More Second Quantization

  • Lecture 1: density correlations in ground state of 1D Fermi gas

$$ \rho_2(x,y) = n^2\left[1 - \left(\frac{\sin[k_\text{F}(x-y)]}{k_\text{F}(x-y)}\right)^2\right]. \label{more_rho2evalFermi} $$

  • How to find this using second quantization?

  • What can these correlations tell us about interactions?

$$ \nonumber \newcommand{\cN}{\mathcal{N}} \newcommand{\br}{\mathbf{r}} \newcommand{\bp}{\mathbf{p}} \newcommand{\bk}{\mathbf{k}} \newcommand{\bq}{\mathbf{q}} \newcommand{\bv}{\mathbf{v}} \newcommand{\pop}{\psi^{\vphantom{\dagger}}} \newcommand{\pdop}{\psi^\dagger} \newcommand{\Pop}{\Psi^{\vphantom{\dagger}}} \newcommand{\Pdop}{\Psi^\dagger} \newcommand{\Phop}{\Phi^{\vphantom{\dagger}}} \newcommand{\Phdop}{\Phi^\dagger} \newcommand{\phop}{\phi^{\vphantom{\dagger}}} \newcommand{\phdop}{\phi^\dagger} \newcommand{\aop}{a^{\vphantom{\dagger}}} \newcommand{\adop}{a^\dagger} \newcommand{\bop}{b^{\vphantom{\dagger}}} \newcommand{\bdop}{b^\dagger} \newcommand{\cop}{c^{\vphantom{\dagger}}} \newcommand{\cdop}{c^\dagger} \newcommand{\Nop}{\mathsf{N}^{\vphantom{\dagger}}} \newcommand{\bra}[1]{\langle{#1}\rvert} \newcommand{\ket}[1]{\lvert{#1}\rangle} \newcommand{\inner}[2]{\langle{#1}\rvert #2 \rangle} \newcommand{\braket}[3]{\langle{#1}\rvert #2 \lvert #3 \rangle} \DeclareMathOperator{\sgn}{sgn} \DeclareMathOperator{\tr}{tr} \newcommand{\abs}[1]{\lvert{#1}\rvert} \newcommand{\brN}{\br_1, \ldots, \br_N} \newcommand{\xN}{x_1, \ldots, x_N} \newcommand{\zN}{z_1, \ldots, z_N} $$

$\rho_2$ from Second Quantization

  • From Lecture 1: pair distribution function $$ \rho_2(x_1,x_2) = N(N-1) \int dx_3\ldots dx_N ,\left|\Psi(x_1,x_2,\ldots,x_N)\right|^2, \label{more_pdf} $$ measures likelihood of finding particles at $x_1$ and $x_2$

  • For 1D Fermi gas ground state we found $$ \rho_2(x,y) = n^2\left[1 - \left(\frac{\sin[k_\text{F}(x-y)]}{k_\text{F}(x-y)}\right)^2\right] $$ recall: Slater determinant, $P$ and $P'$ differ by transposition, etc.

  • Let’s calculate using second quantization!

$$ \begin{align} \rho_2(x,y) &= N(N-1) \int dx_3\ldots dx_N \,\left|\Psi(x,y,\ldots,x_N)\right|^2,\\ \end{align} $$

$$ \begin{align} \ket{\Psi}&\longleftrightarrow \Psi(x_1,\ldots, x_N)\nonumber\\ \psi(X)\ket{\Psi}&\longleftrightarrow \sqrt{N}\Psi(X,x_1,\ldots, x_{N-1})\\ \end{align} $$

  • Second quantized form

$$ \rho_2(x,y) =\braket{\Psi}{\pdop(x)\pdop(y)\pop(y)\pop(x)}{\Psi}. \label{more_rho22ndquant} $$

$$ \rho_2(x,y) =\braket{\Psi}{\pdop(x)\pdop(y)\pop(y)\pop(x)}{\Psi}. $$

  • Operators in which all annihilation operators stand to the right of all creation operators are said to be normal ordered

  • Two particle terms in the Hamiltonian are normal ordered to prevent particle interacting with itself!

