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

Fermion. Antisymmetric under exchange. $n_k = 0, 1$
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

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

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

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.

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

Posted in Math and Physics Learning. | Tagged: Boltzmann distribution, Boson, chemical potential, density of states, Fermion, fugacity, grand canonical ensemble, grand partition function, hamiltonian, number operator, occupation number, particle in a box, PHY452H1S, statistical mechanics, thermal de Broglie wavelength | Leave a Comment »

Addition of two one half spins

Posted by peeterjoot on March 10, 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.)]

In class an example of interacting spin was given where the Hamiltonian included a two spins dot product

$\begin{aligned}H = \mathbf{S}_1 \cdot \mathbf{S}_2.\end{aligned} \hspace{\stretch{1}}(1.0.1)$

The energy eigenvalues for this Hamiltonian were derived by using the trick to rewrite this in terms of just squared spin operators

$\begin{aligned}H = \frac{(\mathbf{S}_1 + \mathbf{S}_2)^2 - \mathbf{S}_1^2 - \mathbf{S}_2^2}{2}.\end{aligned} \hspace{\stretch{1}}(1.0.2)$

For each of these terms we can calculate the total energy eigenvalues from

$\begin{aligned}\mathbf{S}^2 \Psi = \hbar^2 S (S + 1) \Psi,\end{aligned} \hspace{\stretch{1}}(1.0.3)$

where $S$ takes on the values of the total spin for the (possibly composite) spin operator. Thinking about the spin operators in their matrix representation, it’s not obvious to me that we can just add the total spins, so that if $\mathbf{S}_1$ and $\mathbf{S}_2$ are the spin operators for two respective particle, then the total system has a spin operator $\mathbf{S} = \mathbf{S}_1 + \mathbf{S}_2$ (really $\mathbf{S} = \mathbf{S}_1 \otimes I_2 + I_2 \otimes \mathbf{S}_2$ , since the respective spin operators only act on their respective particles).

Let’s develop a bit of intuition on this, by calculating the energy eigenvalues of $\mathbf{S}_1 \cdot \mathbf{S}_2$ using Pauli matrices.

First lets look at how each of the Pauli matrices operate on the $S_z$ eigenvectors

$\begin{aligned}\sigma_x {\left\lvert {+} \right\rangle} = \begin{bmatrix} 0 & 1 \\ 1 & 0 \\ \end{bmatrix} \begin{bmatrix}1 \\ 0\end{bmatrix}=\begin{bmatrix}0 \\ 1 \end{bmatrix}= {\left\lvert {-} \right\rangle}\end{aligned} \hspace{\stretch{1}}(1.0.4a)$

$\begin{aligned}\sigma_x {\left\lvert {-} \right\rangle} = \begin{bmatrix} 0 & 1 \\ 1 & 0 \\ \end{bmatrix} \begin{bmatrix}0 \\ 1\end{bmatrix}=\begin{bmatrix}1 \\ 0 \end{bmatrix}= {\left\lvert {+} \right\rangle}\end{aligned} \hspace{\stretch{1}}(1.0.4b)$

$\begin{aligned}\sigma_y {\left\lvert {+} \right\rangle} = \begin{bmatrix} 0 & -i \\ i & 0 \\ \end{bmatrix} \begin{bmatrix}1 \\ 0\end{bmatrix}=\begin{bmatrix}0 \\ i \end{bmatrix}= i {\left\lvert {-} \right\rangle}\end{aligned} \hspace{\stretch{1}}(1.0.4c)$

$\begin{aligned}\sigma_y {\left\lvert {-} \right\rangle} = \begin{bmatrix} 0 & -i \\ i & 0 \\ \end{bmatrix} \begin{bmatrix}0 \\ 1\end{bmatrix}=\begin{bmatrix}-i \\ 0 \end{bmatrix}= -i {\left\lvert {+} \right\rangle}\end{aligned} \hspace{\stretch{1}}(1.0.4d)$

