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# Posts Tagged ‘particle in a box’

## An updated compilation of notes, for ‘PHY452H1S Basic Statistical Mechanics’, Taught by Prof. Arun Paramekanti

Posted by peeterjoot on March 27, 2013

Here’s my second update of my notes compilation for this course, including all of the following:

March 27, 2013 Fermi gas

March 26, 2013 Fermi gas thermodynamics

March 26, 2013 Fermi gas thermodynamics

March 23, 2013 Relativisitic generalization of statistical mechanics

March 21, 2013 Kittel Zipper problem

March 18, 2013 Pathria chapter 4 diatomic molecule problem

March 17, 2013 Gibbs sum for a two level system

March 16, 2013 open system variance of N

March 16, 2013 probability forms of entropy

March 14, 2013 Grand Canonical/Fermion-Bosons

March 13, 2013 Quantum anharmonic oscillator

March 12, 2013 Grand canonical ensemble

March 11, 2013 Heat capacity of perturbed harmonic oscillator

March 10, 2013 Langevin small approximation

March 10, 2013 Addition of two one half spins

March 10, 2013 Midterm II reflection

March 07, 2013 Thermodynamic identities

March 06, 2013 Temperature

March 05, 2013 Interacting spin

plus everything detailed in the description of my first update and before.

## PHY452H1S Basic Statistical Mechanics. Lecture 15: Grand Canonical/Fermion-Bosons. Taught by Prof. Arun Paramekanti

Posted by peeterjoot on March 14, 2013

[Click here for a PDF of this post with nicer formatting (especially if my latex to wordpress script has left FORMULA DOES NOT PARSE errors.)]

# Disclaimer

Peeter’s lecture notes from class. May not be entirely coherent.

# Grand Canonical/Fermion-Bosons

Was mentioned that three dimensions confines us to looking at either Fermions or Bosons, and that two dimensions is a rich subject (interchange of two particles isn’t the same as one particle cycling around the other ending up in the same place — how is that different than a particle cycling around another in a two dimensional space?)

Definitions

1. Fermion. Antisymmetric under exchange. $n_k = 0, 1$
2. Boson. Symmetric under exchange. $n_k = 0, 1, 2, \cdots$

In either case our energies are

\begin{aligned}\epsilon_k = \frac{\hbar^2 k^2}{2m},\end{aligned} \hspace{\stretch{1}}(1.2.1)

For Fermions we’ll have occupation filling of the form fig. 1.1, where there can be only one particle at any given site (an energy level for that value of momentum). For Bosonic systems as in fig. 1.2, we don’t have a restriction of only one particle for each state, and can have any given number of particles for each value of momentum.

Fig 1.1: Fermionic energy level filling for free particle in a box

Fig 1.2: Bosonic free particle in a box energy level filling

Our Hamiltonian is

\begin{aligned}H = \sum_k \hat{n}_k \epsilon_k,\end{aligned} \hspace{\stretch{1}}(1.2.2)

where we have a number operator

\begin{aligned}N = \sum \hat{n}_k,\end{aligned} \hspace{\stretch{1}}(1.2.3)

such that

\begin{aligned}\left[{N},{H}\right] = 0.\end{aligned} \hspace{\stretch{1}}(1.2.4)

\begin{aligned}Z_{\mathrm{G}} = \sum_{N=0}^\infty e^{\beta \mu N}\sum_{n_k, \sum n_k = N} e^{-\beta \sum_k n_k \epsilon_k}.\end{aligned} \hspace{\stretch{1}}(1.2.5)

While the second sum is constrained, because we are summing over all $n_k$, this is essentially an unconstrained sum, so we can write

\begin{aligned}Z_{\mathrm{G}} &= \sum_{n_k}e^{\beta \mu \sum_k n_k}e^{-\beta \sum_k n_k \epsilon_k} \\ &= \sum_{n_k} \left( \prod_k e^{-\beta(\epsilon_k - \mu) n_k} \right) \\ &= \prod_{n} \left( \sum_{n_k} e^{-\beta(\epsilon_k - \mu) n_k} \right).\end{aligned} \hspace{\stretch{1}}(1.2.6)

Fermions

\begin{aligned}\sum_{n_k = 0}^1 e^{-\beta(\epsilon_k - \mu) n_k} = 1 + e^{-\beta(\epsilon_k - \mu)}\end{aligned} \hspace{\stretch{1}}(1.2.7)

Bosons

\begin{aligned}\sum_{n_k = 0}^\infty e^{-\beta(\epsilon_k - \mu) n_k} = \frac{1}{{1 - e^{-\beta(\epsilon_k - \mu)}}}\end{aligned} \hspace{\stretch{1}}(1.2.8)

($\epsilon_k - \mu \ge 0$).