$$ \rho_2(x,y) =\braket{\Psi}{\pdop(x)\pdop(y)\pop(y)\pop(x)}{\Psi}. $$

  • Insert expansion

$$ \begin{align} \pop(x)=\sum_{\beta} \varphi^{}_{\beta}(x)\aop_{\beta},\\ \pdop(x)=\sum_{\beta} \varphi^*_{\beta}(x)\adop_{\beta}. \end{align} $$

  • This gives

$$ \label{2nd_quant_CEval} \rho_2(x,y)=\sum_{\alpha, \beta, \gamma, \delta}\varphi^{*}_{\alpha}(x)\varphi^{*}_{\beta}(y)\varphi^{}_{\gamma}(y)\varphi^{}_{\delta}(x)\braket{\Psi}{\adop_{\alpha}\adop_{\beta}\aop_{\gamma}\aop_{\delta}}{\Psi}. $$

$$ \braket{\Psi}{\adop_{\alpha}\adop_{\beta}\aop_{\gamma}\aop_{\delta}}{\Psi} $$

  • When $\ket{\Psi}=\ket{\mathbf{N}}$ (product state using $\adop_\alpha$) have two possibilities $$ \begin{align} &\alpha =\delta,\, \beta=\gamma, \text{ or }\\ &\alpha=\gamma,\, \beta=\delta, \end{align} $$ which give rise to two groups of terms $$ \begin{align} \braket{\mathbf{N}}{\adop_{\alpha}\adop_{\gamma}\aop_{\gamma}\aop_{\alpha}}{\mathbf{N}}&=N_{\alpha}N_{\gamma}\nonumber\\ \braket{\mathbf{N}}{\adop_{\alpha}\adop_{\gamma}\aop_{\alpha}\aop_{\gamma}}{\mathbf{N}}&=\pm N_{\alpha}N_{\gamma}\qquad\text{if }\alpha\neq\gamma, \end{align} $$
    $\pm$ corresponding to bosons and fermions

$$ \begin{align} \rho_2(x,y)&=\sum_{\alpha, \beta, \gamma, \delta}\varphi^{*}_{\alpha}(x)\varphi^{*}_{\beta}(y)\varphi^{}_{\gamma}(y)\varphi^{}_{\delta}(x)\braket{\Psi}{\adop_{\alpha}\adop_{\beta}\aop_{\gamma}\aop_{\delta}}{\Psi}\\ &=\sum_{\alpha, \beta}N_\alpha N_\beta\left[\abs{\varphi_{\alpha}(x)}^2\abs{\varphi_{\beta}(y)}^2 \pm \varphi^*_\alpha(x)\varphi^{}_\alpha(y)\varphi^*_\beta(y)\varphi^{}_\beta(x) \right]. \end{align} $$

  • Notice $\alpha=\beta=\gamma=\delta$ has a factor $2N_\alpha^2$

  • Should have $N_\alpha(N_\alpha-1)$ for bosons; zero for fermions

  • Involves sum over a single index, general case is sum over two indices. Often not important in thermodynamic limit.

  • For plane wave states we have the usual prescription $$ \sum_\alpha(\cdots) \longrightarrow L\int (\cdots)\frac{dk}{2\pi} $$ assuming integrand smooth

  • Error from “wrong” terms has extra $L^{-1}$

  • Careful with Bose condensates, where one state has finite fraction of particles

$$ \begin{align} \rho_2(x,y)&=\sum_{\alpha, \beta, \gamma, \delta}\varphi^{*}_{\alpha}(x)\varphi^{*}_{\beta}(y)\varphi^{}_{\gamma}(y)\varphi^{}_{\delta}(x)\braket{\Psi}{\adop_{\alpha}\adop_{\beta}\aop_{\gamma}\aop_{\delta}}{\Psi}\\ &=\sum_{\alpha, \beta}N_\alpha N_\beta\left[\abs{\varphi_{\alpha}(x)}^2\abs{\varphi_{\beta}(y)}^2 \pm \varphi^*_\alpha(x)\varphi^{}_\alpha(y)\varphi^*_\beta(y)\varphi^{}_\beta(x) \right]. \end{align} $$

