advanced33 min studyLesson 370 in Physics

Wigner functions and quasiprobability

Represent a continuous quantum state in phase space while keeping true marginals, normalization conventions, and negative quasiprobability distinct.

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What you’ll understand

  • Define a one-dimensional Wigner function with a declared Fourier convention.
  • Recover position and momentum marginal densities.
  • Explain why negativity does not mean negative detector probability and why positivity does not make a state classical.

The Wigner function places information from a quantum density operator on a position–momentum plane. Integrating along either full axis returns an ordinary measurement density, but values within that plane need not be positive. This representation is useful because it exposes phase-space geometry and interference while preserving the state information. Its convenience does not turn incompatible sharp position and momentum into two simultaneously sampled hidden coordinates of a particle.

Fourier transform the spatial off-diagonal kernel

The density operator has a two-position kernel. At fixed midpoint x, Fourier transforming its dependence on the separation y introduces momentum. Declare both the half-separation convention and the normalization: changing either without adjusting the prefactor causes factor-of-two mistakes. For a pure state, the kernel is a product of the wavefunction at one endpoint and its complex conjugate at the other. A mixed state's kernel enters the same linear transformation.

Hermiticity makes the Wigner function real: conjugation exchanges the two endpoints, and reversing the separation variable restores the original integral. Integrating over momentum collapses the separation to zero through the Fourier delta identity, recovering the position density. Integrating over position recovers the momentum density in the matching Fourier convention. Normalization over both axes follows, and its units are inverse action rather than inverse position alone.

W(x,p)=12πℏ∫Re−ipy/ℏ⟨x+y/2∣ρ∣x−y/2⟩ dy,∫W(x,p) dp=⟨x∣ρ∣x⟩W(x,p)=\frac{1}{2\pi\hbar}\int_{\mathbb R}e^{-ipy/\hbar}\langle x+y/2|\rho|x-y/2\rangle\,dy,\qquad \int W(x,p)\,dp=\langle x|\rho|x\rangle
x and y are lengths, p is momentum, and W has units (position×momentum)⁻¹=(J s)⁻¹. The one-dimensional full-line convention matches the earlier momentum transform. The identity holds with an appropriate integral or distributional interpretation for the state under consideration.↔ Scroll wide equations horizontally.

Marginals are probabilities; arbitrary phase-space cells are not

The position and momentum marginals are nonnegative because they are Born densities. The Wigner function itself can have negative regions while both marginals remain valid. Integrating an arbitrary tiny phase-space cell is therefore not generally the probability that an ideal joint sharp detector finds x and p there. A physically implemented approximate joint measurement uses its own positive effects and resolution, often smoothing or otherwise modifying the phase-space description.

Expectation values can be computed with phase-space symbols when the operator-to-symbol convention is handled consistently. Symmetric operator ordering is built into the Weyl–Wigner correspondence. One cannot replace every operator expression by an ordinary classical polynomial while ignoring ordering; position and momentum do not commute. For example, the symmetric combination of XP and PX has the ordinary xp symbol, whereas the ordered product XP has symbol xp+iℏ/2xp+i\hbar/2 under this convention.

∫W(x,p) dx=⟨p∣ρ∣p⟩,∬W dx dp=1,⟨XP+PX2⟩=∬xp W(x,p) dx dp\int W(x,p)\,dx=\langle p|\rho|p\rangle,\qquad \iint W\,dx\,dp=1,\qquad \left\langle\frac{XP+PX}{2}\right\rangle=\iint xp\,W(x,p)\,dx\,dp
The first two relations give true marginal density and total normalization. The final moment formula assumes the symmetric moment exists and uses Weyl ordering. XP has action units; it is not a joint sharp position–momentum probability assignment.↔ Scroll wide equations horizontally.

Gaussian positivity and non-Gaussian negativity

A normalized unchirped Gaussian has a positive Gaussian Wigner function, with reciprocal widths in x and p. Increasing its position scale squeezes the momentum direction without changing total phase-space normalization. Positive Wigner values alone do not erase uncertainty or make all quantum experiments classically equivalent. Squeezed Gaussian states can be positive in this representation while retaining quantum properties relative to an optical reference or an allowed operation set.

