research34 min studyLesson 486 in Physics

Gross–Pitaevskii equation and healing length

Normalize the condensate field, derive a mean-field interaction scale, and identify when a healing length is controlled by dilute-gas physics.

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What you’ll understand

  • Use a condensate wavefunction normalized to atom number.
  • State the diluteness and depletion assumptions behind three-dimensional GP theory.
  • Relate chemical potential, sound speed, and healing length with consistent factors.

A weakly interacting Bose condensate can be described by a macroscopic complex field whose squared magnitude is condensate density. The Gross–Pitaevskii equation evolves that field with kinetic energy, external trapping, and a density-dependent contact interaction. It is a controlled mean-field approximation in an appropriate dilute regime, rather than an exact equation for every ultracold gas. Its healing length measures how rapidly the order parameter can adjust when density is disturbed.

Normalization fixes the nonlinear coefficient

Choose the field normalized so its spatial density integrates to condensate atom number. The interaction term is then g times its squared magnitude without another factor of atom number. If a unit-normalized orbital is used instead, the nonlinear coefficient contains the appropriate particle-number factor. Both conventions are common, but combining their coefficients overcounts or undercounts interactions. The field has units of inverse square root volume, not the units of an ordinary unit-amplitude optical wave.

The contact coupling is fixed by the low-energy scattering length for identical equal-mass bosons in three dimensions. The cubic term represents the leading mean field and assumes short-range s-wave interactions. Dipolar interactions, spin-dependent channels, strong confinement into reduced dimensions, and narrow-resonance range effects require modifications. Tuning scattering length to very large magnitude can leave the weak-coupling regime even when the physical atomic gas still looks extremely dilute on laboratory length scales.

iℏ∂tψ=[−ℏ2∇22m+V+g∣ψ∣2]ψ,∫∣ψ∣2d3r=N0,g=4πℏ2ami\hbar\partial_t\psi=\left[-\frac{\hbar^2\nabla^2}{2m}+V+g|\psi|^2\right]\psi,\qquad \int|\psi|^2d^3r=N_0,\qquad g=\frac{4\pi\hbar^2a}{m}
ψ has units m⁻³ᐟ², |ψ|² is condensate density in m⁻³, V and g|ψ|² are energies in joules, and g is in J m³. N₀ is condensate atom number. The equation neglects depletion, thermal-cloud dynamics, and losses at leading order.↔ Scroll wide equations horizontally.

Weak interaction means a dimensionless condition

For a homogeneous repulsive gas, the small parameter is density times scattering length cubed. The quantum depletion fraction is of order its square root, so it can remain small even while interaction energy changes a trapped density profile substantially. Diluteness therefore does not mean interactions are experimentally negligible. Low temperature and a large condensate fraction further support a single-field description; near a thermal transition the noncondensed cloud and fluctuations need separate treatment.

Positive scattering length gives a stable homogeneous mean-field compressibility. Negative scattering length makes the homogeneous contact mean field attractive and unstable at sufficiently long wavelength. A finite harmonic trap can support a metastable attractive condensate below a particle-number threshold because kinetic energy resists compression. That does not make its healing length real under the repulsive uniform formula. Stability and the validity of the approximation must be assessed before interpreting a computed length as a physical density-response scale.

na3≪1,μ=gn,c=gnm(a>0)na^3\ll1,\qquad \mu=gn,\qquad c=\sqrt{\frac{gn}{m}}\quad(a>0)
n is uniform condensate density, approximately total density when depletion is small. μ is the uniform zero-temperature mean-field chemical potential in joules, and c is sound speed in m/s. These expressions require a positive contact coupling and a controlled weakly depleted three-dimensional gas.↔ Scroll wide equations horizontally.

Balance kinetic and interaction energy to find a length

A spatial variation over a length ξ costs a kinetic energy of order hbar squared divided by twice mass times ξ squared. Equating that scale to the mean-field energy gives the conventional healing length. Its numerical factor is a convention tied to that balance; related definitions in other fields may differ by square root of two. A vortex core or a boundary profile varies over this scale, but the complete shape requires solving the stationary field equation and its boundary conditions.