  • Can express using density $\rho_1(x)$ and density matrix as $g(x,y)$ $$ \rho_2(x,y) = \rho_1(x)\rho_1(y) \pm g(x,y)g(y,x), \label{more_rho2compact} $$ for ground state of the Fermi gas, reproduces $$ \rho_2(x,y) = n^2\left[1 - \left(\frac{\sin[k_\text{F}(x-y)]}{k_\text{F}(x-y)}\right)^2\right]. $$


  • ‘Hole’ set by $\lambda_\text{F}$ or mean particle separation

  • Decaying Friedel oscillations, indicating liquid-like correlations

  • For bosons $$ \rho_2(x,y) = \rho_1(x)\rho_1(y) + g(x,y)g(y,x), $$

  • If $g(x,y)\to 0$ as $\abs{x-y}\to\infty$ $\rho_2(x,x)$ is twice the value at $\abs{x-y}\to\infty$.

$$ \rho_2(x,y) = \rho_1(x)\rho_1(y) \pm g(x,y)g(y,x), $$

  • For different occupations can recalculate $\rho_1(x)$ and $g(x,y)$

  • Applies to product states only

Hanbury Brown and Twiss Effect

  • $\rho_2(x,y)$ can show interference, as can $\rho_1(x)$ (‘intensity’)

  • Classic BEC experiment of Andrews et al.

  • Earlier demonstrations of HBT in astro, see Wikipedia for more

  • $N$ noninteracting bosons occupying in ground state $\varphi_{0}(\br)$ of some potential: a __Bose condensate__

  • $N$-body wavefunction is $$ \Psi(\br_1,\br_2,\ldots,\br_N)=\prod_i^N \varphi_0(\br_i), \label{2nd_quant_BoseGroundState} $$ $$ \ket{\Psi}=\frac{1}{\sqrt{N!}}\left(\adop_0\right)^N\ket{\text{VAC}}, $$ where $\adop_0$ creates particle in state $\varphi_0(\br)$

Experiment of Andrews et al.

  • Two identical BECs side by side

  • Switch off potentials: particles fly out

  • After some time wavefunctions overlap. What do we see?

A simpler situation..

  • One BEC, each atom in superposition of $\varphi_L(\br)$ and $\varphi_R(\br)$ $$ \ket{\bar N_L,\bar N_R}_\theta\equiv\frac{1}{\sqrt{N!}}\left[\sqrt{\frac{\bar N_L}{N}}e^{-i\theta/2} \adop_L+\sqrt{\frac{\bar N_R}{N}}e^{i\theta/2}\adop_R\right]^N\ket{\text{VAC}}, \label{more_two} $$ $\bar N_{L,R}$ are average particle number ($N=\bar N_L+\bar N_R$)

  • System evolves for time $t$. Field operator obeys $$ i\frac{\partial \pop(\br,t)}{\partial t} = -\frac{1}{2m}\nabla^2\pop(\br,t) $$ (no potential)

  • Write the field operator $$ \pop(\br)=\varphi_L(\br,t)\aop_L+\varphi_R(\br,t)\aop_R+\cdots, $$ where wavefunctions $\varphi_{L/R}(\br,t)$ obey free particle Schrödinger

$$ \begin{align} \rho_1(\br,t)=\bar N_L|\varphi_L(\br,t)|^2+\bar N_R|\varphi_R(\br,t)|^2+\overbrace{2\sqrt{\bar N_L \bar N_R}\mathrm{Re}\,e^{i\theta}\,\varphi^*_L(\br,t)\varphi_R(\br,t)}^{\equiv\rho_{\mathrm{int}( \br,t)}}. \label{dens_int} \end{align} $$

  • If clouds begin overlap, last term can give interference fringes

  • Relative phase has real physical effect.

