Reminder | Notion

<aside> <img src="https://prod-files-secure.s3.us-west-2.amazonaws.com/369dfa6b-d4d9-4cf2-a446-e369553b6347/048cd771-98e8-49a8-b047-5c7b36b813c4/Eigenvalues_and_eigenvectors.png" alt="https://prod-files-secure.s3.us-west-2.amazonaws.com/369dfa6b-d4d9-4cf2-a446-e369553b6347/048cd771-98e8-49a8-b047-5c7b36b813c4/Eigenvalues_and_eigenvectors.png" width="40px" /> Eigenvalues and eigenvectors:

$$ \bold M \vec v=\lambda \vec v $$

Solution is finding all the solutions of $\lambda$ where

$$ \text{det}(\bold M-\lambda \bold 1)=0 $$

To find $\vec v$ plug in your value of $\lambda$ and solve for $\vec v \ne \bold 0$

$$ (\bold M-\lambda \bold 1)\vec v=0 $$

🗒️ Note: for more info go to 🏹 Eigenvalues, eigenvectors:

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<aside> <img src="https://prod-files-secure.s3.us-west-2.amazonaws.com/369dfa6b-d4d9-4cf2-a446-e369553b6347/deba4abd-273a-4ba4-8695-29a189a59d71/Taylor_series.png" alt="https://prod-files-secure.s3.us-west-2.amazonaws.com/369dfa6b-d4d9-4cf2-a446-e369553b6347/deba4abd-273a-4ba4-8695-29a189a59d71/Taylor_series.png" width="40px" /> Taylor series:

$$ f(x)=\sum_{n=0}^\infin \frac{1}{n!} \left ( \frac{\text d^n f(x)}{\text d x^n} \right )_{x=x_0} (x-x_0)^n $$

🗒️ Note: for more info go to Taylor Series:

💃 Examples: around $x=0$:

\small{\begin{aligned} (1+x)^\alpha &\simeq 1+\alpha x \\ \frac{1}{1+x} &=\sum^\infin_{n=0} (-1)^n x^n \\ \frac{1}{1-x} & = \sum^\infin_{n=0} x^n \\ e^x&= \sum_{n=0}^\infin \frac{x^n}{n!} \end{aligned}} \qquad \small{\begin{aligned} \tanh(x)&\simeq x\left ( 1-\frac{x^2}{3} \right ) \\ \cos(x)&\simeq 1-\frac{x^2}{2} \\ \sin(x)&\simeq x \end{aligned}} $$

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Fermions and bosons: if we have $N$ indistinguishable and non-interacting particles that we denote $x_i$ where $i\in[1, N]\cap \N$ we can write $\forall ij$

$$ H(x_1,\ldots,x_i,\ldots ,x_j,\ldots,x_N)=H(x_1,\ldots,x_j,\ldots ,x_i,\ldots,x_N) $$

We define $\widehat P_{ij} \psi(x_1,\ldots,x_i,\ldots ,x_j,\ldots,x_N)=\psi(x_1,\ldots,x_j,\ldots ,x_i,\ldots,x_N)$ and since $[ \widehat P_{ij},H ]=0$ we can write the following

$$ \begin{aligned} \widehat P_{ij}^2 \psi(\ldots,x_i,\ldots ,x_j,\ldots)=\widehat P_{ij}\psi(\ldots,x_j,\ldots ,x_i,\ldots)=\psi(\ldots,x_i,\ldots ,x_j,\ldots) \\ \text{thus the solution to the eigen equation } \widehat P \psi=\lambda \psi \text{ is } \lambda^2=1 \text{ or } \lambda = \pm 1 \text{ thus:}\end{aligned} \\\begin{aligned} \underbrace{\widehat P\psi(\ldots,x_i,\ldots ,x_j,\ldots)}&=\pm 1\psi( \ldots,x_i,\ldots ,x_j,\ldots) \\ \psi(\ldots,x_j,\ldots ,x_i,\ldots)&=\pm 1\psi( \ldots,x_i,\ldots ,x_j,\ldots) \end{aligned} $$

this gives us:

Boson	$+$	Arbitrary number of boson per state
Fermion	$-$	One fermion per state

🗒️ Note: for more info go to 🤖 Bosons: particles with $\Z$ spin

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Separable Hamiltonian: For non-interacting indistinguishable particles

$$ H(x_1,\ldots,x_n)=H_{1P}(x_1)+\ldots + H_{1P}(x_N) $$

where each single particle Hamiltonian is identical because they are indistinguishable

💃 Example: for a 1D free particle $H_{1P}(x)=\widehat p^2/(2m)$

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Properties of particle systems from eigenstates: considering the same system:

$$ H_{1P}(x)\psi_\lambda ^{(1P)}(x)=\epsilon_\lambda \psi _\lambda^{(1p)} (x) $$

We assign a set of integers $n_\lambda$ which represent the number of particles in a state $\epsilon_\lambda$

$$ \begin{aligned} \text{Total number of particles:} \qquad N_i&= \sum_{\lambda} n_\lambda \\ \text{Total energy:} \qquad \qquad \quad \; \; \qquad E_i&= \sum_\lambda n_\lambda \epsilon_\lambda

