_{2}}$ the wave function ${\psi_{f}=0}$, so for fermions it is impossible to have two particles in the same state. This is a quite general result and is known as the Pauli exclusion principle. For the infinite square well, the ground state for fermions is therefore ${n_{1}=1;\; n_{2}=2}$, with energy ${5K}$ and degeneracy 1. In this case, using ${n_{1}=2;\; n_{2}=1}$ does give a different wave function, but it is simply the negative of the original, so differs from the original only by a phase factor which disappears when taking the square modulus.

The next state occurs at ${n_{1}=1;\; n_{2}=3}$ (degeneracy 1, energy ${10K}$), then ${n_{1}=2;\; n_{2}=3}$ (degeneracy 1, energy ${13K}$).

# Identical particles: fermions and bosons

Required math: calculus

Required physics: 3-d Schrödinger equation

Reference: Griffiths, David J. (2005), Introduction to Quantum Mechanics, 2nd Edition; Pearson Education – Problem 5.4.

One of the more unusual postulates of quantum mechanics is that particles of the same type (for example, 2 electrons) are actually identical, in the sense that we cannot distinguish one from the other by any means at all. This is a radical departure from classical mechanics, where all particles are, in principle, different, even if they have the same physical properties. In classical mechanics, it makes sense to talk about ‘electron #1’ and ‘electron #2’, but in quantum mechanics, all electrons (or particles of other types such as protons and neutrons) are exactly the same, and it is impossible to label them in any way.

Up to now, we’ve been considering mainly one-particle systems moving in potentials, so the issue of identifying one of a group of particles hasn’t arisen. We’ve essentially assumed that the one particle (usually an electron) is the only particle in the universe, so we don’t need to think about labelling it.

For two (or more) particles, though, we do need to think about how to describe the quantum state in such a way that, even from the mathematics, we can’t tell which particle is which. For a single particle moving in a potential, we can solve the time-independent Schrödinger equation and get a set of stationary states. Once we’ve done that, we can generate the time-dependent solution in the usual manner by constructing a series:

$\displaystyle \Psi(\mathbf{r},t)=\sum_{n=1}^{\infty}c_{n}\psi_{n}\left(\mathbf{r}\right)e^{-iE_{n}t/\hbar} \ \ \ \ \ (1)$

If we have 2 particles, we need to introduce two spatial coordinates, one for each particle. Now suppose that each particle is in one of the stationary states, say ${\psi_{1}}$ and ${\psi_{2}}$. If the particles are really indistinguishable, then it shouldn’t be possible to tell which particle is in which state. That is, trying a total wave function of the form ${\psi\left(\mathbf{r}_{a},\mathbf{r}_{b}\right)=\psi_{1}\left(\mathbf{r}_{a}\right)\psi_{2}\left(\mathbf{r}_{b}\right)}$ won’t work, because we’ve associated particle ${a}$ with state ${\psi_{1}}$ and ${b}$ with state ${\psi_{2}}$.

One way of mixing things up is to take a symmetric or anti-symmetric combination of the two states. That is

$\displaystyle \psi_{\pm}\left(\mathbf{r}_{a},\mathbf{r}_{b}\right)=A\left[\psi_{1}\left(\mathbf{r}_{a}\right)\psi_{2}\left(\mathbf{r}_{b}\right)\pm\psi_{2}\left(\mathbf{r}_{a}\right)\psi_{1}\left(\mathbf{r}_{b}\right)\right] \ \ \ \ \ (2)$

where ${A}$ is a normalization constant. Assuming the individual ${\psi}$ functions are orthonormal, we can work out ${A}$. Using the orthonormal properties of the ${\psi_{1,2}}$ functions, we get, assuming ${\psi_{1}\ne\psi_{2}}$:

 $\displaystyle \int\psi_{\pm}^*\psi_{\pm}d^{3}\mathbf{r}_{a}d^{3}\mathbf{r}_{b}$ $\displaystyle =$ $\displaystyle \left|A\right|^{2}\int\left[\psi_{1}\left(\mathbf{r}_{a}\right)\psi_{2}\left(\mathbf{r}_{b}\right)\pm\psi_{2}\left(\mathbf{r}_{a}\right)\psi_{1}\left(\mathbf{r}_{b}\right)\right]^*\times\ \ \ \ \ (3)$ $\displaystyle$ $\displaystyle$ $\displaystyle \left[\psi_{1}\left(\mathbf{r}_{a}\right)\psi_{2}\left(\mathbf{r}_{b}\right)\pm\psi_{2}\left(\mathbf{r}_{a}\right)\psi_{1}\left(\mathbf{r}_{b}\right)\right]d^{3}\mathbf{r}_{a}d^{3}\mathbf{r}_{b}\ \ \ \ \ (4)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 2\left|A\right|^{2}\ \ \ \ \ (5)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 1 \ \ \ \ \ (6)$

so ${A=1/\sqrt{2}}$.

In the case where ${\psi_{1}=\psi_{2}=\psi}$, the minus sign results in the total wave function being zero, so we have only the case with the plus sign. In this case, we get

 $\displaystyle \int\psi_{\pm}^*\psi_{\pm}d^{3}\mathbf{r}_{1}d^{3}\mathbf{r}_{2}$ $\displaystyle =$ $\displaystyle \left|A\right|^{2}\int\left[2\psi\left(\mathbf{r}_{a}\right)\psi\left(\mathbf{r}_{b}\right)\right]^*\left[2\psi\left(\mathbf{r}_{a}\right)\psi\left(\mathbf{r}_{b}\right)\right]d^{3}\mathbf{r}_{a}d^{3}\mathbf{r}_{b}\ \ \ \ \ (7)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 4\left|A\right|^{2}=1 \ \ \ \ \ (8)$

so here ${A=1/2}$.

Particles for which we take the plus sign above are known as bosons, and particles using the minus sign are known as fermions.