In a trap, density changes with position, so a local healing length changes too. Near the dilute edge the local expression grows and a simple bulk approximation eventually fails. Dropping kinetic energy in the Thomas–Fermi bulk can be justified when density varies over much longer scales, but the kinetic term remains essential near surfaces and defects. A GP calculation should be checked against grid resolution, particle-number normalization, and boundary conditions as well as the gas parameter.

ξ=ℏ2mgn=18πan,cξ=ℏ2m\xi=\frac{\hbar}{\sqrt{2mgn}}=\frac1{\sqrt{8\pi an}},\qquad c\xi=\frac{\hbar}{\sqrt2m}
ξ is in metres for a > 0. The formula uses the convention ℏ²/(2mξ²) = gn. It applies to the repulsive uniform mean-field gas or a slowly varying local bulk approximation; it is not a universal coherence length for strongly interacting or low-dimensional matter.↔ Scroll wide equations horizontally.

Check diluteness before quoting a healing length

A uniform repulsive condensate has n = 10²⁰ m⁻³ and a = 5.0 nm. Compute na³ and ξ using π ≈ 3.14, and decide whether the gas parameter supports a weakly depleted treatment.

  1. The scattering length cubed is (5×10−9)3=1.25×10−25 m3(5\times10^{-9})^3=1.25\times10^{-25}\,\mathrm{m^3}. Thus na3=1.25×10−5na^3=1.25\times10^{-5}, much smaller than one.

  2. The inverse squared healing length is 8πan≃8(3.14)(5×1011)=1.256×1013 m−28\pi an\simeq8(3.14)(5\times10^{11})=1.256\times10^{13}\,\mathrm{m^{-2}}.

  3. Taking the inverse square root gives ξ≃2.82×10−7 m=0.282 μm\xi\simeq2.82\times10^{-7}\,\mathrm m=0.282\,\mu\mathrm m. This is much larger than a, consistent with the separation of scales.

The result

The gas parameter is 1.25 × 10⁻⁵ and the healing length approximately 0.282 μm. Low temperature and suitable geometry are additional assumptions, not consequences of this one check.

Explore the model

Interactive lab

Interactive lab · function

A local energy curvature is an approximation

Reset and set BOTH Coefficient a to 0.5 and Position x to 1; predict slope 1. Then Reset and set BOTH a to 1 and x to 1; predict slope 2. The exact scalar parabola illustrates an energy derivative only. It has no complex condensate field, spatial kinetic term, density profile, healing length, or GP solver.

A local energy curvature is an approximationThis lab uses only y = ax². Playback traces the curve within the fixed integration interval from 0 to x; it does not estimate numerical error or change the chosen point. Current readings: y at chosen x: 0.608; Tangent slope · 2ax: 1.35; Signed area · ax³/3: 0.182. All parameters are adjustable with the labeled controls below.−2.4−1.201.22.4−9−4.504.59xy(x, ax²)
y = ax²TangentSigned integral
y at chosen x
0.608
Tangent slope · 2ax
1.35
Signed area · ax³/3
0.182
t = 0 s

This lab uses only y = ax². Playback traces the curve within the fixed integration interval from 0 to x; it does not estimate numerical error or change the chosen point. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forAn energy slope can explain a restoring tendency, but a condensate profile also needs its gradient energy and normalization.

Check your understanding

With ∫|ψ|² d³r = N₀, which nonlinear term appears in the leading GP equation?

Try it yourself

Keep positive a and mass fixed while increasing uniform condensate density by four. Find the changes in μ, c, ξ, and na³. If the initial gas parameter is 10⁻⁵, does this scaling alone invalidate diluteness?

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

Use the density powers in the uniform mean-field formulas and then test the new dimensionless parameter.

Compare with the solution

Chemical potential gn becomes four times larger, sound speed increases by two, and healing length halves. The gas parameter becomes 4×10−54\times10^{-5}, still much smaller than one. Thus this density change alone does not invalidate diluteness, although temperature, depletion corrections, three-body loss, and trap geometry still require attention. The product cξ remains ℏ/(√2m), providing a factor and scaling check.

Take it with you

  • Condensate-field normalization determines where atom-number factors enter the GP nonlinearity.
  • A small gas parameter controls weak depletion but does not eliminate measurable interactions.
  • Healing length balances kinetic and interaction energies within a declared repulsive mean-field regime.

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