$\begin{aligned}\sigma_z {\left\lvert {+} \right\rangle} = \begin{bmatrix} 1 & 0 \\ 0 & -1 \\ \end{bmatrix} \begin{bmatrix}1 \\ 0\end{bmatrix}=\begin{bmatrix}1 \\ 0 \end{bmatrix}= {\left\lvert {+} \right\rangle}\end{aligned} \hspace{\stretch{1}}(1.0.4e)$

$\begin{aligned}\sigma_z {\left\lvert {-} \right\rangle} = \begin{bmatrix} 1 & 0 \\ 0 & -1 \\ \end{bmatrix} \begin{bmatrix}0 \\ 1\end{bmatrix}=-\begin{bmatrix}0 \\ 1 \end{bmatrix}= -{\left\lvert {-} \right\rangle}\end{aligned} \hspace{\stretch{1}}(1.0.4f)$

Summarizing, these are

$\begin{aligned}\sigma_x {\left\lvert {\pm} \right\rangle} = {\left\lvert {\mp} \right\rangle}\end{aligned} \hspace{\stretch{1}}(1.0.5a)$

$\begin{aligned}\sigma_y {\left\lvert {\pm} \right\rangle} = \pm i {\left\lvert {\mp} \right\rangle}\end{aligned} \hspace{\stretch{1}}(1.0.5b)$

$\begin{aligned}\sigma_z {\left\lvert {\pm} \right\rangle} = \pm {\left\lvert {\pm} \right\rangle}\end{aligned} \hspace{\stretch{1}}(1.0.5c)$

For convienience let’s avoid any sort of direct product notation, with the composite operations defined implicitly by

$\begin{aligned}\left( S_{1k} \otimes S_{2k} \right)\left( {\left\lvert {\alpha} \right\rangle} \otimes {\left\lvert {\beta} \right\rangle} \right)=S_{1k} S_{2k} {\left\lvert {\alpha \beta} \right\rangle}=\left( S_{1k} {\left\lvert {\alpha} \right\rangle} \right) \otimes\left( S_{2k} {\left\lvert {\beta} \right\rangle} \right).\end{aligned} \hspace{\stretch{1}}(1.0.6)$

Now let’s compute all the various operations

$\begin{aligned}\begin{aligned}\sigma_{1x} \sigma_{2x} {\left\lvert {++} \right\rangle} &= {\left\lvert {--} \right\rangle} \\ \sigma_{1x} \sigma_{2x} {\left\lvert {--} \right\rangle} &= {\left\lvert {++} \right\rangle} \\ \sigma_{1x} \sigma_{2x} {\left\lvert {+-} \right\rangle} &= {\left\lvert {-+} \right\rangle} \\ \sigma_{1x} \sigma_{2x} {\left\lvert {-+} \right\rangle} &= {\left\lvert {+-} \right\rangle}\end{aligned}\end{aligned} \hspace{\stretch{1}}(1.0.7a)$

$\begin{aligned}\begin{aligned}\sigma_{1y} \sigma_{2y} {\left\lvert {++} \right\rangle} &= i^2 {\left\lvert {--} \right\rangle} \\ \sigma_{1y} \sigma_{2y} {\left\lvert {--} \right\rangle} &= (-i)^2 {\left\lvert {++} \right\rangle} \\ \sigma_{1y} \sigma_{2y} {\left\lvert {+-} \right\rangle} &= i (-i) {\left\lvert {-+} \right\rangle} \\ \sigma_{1y} \sigma_{2y} {\left\lvert {-+} \right\rangle} &= (-i) i {\left\lvert {+-} \right\rangle}\end{aligned}\end{aligned} \hspace{\stretch{1}}(1.0.7b)$

$\begin{aligned}\begin{aligned}\sigma_{1z} \sigma_{2z} {\left\lvert {++} \right\rangle} &= (-1)^2 {\left\lvert {--} \right\rangle} \\ \sigma_{1z} \sigma_{2z} {\left\lvert {--} \right\rangle} &= {\left\lvert {++} \right\rangle} \\ \sigma_{1z} \sigma_{2z} {\left\lvert {+-} \right\rangle} &= -{\left\lvert {-+} \right\rangle} \\ \sigma_{1z} \sigma_{2z} {\left\lvert {-+} \right\rangle} &= -{\left\lvert {+-} \right\rangle}\end{aligned}\end{aligned} \hspace{\stretch{1}}(1.0.7c)$

