Position-Space Two-Particle States

A two-particle wavefunction is the position representation of a vector in a tensor-product Hilbert space. For two distinguishable particles moving on a line,

$$\mathcal H_{12} = L^2(\mathbb R)\otimes L^2(\mathbb R) \cong L^2(\mathbb R^2).$$

The wavefunction

$$\Psi(x_1,x_2) = \langle x_1,x_2\vert\Psi\rangle$$

is a probability amplitude on configuration space. The variables $x_1$ and $x_2$ label the position outcomes for the two tensor factors. They are not automatically independent random variables; independence is a property of special states.

This page is the bridge from wave mechanics to composite-system language. The general theory of wavefunction probability densities belongs to Wavefunctions and Probability Density; the general tensor-product rule belongs to Tensor Products of Hilbert Spaces. Here the focus is how $\Psi(x_1,x_2)$ encodes product states, entanglement, partial traces, and exchange symmetry.

Two-Particle Wavefunction

For a normalized two-particle pure state,

$$\int_{-\infty}^{\infty}dx_1 \int_{-\infty}^{\infty}dx_2\, \lvert\Psi(x_1,x_2)\rvert^2 = 1.$$

The joint probability density for position measurements is

$$\rho(x_1,x_2) = \lvert\Psi(x_1,x_2)\rvert^2.$$

Thus the probability that particle 1 is found in region $A$ and particle 2 in region $B$ is

$$P(A,B) = \int_A dx_1 \int_B dx_2\, \lvert\Psi(x_1,x_2)\rvert^2.$$

The marginal density for particle 1 is

$$\rho_1(x_1) = \int_{-\infty}^{\infty}dx_2\, \lvert\Psi(x_1,x_2)\rvert^2,$$

and similarly for particle 2. These marginals describe position statistics only. They do not contain the full reduced quantum state, because phase coherence between different positions is carried by a density-matrix kernel.

In three spatial dimensions, replace $x_i$ by $\mathbf r_i$ and $dx_i$ by $d^3r_i$. The configuration space of two particles is six-dimensional, even when ordinary physical space is three-dimensional.

Product Wavefunctions

A pure product state of two distinguishable particles has the form

$$\lvert\Psi\rangle = \lvert\psi\rangle_1\otimes\lvert\phi\rangle_2.$$

In position representation this becomes

$$\Psi(x_1,x_2) = \psi(x_1)\phi(x_2).$$

If $\psi$ and $\phi$ are normalized, then $\Psi$ is normalized:

$$\int dx_1\,dx_2\, \lvert\psi(x_1)\phi(x_2)\rvert^2 = \left( \int dx_1\,\lvert\psi(x_1)\rvert^2 \right) \left( \int dx_2\,\lvert\phi(x_2)\rvert^2 \right) = 1.$$

The joint density factorizes:

$$\lvert\Psi(x_1,x_2)\rvert^2 = \lvert\psi(x_1)\rvert^2 \lvert\phi(x_2)\rvert^2.$$

This is stronger than saying the particles have no position correlation. The wavefunction itself factorizes, including phases. A state can have a factorized probability density but still fail to be a product state if the phase contains nonlocal dependence on both variables.

For example,

$$\Psi(x_1,x_2) = \psi(x_1)\phi(x_2)e^{i\alpha x_1x_2}$$

has the same position density as $\psi(x_1)\phi(x_2)$, but it is not generally a product wavefunction because the phase does not split into a sum of a function of $x_1$ and a function of $x_2$.

Entangled Wavefunctions

A two-particle pure state is entangled across the particle split when its wavefunction cannot be written as one factor depending only on $x_1$ times one factor depending only on $x_2$.

A transparent example uses orthonormal one-particle wavefunctions $u_0,u_1$ for particle 1 and $v_0,v_1$ for particle 2:

$$\Psi(x_1,x_2) = \frac{1}{\sqrt2} \left[ u_0(x_1)v_0(x_2) + u_1(x_1)v_1(x_2) \right].$$

This is the continuous-variable analogue of a two-term Schmidt state. It is entangled because two nonzero matched product terms appear in orthonormal local bases.

More generally, a normalizable bipartite wavefunction may have a Schmidt expansion

$$\Psi(x_1,x_2) = \sum_{n} \sqrt{p_n}\, u_n(x_1)v_n(x_2),$$

where

$$\int dx\,u_m^*(x)u_n(x) = \delta_{mn}, \qquad \int dx\,v_m^*(x)v_n(x) = \delta_{mn},$$

and

$$p_n\ge0, \qquad \sum_n p_n=1.$$

The state is product exactly when only one Schmidt probability is nonzero. If two or more are nonzero, the state is entangled.

