Hartree Approximation

The Hartree approximation minimizes the expectation value of an interacting many-particle Hamiltonian over uncorrelated product states. Each particle moves in a one-body potential generated by the probability densities of all the others, so the one-body equations and the fields appearing in them must be solved self-consistently.

For distinguishable particles, the trial state has the form

$$\Psi_{\mathrm H}(q_1,\ldots,q_N) = \prod_{i=1}^{N}\phi_i(q_i).$$

For identical bosons in a simple condensate, every factor is the same normalized orbital. A raw product of labeled orbitals is not an admissible state of identical fermions; antisymmetry leads instead to a Slater-determinant variational family and exchange terms.

Hartree theory is therefore more specific than the general instruction to replace fluctuations by averages. It is a restricted variational theory with a precisely stated trial manifold. That precision gives it three important properties:

• the optimized energy is an upper bound when the product is an admissible trial state;
• the direct potential and its double-counting correction follow from one energy functional;
• every omitted effect can be traced to structure absent from the product manifold, especially exchange and connected interparticle correlations.

Helpful background. Product states explains factorization, and the variational principle gives the energy bound. The atomic Hartree method specializes these equations to radial orbitals.

Pair potentials and Hartree product states

Consider a first-quantized Hamiltonian

$$H = \sum_{i=1}^{N}h_i + \frac{1}{2} \sum_{i\ne j}v_{ij}.$$

Here $q_i$ denotes all one-particle coordinates needed for particle $i$, possibly including position and internal labels. The one-body operator $h_i$ contains kinetic energy and external fields. For the main derivation, assume that the interaction is a real, symmetric, local pair potential,

$$v_{ij} = v_{ji}, \qquad v_{ij}(q,q') = v_{ji}(q',q).$$

The factor $1/2$ prevents counting an unordered pair twice. Equivalently,

$$\frac{1}{2} \sum_{i\ne j}v_{ij} = \sum_{i<j}v_{ij}.$$

This convention must remain fixed throughout a calculation. Many apparent factors-of-two disagreements in Hartree formulas come from switching silently between ordered and unordered pair sums.

The same variational logic extends to nonlocal pair kernels. The effective Hartree operator may then be nonlocal rather than multiplication by a function, but it is still obtained by contracting one particle line with the one-body state of another particle.

Product-State Trial Manifold

For distinguishable particles, choose independently normalized orbitals,

$$\langle\phi_i|\phi_i\rangle = 1, \qquad i=1,\ldots,N,$$

and form

$$|\Psi_{\mathrm H}\rangle = |\phi_1\rangle \otimes\cdots\otimes |\phi_N\rangle.$$

No orthogonality condition is required. The factors live in different labeled one-particle Hilbert spaces, so an overlap such as $\langle\phi_i|\phi_j\rangle$ need not even be defined when the species differ.

For a simple condensate of $N$ identical bosons, the state is instead

$$|\Psi_{\mathrm H}^{(B)}\rangle = |\phi\rangle^{\otimes N}, \qquad \langle\phi|\phi\rangle=1.$$

Because every factor is identical, this state is already symmetric. More general bosonic mean-field families can use several orbitals and symmetrized permanents, but they describe fragmentation and introduce additional occupation and orthogonality structure. They are not the elementary Hartree ansatz considered here.

What factorization asserts

For operators $A_i$ and $B_j$ acting on different factors,

$$\langle A_iB_j\rangle_{\mathrm H} = \langle A_i\rangle_{\mathrm H} \langle B_j\rangle_{\mathrm H}, \qquad i\ne j.$$

Thus every connected correlator between distinct particle slots vanishes:

$$\langle A_iB_j\rangle_{\mathrm H,c} \equiv \langle A_iB_j\rangle_{\mathrm H} - \langle A_i\rangle_{\mathrm H} \langle B_j\rangle_{\mathrm H} =0.$$

For identical bosons, individually labeled slot operators are not separate physical particle addresses; symmetric combinations give physical observables. A fixed-$N$ condensate can still have correlated occupations of spatial modes, despite this particle-coordinate factorization.

The orbitals may be strongly distorted by the average interaction field, so Hartree theory is not generally a weak perturbation of the noninteracting orbitals. What remains absent is joint dependence on two or more coordinates beyond the product.