Our grand partition functions are then

\begin{aligned}Z_{\mathrm{G}}^f = \prod_k \left( 1 + e^{-\beta(\epsilon_k - \mu)} \right)\end{aligned} \hspace{\stretch{1}}(1.0.9a)

\begin{aligned}Z_{\mathrm{G}}^b = \prod_k \frac{1}{{ 1 - e^{-\beta(\epsilon_k - \mu)} }}\end{aligned} \hspace{\stretch{1}}(1.0.9b)

We can use these to compute the average number of particles

\begin{aligned}\left\langle{{n_k^f}}\right\rangle = \frac{1 \times 0 + e^{-\beta(\epsilon_k - \mu)} \times 1}{ 1 + e^{-\beta(\epsilon_k - \mu)} }=\frac{1}{{ 1 + e^{-\beta(\epsilon_k - \mu)} }}\end{aligned} \hspace{\stretch{1}}(1.0.10)

\begin{aligned}\left\langle{{n_k^b}}\right\rangle = \frac{1 \times 0 + e^{-\beta(\epsilon_k - \mu)} \times 1+e^{-2 \beta(\epsilon_k - \mu)} \times 2 + \cdots}{ 1+e^{-\beta(\epsilon_k - \mu)} +e^{-2 \beta(\epsilon_k - \mu)} }\end{aligned} \hspace{\stretch{1}}(1.0.11)

This chemical potential over temperature exponential

\begin{aligned}e^{\beta \mu} \equiv z,\end{aligned} \hspace{\stretch{1}}(1.0.12)

is called the fugacity. The denominator has the form

\begin{aligned}D = 1 + z e^{-\beta \epsilon_k}+ z^2 e^{-2 \beta \epsilon_k},\end{aligned} \hspace{\stretch{1}}(1.0.13)

so we see that

\begin{aligned}z \frac{\partial {D}}{\partial {z}} = z e^{-\beta \epsilon_k}+ 2 z^2 e^{-2 \beta \epsilon_k}+ 3 z^3 e^{-3 \beta \epsilon_k}+ \cdots\end{aligned} \hspace{\stretch{1}}(1.0.14)

Thus the numerator is

\begin{aligned}N = z \frac{\partial {D}}{\partial {z}},\end{aligned} \hspace{\stretch{1}}(1.0.15)

and

\begin{aligned}\left\langle{{n_k^b}}\right\rangle &= \frac{z \frac{\partial {D_k}}{\partial {z}} }{D_k} \\ &= z \frac{\partial {}}{\partial {z}} \ln D_k \\ &= \cdots \\ &= \frac{1}{{ e^{\beta(\epsilon_k - \mu)} - 1}}\end{aligned} \hspace{\stretch{1}}(1.0.16)

What is the density $\rho$?

For Fermions

\begin{aligned}\rho = \frac{N}{V} =\frac{1}{{V}} \sum_{\mathbf{k}}\frac{1}{{ e^{\beta(\epsilon_\mathbf{k} - \mu)} + 1}}\end{aligned} \hspace{\stretch{1}}(1.0.17)

Using a “particle in a box” quantization where $k_\alpha = 2 \pi m_\alpha/L$, in a $d$-dimensional space, we can approximate this as

\begin{aligned}\boxed{\rho = \int \frac{d^d k}{(2 \pi)^d}\frac{1}{{ e^{\beta(\epsilon_k - \mu)} - 1}}.}\end{aligned} \hspace{\stretch{1}}(1.0.18)

This integral is actually difficult to evaluate. For $T \rightarrow 0$ ($\beta \rightarrow \infty$, where

\begin{aligned}n_k = \Theta(\mu - \epsilon_k).\end{aligned} \hspace{\stretch{1}}(1.0.19)

This is illustrated in, where we also show the smearing that occurs as temperature increases fig. 1.3.

