Quantum distributions

In the previous section, and in Sec. , we discussed the partition function for a system of independent classical particles and, we argued, that its validity required $\bar{n}_r \ll 1$. This condition is satisfied in a regime for which the average distance between the particles is much larger than their de Broglie wavelength, which represents the extent to which the particle position is de-localised because of its quantum nature. According to quantum mechanics, the position of a particle in space is described by a wavefunction, whose square modulus gives the probability density of being found in a particular position in space. When the distance between particles becomes comparable to their de Broglie wavelengths the wavefunctions start to overlap, and the particles lose their identity. They cannot be treated as separate objects anymore and one needs to construct a many-particle wavefunction that describes the system as a whole. ^5.2 This wavefunction is a solution of the Schrödinger equation:

$$\hat H \Psi = E\Psi,$$ (5.18)

where $\hat H = \hat K + \hat U$ is the hamiltonian operator, being the sum of the kinetic energy and potential energy operators $\hat K$ and $\hat U$, respectively, and $E$ is the eigenvalue. For a system containing $N$ identical particles $\hat H$ is invariant w.r.t. any permutation of the particles and so if $\Psi(1,2,\dots,N)$ is a solution so is $\hat P \Psi(1,2,\dots,N)$, where $\hat P$ is any operator that causes any permutation or particles in the list. Any linear combination of wavefunctions with permutated arguments is also acceptable. Out of all possibilities, we can build two special combinations:

$$\Psi_S = \sum_{\hat P} \hat P\psi(1,2,\dots,N)$$ (5.19)

and

$$\Psi_A = \sum_{\hat P} (-1)^P \hat P\psi(1,2,\dots,N).$$ (5.20)

where $P$ is the number of two-particle permutations to obtain the total permutation $\hat P$, also known as the order of the permutation, and the sum runs over all possible permutations. The wavefunction $\Psi_S$ is called symmetric, and $\Psi_A$ anti-symmetric. They live in two portions of the Hilbert space that are disjoint and cannot be turned into one another. It turns out that different particles obey either one or the other condition, depending on the value of their spin. In particular, particles with integer spin – such as alpha particles, photons etc. – have symmetric wavefunctions and particle with half integer spin – electrons, protons, neutrons, etc.. – have anti-symmetric wavefunctions.

Note that we are dealing with a single state here, $\Psi_A$ or $\Psi_S$, despite these being realised as a linear combination of up to $N!$ of them. If we compare with the situation in the classical limit, we see that the multiplicity due to particle exchange is dealt with from the outset: there is just a single state. We do not treat states that differ by a permutation of the (identical) particles as potentially distinct and therefore we do not run into the issue of the Gibbs paradox.

The hamiltonian in  is the sum of the kinetic energy and the potential energy operators. The kinetic energy operator is $\hat K = -\hbar^2/2m \sum_i \nabla_i^2 = \sum_i \hat K_i$. If the potential energy operator is also the sum of single particle operators $\hat U =\sum_i \hat U_i$, meaning that the particles do not interact amongst themselves, then the solution of  factorises into a product of single particle wavefunctions. For example, let us consider again the example of Sec.  in which we have two particles that can occupy two different states, $\psi_a$ and $\psi_b$. We can build 4 separate wavefunctions:

$$\psi_a(1)\psi_a(2)$$

$$\psi_b(1)\psi_b(2)$$

$$\psi_a(1)\psi_b(2)$$

$$\psi_b(1)\psi_a(2)$$

From these we can produce either symmetric or antisymmetric wavefunctions. We obtain:

$$\Psi_1 = ~\psi_a(1)\psi_a(2)$$

$$\Psi_2 = ~\psi_b(1)\psi_b(2)$$

$$\Psi_3 = ~\psi_a(1)\psi_b(2) + \psi_b(1)\psi_a(2)$$

$$\Psi_4 = ~\psi_a(1)\psi_b(2) - \psi_b(1)\psi_a(2).$$ (5.21)

The first three of these wavefunctions are symmetric and the fourth is antisymmetric. Let us consider the symmetric case. We see that we can build three different two-particle states out of these two single-particle states. This is the same number of states that we argued we should have had in the example described in Sec. , but in that case we ended up with the problem of underweighting the two states with both particles in them, as we divided all terms in the sum  by 2. We had to assume then that that particular expression of the partition function could only be justified if the probability of occupying one state with more than one particle were negligible, so that it wouldn't matter if those states were weighted wrongly. We found that the condition for this assumption to be valid was that the average distance between the particles should be much larger than their de Broglie wavelength, i.e that the particles behaved classically. In the quantum mechanics picture we do not run into the problem of underweighting any state. Any distribution of the $N$ particles with occupation numbers $n_1, n_2, \dots, n_r, \dots$ is a single microstate, obtained by forming a linear combination of all possible distributions of the particles in the single particle states, this linear combination being either symmetric or anti-symmetric. The probability therefore is:

$$p(n_1,n_2,\dots n_r,\dots) = \frac{\Omega(n_1,n_2,\dots n_r,\dots)}{\mathcal{Z} }e^{\beta \sum_r n_r (\mu - \epsilon_r)},$$ (5.22)

with $\Omega(n_1,n_2,\dots n_r,\dots) = 1$, and we have:

$${\mathcal{Z} } = {\mathcal{Z}_1 }{\mathcal{Z}_2 } \dots {\mathcal{Z}_r } \dots,$$ (5.23)

with

$${\mathcal{Z}_r } = \sum_{n_r=0}^\infty e^{\beta n_r (\mu - \epsilon_r)}.$$ (5.24)

The total probability factorises:

$$p(n_1,n_2,\dots n_r,\dots) = p_1(n_1) p_2(n_2) \dots p_r(n_r)\dots,$$ (5.25)

with

$$p_r(n_r) = \frac{1}{{\mathcal{Z}_r }}e^{\beta n_r (\mu - \epsilon_r)}.$$ (5.26)

Note that the probability  differ from  only by a factor $n_r!$ in the denominator, therefore the two expressions are identical for $n_r = 0,1$. This means that if the probabilities for occupation numbers $n_r \ge 2$ are small, these states contribute little to the average occupation number ${\bar{n}_r}$, which is therefore the same when computed with either  or . This condition is verified when ${\bar{n}_r} \ll 1$, which is once again the limit in which the average distance between the particles is much larger than their de Broglie wavelength, and tells us that the MB distribution only becomes a good approximation for the statistics of the occupation numbers $n_r$ when the particles behave classically.