Welcome!

About this course

• What can we solve in quantum mechanics?

  • Particle in a box
  • Harmonic oscillator $$ H = -\frac{1}{2m}\left(\frac{d}{dx}\right)^2 +\frac{1}{2}m\omega^2 x^2 $$
  • Hydrogen atom $$ H = -\frac{1}{2m}\nabla^2 - \frac{e^2}{r} $$

• What would we like to solve?

$$ \nonumber \newcommand{\br}{\mathbf{r}} \newcommand{\bp}{\mathbf{p}} \newcommand{\bk}{\mathbf{k}} \newcommand{\bq}{\mathbf{q}} \newcommand{\bv}{\mathbf{v}} \newcommand{\pop}{\psi^{\vphantom{\dagger}}} \newcommand{\pdop}{\psi^\dagger} \newcommand{\Pop}{\Psi^{\vphantom{\dagger}}} \newcommand{\Pdop}{\Psi^\dagger} \newcommand{\Phop}{\Phi^{\vphantom{\dagger}}} \newcommand{\Phdop}{\Phi^\dagger} \newcommand{\phop}{\phi^{\vphantom{\dagger}}} \newcommand{\phdop}{\phi^\dagger} \newcommand{\aop}{a^{\vphantom{\dagger}}} \newcommand{\adop}{a^\dagger} \newcommand{\bop}{b^{\vphantom{\dagger}}} \newcommand{\bdop}{b^\dagger} \newcommand{\cop}{c^{\vphantom{\dagger}}} \newcommand{\cdop}{c^\dagger} \newcommand{\bra}[1]{\langle{#1}\rvert} \newcommand{\ket}[1]{\lvert{#1}\rangle} \newcommand{\inner}[2]{\langle{#1}\rvert #2 \rangle} \newcommand{\braket}[3]{\langle{#1}\rvert #2 \lvert #3 \rangle} \newcommand{\sgn}{\mathrm{sgn}} \DeclareMathOperator{\tr}{tr} \DeclareMathOperator{\E}{\mathbb{E}} $$

Everything…

• Given ion positions $\mathbf{R}_k$, charge $+Z_k e$

$$ H = -\frac{1}{2m}\sum_{j=1}^M\nabla^2_j + e^2\sum_{j<k}\frac{1}{|\br_j-\br_k|} - e^2\sum_{j<k}\frac{Z_k}{|\br_j-\mathbf{R}_k|} $$

• Want many-electron wavefunction $\Psi(\br_1,\ldots \br_N)$

• Hard for atoms / molecules: impossible for matter $N\sim N_A$!

• Goal: what is a solid / metal / superconductor / superfluid?

So what do we do?

• Use simple models that we can analyse in $N\to\infty$ limit

$$ \label{coll_Hchain} H = \sum_{j=1}^N \left[\frac{p_j^2}{2m} + \frac{k}{2} (u_j-u_{j+1})^2 \right]. $$

• New techniques to represent wavefunctions and Hamiltonians (second quantization aka quantum field theory)

Housekeeping

• Course website austen.uk/courses/tqm-lectures (updated with slides)

• Lecture capture (all three screens)

• Lectures are 90 minutes with 5 minute break roughly halfway

• Supervision details to follow

$\hbar=1$ !

Many Body Wavefunctions

This lecture is about the general character of many body wavefunctions and what they tell us, focusing on the simplest type: product states

Systems of indistinguishable particles are described by totally symmetric or totally antisymmetric wavefunctions.

• The first kind are bosons, the second fermions

Two Particles

• In general described by $\Psi(\mathbf{x},\mathbf{y})$

• A pair of particles in states $\ket{\varphi_1}$ and $\ket{\varphi_2}$

$$ \Psi(\br_1,\br_2)\stackrel{?}{=} \varphi_1(\br_1)\varphi_2(\br_2). \label{quantum_statistics_ProductWavefunction} $$

• This would be fine for distinguishable particles

Accounting for symmetry

$$ \label{quantum_statistics_sym} \Psi(\br_1,\br_2)=\frac{1}{\sqrt{2}}[\varphi_1(\br_1)\varphi_2(\br_2)\pm \varphi_2(\br_1)\varphi_1(\br_2)] $$

• upper sign for bosons and the lower for fermions

• $1/\sqrt{2}$ for normalization if $\varphi_{1,2}(\br)$ are orthonormal

• When $\varphi_1=\varphi_2$ fermion $\Psi(\br_1,\br_2)$ vanishes: exclusion principle

Independent Particles?