Hartree Energy Functional

The one-body contribution factorizes immediately:

$$\langle\Psi_{\mathrm H}|h_i|\Psi_{\mathrm H}\rangle = \langle\phi_i|h_i|\phi_i\rangle.$$

Define the direct pair integral

$$J_{ij} = \int dq\,dq'\, |\phi_i(q)|^2 v_{ij}(q,q') |\phi_j(q')|^2.$$

For a symmetric interaction, $J_{ij}=J_{ji}$. The product-state energy is

$$E_{\mathrm H}[\{\phi_i\}] ={} \sum_i \langle\phi_i|h_i|\phi_i\rangle + \frac{1}{2} \sum_{i\ne j}J_{ij}.$$

In coordinate form,

$$E_{\mathrm H} ={} \sum_i \int dq\, \phi_i^*(q)h_i\phi_i(q) + \frac{1}{2} \sum_{i\ne j} \int dq\,dq'\, |\phi_i(q)|^2 v_{ij}(q,q') |\phi_j(q')|^2.$$

The interaction energy depends on the full set of orbital densities. Consequently, varying one orbital changes both its own kinetic and external energy and every pair term containing that orbital.

Constrained Variational Derivation

Minimize $E_{\mathrm H}$ subject to independent normalization constraints. Introduce real Lagrange multipliers $\epsilon_i$ and the functional

$$\mathcal L = E_{\mathrm H} - \sum_i \epsilon_i \left( \langle\phi_i|\phi_i\rangle-1 \right).$$

Treat $\phi_i$ and $\phi_i^*$ as independent variables during variation. The one-body term gives

$$\frac{\delta}{\delta\phi_i^*(q)} \langle\phi_i|h_i|\phi_i\rangle = h_i\phi_i(q).$$

Particle $i$ appears in both ordered versions of each pair integral. The prefactor $1/2$ cancels that duplication:

$$\begin{aligned} \frac{\delta}{\delta\phi_i^*(q)} \left[ \frac{1}{2} \sum_{k\ne l}J_{kl} \right] ={}& \sum_{j\ne i} \left[ \int dq'\, |\phi_j(q')|^2 v_{ij}(q,q') \right] \phi_i(q). \end{aligned}$$

This identifies the direct Hartree potential seen by particle $i$:

$$V_i^{\mathrm H}(q) = \sum_{j\ne i} \int dq'\, |\phi_j(q')|^2 v_{ij}(q,q').$$

Stationarity, $\delta\mathcal L/\delta\phi_i^*=0$, gives the coupled Hartree equations

$$\left[ h_i+V_i^{\mathrm H}[\{\phi_j\}] \right] \phi_i = \epsilon_i\phi_i.$$

The notation emphasizes that $V_i^{\mathrm H}$ is a functional of all the other orbitals. The equation is linear in $\phi_i$ when the other orbitals are held fixed, but the full system is nonlinear.

Physical interpretation

The quantity $|\phi_j(q')|^2dq'$ is the probability of finding particle $j$ near $q'$. Hartree theory replaces the pair interaction with particle $j$ by its average over this distribution:

$$v_{ij}(q,q_j) \longrightarrow \int dq'\, |\phi_j(q')|^2v_{ij}(q,q').$$

Particle $i$ then moves in the external field plus the sum of these averaged fields. Its new orbital changes its density, which changes the fields seen by the other particles. A Hartree solution is a fixed point of this feedback loop.

Self-Consistency and Stationarity

A standard static iteration is:

1. choose normalized initial orbitals $\{\phi_i^{(0)}\}$;
2. construct every $V_i^{\mathrm H}$ from the current densities;
3. solve the effective one-body eigenproblems;
4. select the desired orbitals and normalize them;
5. mix old and new densities or orbitals if necessary;
6. repeat until densities, energy, and residuals converge.

An orbital residual is

$$R_i = \left( h_i+V_i^{\mathrm H}-\epsilon_i \right) \phi_i.$$

Small energy changes alone are not enough: cancellation can make the total energy appear stationary while one or more orbital equations remain inaccurate. A reliable calculation monitors norms such as $\|R_i\|$, density changes, interaction-energy consistency, and the original variational functional.

