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.

Helpful background. Wavefunction probability density supplies the normalization and integration rules, while tensor products identifies the particle factors. The first calculations concern distinguishable particles; exchange symmetry is discussed separately below.

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 the phase does not generally split into a sum of one-coordinate phases. Here $\alpha$ is real and has units of inverse length squared. For smooth wavefunctions nonzero on intervals, $\partial_{x_1}\partial_{x_2}\log\Psi=i\alpha$; a product would give zero. Thus $\alpha\ne0$ produces entanglement even with statistically independent position outcomes.

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.

For every normalized square-integrable bipartite wavefunction, the reduced density operator is automatically positive and trace class, with trace one. Its nonzero spectrum is countable even though the position observable has a continuous spectrum. Trace class does not guarantee finite entropy: the sum $-\sum_n p_n\log_2p_n$ can diverge. Ideal EPR distributions are outside this normalized-state setting.

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 $1$ bit.

For a more revealing example, choose centered Gaussian $\psi$ and $\phi$ whose position variances are $\sigma_1^2$ and $\sigma_2^2$, and retain the phase $e^{i\alpha xy}$. The Gaussian characteristic function gives

$$\rho_1(x,x')=\psi(x)\psi^*(x') \exp\left[-\frac{\alpha^2\sigma_2^2}{2}(x-x')^2\right].$$

The diagonal is unchanged, but off-diagonal coherence is suppressed. Integrating $|\rho_1(x,x')|^2$ gives

$$\operatorname{Tr}\rho_1^2 =\frac{1}{\sqrt{1+4\alpha^2\sigma_1^2\sigma_2^2}}.$$

It is below one for nonzero $\alpha$. Position measurements alone miss this entanglement; momentum statistics or other coherence-sensitive measurements can reveal it.

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 writes the same state as a superposition over slots. That slot nonfactorization is a real property of the representation, but is not by itself an operational resource between separate laboratories. 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:

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

For example, let $\Phi(R)\propto e^{-R^2/(4a^2)}$ and $\chi(r)\propto e^{-r^2/(4b^2)}$, where $a,b>0$ are their probability standard deviations. Expanding the exponent in particle coordinates produces an $x_1x_2$ coefficient

$$-\frac{1}{8a^2}+\frac{1}{2b^2}.$$

The particle state is product exactly when $b=2a$. All other width ratios in this family give particle entanglement, despite perfect factorization in $R,r$.

The canonical momenta are $P=P_1+P_2$ and $p=(P_1-P_2)/2$, satisfying $[R,P]=[r,p]=i\hbar$ with crossed commutators zero. For equal masses, the kinetic energy is $P^2/(4m)+p^2/m$. These factors explain why widths in the unscaled $R,r$ coordinates cannot be compared as if they were identical particle coordinates.

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 $\mathcal H_1\otimes\mathcal H_2$.

References

• L. E. Ballentine, Quantum Mechanics: A Modern Development, 2nd ed., World Scientific (2014).
• B. C. Hall, Quantum Theory for Mathematicians, Springer (2013), doi:10.1007/978-1-4614-7116-5.
• M. A. Nielsen and I. L. Chuang, Quantum Computation and Quantum Information, 10th anniversary ed., Cambridge University Press (2010).
• R. Shankar, Principles of Quantum Mechanics, 2nd ed., Springer (1994).

Exercises

1. Entanglement hidden in a phase. For the Gaussian phase-correlated state above, take $\alpha=1/(2\sigma_1\sigma_2)$. Find the reduced purity and explain what the position density alone would show.

Solution

The purity is $1/\sqrt{1+1}=1/\sqrt2<1$, so the globally pure state is entangled. Its position density is still $|\psi(x)|^2|\phi(y)|^2$, with no position correlations. The phase changes the full reduced kernel even though its diagonal is unchanged.

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 $1$ bit.

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 $x_2$.