... etc. ^1.1

We refer here to quantities such as pressure and temperature in their intuitive meaning. We will define them more precicely in due course.

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... systems ^1.2

This statement can be violated for systems close to their critical point, but we will not discuss these conditions in this notes.

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... process^2.1

We are also implicitly assuming that the internal and the external pressures are equal. This issue is discussed further below.

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... changes ^2.2

Infinitesimal changes are of course only an idealisation, for every system is finite and its energy changes in steps, according to quantum theory. However, in those circumstances where the separation between energy levels is much smaller than the typical energies at play (i.e. $\delta \epsilon \ll k_{\rm B}T$), switching to a continuum description is usually an acceptable approximation.

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... grows ^3.1

Exercise: work out the fraction $\omega(0,N)/\omega(N/2,N)$ as function of $N$

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... ways ^3.2

Since we assumed that the single particle energies are non-degenerate, $\epsilon_2 > \epsilon_1$ and therefore $E_2 > E_1$, i.e. the state with energy $E_2$ is distinct from the state with energy $E_1$.

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... factorial: ^3.3

In fact, a more accurate expression for the Stirling formula is $N! \simeq e^{-N} N^N\sqrt{2 \pi N}$. Take the logarithm of this quantity: $\ln N! \simeq N \ln N - N + \ln \sqrt{2 \pi N}$. For macroscopic systems $N$ is of order $10^{23}$, and therefore the term $\ln \sqrt{2 \pi N}$ is completely negligible compared to $N \ln N - N$.

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... have ^3.4

We assume here that we can ignore the extra energy and entropy due to the presence of the surface.

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... increases, ^3.5

This may not necessarily be the case. For example, systems that have an upper bound to the energy (e.g. spin systems) may have a maximum number of microstates for a value of the energy inside the energy spectrum.

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... increases ^3.6

Note that the number of microstates reduction in the hot system is larger than the increase in the cold system. The sum of the number of microstates in each system is reduced as the system tends to equalisation of temperature. However, the total number of microstates is the product of the two, not the sum.

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... system ^3.7

Even in the more general definition  (see below), it is clear that the entropy must be finite.

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... bath ^3.8

This statement needs some care, as in principle the energy $E_r$ could be as large as wanted if the energy of the system is not bound. However, our following result will show that the probability for this to happen decreases exponentially with the magnitude of $E_r$ and so the assumption of $E_r$ being small compared to the energy of the heat bath becomes justified.

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... ^3.9

Since we have assumed a discrete energy spectrum, we need to clarify what we mean by $\left ( \frac{\partial S_2}{\partial E}\right )$. One approach is to assume that the energy levels are so finely spaced that we can consider the energy to be continuous. A different approach is to define extended functions on the continuum energy spectrum, and recognise that they only represent real physical quantities when the energy coincide with one of the available energy levels. The latter approach would seem to have a better justification.

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... integral ^3.10

We should keep in mind, however, that this equality breaks down when the number of states in the energy range $k_{\rm B}T$ becomes small. A notable case is the Bose-Einstein condensation, where at sufficiently low temperature a large fraction of particles is found to be in the ground state at $E = 0$. These states contribute nothing to the integral, because the volume of the point $E = 0$ is zero, but they obviously give a finite contribution to the sum.

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... probability ^3.11

We postulated that.

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...\space ^4.1

Note that we are allowing the possibility of the energy levels to be degenerate. This is clearly the case for particles in a box, as for example the energy of a particle at position r with velocity vector (v,0,0) is the same as that with velocity (0,v,0), but the two states are different.

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... configuration.^4.2

Unless of course there were a label on each particle, or some physical mechanism to tell them a part, in which case they would not be identical.

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... rest ^4.3

Strictly speaking, the momentum of a particle in an enclosure of volume $V$ cannot be exactly zero, because of the Heisemberg indeterminacy principle. However, for the development of our argument we may assume that the volume is so large that the momentum is negligibly small.

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... freedom ^4.4

In classical non-relativistic physics this is correct, but relativistic effects for the internal degrees of freedom may affect the mass of the particles, which in turn would affect the motion of the centre of mass.

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..., ^4.5

This simply follows from the property of the logarithm $\ln \alpha x = \ln \alpha + \ln x$. If $\alpha$ does not depend on $x$ then $d \ln \alpha/ d x = 0$ and one is left with just the derivative of $\ln x$, which is equal to $1/x$.

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... with ^5.1

$\sum_{n=0}^\infty \frac{a^n}{n!} = 1 + a + \frac{a^2}{2!} + \frac{a^3}{3!} + \dots = e^a$

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... whole. ^5.2

In fact, in a quantum mechanical picture this is the case also if the particles are very far from each other, and indeed it is possible to realise this condition with very careful experiments, but their entanglement is fragile and easily destroyed by unavoidable interactions with the environment, so that they would eventually behave as two independent particles.

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...: ^5.3

Note that the chemical potential has to be negative, and so will never actually be zero, but $\mu=0$ is a useful limiting case to consider to find the lowest possible temperature that the system may take for the integral in  to be equal to $N$. Indeed, the integral in Eq.  does not change much if the chemical potential is set equal to zero or simply to a very small number.

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... equilibrium ^6.1

In fact, since equilibrium is dynamic, there would be some evaporation also in a saturated atmosphere, but this would be matched by condensation.