Consider a Gaussian wavefunction of width $R_0$ at time $t=0$. Show (by substitution into the Schrödinger equation is fine) that this function evolves as

$$ \varphi(\br,t)=\frac{1}{\left(\pi R_t^{2}\right)^{3/4}}\exp\left[-\frac{\br^2\left(1+i t/m R_0^2)\right)}{2R_t^2}\right], \label{Gaussian} $$


$$ R_t^2=R_0^2+\left(\frac{ t}{mR_0}\right)^2. $$

Back to Andrews…

  • Do same thing with two condensates of fixed particle number

  • System is in product state (often Fock state in this context) $$ \ket{N_L,N_R}\equiv\frac{1}{\sqrt{N_L! N_R!}}\left(\adop_L\right)^{N_L}\left(\adop_R\right)^{N_R}\ket{\text{VAC}} $$

  • Compute density as before $$ \rho_1(\br,t)=N_L|\varphi_L(\br,t)|^2+N_R|\varphi_R(\br,t)|^2, \label{dens_fock} $$ which differs from the previous result by the absence of the interference term.

  • But a single image is not an expectation value!

  • Correlation function can tell us about fluctuations between images

  • Our result for $\rho_2$ gives

$$ \begin{align} \rho_2(\br,\br')&=\rho_1(\br)\rho_1(\br') +N_LN_{R}\varphi_L^*(\br)\varphi_R^*(\br')\varphi_L(\br')\varphi_R(\br) \nonumber\\ &\qquad+N_{L}N_R\varphi_R^*(\br)\varphi_L^*(\br') \varphi_R(\br')\varphi_L(\br). \label{dens_corr} \end{align} $$

  • Second line contains interference fringes!
  • $\rho_2$ gives probability of finding an atom at $\br'$ if there is one at $\br$

  • Conclusion: in each measurement of density, fringes are present but with random phase


  • Expectations in Fock state can be obtained using relative phase state, but with subsequent averaging over phase.

Prove this by showing that the density matrix

$$ \rho=\int_0^{2\pi}\frac{d\theta}{2\pi}\ket{\bar N_L,\bar N_R}_\theta\bra{\bar N_R,\bar N_L}_\theta $$

coincides with that of a mixture of Fock states with binomial distribution of atoms into states $\varphi_{L}$, $\varphi_{r}$. At large $N$ this distribution becomes sharply peaked at occupations $\bar N_L$, $\bar N_R$.

Hartree–Fock Theory

$$ \hat H_\text{int.} = \sum_{j<k} U(\br_j-\br_k)=\frac{1}{2}\int d\br_1 d\br_2\, U(\br_1-\br_2)\pdop(\br_1)\pdop(\br_2)\pop(\br_2)\pop(\br_1). $$

  • Since

$$ \sum_{j<k} U(\br_j-\br_k) = \frac{1}{2}\int \sum_{j\neq k}\delta(\br_1-\br_j)\delta(\br_2-\br_k)U(\br_1-\br_2) d\br_1 d\br_2, $$

  • Find expectation value of interaction energy in a product state

$$ \begin{align} \label{2nd_quant_HartreeFock} \langle \hat V\rangle &= \overbrace{\frac{1}{2}\int d\br\, d\br'\, \rho_1(\br) U(\br-\br')\rho_1(\br')}^{\equiv E_\text{Hartree}} \\ &\qquad\overbrace{\pm \frac{1}{2}\int d\br\, d\br'\, U(\br-\br')g(\br,\br')g(\br',\br)}^{\equiv E_\text{Fock}}. \end{align} $$

  • The two terms are Hartree and Fock (or exchange) contributions

  • Basis of variational Hartree–Fock method for many body systems: approximate ground state by product state

Hartree–Fock for Electron Gas

  • What about spin?

  • Use field operators $\pop_\sigma(\br)$, $\pdop_\sigma(\br’)$ satisfying

$$ \begin{gather} \left\{\pop_{\sigma_1}(\br_1),\pdop_{\sigma_2}(\br_2)\right\}=\delta_{\sigma_1\sigma_2}\delta(\br_1-\br_2)\nonumber\\ \left\{\pop_{\sigma_1}(\br_1),\pop_{\sigma_2}(\br_2)\right\}=\left\{\pdop_{\sigma_1}(\br_1),\pdop_{\sigma_2}(\br_2)\right\}=0. \label{2nd_quant_PositionRelationsAnti} \end{gather} $$