• Probability density is $\rho_{12}(\br_1,\br_2)=|\Psi(\br_1,\br_2)|^2$

• For distinguishable particles

$$ \rho_{12}(\br_1,\br_2)=|\varphi_1(\br_1)\varphi_2(\br_2)|^2=\rho_1(\br_1)\rho_2(\br_2) \label{eq:classicaljoint} $$

Work it out for identical bosons and fermions Show that $\rho_{12}(\br,\br) = 0$ for fermions, and $\rho_{12}(\br,\br) = 2\rho_1(\br)\rho_2(\br)$ for bosons.

• For identical particles

$$ \begin{align} \rho_{12}(\br_1,\br_2) &= \frac{1}{2}\left[\rho_1(\br_1)\rho_2(\br_2)+\rho_1(\br_2)\rho_2(\br_1)\right] \\ &\pm\frac{1}{2}\left[\varphi^{}_1(\br_1)\varphi^*_2(\br_1)\varphi^{}_2(\br_2)\varphi^*_1(\br_2)+\varphi^{}_1(\br_2)\varphi^*_2(\br_2)\varphi^{}_2(\br_1)\varphi^*_1(\br_1)\right] \end{align} $$

• Identical particles in quantum mechanics are never truly independent.

Hong–Ou–Mandel effect

• Two photons (bosons) approach a beam splitter from either side.

• Start in orthogonal states, end up in orthogonal states e.g.

$$ \begin{align} \ket{\text{Left}}\to\frac{1}{\sqrt{2}}\left(\ket{\text{Left}}+ \ket{\text{Right}}\right)\\ \ket{\text{Right}}\to\frac{1}{\sqrt{2}}\left(\ket{\text{Left}}- \ket{\text{Right}}\right) \end{align} $$

If we start in $$\frac{1}{\sqrt{2}}[\varphi_\text{L}(\br_1)\varphi_\text{R}(\br_2)\pm \varphi_\text{R}(\br_1)\varphi_\text{L}(\br_2)]$$ What state do we end up in?

Noninteracting particles

$$ H = \sum_{i=1}^{N} \overbrace{\left[-\frac{\nabla_i^{2}}{2m}+V(\br_i)\right]}^{\equiv H_\text{sp}\text{, single particle Hamiltonian}} \label{quantum_statistics_SPHamiltonian} $$

• $H_\text{sp}$ has eigenstates ${\varphi_\alpha(\br)}$ and eigenenergies ${E_\alpha}$

• Eigenstatate of $N$ distinguishable particles $$ \label{quantum_statistics_disting} \ket{\Psi_{\alpha_{1}\alpha_{2}\cdots \alpha_{N}}}=\varphi_{\alpha_{1}}(\mathbf{r_{1}})\varphi_{\alpha_{2}}(\mathbf{r_{2}})\cdots\varphi_{\alpha_{N}}(\mathbf{r_{N}}) $$

What’s the energy?

$$ E = \sum_{j=1}^N E_{\alpha_j} $$

What are the energy eigenstates?

Symmetrize / Antisymmetrize

$$ \label{quantum_statistics_SymAntisym} \mathcal{S}=\frac{1}{N!}\sum_{\pi} P_\pi, \qquad \mathcal{A}=\frac{1}{N!}\sum_{\pi} \sgn(\pi)P_\pi $$

• $\pi$ is one of $N!$ permutations of $N$ objects

• $P_\pi$ is permutation operator for permutation $\pi$

• For two particles there are two permutations: $\pi=(12)$ and $\pi=(21)$

$$ P_{(21)}\Psi(\br_1,\br_2) = \Psi(\br_2,\br_1) $$

Signature of Permutation

• $\sgn(\pi)$ is the signature of the permutation

• $\pm1$ for $\pi$ involving an even (odd) number of exchanges

Symmetric and antisymmetric product states

$$ \begin{align} \ket{\Psi^{S}_{\alpha_{1}\alpha_{2}\cdots\alpha_{N}}}&=\sqrt{N!\prod_{\alpha}N_{\alpha}!}\mathcal{S}\,\varphi_{\alpha_{1}}(\mathbf{r_{1}})\varphi_{\alpha_{2}}(\mathbf{r_{2}})\cdots\varphi_{\alpha_{N}}(\mathbf{r_{N}}) \nonumber \\ \ket{\Psi^{A}_{\alpha_{1}\alpha_{2}\cdots\alpha_{N}}}&=\sqrt{N!}\mathcal{A}\,\varphi_{\alpha_{1}}(\mathbf{r_{1}})\varphi_{\alpha_{2}}(\mathbf{r_{2}})\cdots\varphi_{\alpha_{N}}(\mathbf{r_{N}}) \label{quantum_statistics_norm} \end{align} $$