Fig 1.3: Occupation numbers for different energies

With

\begin{aligned}E_{\mathrm{F}} = \mu(T = 0),\end{aligned} \hspace{\stretch{1}}(1.0.20)

we want to ask what is the radius of the ball for which

\begin{aligned}\epsilon_k = E_{\mathrm{F}}\end{aligned} \hspace{\stretch{1}}(1.0.21)

or

\begin{aligned}E_{\mathrm{F}} = \frac{\hbar^2 k_{\mathrm{F}}^2}{2m},\end{aligned} \hspace{\stretch{1}}(1.0.22)

so that

\begin{aligned}k_{\mathrm{F}} = \sqrt{\frac{2 m E_{\mathrm{F}}}{\hbar^2}},\end{aligned} \hspace{\stretch{1}}(1.0.23)

so that our density where $\epsilon_k = \mu$ is

\begin{aligned}\rho &= \int_{k \le k_{\mathrm{F}}} \frac{d^3 k}{(2 \pi)^3} \times 1 \\ &= \frac{1}{{(2\pi)^3}} 4 \pi \int^{k_{\mathrm{F}}} k^2 dk \\ &= \frac{4 \pi}{3} k_{\mathrm{F}}^3 \frac{1}{{(2 \pi)^3}},\end{aligned} \hspace{\stretch{1}}(1.0.24)

so that

\begin{aligned}k_{\mathrm{F}} = (6 \pi^2 \rho)^{1/3},\end{aligned} \hspace{\stretch{1}}(1.0.25)

Our chemical potential at zero temperature is then

\begin{aligned}\mu(T = 0) = \frac{\hbar^2}{2m} (6 \pi^2 \rho)^{2/3}.\end{aligned} \hspace{\stretch{1}}(1.0.26)

\begin{aligned}\rho^{-1/3} = \mbox{interparticle spacing}.\end{aligned} \hspace{\stretch{1}}(1.0.27)

We can convince ourself that the chemical potential must have the form fig. 1.4.

Fig 1.4: Large negative chemical potential at high temperatures

Given large negative chemical potential at high temperatures our number distribution will have the form

\begin{aligned}\left\langle{{n_k}}\right\rangle = e^{-\beta (\epsilon_k - \mu)} \propto e^{-\beta \epsilon_k}\end{aligned} \hspace{\stretch{1}}(1.0.28)

We see that the classical Boltzmann distribution is recovered for high temperatures.

We can also calculate the chemical potential at high temperatures. We’ll find that this has the form

\begin{aligned}e^{\beta \mu} = \frac{4}{3} \rho \lambda_T^3,\end{aligned} \hspace{\stretch{1}}(1.0.29)

where this quantity $\lambda_T$ is called the Thermal de Broglie wavelength.

\begin{aligned}\lambda_T = \sqrt{\frac{ 2 \pi \hbar^2}{m k_{\mathrm{B}} T}}.\end{aligned} \hspace{\stretch{1}}(1.0.30)

## An updated compilation of notes, for ‘PHY452H1S Basic Statistical Mechanics’, Taught by Prof. Arun Paramekanti

Posted by peeterjoot on March 3, 2013

That compilation now all of the following too (no further updates will be made to any of these) :

February 28, 2013 Rotation of diatomic molecules

February 28, 2013 Helmholtz free energy

February 26, 2013 Statistical and thermodynamic connection

February 24, 2013 Ideal gas

February 16, 2013 One dimensional well problem from Pathria chapter II

February 15, 2013 1D pendulum problem in phase space

February 14, 2013 Continuing review of thermodynamics

February 13, 2013 Lightning review of thermodynamics

February 11, 2013 Cartesian to spherical change of variables in 3d phase space

February 10, 2013 n SHO particle phase space volume

February 10, 2013 Change of variables in 2d phase space

February 10, 2013 Some problems from Kittel chapter 3

February 07, 2013 Midterm review, thermodynamics

February 06, 2013 Limit of unfair coin distribution, the hard way

February 05, 2013 Ideal gas and SHO phase space volume calculations

February 03, 2013 One dimensional random walk

February 02, 2013 1D SHO phase space

February 02, 2013 Application of the central limit theorem to a product of random vars

January 31, 2013 Liouville’s theorem questions on density and current

January 30, 2013 State counting

## PHY452H1S Basic Statistical Mechanics. Lecture 7: Ideal gas and SHO phase space volume calculations. Taught by Prof. Arun Paramekanti

Posted by peeterjoot on February 5, 2013

# Disclaimer

Peeter’s lecture notes from class. May not be entirely coherent.

# Review. Classical phase space calculation

\begin{aligned}E_{\mathrm{ideal}} = \sum_i \frac{\mathbf{p}_i^2}{2 m}\end{aligned} \hspace{\stretch{1}}(1.2.1)

From this we calculated $\gamma(E)$, and

\begin{aligned}\frac{d\gamma(E)}{dE} = \Omega_{\mathrm{classical}}(E)\end{aligned} \hspace{\stretch{1}}(1.2.2)

Fudging with a requirement that $\Delta x \Delta p \sim h$, we corrected this as

\begin{aligned}\Omega_{\mathrm{quantum}}(E) = \frac{\Omega_{\mathrm{classical}}(E)}{N! h^{3N}}\end{aligned} \hspace{\stretch{1}}(1.2.3)

Now let’s do the quantum calculation.

