B is for Bunching

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*} \lvert{\Psi}\rangle&\longleftrightarrow \Psi(x_1,\ldots, x_N)\nonumber\\ \psi(X)\lvert{\Psi}\rangle&\longleftrightarrow \sqrt{N}\Psi(X,x_1,\ldots, x_{N-1})\\ \end{align*}

  • Second quantized form

\rho_2(x,y) =\braket{\Psi|\psi^\dagger(x)\psi^\dagger(y)\psi^{\vphantom{\dagger}}(y)\psi^{\vphantom{\dagger}}(x)|\Psi}

\rho_2(x,y) =\braket{\Psi|\psi^\dagger(x)\psi^\dagger(y)\psi^{\vphantom{\dagger}}(y)\psi^{\vphantom{\dagger}}(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|\psi^\dagger(x)\psi^\dagger(y)\psi^{\vphantom{\dagger}}(y)\psi^{\vphantom{\dagger}}(x)|\Psi}

  • Insert expansion

\begin{align*} \psi^{\vphantom{\dagger}}(x)=\sum_{\beta} \varphi^{}_{\beta}(x)a^{\vphantom{\dagger}}_{\beta},\\ \psi^\dagger(x)=\sum_{\beta} \varphi^*_{\beta}(x)a^\dagger_{\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|a^\dagger_{\alpha}a^\dagger_{\beta}a^{\vphantom{\dagger}}_{\gamma}a^{\vphantom{\dagger}}_{\delta}|\Psi}.

\braket{\Psi|a^\dagger_{\alpha}a^\dagger_{\beta}a^{\vphantom{\dagger}}_{\gamma}a^{\vphantom{\dagger}}_{\delta}|\Psi}

  • When \lvert{\Psi}\rangle=\lvert{\mathbf{N}}\rangle (product state using a^\dagger_\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}|a^\dagger_{\alpha}a^\dagger_{\gamma}a^{\vphantom{\dagger}}_{\gamma}a^{\vphantom{\dagger}}_{\alpha}|\mathbf{N}}&=N_{\alpha}N_{\gamma}\nonumber\\ \braket{\mathbf{N}|a^\dagger_{\alpha}a^\dagger_{\gamma}a^{\vphantom{\dagger}}_{\alpha}a^{\vphantom{\dagger}}_{\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|a^\dagger_{\alpha}a^\dagger_{\beta}a^{\vphantom{\dagger}}_{\gamma}a^{\vphantom{\dagger}}_{\delta}|\Psi}\\ &=\sum_{\alpha, \beta}N_\alpha N_\beta\left[\lvert{\varphi_{\alpha}(x)}\rvert^2\lvert{\varphi_{\beta}(y)}\rvert^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|a^\dagger_{\alpha}a^\dagger_{\beta}a^{\vphantom{\dagger}}_{\gamma}a^{\vphantom{\dagger}}_{\delta}|\Psi}\\ &=\sum_{\alpha, \beta}N_\alpha N_\beta\left[\lvert{\varphi_{\alpha}(x)}\rvert^2\lvert{\varphi_{\beta}(y)}\rvert^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 \lvert{x-y}\rvert\to\infty \rho_2(x,x) is twice the value at \lvert{x-y}\rvert\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 ground state \varphi_{0}(\mathbf{r}) of some potential: a Bose condensate

  • N-body wavefunction is \Psi(\mathbf{r}_1,\mathbf{r}_2,\ldots,\mathbf{r}_N)=\prod_i^N \varphi_0(\mathbf{r}_i) \lvert{\Psi}\rangle=\frac{1}{\sqrt{N!}}\left(a^\dagger_0\right)^N\lvert{\text{VAC}}\rangle where a^\dagger_0 creates particle in state \varphi_0(\mathbf{r})