A vanishing orbital residual establishes a stationary point; a stalled iteration need not do even that. Different seeds can converge to different local minima, excited self-consistent solutions, or symmetry-related branches. The expectation of the original Hamiltonian in every admissible normalized trial state is an upper bound to its ground energy, including an excited stationary product. Finding the global product-state minimum tightens that bound; convergence alone does not establish that this minimum was found.

Meaning of the Orbital Multipliers

Multiplying the $i$th Hartree equation by $\phi_i^*$ and integrating gives

$$\epsilon_i = \langle\phi_i|h_i|\phi_i\rangle + \sum_{j\ne i}J_{ij}.$$

The multiplier $\epsilon_i$ enforces normalization and labels an effective one-body eigenvalue. It is not generally a literal share of the total energy. Summing over $i$ counts every pair interaction twice:

$$\sum_i\epsilon_i = \sum_i \langle\phi_i|h_i|\phi_i\rangle + \sum_{i\ne j}J_{ij}.$$

Therefore

$$E_{\mathrm H} = \sum_i\epsilon_i - \frac{1}{2} \sum_{i\ne j}J_{ij}.$$

The subtraction is the direct-interaction double-counting correction. Simply adding the occupied Hartree eigenvalues overcounts the interaction contribution. It overestimates the energy for repulsive pair integrals, but the direction reverses for attractive ones.

Distinguishable Particles and Mixtures

The labeled-orbital derivation applies directly to distinguishable particles. It also generalizes naturally to mixtures in which each bosonic species occupies one orbital.

Let species $a$ contain $N_a$ particles in a normalized orbital $\phi_a$. Define

$$U_{ab} = \int dq\,dq'\, |\phi_a(q)|^2 v_{ab}(q,q') |\phi_b(q')|^2.$$

Then

$$E_{\mathrm H} ={} \sum_a N_a \langle\phi_a|h_a|\phi_a\rangle + \frac{1}{2} \sum_a N_a(N_a-1)U_{aa} + \sum_{a<b}N_aN_bU_{ab}.$$

Variation with respect to $\phi_a^*$ gives

$$\bigg[ h_a + (N_a-1)V_{aa}^{\mathrm H} + \sum_{b\ne a}N_bV_{ab}^{\mathrm H} \bigg]\phi_a = \mu_a\phi_a,$$

where

$$V_{ab}^{\mathrm H}(q) = \int dq'\, |\phi_b(q')|^2v_{ab}(q,q').$$

The same-species coefficient is $N_a-1$ because a particle does not interact with itself. The other-species coefficient is $N_b$ because all particles of species $b$ are genuine partners.

Identical Bosons in One Orbital

For $N$ identical bosons with the same one-body operator $h$ and symmetric pair potential $v$, use

$$\Psi_{\mathrm H}^{(B)}(q_1,\ldots,q_N) = \prod_{i=1}^{N}\phi(q_i).$$

There are $N$ one-body terms and $N(N-1)/2$ unordered pairs. Define

$$U[\phi] = \int dq\,dq'\, |\phi(q)|^2 v(q,q') |\phi(q')|^2.$$

The energy functional is

$$E_{\mathrm H}^{(B)}[\phi] = N\langle\phi|h|\phi\rangle + \frac{N(N-1)}{2}U[\phi].$$

Varying under $\langle\phi|\phi\rangle=1$ gives

$$\left[ h + (N-1)V_{\phi}^{\mathrm H}(q) \right] \phi(q) = \mu_{\mathrm H}\phi(q),$$

with

$$V_{\phi}^{\mathrm H}(q) = \int dq'\, |\phi(q')|^2v(q,q').$$

The factor $N-1$, rather than $N$, is exact within this finite-$N$ product ansatz. It encodes self-exclusion at the level of particle counting, even though the state has no correlation hole.