Quantum calculation

Recall that for the solutions of the Quantum free particle in a box, as in (Fig 1), our solutions are

Fig 1: 1D Quantum free particle in a box

\begin{aligned}\Psi_n(x) = \sqrt{\frac{2}{L}} \sin\left( \frac{ n \pi x}{L} \right),\end{aligned} \hspace{\stretch{1}}(1.2.4)

where $n = 1, 2, \cdots$, and

\begin{aligned}\epsilon_n = \frac{n^2 h^2}{8 m L^2}\end{aligned} \hspace{\stretch{1}}(1.2.5)

.

In three dimensions, with $n_i = 1, 2, \cdots$ we have

\begin{aligned}\Psi_{n_1, n_2, n_3}(x, y, z) = \left( \frac{2}{L} \right)^{3/2} \sin\left( \frac{ n_1 \pi x}{L} \right)\sin\left( \frac{ n_2 \pi x}{L} \right)\sin\left( \frac{ n_3 \pi x}{L} \right)\end{aligned} \hspace{\stretch{1}}(1.2.6)

and

\begin{aligned}\epsilon_{n_1, n_2, n_3} = \frac{h^2}{8 m L^2} \left( n_1^2 + n_2^2 + n_3^2 \right)\end{aligned} \hspace{\stretch{1}}(1.2.7)

\begin{aligned}\gamma^{3d}_{\mathrm{classical}}(E) = \underbrace{V}_{L^3}\int d^3 p \theta \left( E - \frac{\mathbf{p}^2}{2m} \right)= V \frac{4 \pi}{3} (2 m E)^{3/2}\end{aligned} \hspace{\stretch{1}}(1.2.8)

so that

\begin{aligned}\gamma^{3d}_{\mathrm{corrected}}(E) = V \frac{4 \pi}{3} \frac{(2 m E)^{3/2}}{h^3}\end{aligned} \hspace{\stretch{1}}(1.2.9)

\begin{aligned}\gamma^{3d}_{\mathrm{quantum}}(E) = \sum_{n_1, n_2, n_3} \Theta(E - \epsilon_{n_1, n_2, n_3} ).\end{aligned} \hspace{\stretch{1}}(1.2.10)

How do the multiplicities scale by energy? We’ll have expect something like (Fig 2).

Provided the energies $E \gg 3h^2/(8 m L)$ are large enough, we can approximate the sum with

\begin{aligned}\sum_{n_1, n_2, n_3} \sim \int_0^\infty dn_1 dn_2 dn_3\end{aligned} \hspace{\stretch{1}}(1.2.11)

So

\begin{aligned}\gamma^{3d}_{\mathrm{quantum}} \left( E \gg \frac{h^2}{8 m L^2} \right) \approx\int_0^\infty dn_1 dn_2 dn_3 \Theta \left( E - \frac{h^2}{8 m L^2} \left( n_1^2 + n_2^2 + n_3^2 \right) \right)=\frac{1}{{8}}\frac{4 \pi}{3} \left( \frac{8 m L^2 E}{h^2} \right)^{3/2}=L^3\frac{4 \pi}{3} \frac{\left( 2 m E \right)^{3/2}}{h^3}\end{aligned} \hspace{\stretch{1}}(1.2.12)

Harmonic oscillator in 1D.

Our phase space is of the form (Fig 3).

Fig 3: 1D classical SHO phase space

Where the number of states in this classical picture are found with:

\begin{aligned}\gamma^{\mathrm{classical}}(E) = \int dx dp \theta\left( E - \left( \frac{1}{{2}} k x^2 + \frac{1}{{2m }} p^2 \right) \right).\end{aligned} \hspace{\stretch{1}}(1.2.13)

Rescale

\begin{aligned}\tilde{x} = x \sqrt{ \frac{k}{2}}\end{aligned} \hspace{\stretch{1}}(1.0.14a)

\begin{aligned}\tilde{p} = \frac{p}{\sqrt{2m}}\end{aligned} \hspace{\stretch{1}}(1.0.14b)

so that we have

\begin{aligned}\gamma^{\mathrm{classical}}(E) = \int d\tilde{x} d \tilde{p} \sqrt{\frac{2 \times 2 m}{k}} \theta\left( E - \tilde{x}^2 - \tilde{p}^2 \right)=2 \sqrt{\frac{m}{k}} \pi E= 2 \pi \sqrt{\frac{m}{k}} E.\end{aligned} \hspace{\stretch{1}}(1.0.15)

\begin{aligned}\gamma^{\mathrm{SHO}}_{\mathrm{corrected}}(E) = 2 \pi \sqrt{\frac{m}{k}} \frac{E}{h}.\end{aligned} \hspace{\stretch{1}}(1.0.16)