Tabulating first the action of the sum of the $x$ and $y$ operators we have

$\begin{aligned}\begin{aligned}\left( \sigma_{1x} \sigma_{2x} + \sigma_{1y} \sigma_{2y} \right) {\left\lvert {++} \right\rangle} &= 0 \\ \left( \sigma_{1x} \sigma_{2x} + \sigma_{1y} \sigma_{2y} \right) {\left\lvert {--} \right\rangle} &= 0 \\ \left( \sigma_{1x} \sigma_{2x} + \sigma_{1y} \sigma_{2y} \right) {\left\lvert {+-} \right\rangle} &= 2 {\left\lvert {-+} \right\rangle} \\ \left( \sigma_{1x} \sigma_{2x} + \sigma_{1y} \sigma_{2y} \right) {\left\lvert {-+} \right\rangle} &= 2 {\left\lvert {+-} \right\rangle}\end{aligned}\end{aligned} \hspace{\stretch{1}}(1.0.8)$

so that

$\begin{aligned}\begin{aligned}\mathbf{S}_1 \cdot \mathbf{S}_2 {\left\lvert {++} \right\rangle} &= {\left\lvert {++} \right\rangle} \\ \mathbf{S}_1 \cdot \mathbf{S}_2 {\left\lvert {--} \right\rangle} &= {\left\lvert {--} \right\rangle} \\ \mathbf{S}_1 \cdot \mathbf{S}_2 {\left\lvert {+-} \right\rangle} &= 2 {\left\lvert {-+} \right\rangle} - {\left\lvert {+-} \right\rangle} \\ \mathbf{S}_1 \cdot \mathbf{S}_2 {\left\lvert {-+} \right\rangle} &= 2 {\left\lvert {+-} \right\rangle} - {\left\lvert {-+} \right\rangle}\end{aligned}\end{aligned} \hspace{\stretch{1}}(1.0.9)$

Now we are set to write out the Hamiltonian matrix. Doing this with respect to the basis $\beta = \{ {\left\lvert {++} \right\rangle}, {\left\lvert {--} \right\rangle}, {\left\lvert {+-} \right\rangle}, {\left\lvert {-+} \right\rangle} \}$ , we have

$\begin{aligned}H &= \mathbf{S}_1 \cdot \mathbf{S}_2 \\ &= \frac{\hbar^2}{4} \begin{bmatrix}\left\langle ++ \right\rvert H \left\lvert ++ \right\rangle & \left\langle ++ \right\rvert H \left\lvert -- \right\rangle & \left\langle ++ \right\rvert H \left\lvert +- \right\rangle & \left\langle ++ \right\rvert H \left\lvert -+ \right\rangle \\ \left\langle -- \right\rvert H \left\lvert ++ \right\rangle & \left\langle -- \right\rvert H \left\lvert -- \right\rangle & \left\langle -- \right\rvert H \left\lvert +- \right\rangle & \left\langle -- \right\rvert H \left\lvert -+ \right\rangle \\ \left\langle +- \right\rvert H \left\lvert ++ \right\rangle & \left\langle +- \right\rvert H \left\lvert -- \right\rangle & \left\langle +- \right\rvert H \left\lvert +- \right\rangle & \left\langle +- \right\rvert H \left\lvert -+ \right\rangle \\ \left\langle -+ \right\rvert H \left\lvert ++ \right\rangle & \left\langle -+ \right\rvert H \left\lvert -- \right\rangle & \left\langle -+ \right\rvert H \left\lvert +- \right\rangle & \left\langle -+ \right\rvert H \left\lvert -+ \right\rangle \end{bmatrix} \\ &= \frac{\hbar^2}{4} \begin{bmatrix}1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & -1 & 2 \\ 0 & 0 & 2 & -1 \end{bmatrix} \end{aligned} \hspace{\stretch{1}}(1.0.10)$

Two of the eigenvalues we can read off by inspection, and for the other two need to solve