Multiplying the orbital equation by $\phi^*$ gives

$$\mu_{\mathrm H} = \langle h\rangle_\phi + (N-1)U[\phi].$$

By contrast, the energy per particle is

$$\frac{E_{\mathrm H}^{(B)}}{N} = \langle h\rangle_\phi + \frac{N-1}{2}U[\phi].$$

Thus the nonlinear orbital eigenvalue is not the energy per particle. Nor is it exactly a finite-$N$ addition energy. At a fixed orbital,

$$E(N+1)-E(N)=\langle h\rangle_\phi+NU[\phi],$$

whereas $\mu_{\mathrm H}=\langle h\rangle_\phi+(N-1)U[\phi]$. Relaxation in the new number sector changes the addition energy further. The orbital multiplier approaches a thermodynamic chemical potential only in an appropriate large-$N$ limit.

Contact Interaction and the Gross–Pitaevskii Boundary

If the effective pair interaction is modeled as

$$v(\mathbf r,\mathbf r') = g\,\delta^{(3)}(\mathbf r-\mathbf r'),$$

then

$$U[\phi] = g\int d^3r\,|\phi(\mathbf r)|^4,$$

and the bosonic Hartree equation becomes

$$\left[ h + g(N-1)|\phi(\mathbf r)|^2 \right] \phi(\mathbf r) = \mu_{\mathrm H}\phi(\mathbf r).$$

Writing a condensate wavefunction $\Phi=\sqrt{N}\phi$ gives

$$\left[ h + g\left(1-\frac{1}{N}\right)|\Phi|^2 \right] \Phi = \mu_{\mathrm H}\Phi.$$

At large $N$, this has the familiar form of the stationary Gross–Pitaevskii equation. The conceptual relation is close, but the theories should not be identified carelessly:

• elementary Hartree variation averages a specified pair potential over a product state;
• Gross–Pitaevskii theory uses a low-energy coupling fixed by the two-body scattering length;
• in the microscopic dilute-gas limit, short-range pair correlations can be essential even when the leading energy and density are described by a one-field functional;
• condensate depletion and Bogoliubov quasiparticles lie outside the pure product ansatz.

The focused Gross–Pitaevskii Equation treatment develops scattering-length matching, healing length, vortices and condensate dynamics.

Why Raw Hartree Is Not a Fermion Theory

For identical fermions, a physical wavefunction must satisfy

$$\Psi(\ldots,q_i,\ldots,q_j,\ldots) = - \Psi(\ldots,q_j,\ldots,q_i,\ldots).$$

A labeled product

$$\phi_1(q_1)\phi_2(q_2)\cdots\phi_N(q_N)$$

does not have this transformation law. It is therefore not an admissible variational state for identical fermions, even if the orbitals are orthogonal.

Replacing the product by a Slater determinant restores antisymmetry. Evaluation of a two-body interaction in that determinant produces both direct and exchange integrals. The resulting Hartree–Fock equations are not merely Hartree equations supplemented by an optional empirical correction: exchange follows from the allowed fermionic trial manifold.

This distinction also clarifies terminology. In electronic-structure contexts, the phrase “Hartree term” often means only the direct density-generated contribution inside Hartree–Fock, density-functional, or Green-function equations. That direct term is meaningful, but a direct-only product theory is not by itself a valid many-electron wavefunction approximation.

Coulomb Hartree Potential

For particles with repulsive Coulomb interaction,

$$v(\mathbf r,\mathbf r') = \frac{e^2}{4\pi\epsilon_0} \frac{1}{|\mathbf r-\mathbf r'|},$$

an orbital density $n_j(\mathbf r)=|\phi_j(\mathbf r)|^2$ generates

$$V_{ij}^{\mathrm H}(\mathbf r) = \frac{e^2}{4\pi\epsilon_0} \int d^3r'\, \frac{n_j(\mathbf r')} {|\mathbf r-\mathbf r'|}.$$

The field seen by particle $i$ is the sum over $j\ne i$. If

$$n(\mathbf r) = \sum_j n_j(\mathbf r),$$

then a common total-density potential is

$$V_{\mathrm H}[n](\mathbf r) = \frac{e^2}{4\pi\epsilon_0} \int d^3r'\, \frac{n(\mathbf r')} {|\mathbf r-\mathbf r'|}.$$

It satisfies

$$\nabla^2V_{\mathrm H}[n](\mathbf r) = -\frac{e^2}{\epsilon_0}n(\mathbf r).$$

This sign corresponds to the positive potential energy of repulsion between two electrons. It should not be confused with the electrostatic scalar potential produced by negative charge, whose sign is opposite before multiplication by the test electron charge.