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... energy ^6.2

We choose as reference the energy of the bottom of the potential energy of the H$_2$ molecule. A different choice $U_0$ for this reference would require adding $U_0$ to the r.h.s of Eq .

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... frequency ^6.3

We are assuming that we can represent the hydrogen molecule as a harmonic oscillator.

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... molecule ^6.4

We assume here that the distance between the atoms in the H$_2$ molecule stays constant. This is usually a good approximation at reasonable temperatures, as the molecule stays in its lowest vibrational state.

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... identity ^6.5

$\sum_{\nu=0}^\infty \phi(\nu) = \int_0^\infty \phi(x)dx + \frac{1}{2}\phi(0)-\frac{1}{12}\phi'(0)+\frac{1}{720}\phi'''(0)+ \dots$

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... occupied ^6.6

At $T = 298$ K the ratio of the probability of occupying the first excited state over the probability of occupying the ground state is $p(\epsilon_1)/p(\epsilon_0) = e^{-\hbar \omega/k_{\rm B}T} \simeq e^{-21} = 7.8 \times 10^{-10}$. The contribution of the excited states to the vibrational free energy of the H$_2$ molecule is therefore negligible at room temperature, and remains less than 1 meV up to $T = 900$ K.

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... phase. ^7.1

Without loss of generality, we are omitting an additional constant and an additional linear velocity, admitted by a general solution of Eq. , i.e. we are assuming that the average position of the oscillator is fixed around $q_i = 0$.

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... ^7.2

Prove it.

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... samples ^7.3

In the extreme case of just one sample ($N=1$) it would be impossible to compute any root mean square fluctuation.

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...Andersen ^7.4

H. C. Andersen, The Journal of Chemical Physics 72, 2384 (1980).

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... by ^8.1

Here we are using atomic units throughout, $\hbar = m_e = e = 1$

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... normalised ^8.2

The wavefunction $\Psi$ is normalised over the volume $V$ if $\int_V d^3{\bf r}_1 \dots d^3{\bf r}_N \Psi^*({\bf r}_1,\dots,{\bf r}_N)\Psi({\bf r}_1,\dots,{\bf r}_N) = 1$.

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... show ^8.3

P. Hohenberg and W. Kohn, Physical Review 136, B864 (1964).

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... different ^8.4

Any physical observable GS property of the system can be obtained from the expectation value of the corresponding operator over the GS wavefunction, $\langle \Psi_0 \vert \hat{O} \vert \Psi_0 \rangle$. If $\Psi_0 = \Psi'_0$ then all physical properties of the two systems are the same, and therefore the two potential energies cannot be different (a part from a trivial constant).

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... electrons ^8.5

W. Kohn and L. J. Sham, Physical Review 140(4A), 1133 (1965).

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... orbital ^8.6

We assume here a spin restricted case, where the orbital part of the wavefunction is the same for two electrons of opposing spin. In the general case this is not necessarily true, and one needs to consider a different orbital for each electron.

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... ^8.7

Note that for the electrostatic term there are two contributions from the double integral, obtained when each of the two integration variables is equal to $\bf {r}$, hence the disappearance of the factor 1/2. The additional factor 2 comes from the chain derivative $\frac{\delta}{\delta\psi_n^*} = \frac{\delta \rho}{\delta \psi_n^*}\frac{\delta}{\delta\rho} = 2 \psi_n\frac{\delta}{\delta\rho}$

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...dft:22 ^8.8

More advanced algorithms involve mixing $\rho_2$ and $\rho_1$, or for higher order iterations mixing several previously obtained charge densities.

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.... ^8.9

For a more in depth discussion of this point see R. G. Parr and W. Yang, "Density Functional Theory of Atoms and Molecules", Oxford university Press (1989), pp. 53 and pp. 148, and references therein.

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... points ^8.10

H. J. Monkhorst and J. D. Pack, Phys. Rev. B 13, 5188 (1976).

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... Baldereschi ^8.11

A. Baldereschi, Phys. Rev. B 7, 5212 (1973).

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... Mermin ^8.12

N. D. Mermin, Phys. Rev. 137, A1441 (1965).

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... one ^8.13

M. Methfessel and A. Paxton, Phys. Rev. B, 40, 3616 (1989).

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... style="font-family:sans-serif">vasp ^9.1

G. Kresse and J. Furthmüller, Physical Review B 54, 11169 (1996)

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... state ^9.2

F. Birch, Physical Review 71 809 (1947)

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... code ^9.3

D. Alfè, Computer Physics Communications 180, 2622 (2009)

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... EXPAL ^9.4

R. Stedman and G Nilsson, Physical Review 145, 492 (1966).

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... description. ^9.5

Elaborate on this failure. Hint, relate the thermal expansion coefficient to the entropy.

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... ^12.1

This is a general result of quantum mechanics and is given here without proof. An equivalent way of expressing it is that the eigenstates of any operator form a complete set in Hilbert space, and we can write $\sum_j \vert\psi_j\rangle \langle \psi_j \vert = \mathds{1}$. This is also known as resolution of the identity. A special case is the position operator $\hat{\bf r}$, which has a continuous set of eigenstates $\vert{\bf r}\rangle$ and we have $\mathds{1}= \int d^3{\bf r} \vert{\bf r} \rangle \langle {\bf r} \vert.$

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