  • Density matrix is a matrix in spin space as well as real space

$$ g_{\sigma_1\sigma_2}(\br_1,\br_2) = \braket{\Psi}{\pdop_{\sigma_1}(\br_1)\pop_{\sigma_2}(\br_2)}{\Psi}. $$

  • From $g_{\sigma_1\sigma_2}(\br_1,\br_2)$ we get density and spin density

$$ \mathbf{\rho}(\br) = \tr\left[g(\br,\br)\right],\quad \mathbf{s}(\br) = \frac{1}{2}\tr\left[\boldsymbol{\sigma}g(\br,\br)\right]. $$

  • Spin-independent interaction

$$ \hat H_\text{int.} = \frac{1}{2}\sum_{\sigma_1,\sigma_2}\int d\br_1 d\br_2, U(\br_1-\br_2)\pdop_{\sigma_1}(\br_1)\pdop_{\sigma_2}(\br_2)\pop_{\sigma_2}(\br_2)\pop_{\sigma_1}(\br_1). $$

  • Hartree–Fock energy then

$$ \begin{align} \langle \hat H_\text{int.}\rangle &=\frac{1}{2}\int d\br\, d\br'\, \rho(\br) U(\br-\br')\rho(\br')\\ &- \frac{1}{2}\int d\br\, d\br'\, U(\br-\br')\tr\left[g(\br,\br')g(\br',\br)\right]. \label{2nd_quant_HFSpin} \end{align} $$

  • Fock term can be rewritten using identity $$ \delta_{ab}\delta_{cd} = \frac{1}{2}\left[\boldsymbol{\sigma}_{a c}\cdot \boldsymbol{\sigma}_{d b} + \delta_{ac}\delta_{db}\right], $$ which gives $$ \begin{align} E_{\text{Fock}} &=-\frac{1}{4} \int d\br\, d\br'\, U(\br-\br')\tr\left[g(\br,\br')\right]\tr\left[g(\br',\br)\right]\\&-\frac{1}{4}\int d\br\, d\br'\, U(\br-\br')\tr\left[\boldsymbol{\sigma}g(\br,\br')\right]\cdot\tr\left[\boldsymbol{\sigma}g(\br',\br)\right]. \end{align} $$

Suppose $U(\br)=U_0 \delta(\br)$

$$ \begin{align} E_{\text{Fock}} =-\frac{V_0}{4} \int d\br\, \rho(\br)^2-V_0\int d\br\, \mathbf{s}(\br)\cdot\mathbf{s}(\br) \end{align} $$

  • Second term favours ferromagnetism for repulsive interactions (c.f. Hund’s rules)

This is most succintly put by the formula

$$ \rho_2(\br,\br) = \frac{1}{2}\rho(\br)^2 - 2\mathbf{s}(\br)\cdot\mathbf{s}(\br) $$

  • Hartree–Fock energy forms the basis of variational method using product states

  • For a Hamiltonian with translational invariance $$ H = \int d\br \frac{1}{2m}\nabla\pdop\cdot\nabla\pop + \frac{1}{2}\int d\br d\br' U(\br-\br')\pdop(\br)\pdop(\br')\pop(\br')\pop(\br), \label{more_H2nd} $$ guaranteed to involve with plane wave single particle states. Only variational parameters are occupancies of these states

  • If translational symmetry is broken, have to allow the states, as well as the occupancies, to vary.

Stoner Criterion for Ferromagnetism

  • Polarizing spins in a Fermi gas is not without cost (otherwise everything would be ferromagnetic!)

  • Price to pay is increased kinetic energy!