Removing the self-generated field

The exact labeled Hartree equation uses $j\ne i$. Replacing that sum by the total density makes orbital $i$ feel its own density unless one subtracts

$$V_{ii}^{\mathrm H}(\mathbf r) = \frac{e^2}{4\pi\epsilon_0} \int d^3r'\, \frac{|\phi_i(\mathbf r')|^2} {|\mathbf r-\mathbf r'|}.$$

For one particle, any nonzero Coulomb Hartree field generated by its own density is manifestly spurious. In a finite bosonic condensate, the analogous correction is the difference between $N$ and $N-1$. In Hartree–Fock, exchange cancels the direct self-interaction of an occupied spin-orbital exactly, although correlation and approximate density functionals raise separate self-interaction questions.

Worked Benchmark: Two Coupled Oscillators

An exactly soluble model isolates what orbital relaxation can and cannot accomplish. Consider two distinguishable particles of equal mass in one-dimensional traps,

$$H ={} \sum_{i=1}^{2} \left( \frac{p_i^2}{2m} + \frac{m\omega^2x_i^2}{2} \right) + \frac{\kappa}{2}(x_1-x_2)^2, \qquad \kappa\ge 0.$$

The interaction favors correlated displacements $x_1\simeq x_2$.

Joint-density contours at $|\Psi|^2/|\Psi(0,0)|^2=e^{-1}$ for $\kappa/(m\omega^2)=2$. Coordinates are in oscillator-length units $\ell=\sqrt{\hbar/(m\omega)}$. The exact ellipse tilts toward $x_1=x_2$, showing correlated displacements. Hartree adjusts both widths but its circular contour has zero cross covariance.

Exact solution

Introduce orthonormal normal coordinates

$$q_+ = \frac{x_1+x_2}{\sqrt{2}}, \qquad q_- = \frac{x_1-x_2}{\sqrt{2}}.$$

The normal-mode frequencies are

$$\omega_+ = \omega, \qquad \omega_- = \sqrt{\omega^2+\frac{2\kappa}{m}}.$$

Hence the exact ground-state energy is

$$E_0 = \frac{\hbar}{2} \left( \omega + \sqrt{\omega^2+\frac{2\kappa}{m}} \right).$$

The exact state is Gaussian in $q_+$ and $q_-$, but their unequal widths make it nonfactorizable in $x_1$ and $x_2$. In particular,

$$\langle x_1x_2\rangle_0 = \frac{\hbar}{4m} \left( \frac{1}{\omega} - \frac{1}{\omega_-} \right) >0$$

for $\kappa>0$.

Hartree minimization

For a product state, the energy of the two means is a positive quadratic form, so it is minimized at zero means. With those means fixed, each factor minimizes an identical harmonic oscillator of frequency $\sqrt{\omega^2+\kappa/m}$. Thus Gaussian factors are the full product-state minimizers here, not merely a guess based on exchange symmetry. To obtain their width variationally, write

$$\Psi_{\mathrm H}(x_1,x_2) = \phi_{\Omega}(x_1) \phi_{\Omega}(x_2).$$

For one factor,

$$\langle x^2\rangle_{\Omega} = \frac{\hbar}{2m\Omega}, \qquad \left\langle\frac{p^2}{2m}\right\rangle_{\Omega} = \frac{\hbar\Omega}{4}.$$

Factorization gives $\langle x_1x_2\rangle_{\mathrm H}=0$. The variational energy is

$$E_{\mathrm H}(\Omega) = \frac{\hbar}{2} \left[ \Omega + \frac{\omega^2+\kappa/m}{\Omega} \right].$$

Stationarity yields

$$\Omega_{\mathrm H} = \sqrt{\omega^2+\frac{\kappa}{m}}, \qquad E_{\mathrm H} = \hbar\Omega_{\mathrm H}.$$

The self-consistent field correctly narrows each one-particle orbital. Nevertheless,

$$E_{\mathrm H} - E_0 \ge 0,$$

with equality only for $\kappa=0$. Defining $\lambda=\kappa/(m\omega^2)$ gives the weak-coupling difference

$$E_{\mathrm H}-E_0 = \frac{\hbar\omega}{8}\lambda^2 + O(\lambda^3).$$

Hartree theory is correct through first order here because first-order perturbation theory needs only the unperturbed product density. The first missing energy appears at second order, where virtual correlated motion matters.