We have for the energy

\begin{aligned}E_n^{\mathrm{SHO}} = \left( n + \frac{1}{{2}} \right) \hbar \omega\end{aligned} \hspace{\stretch{1}}(1.0.17a)

\begin{aligned}\omega = \sqrt{\frac{k}{m}}\end{aligned} \hspace{\stretch{1}}(1.0.17b)

\begin{aligned}\hbar = \frac{h}{2 \pi}\end{aligned} \hspace{\stretch{1}}(1.0.17c)

graphing the counts (Fig 4), we again have stepping as a function of energy, but no multiplicities this time

Fig 4: 1D quantum SHO states per energy level

\begin{aligned}\gamma_{\mathrm{quantum}}(E) = \sum_{n = 0}^\infty \Theta\left( E - \left( n + \frac{1}{{2}} \hbar \omega \right) \right)\end{aligned} \hspace{\stretch{1}}(1.0.18)

we make the continuous approximation for the summation again, and throwing away the zero point energy, we have

\begin{aligned}\gamma_{\mathrm{quantum}}(E \gg \hbar \omega) \approx\int_{0}^\infty dn \Theta\left( E - n \hbar \omega \right)= 2 \pi \frac{E}{h} \sqrt{\frac{m}{k}}\end{aligned} \hspace{\stretch{1}}(1.0.19)

Why $N!$?

We have a problem with out counting here. Consider some particles in a box as in (Fig 5).

Fig 5: Three particles in a box

1. particle $1$ at $\mathbf{x}_1$
2. particle $2$ at $\mathbf{x}_2$
3. particle $3$ at $\mathbf{x}_3$

or

1. particle $1$ at $\mathbf{x}_2$
2. particle $2$ at $\mathbf{x}_3$
3. particle $3$ at $\mathbf{x}_1$

This is fine in the classical picture, but in the quantum picture with an assumption of indistinguishability, no two particles (say electrons) cannot be labelled in this fashion.

\begin{aligned}\underbrace{S_{\mathrm{ideal}}^{(\mathrm{E}, \mathrm{N}, \mathrm{V})}}_{\text{Statistical entropy}}= k_{\mathrm{B}} \ln \left( \frac{\Omega_{\mathrm{classical}}}{h^{3N}} \right)\underbrace{\approx}_{N \gg 1} k_{\mathrm{B}} \left( N \ln V + \frac{3 N}{2} \ln \left( \frac{4 \pi m E }{3 N h^2} \right) + \frac{3 N}{2} \right)\end{aligned} \hspace{\stretch{1}}(1.0.20)

Suppose we double the volume as in (Fig 6), then our total entropy for the bigger system would be

Fig 6: Gibbs volume doubling argument. Two identical systems allowed to mix

\begin{aligned}S_{\mathrm{total}}^{(\mathrm{E}, \mathrm{N}, \mathrm{V})}= k_{\mathrm{B}} \ln \left( \frac{\Omega_{\mathrm{classical}}}{h^{3N}} \right)\approx k_{\mathrm{B}} \left( (2 N) \ln (2 V) + \frac{3 (2 N)}{2} \ln \left( \frac{4 \pi m (2 E) }{2 ( 2 N) h^2} \right) + \frac{3 (2 N)}{2} \right).\end{aligned} \hspace{\stretch{1}}(1.0.21)

We have

\begin{aligned}S_{\mathrm{total}} = S_1 + S_2 + k_{\mathrm{B}} (2 N) \ln 2= S_1 + S_2 + k_{\mathrm{B}} \ln 2^{2 N}.\end{aligned} \hspace{\stretch{1}}(1.0.22)

This is telling us that each particle could be in either the left or the right side, but we know that this uncertainty shouldn’t be in the final answer. We must drop this $k_{\mathrm{B}}$ term.

So, if we assume that these particles are identical, and divide $\Omega$ by $N!$, then we find

\begin{aligned}S_{\mathrm{ideal}} = k_{\mathrm{B}} \left( N \ln \frac{V}{N} + \frac{3 N}{2} \ln \left( \frac{4 \pi m E }{3 N h^2} \right) + \frac{5 N}{2} \right)\end{aligned} \hspace{\stretch{1}}(1.0.23)