The first excited oscillator state instead has a negative Wigner value at the origin. A classical mixture transforms linearly into the same weighted mixture of Wigner functions, while a coherent superposition introduces cross terms before transformation. Negativity is a useful nonclassicality witness in this chosen continuous-variable representation, but it is not a universal measure of every quantum resource. Distinguish this established representation from stronger claims about computational power, Bell nonlocality, or literal negative event frequencies.

Check a Gaussian phase-space normalization

For ψ(x)=(πa2)−1/4e−x2/(2a2)\psi(x)=(\pi a^2)^{-1/4}e^{-x^2/(2a^2)}, the stated convention gives W=(πℏ)−1e−x2/a2−a2p2/ℏ2W=(\pi\hbar)^{-1}e^{-x^2/a^2-a^2p^2/\hbar^2}. Find the position marginal, W(a,0)/W(0,0), and the two variances.

  1. Integrate the momentum Gaussian using ∫e−a2p2/ℏ2dp=πℏ/a\int e^{-a^2p^2/\hbar^2}dp=\sqrt\pi\hbar/a. The marginal is (πa)−1e−x2/a2(\sqrt\pi a)^{-1}e^{-x^2/a^2}, exactly the squared wavefunction and correctly normalized over x.

  2. At (a,0) the exponent is −1, while at the origin it is zero. Their ratio is therefore e−1≈0.3679e^{-1}\approx0.3679, independent of the action-unit prefactor and of the chosen positive length a.

  3. The position density gives Var⁡X=a2/2\operatorname{Var}X=a^2/2, and the momentum marginal gives Var⁡P=ℏ2/(2a2)\operatorname{Var}P=\hbar^2/(2a^2). Their product is ℏ²/4, so the standard-deviation product is ℏ/2. Integrating both normalized marginals also checks the total Wigner normalization.

The result

The Wigner function has the correct Gaussian position marginal, relative value e⁻¹ at (a,0), and minimum uncertainty variances a²/2 and ℏ²/(2a²). Its positivity does not turn it into an unrestricted joint sharp-measurement density.

Explore the model

Interactive lab

Interactive lab · function

Read a Gaussian cut through its exponent

For the Gaussian cut p=0, write q=x/a and W/W(0,0)=exp(−q²). Reset and set both controls to Coefficient a=1 and Position x=1 in the mathematical lab; its y readout is 1, giving a separately calculated ratio e⁻¹. For a fresh comparison, Reset and set both controls to a=1 and x=2; y=4 gives e⁻⁴. The control names here belong to the quadratic lab, not to the physical Gaussian width.

Read a Gaussian cut through its exponentThis 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 forThe quadratic readout checks the exponent of a Gaussian phase-space cut. The lab plots y=a x², not W itself, and has no momentum control, density operator, Wigner negativity, or quantum evolution.

Check your understanding

What does a negative region of a valid Wigner function imply for ordinary position-measurement probabilities?

Try it yourself

For one oscillator let W0(0,0)=1/(πℏ)W_0(0,0)=1/(\pi\hbar) and W1(0,0)=−1/(πℏ)W_1(0,0)=-1/(\pi\hbar). A classical preparation mixture has weight 1/4 in |0⟩ and 3/4 in |1⟩. Find its Wigner value at the origin and explain why its position marginal remains a valid density.

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

Give me a hint

The Wigner transformation is linear in the density operator. Average the two phase-space functions with preparation weights rather than superposing wavefunctions.

Compare with the solution

The origin value is [(1/4)−(3/4)]/(πℏ)=−1/(2πℏ)[(1/4)-(3/4)]/(\pi\hbar)=-1/(2\pi\hbar). The negative value is consistent with a normalized physical density operator. Its position marginal is (1/4)∣ψ0(x)∣2+(3/4)∣ψ1(x)∣2(1/4)|\psi_0(x)|^2+(3/4)|\psi_1(x)|^2, a nonnegative normalized mixture of two Born densities. Adding wavefunctions instead would describe a different coherent state with additional cross terms, and it would not be the stipulated preparation mixture.

Take it with you

  • A declared Fourier and separation convention fixes Wigner normalization and units.
  • Full position and momentum marginals are true probabilities despite possible Wigner negativity.
  • Positive phase-space appearance does not remove quantum uncertainty or establish a general classical model.

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