$\begin{aligned}0 =\begin{vmatrix}-\hbar^2/4 - \lambda & \hbar^2/2 \\ \hbar^2/2 & -\hbar^2/4 - \lambda\end{vmatrix}= (\hbar^2/4 + \lambda)^2 - (\hbar^2/2)^2\end{aligned} \hspace{\stretch{1}}(1.0.11)$

$\begin{aligned}\lambda = -\frac{\hbar^2}{4} \pm \frac{\hbar^2}{2} = \frac{\hbar^2}{4}, -\frac{3 \hbar^2}{4}.\end{aligned} \hspace{\stretch{1}}(1.0.12)$

These are the last of the triplet energy eigenvalues and the singlet value that we expected from the spin addition method. The eigenvectors for the $\hbar^2/4$ eigenvalue is given by the solution of

$\begin{aligned}0 =\frac{\hbar^2}{2}\begin{bmatrix}-1 & 1 \\ 1 & -1\end{bmatrix}\begin{bmatrix}a \\ b\end{bmatrix}\end{aligned} \hspace{\stretch{1}}(1.0.13)$

So the eigenvector is

$\begin{aligned}\frac{1}{{\sqrt{2}}} \left( {\left\lvert {+-} \right\rangle} + {\left\lvert {-+} \right\rangle} \right)\end{aligned} \hspace{\stretch{1}}(1.0.14)$

For our $-3\hbar^2/4$ eigenvalue we seek

$\begin{aligned}0 =\frac{\hbar^2}{2}\begin{bmatrix}1 & 1 \\ 1 & 1\end{bmatrix}\begin{bmatrix}a \\ b\end{bmatrix}\end{aligned} \hspace{\stretch{1}}(1.0.15)$

So the eigenvector is

$\begin{aligned}\frac{1}{{\sqrt{2}}} \left( {\left\lvert {+-} \right\rangle} - {\left\lvert {-+} \right\rangle} \right)\end{aligned} \hspace{\stretch{1}}(1.0.16)$

An orthonormal basis with respective eigenvalues $\hbar^2/4 (\times 3), -3\hbar^2/4$ is thus given by

$\begin{aligned}\beta' = \left\{{\left\lvert {++} \right\rangle},{\left\lvert {--} \right\rangle},\frac{1}{{\sqrt{2}}} \left( {\left\lvert {+-} \right\rangle} + {\left\lvert {-+} \right\rangle} \right),\frac{1}{{\sqrt{2}}} \left( {\left\lvert {+-} \right\rangle} - {\left\lvert {-+} \right\rangle} \right)\right\}.\end{aligned} \hspace{\stretch{1}}(1.0.17)$

Confirmation of spin additivity.

Let’s use this to confirm that for $H = (\mathbf{S}_1 + \mathbf{S}_2)^2$ , the two spin $1/2$ particles have a combined spin given by

$\begin{aligned}S(S + 1) \hbar^2.\end{aligned} \hspace{\stretch{1}}(1.0.18)$

With

$\begin{aligned}(\mathbf{S}_1 + \mathbf{S}_2)^2 = \mathbf{S}_1^2 + \mathbf{S}_2^2 + 2 \mathbf{S}_1 \cdot \mathbf{S}_2,\end{aligned} \hspace{\stretch{1}}(1.0.19)$

we have for the $\hbar^2/4$ energy eigenstate of $\mathbf{S}_1 \cdot \mathbf{S}_2$

$\begin{aligned}2 \hbar^2 \frac{1}{{2}} \left( 1 + \frac{1}{{2}} \right) + 2 \frac{\hbar^2}{4} = 2 \hbar^2,\end{aligned} \hspace{\stretch{1}}(1.0.20)$

and for the $-3\hbar^2/4$ energy eigenstate of $\mathbf{S}_1 \cdot \mathbf{S}_2$

$\begin{aligned}2 \hbar^2 \frac{1}{{2}} \left( 1 + \frac{1}{{2}} \right) + 2 \left( - \frac{3 \hbar^2}{4} \right) = 0.\end{aligned} \hspace{\stretch{1}}(1.0.21)$

We get the $2 \hbar^2$ and $0$ eigenvalues respectively as expected.