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(\mathbf{r}) and \varphi_R(\mathbf{r}) \lvert{\bar N_L,\bar N_R}\rangle_\theta\equiv\frac{1}{\sqrt{N!}}\left[\sqrt{\frac{\bar N_L}{N}}e^{-i\theta/2} a^\dagger_L+\sqrt{\frac{\bar N_R}{N}}e^{i\theta/2}a^\dagger_R\right]^N\lvert{\text{VAC}}\rangle \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 \psi^{\vphantom{\dagger}}(\mathbf{r},t)}{\partial t} = -\frac{1}{2m}\nabla^2\psi^{\vphantom{\dagger}}(\mathbf{r},t) (no potential)

  • Write the field operator \psi^{\vphantom{\dagger}}(\mathbf{r})=\varphi_L(\mathbf{r},t)a^{\vphantom{\dagger}}_L+\varphi_R(\mathbf{r},t)a^{\vphantom{\dagger}}_R+\cdots, where wavefunctions \varphi_{L/R}(\mathbf{r},t) obey free particle Schrödinger

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

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

  • Relative phase has real physical effect.

Check

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(\mathbf{r},t)=\frac{1}{\left(\pi R_t^{2}\right)^{3/4}}\exp\left[-\frac{\mathbf{r}^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) \lvert{N_L,N_R}\rangle\equiv\frac{1}{\sqrt{N_L! N_R!}}\left(a^\dagger_L\right)^{N_L}\left(a^\dagger_R\right)^{N_R}\lvert{\text{VAC}}\rangle

  • Compute density as before \rho_1(\mathbf{r},t)=N_L|\varphi_L(\mathbf{r},t)|^2+N_R|\varphi_R(\mathbf{r},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(\mathbf{r},\mathbf{r}')&=\rho_1(\mathbf{r})\rho_1(\mathbf{r}') +N_LN_{R}\varphi_L^*(\mathbf{r})\varphi_R^*(\mathbf{r}')\varphi_L(\mathbf{r}')\varphi_R(\mathbf{r}) \nonumber\\ &\qquad+N_{L}N_R\varphi_R^*(\mathbf{r})\varphi_L^*(\mathbf{r}') \varphi_R(\mathbf{r}')\varphi_L(\mathbf{r}) \end{align*}

  • Second line contains interference fringes!

  • \rho_2 gives probability of finding an atom at \mathbf{r}' if there is one at \mathbf{r}

  • 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.

Check

Prove this by showing that the density matrix

\rho=\int_0^{2\pi}\frac{d\theta}{2\pi}\lvert{\bar N_L,\bar N_R}\rangle_\theta\langle{\bar N_R,\bar N_L}\rvert_\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(\mathbf{r}_j-\mathbf{r}_k)=\frac{1}{2}\int d\mathbf{r}_1 d\mathbf{r}_2\, U(\mathbf{r}_1-\mathbf{r}_2)\psi^\dagger(\mathbf{r}_1)\psi^\dagger(\mathbf{r}_2)\psi^{\vphantom{\dagger}}(\mathbf{r}_2)\psi^{\vphantom{\dagger}}(\mathbf{r}_1).

  • Since

\sum_{j<k} U(\mathbf{r}_j-\mathbf{r}_k) = \frac{1}{2}\int \sum_{j\neq k}\delta(\mathbf{r}_1-\mathbf{r}_j)\delta(\mathbf{r}_2-\mathbf{r}_k)U(\mathbf{r}_1-\mathbf{r}_2) d\mathbf{r}_1 d\mathbf{r}_2,

  • Find expectation value of interaction energy in a product state

\begin{align*} \langle \hat V\rangle &= \overbrace{\frac{1}{2}\int d\mathbf{r}\, d\mathbf{r}'\, \rho_1(\mathbf{r}) U(\mathbf{r}-\mathbf{r}')\rho_1(\mathbf{r}')}^{\equiv E_\text{Hartree}} \\ &\qquad\overbrace{\pm \frac{1}{2}\int d\mathbf{r}\, d\mathbf{r}'\, U(\mathbf{r}-\mathbf{r}')g(\mathbf{r},\mathbf{r}')g(\mathbf{r}',\mathbf{r})}^{\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 \psi^{\vphantom{\dagger}}_\sigma(\mathbf{r}), \psi^\dagger_\sigma(\mathbf{r}') satisfying

