# More Second Quantization

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

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

• How to find this using second quantization?

• What can these correlations tell us about interactions?

## $\rho_2$ from Second Quantization

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

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

• Let’s calculate using second quantization!

$$\rho_2(x,y) = N(N-1) \int dx_3\ldots dx_N \left|\Psi(x,y,\ldots,x_N)\right|^2$$

\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

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

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

• 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

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

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

• 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 $g(x,y)$ $$\rho_2(x,y) = \rho_1(x)\rho_1(y) \pm g(x,y)g(y,x)$$ for ground state of the Fermi gas, reproduces $$\rho_2(x,y) = n^2\left[1 - \left(\frac{\sin\left[k_\text{F}(x-y)\right]}{k_\text{F}(x-y)}\right)^2\right]$$

• ‘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 $$\Psi(\br_1,\br_2,\ldots,\br_N)=\prod_i^N \varphi_0(\br_i)$$ $$\ket{\Psi}=\frac{1}{\sqrt{N!}}\left(\adop_0\right)^N\ket{\text{VAC}}$$ 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 $\varphi_R(\br)$ $$\ket{\bar N_L,\bar N_R}_\theta\equiv\frac{1}{\sqrt{N!}}\left[\sqrt{\frac{\bar N_L}{N}}e^{-i\theta/2} \adop_L+\sqrt{\frac{\bar N_R}{N}}e^{i\theta/2}\adop_R\right]^N\ket{\text{VAC}}$$ $\bar N_{L,R}$ are average particle number ($N=\bar N_L+\bar N_R$)

• System evolves for time $t$. Recall that 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)}} \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

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

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

### 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 $$\rho_1(\br,t)=N_L|\varphi_L(\br,t)|^2+N_R|\varphi_R(\br,t)|^2$$ 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) \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*} \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

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

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

• 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] \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 $U(\br)=V_0 \delta(\br)$

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

• Second term favours ferromagnetism for $V_0>0$ (c.f. Hund’s rules)

This is most succintly put by the formula $$\rho_2(\br,\br) = \frac{1}{2}\rho(\br)^2 - 2\mathbf{s}(\br)\cdot\mathbf{s}(\br)$$ So $\abs{\mathbf{s}(\br)}=\rho(\br)/2\longrightarrow\rho_2(\br,\br)=0$

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

• For a Hamiltonian with translational invariance $$H = \int d\br \frac{1}{2m}\nabla\pdop\cdot\nabla\pop + \frac{1}{2}\int d\br d\br' U(\br-\br')\pdop(\br)\pdop(\br')\pop(\br')\pop(\br)$$ guaranteed to involve with plane wave single particle states. Only variational parameters are occupancies of 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}}1\longrightarrow L^3 \int_{|\bk|<k_\text{F}} \frac{d\bk}{(2\pi)^3} = \frac{k_\text{F}^3L^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}^5L^3}{20\pi^2 m} = L^3 \frac{3}{10m}(6\pi^2)^{2/3} n^{5/3} \end{align*}

• Total energy is $E_\text{kin}(n_\uparrow,n_\downarrow) = (\text{const.})\frac{L^3}{m}\left(n_\uparrow^{5/3}+n_\downarrow^{5/3}\right)$
• 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*}

• Express interaction Hamiltonian in Fourier components

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

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

• Graphical representation

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

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

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

• 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

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