Posted in Math and Physics Learning. | Tagged: addition of spin, eigenvalue, eigenvector, hamiltonian, orthonormal basis, pauli matrix, PHY452H1S, spin one half, statistical mechanics | Leave a Comment »

PHY452H1S Basic Statistical Mechanics. Problem Set 5: Temperature

Posted by peeterjoot on March 10, 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

This is an ungraded set of answers to the problems posed.

Question: Polymer stretching – “entropic forces” (2013 problem set 5, p1)

Consider a toy model of a polymer in one dimension which is made of $N$ steps (amino acids) of unit length, going left or right like a random walk. Let one end of this polymer be at the origin and the other end be at a point $X = \sqrt{N}$ (viz. the rms size of the polymer) , so $1 \ll X \ll N$ . We have previously calculated the number of configurations corresponding to this condition (approximate the binomial distribution by a Gaussian).

Part a

Using this, find the entropy of this polymer as $S = k_{\mathrm{B}} \ln \Omega$ . The free energy of this polymer, even in the absence of any other interactions, thus has an entropic contribution, $F = -T S$ . If we stretch this polymer, we expect to have fewer available configurations, and thus a smaller entropy and a higher free energy.

Part b

Find the change in free energy of this polymer if we stretch this polymer from its end being at $X$ to a larger distance $X + \Delta X$ .

Part c

Show that the change in free energy is linear in the displacement for small $\Delta X$ , and hence find the temperature dependent “entropic spring constant” of this polymer. (This entropic force is important to overcome for packing DNA into the nucleus, and in many biological processes.)

Typo correction (via email):
You need to show that the change in free energy is quadratic in the displacement $\Delta X$ , not linear in $\Delta X$ . The force is linear in $\Delta X$ . (Exactly as for a “spring”.)

Answer

Entropy.

In lecture 2 probabilities for the sums of fair coin tosses were considered. Assigning $\pm 1$ to the events $Y_k$ for heads and tails coin tosses respectively, a random variable $Y = \sum_k Y_k$ for the total of $N$ such events was found to have the form

$\begin{aligned}P_N(Y) = \left\{\begin{array}{l l}\left(\frac{1}{{2}}\right)^N \frac{N!}{\left(\frac{N-Y}{2}\right)!\left(\frac{N+Y}{2}\right)!}& \quad \mbox{if Y and N have same parity} \\ 0& \quad \mbox{otherwise} \end{array}\right.\end{aligned} \hspace{\stretch{1}}(1.1.1)$

For an individual coin tosses we have averages $\left\langle{{Y_1}}\right\rangle = 0$ , and $\left\langle{{Y_1^2}}\right\rangle = 1$ , so the central limit theorem provides us with a large $N$ Gaussian approximation for this distribution

$\begin{aligned}P_N(Y) \approx\frac{2}{\sqrt{2 \pi N}} \exp\left( -\frac{Y^2}{2N} \right).\end{aligned} \hspace{\stretch{1}}(1.1.2)$

This fair coin toss problem can also be thought of as describing the coordinate of the end point of a one dimensional polymer with the beginning point of the polymer is fixed at the origin. Writing $\Omega(N, Y)$ for the total number of configurations that have an end point at coordinate $Y$ we have

$\begin{aligned}P_N(Y) = \frac{\Omega(N, Y)}{2^N},\end{aligned} \hspace{\stretch{1}}(1.1.3)$

From this, the total number of configurations that have, say, length $X = \left\lvert {Y} \right\rvert$ , in the large $N$ Gaussian approximation, is

$\begin{aligned}\Omega(N, X) &= 2^N \left( P_N(+X) +P_N(-X) \right) \\ &= \frac{2^{N + 2}}{\sqrt{2 \pi N}} \exp\left( -\frac{X^2}{2N} \right).\end{aligned} \hspace{\stretch{1}}(1.1.4)$

The entropy associated with a one dimensional polymer of length $X$ is therefore

$\begin{aligned}S_N(X) &= - k_{\mathrm{B}} \frac{X^2}{2N} + k_{\mathrm{B}} \ln \frac{2^{N + 2}}{\sqrt{2 \pi N}} \\ &= - k_{\mathrm{B}} \frac{X^2}{2N} + \text{constant}.\end{aligned} \hspace{\stretch{1}}(1.1.5)$

Writing $S_0$ for this constant the free energy is

$\begin{aligned}\boxed{F = U - T S = U + k_{\mathrm{B}} T \frac{X^2}{2N} + S_0 T.}\end{aligned} \hspace{\stretch{1}}(1.1.6)$

Change in free energy.