Show that these are normalized wavefunctions if the single particle state $\ket{\varphi_\alpha}$ are orthonormal

Occupation Numbers

• State is characterized only by occupation numbers ${N_{\alpha}}$. Use these as a label instead

• Caution: To fix the sign of the fermion wavefunction, have to choose an order e.g. $\varphi_{\alpha_{1}}(\mathbf{r_{1}})\varphi_{\alpha_{2}}(\mathbf{r_{2}})\cdots\varphi_{\alpha_{N}}(\mathbf{r_{N}})$

• The total energy is

$$ \label{quantum_statistics_TotalEnergy} E=\sum_{i=1}^{N}E_{\alpha_{i}}=\sum_{\alpha}N_{\alpha} E_{\alpha} $$

Slater Determinant

• The fermion product state is a determinant!

$$ \label{quantum_statistics_slater} \ket{\Psi^{A}_{\alpha_{1}\alpha_{2}\cdots\alpha_{N}}}=\frac{1}{\sqrt{N!}}\begin{vmatrix} \varphi_{\alpha_{1}}(\mathbf{r_{1}}) & \varphi_{\alpha_{1}}(\mathbf{r_{2}}) & \cdots & \varphi_{\alpha_{1}}(\mathbf{r_{N}}) \\ \varphi_{\alpha_{2}}(\mathbf{r_{1}}) & \cdots & \cdots & \cdots \\ \cdots & \cdots & \cdots & \cdots \\ \varphi_{\alpha_{N}}(\mathbf{r_{1}}) & \cdots & \cdots & \varphi_{\alpha_{N}}(\mathbf{r_{N}}) \end{vmatrix} $$

• Vanishes when two rows or two columns are identical
  • Zero if two particle coordinates coincide ($\br_{i}=\br_{j}$)
  • Zero if two particles occupy the same state ($\alpha_{i}=\alpha_{j}$)

Noninteracting particles on a ring

• Ring has circumference $L$. Single particle eigenstates $$ \label{quantum_statistics_spstates} \varphi_{n}(x)=\frac{1}{\sqrt{L}}\exp\left(ik_n x\right), $$

• $k_n=\frac{2\pi n}{L}$, $n\in\mathbb{Z}$. Energies are $E_{n}=\frac{k_n^2}{2m}$.

Ground State of Bosons

• Every particle in the state $\ket{\varphi_{0}}$ with zero energy: $N_{0}=N$

$$ \Psi^{S}(x_{1},x_{2},\ldots x_{N})=\frac{1}{L^{N/2}} $$

• That was easy! The fermion case is harder.

Ground State of Fermions

• Fill each level with one particle, starting at the bottom (Fermi sea)

  • For $N$ odd $n=-(N-1)/2, -(N-3)/2,\ldots, -1, 0, 1 \ldots (N-1)/2$

  • For $N$ even have to decide whether to put the last particle at $n=\pm N/2$

Complex notation

$$ z_{i}=\exp(2 \pi i x_{i}/L), $$

• The Slater determinant is

$$ \label{quantum_statistics_1ddet} \Psi_0(x_1,\ldots, x_N)=\begin{vmatrix} z_{1}^{-(N-1)/2} & z_{2}^{-(N-1)/2} & \cdots & z_{N}^{-(N-1)/2} \\ z_{1}^{-(N-3)/2} & \cdots & \cdots & \cdots \\ \cdots & \cdots & \cdots & \cdots \\ z_{1}^{(N-1)/2} & \cdots & \cdots & z_{N}^{(N-1)/2} \end{vmatrix}. $$

Simple example $N=3$

$$ \begin{align} \Psi_0(x_1,x_2,x_3)&=\begin{vmatrix} z_{1}^{-1} & z_{2}^{-1} & z_{3}^{-1} \\ 1 & 1 & 1 \\ z_{1} & z_{2} & z_{3} \end{vmatrix} = \frac{z_{1}}{z_{2}}-\frac{z_{2}}{z_{1}}+\frac{z_{3}}{z_{1}}-\frac{z_{1}}{z_{3}}+\frac{z_{2}}{z_{3}}-\frac{z_{3}}{z_{2}}\nonumber\\ &=\left(\sqrt{\frac{z_{3}}{z_{1}}}-\sqrt{\frac{z_{1}}{z_{3}}}\right)\left(\sqrt{\frac{z_{1}}{z_{2}}}-\sqrt{\frac{z_{2}}{z_{1}}}\right)\left(\sqrt{\frac{z_{2}}{z_{3}}}-\sqrt{\frac{z_{3}}{z_{2}}}\right) \nonumber\\ &\propto \sin\left(\frac{ \pi[x_{1}-x_{2}]}{L}\right)\sin\left(\frac{ \pi[x_{3}-x_{1}]}{L}\right)\sin\left(\frac{ \pi[x_{2}-x_{3}]}{L}\right) \label{3particle} \end{align} $$

• Wavefunction vanishes $x_{i}=x_{j}$

Check it is periodic and totally antisymmetric.