Continuous variables allow infinite Schmidt rank and continuous-spectrum subtleties. For ordinary square-integrable wavefunctions, the reduced density operator should be trace class before entropies are used. Ideal distributions, such as exact EPR states, must be treated as limiting models rather than as ordinary vectors.

Reduced Density Kernel

The density operator of a pure two-particle state is

$$\rho_{12} = \lvert\Psi\rangle\langle\Psi\rvert.$$

Its position-space kernel is

$$\rho_{12}(x_1,x_2;x_1',x_2') = \Psi(x_1,x_2)\Psi^*(x_1',x_2').$$

Tracing out particle 2 gives the reduced density operator of particle 1. In position representation,

$$\rho_1(x,x') = \int_{-\infty}^{\infty}dy\, \Psi(x,y)\Psi^*(x',y).$$

The diagonal of this kernel is the marginal position density:

$$\rho_1(x,x) = \int dy\, \lvert\Psi(x,y)\rvert^2.$$

The off-diagonal entries carry coherence between different positions of particle 1. They are why the reduced density operator is more informative than the marginal density alone.

For a product wavefunction $\Psi(x,y)=\psi(x)\phi(y)$,

$$\rho_1(x,x') = \psi(x)\psi^*(x') \int dy\, \lvert\phi(y)\rvert^2 = \psi(x)\psi^*(x').$$

Particle 1 remains in a pure state. For the two-term Schmidt example,

$$\rho_1(x,x') = \frac12 u_0(x)u_0^*(x') + \frac12 u_1(x)u_1^*(x'),$$

so particle 1 is mixed and the entanglement entropy is $\log2$.

Identical-Particle Symmetry

For identical particles without spin, physical two-particle wavefunctions must have definite exchange symmetry:

$$\Psi(x_2,x_1) = +\Psi(x_1,x_2) \quad \text{for bosons},$$

and

$$\Psi(x_2,x_1) = -\Psi(x_1,x_2) \quad \text{for fermions}.$$

For particles with spin or other internal labels, exchange acts on the complete one-particle label

$$q=(x,s,\ldots),$$

not just on position. The symmetry condition is imposed on $\Psi(q_1,q_2)$.

The canonical construction is developed in Symmetric and Antisymmetric Wavefunctions. The main caution here is conceptual: the slot labels $1$ and $2$ used in first-quantized notation are not directly observable particle identities for identical particles. Exchange symmetry by itself should not be confused with operational entanglement between two addressable subsystems.

For identical particles, mode language is often cleaner. A two-boson state with one excitation in each of two orthogonal modes is naturally described in Fock space, while a first-quantized symmetrized wavefunction can make the same state look artificially like a superposition over particle labels. The physical question should specify whether the subsystems are particles, modes, spatial regions, or internal degrees of freedom.

Center-of-Mass and Relative Coordinates

For equal masses, define

$$R = \frac{x_1+x_2}{2}, \qquad r = x_1-x_2.$$

The inverse transformation is

$$x_1 = R+\frac r2, \qquad x_2 = R-\frac r2,$$

and the measure is

$$dx_1\,dx_2 = dR\,dr.$$

Many two-body Hamiltonians separate in $R$ and $r$, so solutions often have the form

$$\Psi(x_1,x_2) = \Phi(R)\chi(r).$$

This is product in the collective variables $R,r$, not necessarily product in the particle variables $x_1,x_2$. The coordinate change mixes the two tensor factors:

$$R = \frac{x_1+x_2}{2}, \qquad r = x_1-x_2.$$

Therefore separability in center-of-mass and relative coordinates is a calculational statement, not automatically a statement about entanglement across the particle split. Conversely, an interaction that is simple in relative coordinates can generate correlations between particle positions.

The safe rule is to name the tensor-product structure before using words such as product, separable, or entangled. A change from $(x_1,x_2)$ to $(R,r)$ is not a local change of basis on $\mathcal H_1\otimes\mathcal H_2$.

Common Mistakes

• Treating a two-variable wavefunction as entangled merely because it depends on two variables.
• Checking only whether the probability density factorizes and ignoring the phase.
• Confusing marginal densities with reduced density operators.
• Using ideal delta-correlated states as if they were normalizable wavefunctions.
• Treating identical-particle slot labels as observable particle identities.
• Forgetting spin or internal labels when imposing exchange symmetry.
• Mistaking center-of-mass separability for particle separability.