\end{aligned} $$

If we minimize the total entropy, with number of particles and energy fixed we get the grand-canonical partition

$$ \begin{aligned} Z&=\sum_i e^{-\beta (E_i - \mu N_i)} \qquad \text{ where: } \beta= (k_B T)^{-1} \\ \text{Subbing in:} \qquad &=\sum_{n_\lambda}e^{-\beta \sum_{\lambda} n_\lambda (\epsilon_\lambda - \mu)} \\ &=\prod_\lambda \left ( \sum_{n_\lambda}e^{-\beta n_\lambda (\epsilon_\lambda - \mu)}\right ) \\ Z&=e^{-\beta \Phi} \\ \text{Where:} \qquad \Phi&= -k_B T \sum_\lambda \ln \left [\sum_{n_\lambda} e^{-\beta n_\lambda (\epsilon_\lambda - \mu)} \right ] \quad \text{is the grand potential}

\end{aligned} $$

🗒️ Note: for more info go to Partition function

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Bosons properties: $n_\lambda \in \N$ so

$$ \sum^\infin_{n_\lambda=0} e^{-\beta n_\lambda (\epsilon \lambda -\mu)}=\sum^\infin_{n_\lambda=0}\left ( e^{-\beta (\epsilon_\lambda - \mu)} \right ) ^{n_\lambda}=\frac{1}{1-e^{-\beta(\epsilon_\lambda-\mu)}} $$

Subbing into the grand potential

$$ \Phi(T,V,\mu)=k_B T \sum_\lambda \ln(1-e^{-\beta (\epsilon_\lambda - \mu)}) $$

The average number of particle $\lang N \rang \equiv N$

$$ N=\left . -\frac{\partial \Phi}{\partial \mu} \right |{T,V} = \sum\lambda \frac{1}{e^{\beta (\epsilon_\lambda - \mu)}-1} $$

The Bose Einstein distribution (how many particles on average whose energy is $\epsilon_\lambda$)

$$ n_\lambda^{(B)}\equiv \frac{1}{e^{\beta (\epsilon_\lambda - \mu)}-1} $$

🗒️ Note: for more info go to The ideal Bose gas

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Fermion properties: $n_\lambda =0,1$ so

$$ \begin{aligned} \sum^1_{n_\lambda=0} e^{-\beta n_{\lambda} (\epsilon_\lambda - \mu)} &=1+e^{-\beta (\epsilon_\lambda-\mu)} \\ \text{Grand potential:} \qquad \Phi(T,V,\mu)&=-k_B T \sum_{\lambda} \ln ( 1+ e^{-\beta (\epsilon_\lambda - \mu)} ) \\ \text{Average number of particles:} \quad N&=\left . -\frac{\partial \Phi}{\partial \mu} \right |{T,V} = \sum\lambda \frac{1}{e^{\beta (\epsilon_\lambda - \mu)}+1} \end{aligned} $$

For non-relativistic fermions in $D$ dimensions, eigenfunctions of $\widehat H$ are plane wave $\sim e^{i\vec k \cdot \vec r}$ each labeled by a wave-vector $\vec k$ with energy $\epsilon_{\vec k} = \hbar ^2 k^2/2m$. The spin $1/2$ introduction a degeneracy $g_s=2s+1 =2$

The Fermi distribution is

$$ n_{\vec k}^{(F)}\equiv \frac{1}{e^{\beta (\epsilon_{\vec k} - \mu)}+1} $$
In the limit where $T\to 0 \degree \rm K$, $n_{\vec k}^{(F)}$ becomes a step function $\to \Theta(\epsilon_F-\epsilon_{\vec k})$ where $\epsilon_F$ is the fermi energy
Fermi momentum (largest occupied momentum)

$$ k_F=\frac{2m \sqrt{\epsilon_F}}{\hbar} $$

🗒️ Note: for more info go to The ideal Fermi gas

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Density of states: Number of of particles fermi gas continuous

$$ N=g_s V \int \frac{\text d D_{\vec k}}{(2\pi)^D}n_{\vec k}^{(F)} $$

Using wavenumber we have

$$ \begin{aligned} N&= \int \text dk\, g(k) \frac{1}{e^{\beta \left ( \frac{\hbar^2 k^2}{2m}-\mu \right )}+1} \\ \text{Density of states:} \quad g(k)&= \left \{ \begin{matrix} \frac{g_sV}{2\pi} & \text{if} & D=1 \\ \frac{g_sVk}{2\pi} & \text{if} & D=2\\ \frac{g_sVk^2}{2\pi} & \text{if} & D=3 \end{matrix} \right . \\ \lim (T\to0\degree {\rm K} ): \quad N&=\int_0^{k_F} \text dk \, g(k) \end{aligned} $$
Using energy we have

$$ \begin{aligned} N&= \int \text dk\, \tilde g(\epsilon) \frac{1}{e^{\beta \left ( \epsilon-\mu \right )}+1} \\ \text{Density of states:} \quad \tilde g(\epsilon)&= \left \{ \begin{matrix} \frac{g_sV}{4\pi\hbar } \sqrt{\frac{2m}{ \epsilon}} & \text{if} & D=1 \\ \frac{g_sV(2m)}{4\pi\hbar^2} & \text{if} & D=2\\ \frac{g_sV(2m)^\frac{3}{2}\epsilon^\frac{1}{2}}{4\pi^2 \hbar^3} & \text{if} & D=3 \end{matrix} \right . \\ \lim (T\to0\degree {\rm K} ): \quad N&=\int_0^{\epsilon_F} \text d\epsilon \, \tilde g(\epsilon) \end{aligned} $$

🗒️ Note: for more info go to The ideal Fermi gas

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