Density-Matrix View

For a distinguishable product state, the reduced state of particle $i$ is pure:

$$\rho_i^{(1)} = |\phi_i\rangle\langle\phi_i|.$$

For the simple $N$-boson product, two common one-body density-matrix conventions are

$$\gamma^{(1)} = |\phi\rangle\langle\phi|, \qquad \operatorname{Tr}\gamma^{(1)}=1,$$

or

$$\Gamma^{(1)} = N|\phi\rangle\langle\phi|, \qquad \operatorname{Tr}\Gamma^{(1)}=N.$$

The normalized two-body reduced state is also a product,

$$\gamma^{(2)} = |\phi\phi\rangle \langle\phi\phi|.$$

Mean-field convergence theorems are often stated as convergence of fixed-order reduced density matrices toward such tensor powers as $N\to\infty$. This is more precise than claiming that the full $N$-body wavefunction becomes close in norm: small correlations distributed across many particles can leave fixed-particle observables asymptotically factorized without making the entire many-body vector a literal product.

Time-Dependent Hartree Theory

Apply the Dirac–Frenkel variational principle to the time-dependent product manifold. Up to time-dependent orbital phase conventions, the distinguishable-particle equations are

$$i\hbar\frac{\partial}{\partial t}\phi_i(q,t) = \left[ h_i(t)+V_i^{\mathrm H}(q,t) \right] \phi_i(q,t),$$

where

$$V_i^{\mathrm H}(q,t) = \sum_{j\ne i} \int dq'\, |\phi_j(q',t)|^2 v_{ij}(q,q';t).$$

For $N$ identical bosons in one orbital,

$$i\hbar\partial_t\phi = \left[ h(t) + (N-1)V_{\phi}^{\mathrm H}(t) \right] \phi.$$

If $h$ and $v$ are time independent and the propagation is exact within the Hartree equations, norms and the Hartree energy are conserved. Independent transformations

$$\phi_i(t) \longrightarrow e^{i\theta_i(t)}\phi_i(t)$$

alter only scalar gauge terms in the orbital equations and multiply the total product by a global phase. Implementations may choose a gauge that removes orbital expectation values from the generators or improves numerical conditioning.

Time-dependent Hartree can describe collective changes in one-body densities, but it cannot generate entanglement between distinguishable particle factors because the trajectory remains on their product manifold. Particle factorization is distinct from factorization into spatial modes.

Mean-Field Scaling and Controlled Limits

For $N$ bosons in a fixed orbital with a nonzero, $N$-independent pair integral, the interaction energy grows as $N^2$. Expanding the occupied volume at fixed density is a different limit and changes that integral. A standard mean-field or Kac scaling uses

$$H_N = \sum_{i=1}^{N}h_i + \frac{1}{N-1} \sum_{i<j}v_{ij}.$$

The Hartree energy per particle is then

$$\frac{E_{\mathrm H}}{N} = \langle\phi|h|\phi\rangle + \frac{1}{2}U[\phi],$$

which remains finite as $N\to\infty$. The stationary equation becomes

$$\left[ h+V_{\phi}^{\mathrm H} \right]\phi = \mu\phi.$$

Under appropriate assumptions on the interaction, initial data, and observables, the many-body dynamics in this scaling converges to time-dependent Hartree dynamics at the level of reduced density matrices. The scaling and hypotheses are part of the theorem; “large $N$” by itself is not a proof that Hartree theory applies.

Other routes to mean-field accuracy include sufficiently long-range weak interactions, high connectivity with suitable coupling rescaling, or observables insensitive to short-distance correlations. The dilute Gross–Pitaevskii limit is different: it retains nontrivial two-body scattering correlations at short range while producing a nonlinear one-field description at leading order.

What the Product Manifold Omits

Exchange

Identical fermions require antisymmetry. Direct Hartree theory omits exchange energy, the exchange hole, and all consequences that follow from determinant structure.

