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Schrödinger Equation in Three Dimensions

The three-dimensional Schrödinger equation is the coordinate-space equation of motion for a nonrelativistic particle moving in ordinary space. It replaces the one-dimensional second derivative with the Laplacian and interprets ∣ψ(r,t)∣2\lvert\psi(\mathbf r,t)\rvert^2 as a probability density per unit volume.

For a particle of mass mm in a scalar potential V(r,t)V(\mathbf r,t),

iℏ∂ψ(r,t)∂t=[−ℏ22m∇2+V(r,t)]ψ(r,t).i\hbar\frac{\partial \psi(\mathbf r,t)}{\partial t} = \left[ -\frac{\hbar^2}{2m}\nabla^2 +V(\mathbf r,t) \right]\psi(\mathbf r,t).

Here r=(x,y,z)\mathbf r=(x,y,z) and ∇2\nabla^2 is the spatial Laplacian.

For a normalized wavefunction on a region Ω⊆R3\Omega\subseteq\mathbb R^3,

∫Ω∣ψ(r,t)∣2 d3r=1.\int_\Omega \lvert\psi(\mathbf r,t)\rvert^2\,d^3r =1.

The probability of finding the particle in a subregion A⊆ΩA\subseteq\Omega is

P(A)=∫A∣ψ(r,t)∣2 d3r.P(A) =\int_A \lvert\psi(\mathbf r,t)\rvert^2\,d^3r.

Thus ∣ψ∣2\lvert\psi\rvert^2 has units of inverse volume, and ψ\psi has units of length−3/2^{-3/2} in ordinary three-dimensional normalization.

The volume element depends on coordinates. In Cartesian coordinates,

d3r=dx dy dz.d^3r=dx\,dy\,dz.

In spherical coordinates,

d3r=r2sin⁡θ dr dθ dϕ.d^3r=r^2\sin\theta\,dr\,d\theta\,d\phi.

Forgetting the coordinate volume element changes the probability distribution.

In Cartesian coordinates,

∇2=∂2∂x2+∂2∂y2+∂2∂z2.\nabla^2 =\frac{\partial^2}{\partial x^2} +\frac{\partial^2}{\partial y^2} +\frac{\partial^2}{\partial z^2}.

The time-dependent equation becomes

iℏ∂ψ∂t=−ℏ22m(∂2ψ∂x2+∂2ψ∂y2+∂2ψ∂z2)+Vψ.i\hbar\frac{\partial\psi}{\partial t} =-\frac{\hbar^2}{2m} \left( \frac{\partial^2\psi}{\partial x^2} +\frac{\partial^2\psi}{\partial y^2} +\frac{\partial^2\psi}{\partial z^2} \right) +V\psi.

In curvilinear coordinates the Laplacian contains metric and coordinate-volume factors. For example, the spherical-coordinate Laplacian is not obtained by replacing xx with rr in the one-dimensional formula. This is one reason central potentials require their own radial equation.

If the potential is time independent, separated stationary solutions have the form

ψ(r,t)=φ(r)e−iEt/ℏ.\psi(\mathbf r,t) =\varphi(\mathbf r)e^{-iEt/\hbar}.

Substitution gives the three-dimensional time-independent Schrödinger equation:

[−ℏ22m∇2+V(r)]φ(r)=Eφ(r).\left[ -\frac{\hbar^2}{2m}\nabla^2 +V(\mathbf r) \right]\varphi(\mathbf r) =E\varphi(\mathbf r).

This is an eigenvalue problem for a differential operator plus boundary conditions. Bound systems usually impose square integrability and surface conditions; scattering systems impose incoming and outgoing behavior at large distances.

The probability density is

ρ(r,t)=∣ψ(r,t)∣2.\rho(\mathbf r,t)=\lvert\psi(\mathbf r,t)\rvert^2.

For a real scalar potential and suitable boundary conditions, it obeys the continuity equation

∂ρ∂t+∇⋅j=0,\frac{\partial\rho}{\partial t} +\nabla\cdot\mathbf j=0,

where the probability-current density is

j=ℏ2mi(ψ∗∇ψ−ψ∇ψ∗).\mathbf j =\frac{\hbar}{2mi} \left( \psi^*\nabla\psi -\psi\nabla\psi^* \right).

