Gaussian Distributions

Gaussian distributions are the most reusable continuous probability distributions in quantum mechanics.

They appear as the limiting distribution in central-limit arguments, as the probability densities of many minimum-uncertainty wave packets, as the tractable family behind continuous-variable Gaussian states, and as the finite-dimensional model for quadratic fluctuations in saddle-point and path-integral calculations.

This page is the probability-theory home for Gaussian densities and Gaussian integrals. The time evolution of Gaussian wave packets belongs to Gaussian Wave Packets, and convention-sensitive Fourier transforms are collected in Fourier Transform Tables for QM.

One-Dimensional Definition

A real random variable $X$ has a Gaussian, or normal, distribution with mean $\mu$ and variance $\sigma^2$ if its density is

$$f_X(x) = \frac{1}{\sqrt{2\pi}\,\sigma} \exp\!\left[ - \frac{(x-\mu)^2}{2\sigma^2} \right], \qquad \sigma>0.$$

This is written

$$X\sim \mathcal N(\mu,\sigma^2).$$

The parameter $\mu$ locates the center of the distribution. The parameter $\sigma$ is the standard deviation, not the variance. The variance is $\sigma^2$.

The standard normal distribution is

$$Z\sim\mathcal N(0,1), \qquad \phi(z) = \frac{1}{\sqrt{2\pi}}e^{-z^2/2}.$$

If $X\sim\mathcal N(\mu,\sigma^2)$, then

$$Z = \frac{X-\mu}{\sigma}$$

is standard normal. Conversely,

$$X=\mu+\sigma Z$$

has distribution $\mathcal N(\mu,\sigma^2)$.

Normalization

The basic Gaussian integral is

$$\int_{-\infty}^{\infty} e^{-a x^2}\,dx = \sqrt{\frac{\pi}{a}}, \qquad a>0.$$

Setting

$$a=\frac{1}{2\sigma^2}$$

gives

$$\int_{-\infty}^{\infty} \exp\!\left[ - \frac{(x-\mu)^2}{2\sigma^2} \right]dx = \sqrt{2\pi}\,\sigma.$$

The prefactor in the normal density is exactly the reciprocal of this integral.

The shift by $\mu$ does not affect the value of the integral. It only moves the center of the graph.

Mean and Variance

Using the standard variable

$$z=\frac{x-\mu}{\sigma}, \qquad dx=\sigma\,dz,$$

the expectation value becomes

$$\begin{aligned} \mathbb E[X] &= \int_{-\infty}^{\infty}x f_X(x)\,dx\\ &= \int_{-\infty}^{\infty} (\mu+\sigma z) \frac{1}{\sqrt{2\pi}}e^{-z^2/2}\,dz\\ &= \mu. \end{aligned}$$

The odd term in $z$ integrates to zero. Similarly,

$$\begin{aligned} \operatorname{Var}(X) &= \mathbb E[(X-\mu)^2]\\ &= \sigma^2 \int_{-\infty}^{\infty} z^2 \frac{1}{\sqrt{2\pi}}e^{-z^2/2}\,dz\\ &= \sigma^2. \end{aligned}$$

Thus the notation $\mathcal N(\mu,\sigma^2)$ names the mean and variance directly.

Gaussian Integrals by Completing the Square

Many quantum calculations reduce to completing the square. For $a>0$,

$$\int_{-\infty}^{\infty} e^{-a x^2+bx+c}\,dx = \sqrt{\frac{\pi}{a}}\, \exp\!\left( \frac{b^2}{4a}+c \right).$$

The reason is

$$-a x^2+bx+c = -a \left( x-\frac{b}{2a} \right)^2 + \frac{b^2}{4a} + c.$$

The shifted square gives the same normalized Gaussian integral, and the remaining terms factor out.

Two useful moment integrals are

$$\int_{-\infty}^{\infty} x\,e^{-a x^2}\,dx = 0,$$

and

$$\int_{-\infty}^{\infty} x^2e^{-a x^2}\,dx = \frac{\sqrt{\pi}}{2a^{3/2}}, \qquad a>0.$$

The first follows from oddness. The second follows by differentiating the basic integral with respect to $a$.

Higher Gaussian moments and radial Gaussian integrals are most compactly written with gamma functions; see Gamma and Beta Functions.

Products and Convolutions

Products of Gaussian functions are Gaussian functions up to an overall constant. For example,

$$\exp \left[ - \frac{(x-a)^2}{2s^2} \right] \exp \left[ - \frac{(x-b)^2}{2t^2} \right] = C \exp \left[ - \frac{(x-m)^2}{2u^2} \right],$$

where

$$\frac{1}{u^2} = \frac{1}{s^2} + \frac{1}{t^2}, \qquad m = u^2 \left( \frac{a}{s^2} + \frac{b}{t^2} \right),$$

and $C$ is independent of $x$.

This is why Gaussian priors and Gaussian likelihoods combine so cleanly in elementary Bayesian inference. It is also why Gaussian trial wavefunctions are algebraically convenient in variational estimates.

Convolutions behave just as simply. If $X$ and $Y$ are independent and

$$X\sim\mathcal N(\mu_X,\sigma_X^2), \qquad Y\sim\mathcal N(\mu_Y,\sigma_Y^2),$$

then

$$X+Y \sim \mathcal N \left( \mu_X+\mu_Y, \sigma_X^2+\sigma_Y^2 \right).$$

The cleanest proof uses characteristic functions.

Characteristic Function

For $X\sim\mathcal N(\mu,\sigma^2)$, the characteristic function is

$$\chi_X(t) = \mathbb E[e^{itX}] = \exp \left( i\mu t-\frac12\sigma^2t^2 \right).$$

The linear term in $t$ encodes the mean. The quadratic term encodes the variance. Since characteristic functions multiply for independent sums, this formula immediately proves that independent Gaussian sums remain Gaussian.