At constant temperature, stretching the polymer from its end being at $X$ to a larger distance $X + \Delta X$ , results in a free energy change of

$\begin{aligned}\Delta F &= F( X + \Delta X ) - F(X) \\ &= \frac{k_{\mathrm{B}} T}{2N} \left( (X + \Delta X)^2 - X^2 \right) \\ &= \frac{k_{\mathrm{B}} T}{2N} \left( 2 X \Delta X + (\Delta X)^2 \right)\end{aligned} \hspace{\stretch{1}}(1.1.7)$

If $\Delta X$ is assumed small, our constant temperature change in free energy $\Delta F \approx (\partial F/\partial X)_T \Delta X$ is

$\begin{aligned}\boxed{\Delta F = \frac{k_{\mathrm{B}} T}{N} X \Delta X.}\end{aligned} \hspace{\stretch{1}}(1.1.8)$

Temperature dependent spring constant.

I found the statement and subsequent correction of the problem statement somewhat confusing. To figure this all out, I thought it was reasonable to step back and relate free energy to the entropic force explicitly.

Consider temporarily a general thermodynamic system, for which we have by definition free energy and thermodynamic identity respectively

$\begin{aligned}F = U - T S,\end{aligned} \hspace{\stretch{1}}(1.0.9a)$

$\begin{aligned}dU = T dS - P dV.\end{aligned} \hspace{\stretch{1}}(1.0.9b)$

The differential of the free energy is

$\begin{aligned}dF &= dU - T dS - S dT \\ &= -P dV - S dT \\ &= \left( \frac{\partial {F}}{\partial {T}} \right)_V dT+\left( \frac{\partial {F}}{\partial {V}} \right)_T dV.\end{aligned} \hspace{\stretch{1}}(1.0.10)$

Forming the wedge product with $dT$ , we arrive at the two form

$\begin{aligned}0 &= \left( \left( P + \left( \frac{\partial {F}}{\partial {V}} \right)_T \right) dV + \left( S + \left( \frac{\partial {F}}{\partial {T}} \right)_V \right) dT \right)\wedge dT \\ &= \left( P + \left( \frac{\partial {F}}{\partial {V}} \right)_T \right) dV \wedge dT,\end{aligned} \hspace{\stretch{1}}(1.0.11)$

This provides the relation between free energy and the “pressure” for the system

$\begin{aligned}P = - \left( \frac{\partial {F}}{\partial {V}} \right)_T.\end{aligned} \hspace{\stretch{1}}(1.0.12)$

For a system with a constant cross section $\Delta A$ , $dV = \Delta A dX$ , so the force associated with the system is

$\begin{aligned}f &= P \Delta A \\ &= - \frac{1}{{\Delta A}} \left( \frac{\partial {F}}{\partial {X}} \right)_T \Delta A,\end{aligned} \hspace{\stretch{1}}(1.0.13)$

$\begin{aligned}f = - \left( \frac{\partial {F}}{\partial {X}} \right)_T.\end{aligned} \hspace{\stretch{1}}(1.0.14)$

Okay, now we have a relation between the force and the rate of change of the free energy

$\begin{aligned}f(X) = -\frac{k_{\mathrm{B}} T}{N} X.\end{aligned} \hspace{\stretch{1}}(1.0.15)$

Our temperature dependent “entropic spring constant” in analogy with $f = -k X$ , is therefore

$\begin{aligned}\boxed{k = \frac{k_{\mathrm{B}} T}{N}.}\end{aligned} \hspace{\stretch{1}}(1.0.16)$