This is the three-dimensional version of the current used in one-dimensional scattering. Probability is conserved by flowing through space or across boundaries, not by being arbitrarily created or destroyed.

A three-dimensional wave-mechanics problem is not specified by the differential equation alone. One must also specify the region and boundary behavior.

For a hard-wall box, the wavefunction vanishes on the boundary:

ψ=0on the wall.\psi=0 \quad\text{on the wall}.

At a finite potential interface, ψ\psi is continuous. For equal mass and no singular surface interaction, the normal derivative is also continuous:

∂ψ∂n∣left=∂ψ∂n∣right.\frac{\partial\psi}{\partial n}\bigg\rvert_{\mathrm{left}} = \frac{\partial\psi}{\partial n}\bigg\rvert_{\mathrm{right}}.

For bound states on all of space, square integrability requires decay at infinity:

∫R3∣ψ(r)∣2 d3r<∞.\int_{\mathbb R^3} \lvert\psi(\mathbf r)\rvert^2\,d^3r \lt \infty.

For scattering states, one instead uses asymptotic incoming and outgoing wave conditions. The boundary conditions encode the physics just as much as the Hamiltonian does.

Some three-dimensional problems reduce to simpler ordinary differential equations by separation of variables. The method works when the potential, coordinate system, and boundary conditions are compatible.

Examples:

  • a rectangular box separates in Cartesian coordinates;
  • a central potential separates into radial and angular equations in spherical coordinates;
  • cylindrical symmetry can separate axial, radial, and angular variables.

Separation is a powerful method, not a universal guarantee. The first wave-mechanics treatment is Separation of Variables, with mathematical background in the Mathematical Toolkit.

  • Forgetting that ∣ψ∣2\lvert\psi\rvert^2 is a probability density per unit volume.
  • Omitting the volume element in spherical or cylindrical coordinates.
  • Treating ∇2\nabla^2 as a one-dimensional second derivative in radial problems.
  • Assuming every three-dimensional potential separates.
  • Specifying a Hamiltonian without specifying the spatial region and boundary conditions.
  • Confusing evanescent decay in one coordinate with full square integrability in three dimensions.
  • D. J. Griffiths and D. F. Schroeter, Introduction to Quantum Mechanics, 3rd ed., Cambridge University Press, 2018.
  • R. Shankar, Principles of Quantum Mechanics, 2nd ed., Springer, 1994.
  • C. Cohen-Tannoudji, B. Diu, and F. Laloë, Quantum Mechanics, Wiley, 1977.
  • E. Merzbacher, Quantum Mechanics, 3rd ed., Wiley, 1998.
  1. What are the units of a normalized three-dimensional wavefunction?
Solution

Normalization requires

∫∣ψ(r)∣2 d3r=1.\int \lvert\psi(\mathbf r)\rvert^2\,d^3r=1.

Since d3rd^3r has units of length3^3, ∣ψ∣2\lvert\psi\rvert^2 has units of length−3^{-3}. Therefore ψ\psi has units of length−3/2^{-3/2}.

  1. Show that if ψ(x,y,z)=X(x)Y(y)Z(z)\psi(x,y,z)=X(x)Y(y)Z(z) is normalized in a rectangular box and each factor is normalized on its interval, then the full wavefunction is normalized.
Solution

The norm factors:

∫∣X(x)Y(y)Z(z)∣2 dx dy dz=(∫∣X(x)∣2 dx)(∫∣Y(y)∣2 dy)(∫∣Z(z)∣2 dz).\int\lvert X(x)Y(y)Z(z)\rvert^2\,dx\,dy\,dz = \left(\int\lvert X(x)\rvert^2\,dx\right) \left(\int\lvert Y(y)\rvert^2\,dy\right) \left(\int\lvert Z(z)\rvert^2\,dz\right).

If each factor is normalized, each parenthesis equals 11, so the product equals 11.

  1. Why is the spherical volume element essential for radial probabilities?
Solution

In spherical coordinates,

d3r=r2sin⁡θ dr dθ dϕ.d^3r=r^2\sin\theta\,dr\,d\theta\,d\phi.

The factor r2sin⁡θr^2\sin\theta accounts for how much physical volume corresponds to a coordinate interval. Omitting it would give the wrong probability assigned to spherical shells and angular regions.