Dynamical and static correlation

The product state cannot adjust the conditional position of one particle after another particle is observed. It misses correlation holes, pair cusps, correlated tunneling, dispersion forces between neutral fragments, and the coupled-oscillator covariance exhibited above.

Bosonic depletion and fragmentation

A single orbital has one macroscopically occupied natural orbital and zero depletion by construction. It cannot represent fragmented condensates or occupation of noncondensed modes.

Critical fluctuations

Near critical points, long-wavelength fluctuations may dominate and invalidate a smooth deterministic field even when a self-consistent solution exists.

Branch superpositions

Nonlinear Hartree equations can have symmetry-broken stationary solutions. The exact finite-system ground state may instead be a symmetric superposition of branches. A single product selects one branch and omits tunneling between them.

Short-range scattering structure

For singular or hard-core interactions, the exact wavefunction can develop rapid pair dependence at separations much shorter than the density-variation scale. Orbital optimization alone cannot build that pair structure.

Exercises

Two bosonic species with contact interactions

Species $a$ and $b$ contain $N_a$ and $N_b$ bosons in normalized orbitals $\phi_a$ and $\phi_b$. Let

$$v_{aa}=g_{aa}\delta, \qquad v_{bb}=g_{bb}\delta, \qquad v_{ab}=g_{ab}\delta.$$

Derive the Hartree equation for species $a$.

Solution

The terms depending on $\phi_a$ are

$$E_a ={} N_a\langle\phi_a|h_a|\phi_a\rangle + \frac{g_{aa}}{2}N_a(N_a-1) \int dq\,|\phi_a|^4 + g_{ab}N_aN_b \int dq\,|\phi_a|^2|\phi_b|^2.$$

Variation and division by $N_a$ give

$$\left[ h_a + g_{aa}(N_a-1)|\phi_a|^2 + g_{ab}N_b|\phi_b|^2 \right] \phi_a = \mu_a\phi_a.$$

The same-species field excludes one $a$ boson; the cross-species field contains all $N_b$ particles.

Diagnose one-particle self-interaction

Suppose a Coulomb Hartree equation is written using the total density $n=|\phi|^2$ for a system with $N=1$. Show why the result is inconsistent and state the correction.

Solution

The total-density formula produces

$$V_{\mathrm H}(\mathbf r) = \frac{e^2}{4\pi\epsilon_0} \int d^3r'\, \frac{|\phi(\mathbf r')|^2} {|\mathbf r-\mathbf r'|},$$

which is nonzero for a normalized orbital. But the original pair Hamiltonian has no terms when $N=1$ because there is no pair. The exact Hartree sum $\sum_{j\ne i}$ is empty. One must therefore subtract the orbital’s own contribution or retain the explicit self-excluding sum from the start.

Minimize the coupled-oscillator product energy

For the two-oscillator benchmark, minimize

$$E_{\mathrm H}(\Omega) = \frac{\hbar}{2} \left[ \Omega + \frac{\omega^2+\kappa/m}{\Omega} \right]$$

and compare the weak-coupling expansion with the exact energy through order $\kappa^2$.

Solution

Differentiation gives

$$\frac{dE_{\mathrm H}}{d\Omega} = \frac{\hbar}{2} \left[ 1 - \frac{\omega^2+\kappa/m}{\Omega^2} \right].$$

The positive stationary point is

$$\Omega_{\mathrm H} = \sqrt{\omega^2+\frac{\kappa}{m}},$$

and it is a minimum. Set $\lambda=\kappa/(m\omega^2)$. Then

$$\frac{E_{\mathrm H}}{\hbar\omega} = \sqrt{1+\lambda} = 1+\frac{\lambda}{2} -\frac{\lambda^2}{8} +O(\lambda^3).$$

The exact result is

$$\begin{aligned} \frac{E_0}{\hbar\omega} &= \frac{1+\sqrt{1+2\lambda}}{2} \\ &= 1+\frac{\lambda}{2} -\frac{\lambda^2}{4} +O(\lambda^3). \end{aligned}$$

Thus

$$E_{\mathrm H}-E_0 = \frac{\hbar\omega}{8}\lambda^2 +O(\lambda^3).$$

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