\begin{gather*} \left\{\psi^{\vphantom{\dagger}}_{\sigma_1}(\mathbf{r}_1),\psi^\dagger_{\sigma_2}(\mathbf{r}_2)\right\}=\delta_{\sigma_1\sigma_2}\delta(\mathbf{r}_1-\mathbf{r}_2)\nonumber\\ \left\{\psi^{\vphantom{\dagger}}_{\sigma_1}(\mathbf{r}_1),\psi^{\vphantom{\dagger}}_{\sigma_2}(\mathbf{r}_2)\right\}=\left\{\psi^\dagger_{\sigma_1}(\mathbf{r}_1),\psi^\dagger_{\sigma_2}(\mathbf{r}_2)\right\}=0 \end{gather*}

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

g_{\sigma_1\sigma_2}(\mathbf{r}_1,\mathbf{r}_2) = \braket{\Psi|\psi^\dagger_{\sigma_1}(\mathbf{r}_1)\psi^{\vphantom{\dagger}}_{\sigma_2}(\mathbf{r}_2)|\Psi}.

  • From g_{\sigma_1\sigma_2}(\mathbf{r}_1,\mathbf{r}_2) we get density and spin density

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

  • Spin-independent interaction

\hat H_\text{int.} = \frac{1}{2}\sum_{\sigma_1,\sigma_2}\int d\mathbf{r}_1 d\mathbf{r}_2\, U(\mathbf{r}_1-\mathbf{r}_2)\psi^\dagger_{\sigma_1}(\mathbf{r}_1)\psi^\dagger_{\sigma_2}(\mathbf{r}_2)\psi^{\vphantom{\dagger}}_{\sigma_2}(\mathbf{r}_2)\psi^{\vphantom{\dagger}}_{\sigma_1}(\mathbf{r}_1)

  • Hartree–Fock energy then

\begin{align*} \langle \hat H_\text{int.}\rangle &=\frac{1}{2}\int d\mathbf{r}\, d\mathbf{r}'\, \rho(\mathbf{r}) U(\mathbf{r}-\mathbf{r}')\rho(\mathbf{r}')\\ &- \frac{1}{2}\int d\mathbf{r}\, d\mathbf{r}'\, U(\mathbf{r}-\mathbf{r}')\operatorname{tr}\left[g(\mathbf{r},\mathbf{r}')g(\mathbf{r}',\mathbf{r})\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\mathbf{r}\, d\mathbf{r}'\, U(\mathbf{r}-\mathbf{r}')\operatorname{tr}\left[g(\mathbf{r},\mathbf{r}')\right]\operatorname{tr}\left[g(\mathbf{r}',\mathbf{r})\right]\\&-\frac{1}{4}\int d\mathbf{r}\, d\mathbf{r}'\, U(\mathbf{r}-\mathbf{r}')\operatorname{tr}\left[\boldsymbol{\sigma}g(\mathbf{r},\mathbf{r}')\right]\cdot\operatorname{tr}\left[\boldsymbol{\sigma}g(\mathbf{r}',\mathbf{r})\right] \end{align*}
  • Suppose U(\mathbf{r})=V_0 \delta(\mathbf{r})

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

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

Check

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

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

  • For a Hamiltonian with translational invariance H = \int d\mathbf{r}\frac{1}{2m}\nabla\psi^\dagger\cdot\nabla\psi^{\vphantom{\dagger}}+ \frac{1}{2}\int d\mathbf{r}d\mathbf{r}' U(\mathbf{r}-\mathbf{r}')\psi^\dagger(\mathbf{r})\psi^\dagger(\mathbf{r}')\psi^{\vphantom{\dagger}}(\mathbf{r}')\psi^{\vphantom{\dagger}}(\mathbf{r}) 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_{|\mathbf{k}|<k_\text{F}}1\longrightarrow L^3 \int_{|\mathbf{k}|<k_\text{F}} \frac{d\mathbf{k}}{(2\pi)^3} = \frac{k_\text{F}^3L^3}{6\pi^2} \\ E_\text{kin} &= \sum_{|\mathbf{k}|<k_\text{F}} \frac{\mathbf{k}^2}{2m} \longrightarrow L^3 \int_{|\mathbf{k}|<k_\text{F}} \frac{d\mathbf{k}}{(2\pi)^3} \frac{\mathbf{k}^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).