Generalization for $N$ odd

$$ \Psi_0(x_1,\ldots, x_N)\propto\prod_{i<j}^{N} \sin\left(\frac{\pi[x_{i}-x_{j}]}{L}\right). \label{quantum_statistics_1dFermiGS} $$

Show this using the Vandermonde determinant $$ \begin{vmatrix} 1 & 1 & \cdots & 1 \\ z_{1} & z_{2} & \cdots & \cdots \\ z_{1}^{2} & z_{2}^{2} & \cdots & \cdots \\ z_{1}^{N-1} & z_{2}^{N-1} & \cdots & z_{N}^{N-1} \end{vmatrix}=\prod_{i<j}^{N}(z_{j}-z_{i}) $$

Terminology

• The wavevector of the last fermion added is called the Fermi wavevector $k_\text{F}$

• Here $k_\text{F}=\frac{(N-1)\pi}{L}$

• Fermi energy $E_{F}=\frac{k_\text{F}^{2}}{2m}$ is the corresponding energy

Probability distribution

• For boson ground state $\Psi$ is constant, so is $|\Psi|^2$

• For fermions

Marginal distributions

• One particle distribution is related to average density of particles $$ \rho_1(x_1) = N \int dx_2\ldots dx_N \lvert\Psi(x_1,x_2,\ldots,x_N)\rvert^2. \label{ave_density} $$

• Note that density is always a number density (not mass density).

• Normalization of the wavefunction implies $$ \int dx \rho_1(x) = N. $$

• In a translationally invariant system like the fermion gas on a ring we expect the average density to be constant.

Density Operator

• Regard $\rho_1(x_1)$ as an expectation of a density operator $$ \rho(x) = \sum_j \delta(x-x_j), \label{many_densityop} $$ so that $\rho_1(x) = \braket{\Psi}{\rho(x)}{\Psi}$.

Single particle density matrix

$$ g(x,y) \equiv N\int dx_2\ldots dx_N \Psi^{}(x,x_2,\ldots,x_N)\Psi^{*}(y,x_2,\ldots,x_N). $$

• Note $g(x,x) = \rho_1(x)$.

Starting from the Slater determinant, show that $g(x,y)$ for the ground state of the Fermi gas is $$ \begin{align} g(x,y) &= \frac{1}{L}\sum_{|k|<k_\text{F}} e^{ik(x-y)} \longrightarrow \int_{-k_\text{F}}^{k_\text{F}} \frac{dk}{2\pi} e^{ik(x-y)} \\ &\longrightarrow n \frac{\sin \left[k_\text{F}(x-y)\right]}{k_\text{F}(x-y)} \end{align} $$ where $n \equiv \frac{k_\text{F}}{\pi}$ is the average density.

Pair Distribution

• Marginal probability distribution of a pair of particles $$ \rho_2(x_1,x_2) = N(N-1) \int dx_3\ldots dx_N\left|\Psi(x_1,x_2,\ldots,x_N)\right|^2. $$
• The prefactor is to account for all pairs of particles.

Starting from the Slater determinant, show that $$ \rho_2(x_1,x_2) = n^2\left[1 - \left(\frac{\sin\left[k_\text{F}(x_1-x_2)\right]}{k_\text{F}(x_1-x_2)}\right)^2\right]. $$ This vanishes at $x_1=x_2$.

Impenetrable Bose Gas

$$ H = -\frac{1}{2m}\sum_j \frac{\partial^2}{\partial x_j^2} + \overbrace{c\sum_{j<k}\delta(x_j-x_k)}^{\equiv H_\text{int}}. \label{many_LL} $$

• The second term is interaction between pairs of particles

• For fermions, the interaction has no effect at all ($\delta$-function)!

• For bosons with $c\to \infty$: eigenenergies coincide with those of free fermions, and eigenstates are modulus of fermion eigenstates

Why?

Ground state

$$ \Psi_0(x_1,\ldots, x_N) = \prod_{i<j}^{N} \left|\sin\left(\frac{\pi[x_{i}-x_{j}]}{L}\right)\right|. $$

• Can calculate any observable as long as it’s insensitive to taking the modulus (e.g. average density and pair distribution)