  • Ground state kinetic energy of $N$ (spinless) fermions in three dimensions, obtained from

$$ \begin{align} N = \sum_{|\bk|<k_\text{F}}&\longrightarrow L^3 \int_{|\bk|<k_\text{F}} \frac{d\bk}{(2\pi)^3} = \frac{k_\text{F}^3}{6\pi^2} \\ E_\text{kin} = \sum_{|\bk|<k_\text{F}} \frac{\bk^2}{2m} &\longrightarrow L^3 \int_{|\bk|<k_\text{F}} \frac{d\bk}{(2\pi)^3} \frac{\bk^2}{2m}\\ &= \frac{k_\text{F}^5}{20\pi^2 m} = L^3 \frac{3}{10m}(6\pi^2)^{2/3} n^{5/3}, \end{align} $$

  • Total energy is $$ \begin{equation} E_\text{kin}(n_\uparrow,n_\downarrow) = \frac{cL^3}{m}\left(n_\uparrow^{5/3}+n_\downarrow^{5/3}\right), \end{equation} $$ $c=\frac{3}{10}(6\pi^2)^{2/3}$
  • Writing $n=n_\uparrow+n_\downarrow$ $\bar s = \left(n_\uparrow-n_\downarrow\right)/2$, we have

$$ E_\text{kin}(n, \bar s) = \frac{cL^3}{m}\left(\left[n/2+\bar s\right]^{5/3}+\left[n/2-\bar s\right]^{5/3}\right). $$

  • In terms of the polarization $P \equiv \frac{n_\uparrow-n_\downarrow}{n}$ in range $[-1,1]$

$$ E_\text{kin}(P) = \frac{E_\text{K}}{2}\left[(1+P)^{5/3}+(1-P)^{5/3}\right]. $$

  • $E^{(0)}_\text{kin}(n, \bar s)$ is minimized for $s=0$.
  • Total Hartree–Fock energy is

$$ E_\text{HF}(n,\bar s) = \frac{V_0L^3}{2} n^2 - \frac{V_0L^3}{2} \left(n_\uparrow^2+n_\downarrow^2\right) = \frac{V_0L^3}{2} \left(\frac{1}{2}n^2 - 2\bar s^2\right). $$

  • In terms of $P$

$$ E_\text{HF}(P) = \frac{E_V}{2}(1-P^2). $$

Minimize the total energy $E(P) = E_\text{kin}(P) + E_\text{HF}(P)$ to show

  1. For $E_V/E_K<10/9$ the ground state is non-magnetic.
  2. As $E_V/E_K$ increases past $10/9$ the magnetization begins to increase.
  3. At $E_V/E_K>\frac{5}{6}2^{2/3}$ is the ground state is fully polarized.

Excited State Energies

$$ \begin{align} \pop(\br)\equiv\frac{1}{L^{3/2}}\sum_{\bk} \exp(i\bk\cdot\br)\aop_{\bk},\
\pdop(\br)\equiv\frac{1}{L^{3/2}}\sum_{\bk} \exp(-i\bk\cdot\br)\adop_{\bk}, \end{align} $$

  • Interaction potential in terms of its Fourier components

$$ U(\br-\br’) = \frac{1}{L^3}\sum_\bq \tilde U(\bq) \exp(i\bq\cdot[\br-\br’]). $$

  • Interaction Hamiltonian for spinless particles can then be written

$$ \hat H_\text{int.} = \frac{1}{2L^3} \sum_{\bk_1+\bk_2=\bk_3+\bk_4} \tilde U(\bk_1-\bk_4) \adop_{\bk_1}\adop_{\bk_2}\aop_{\bk_3}\aop_{\bk_4}. \label{more_vertex} $$

  • Graphical representation


  • Expectation value in a product state of momentum eigenstates gives two terms


  • Evaluating the two contributions

$$ \braket{\mathbf{N}}{\hat H_\text{int.}}{\mathbf{N}} = \frac{1}{2V}\tilde U(0) \sum_{\bk_1,\bk_2} N_{\bk_1}N_{\bk_2} - \frac{1}{2V} \sum_{\bk_1,\bk_2} \tilde U(\bk_1-\bk_2) N_{\bk_1}N_{\bk_2} $$

  • Hartree term just depends on total number

  • Fock term depends on the individual occupations

  • Interaction energy to add a single particle to state $\bk$ is

$$ \Delta U_{\bk} = \frac{\tilde U(0)}{V} \sum_{\bk’} N_{\bk’} - \frac{1}{V}\sum_{\bk’} \tilde U(\bk-\bk’) N_{\bk’} $$

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