Check

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*} \psi^{\vphantom{\dagger}}(\mathbf{r})\equiv\frac{1}{L^{3/2}}\sum_{\mathbf{k}} \exp(i\mathbf{k}\cdot\mathbf{r})a^{\vphantom{\dagger}}_{\mathbf{k}} \\ \psi^\dagger(\mathbf{r})\equiv\frac{1}{L^{3/2}}\sum_{\mathbf{k}} \exp(-i\mathbf{k}\cdot\mathbf{r})a^\dagger_{\mathbf{k}} \end{align*}

  • Express interaction Hamiltonian in Fourier components

U(\mathbf{r}-\mathbf{r}') = \frac{1}{L^3}\sum_\mathbf{q}\tilde U(\mathbf{q}) \exp(i\mathbf{q}\cdot[\mathbf{r}-\mathbf{r}']).

\hat H_\text{int.} = \frac{1}{2L^3} \sum_{\mathbf{k}_1+\mathbf{k}_2=\mathbf{k}_3+\mathbf{k}_4} \tilde U(\mathbf{k}_1-\mathbf{k}_4) a^\dagger_{\mathbf{k}_1}a^\dagger_{\mathbf{k}_2}a^{\vphantom{\dagger}}_{\mathbf{k}_3}a^{\vphantom{\dagger}}_{\mathbf{k}_4}

  • Graphical representation

\hat H_\text{int.} = \frac{1}{2L^3} \sum_{\mathbf{k}_1+\mathbf{k}_2=\mathbf{k}_3+\mathbf{k}_4} \tilde U(\mathbf{k}_1-\mathbf{k}_4) a^\dagger_{\mathbf{k}_1}a^\dagger_{\mathbf{k}_2}a^{\vphantom{\dagger}}_{\mathbf{k}_3}a^{\vphantom{\dagger}}_{\mathbf{k}_4}

\langle{\mathbf{N}}\rvert \hat H_\text{int.} \lvert \mathbf{N} \rangle = \frac{1}{2V}\tilde U(0) \sum_{\mathbf{k}_1,\mathbf{k}_2} N_{\mathbf{k}_1}N_{\mathbf{k}_2} - \frac{1}{2V} \sum_{\mathbf{k}_1,\mathbf{k}_2} \tilde U(\mathbf{k}_1-\mathbf{k}_2) N_{\mathbf{k}_1}N_{\mathbf{k}_2}

\langle{\mathbf{N}}\rvert \hat H_\text{int.} \lvert \mathbf{N} \rangle = \frac{1}{2V}\tilde U(0) \sum_{\mathbf{k}_1,\mathbf{k}_2} N_{\mathbf{k}_1}N_{\mathbf{k}_2} - \frac{1}{2V} \sum_{\mathbf{k}_1,\mathbf{k}_2} \tilde U(\mathbf{k}_1-\mathbf{k}_2) N_{\mathbf{k}_1}N_{\mathbf{k}_2}

  • Hartree term just depends on total number

  • Fock term depends on the individual occupations

  • Interaction energy to add a single particle to state \mathbf{k} is

\Delta U_{\mathbf{k}} = \frac{\tilde U(0)}{V} \sum_{\mathbf{k}'} N_{\mathbf{k}'} - \frac{1}{V}\sum_{\mathbf{k}'} \tilde U(\mathbf{k}-\mathbf{k}') N_{\mathbf{k}'}