Cross-Links

• Continuous-Variable Systems
• EPR State Preview
• Gaussian States Preview
• Partial Trace
• Reduced Density Operators
• Subsystem Entropy
• Schmidt Decomposition
• Symmetric and Antisymmetric Wavefunctions
• Spin and Spatial Wavefunctions
• Formula Sheet
• Wavefunctions and Probability Density
• Normalization Conventions

References

• R. Shankar, Principles of Quantum Mechanics, 2nd ed., Springer, 1994.
• L. E. Ballentine, Quantum Mechanics: A Modern Development, 2nd ed., World Scientific, 2014.
• C. Cohen-Tannoudji, B. Diu, and F. Laloë, Quantum Mechanics, Wiley, 1977.
• B. C. Hall, Quantum Theory for Mathematicians, Springer, 2013.
• A. Peres, Quantum Theory: Concepts and Methods, Kluwer, 1995.
• M. A. Nielsen and I. L. Chuang, Quantum Computation and Quantum Information, Cambridge University Press, 2010.

Exercises

1. Product normalization. Let $\Psi(x_1,x_2)=\psi(x_1)\phi(x_2)$, with $\psi$ and $\phi$ normalized. Show that $\Psi$ is normalized.

Solution

Compute

$$\int dx_1\,dx_2\, \lvert\Psi(x_1,x_2)\rvert^2 = \int dx_1\,\lvert\psi(x_1)\rvert^2 \int dx_2\,\lvert\phi(x_2)\rvert^2.$$

Each factor is $1$, so the product is $1$.

2. Reduced kernel of a product state. For $\Psi(x,y)=\psi(x)\phi(y)$, compute $\rho_1(x,x')$.

Solution

By definition,

$$\rho_1(x,x') = \int dy\, \Psi(x,y)\Psi^*(x',y).$$

Substitute the product form:

$$\rho_1(x,x') = \psi(x)\psi^*(x') \int dy\,\lvert\phi(y)\rvert^2 = \psi(x)\psi^*(x').$$

3. Two-term Schmidt wavefunction. For

$$\Psi(x_1,x_2) = \frac{1}{\sqrt2} \left[ u_0(x_1)v_0(x_2) + u_1(x_1)v_1(x_2) \right],$$

with orthonormal $u$ and $v$ functions, find the nonzero eigenvalues of $\rho_1$.

Solution

The wavefunction is already in Schmidt form with probabilities $1/2$ and $1/2$. Therefore the reduced density operator has two nonzero eigenvalues:

$$\lambda_0 = \lambda_1 = \frac12.$$

The entanglement entropy is $\log2$.

4. Exchange symmetry. Let

$$\Psi_\pm(x_1,x_2) = \frac{1}{\sqrt2} \left[ u(x_1)v(x_2) \pm v(x_1)u(x_2) \right],$$

where $u$ and $v$ are orthonormal. Show that $\Psi_+$ is symmetric and $\Psi_-$ is antisymmetric.

Solution

Exchange the coordinates:

$$\Psi_\pm(x_2,x_1) = \frac{1}{\sqrt2} \left[ u(x_2)v(x_1) \pm v(x_2)u(x_1) \right].$$

Reordering scalar factors gives

$$\Psi_\pm(x_2,x_1) = \frac{1}{\sqrt2} \left[ v(x_1)u(x_2) \pm u(x_1)v(x_2) \right].$$

For the plus sign this equals $\Psi_+(x_1,x_2)$. For the minus sign it equals $-\Psi_-(x_1,x_2)$.

5. Center-of-mass caution. Explain why $\Psi(x_1,x_2)=\Phi((x_1+x_2)/2)\chi(x_1-x_2)$ is not necessarily a product state of particle 1 and particle 2.

Solution

A product state across the particle split must have the form

$$\Psi(x_1,x_2) = \psi(x_1)\phi(x_2).$$

The expression $\Phi((x_1+x_2)/2)\chi(x_1-x_2)$ factors in the collective variables $R=(x_1+x_2)/2$ and $r=x_1-x_2$. Those variables each depend on both particle coordinates. Unless the special functions $\Phi$ and $\chi$ combine so that all cross-dependence cancels, the result cannot be written as one function of $x_1$ times one function of $x_2$.