The same formula also explains why Gaussians are stable under Fourier transforms. Up to convention-dependent constants, the Fourier transform of a Gaussian is another Gaussian. In wave mechanics, that stability makes Gaussian packets the canonical bridge between position-space width and momentum-space width.

Multivariate Gaussian Distributions

Let $\mathbf X$ be an $\mathbb R^n$-valued random vector. A nondegenerate multivariate Gaussian distribution with mean vector $\boldsymbol\mu$ and covariance matrix $\Sigma$ has density

$$f_{\mathbf X}(\mathbf x) = \frac{1}{ \sqrt{(2\pi)^n\det\Sigma} } \exp \left[ - \frac12 (\mathbf x-\boldsymbol\mu)^\mathsf T \Sigma^{-1} (\mathbf x-\boldsymbol\mu) \right].$$

Here $\Sigma$ must be real, symmetric, and positive definite. Its entries are

$$\Sigma_{ij} = \operatorname{Cov}(X_i,X_j).$$

The quadratic form

$$(\mathbf x-\boldsymbol\mu)^\mathsf T \Sigma^{-1} (\mathbf x-\boldsymbol\mu)$$

measures squared distance from the mean in covariance-scaled coordinates. The level surfaces of the density are ellipsoids. Their principal axes are the eigenvectors of $\Sigma$, and their squared widths are the eigenvalues of $\Sigma$.

If $\Sigma$ is diagonal,

$$\Sigma = \operatorname{diag} (\sigma_1^2,\ldots,\sigma_n^2),$$

then the density factorizes:

$$f_{\mathbf X}(\mathbf x) = \prod_{j=1}^n \frac{1}{\sqrt{2\pi}\,\sigma_j} \exp \left[ - \frac{(x_j-\mu_j)^2}{2\sigma_j^2} \right].$$

For a multivariate Gaussian, uncorrelated components are independent. This is a special Gaussian property, not a general fact about arbitrary distributions.

Multivariate Gaussian Integral

The core multidimensional identity is

$$\int_{\mathbb R^n} \exp \left( - \frac12\mathbf x^\mathsf T A\mathbf x \right)d^n x = \frac{(2\pi)^{n/2}}{\sqrt{\det A}},$$

where $A$ is real, symmetric, and positive definite.

Diagonalize

$$A=O^\mathsf T D O,$$

with $O$ orthogonal and

$$D=\operatorname{diag}(\lambda_1,\ldots,\lambda_n), \qquad \lambda_j>0.$$

The change of variables $\mathbf y=O\mathbf x$ has unit Jacobian, so the integral factorizes:

$$\prod_{j=1}^n \int_{-\infty}^{\infty} \exp \left( - \frac12\lambda_j y_j^2 \right)dy_j = \prod_{j=1}^n \sqrt{\frac{2\pi}{\lambda_j}} = \frac{(2\pi)^{n/2}}{\sqrt{\det A}}.$$

Taking $A=\Sigma^{-1}$ gives the normalization of the multivariate Gaussian density.

Marginals and Conditionals

Marginals of a multivariate Gaussian are Gaussian. If

$$\mathbf X = \begin{pmatrix} \mathbf X_1\\ \mathbf X_2 \end{pmatrix}, \qquad \boldsymbol\mu = \begin{pmatrix} \boldsymbol\mu_1\\ \boldsymbol\mu_2 \end{pmatrix},$$

and

$$\Sigma = \begin{pmatrix} \Sigma_{11} & \Sigma_{12}\\ \Sigma_{21} & \Sigma_{22} \end{pmatrix},$$

then the marginal distribution of $\mathbf X_1$ has mean $\boldsymbol\mu_1$ and covariance $\Sigma_{11}$.

Conditionals are also Gaussian. Assuming $\Sigma_{22}$ is invertible,

$$\mathbf X_1\mid \mathbf X_2=\mathbf y$$

has mean

$$\boldsymbol\mu_{1\mid2} = \boldsymbol\mu_1 + \Sigma_{12}\Sigma_{22}^{-1} (\mathbf y-\boldsymbol\mu_2)$$

and covariance

$$\Sigma_{1\mid2} = \Sigma_{11} - \Sigma_{12}\Sigma_{22}^{-1}\Sigma_{21}.$$

This Schur-complement formula is the finite-dimensional algebra behind many Gaussian update rules. It also foreshadows covariance-matrix descriptions of continuous-variable quantum systems.

Degenerate Gaussians

If the covariance matrix is positive semidefinite but not invertible, the distribution can still be Gaussian in a broader sense, but it no longer has an ordinary density on all of $\mathbb R^n$.

Instead, its probability is supported on a lower-dimensional affine subspace. For example, if

$$Y=X$$

with $X$ Gaussian, then $(X,Y)$ lives on the line $y=x$ in the plane. Writing a two-dimensional density with $\det\Sigma=0$ is not legitimate. One must either use a lower-dimensional density on the support or treat the distribution with delta functions.

This warning is important in constrained systems and in idealized quantum calculations where some variables are perfectly correlated.

Quantum-Mechanics Uses

Gaussian probability densities enter quantum mechanics in several distinct roles.

• A normalized Gaussian wavefunction can produce a Gaussian position density, but the wavefunction itself is an amplitude. The probability density is $\lvert\psi(x)\rvert^2$.
• The harmonic-oscillator ground state has a Gaussian wavefunction. Coherent states are displaced minimum-uncertainty packets whose first and second moments remain especially simple.
• Free-particle Gaussian packets are analytically tractable because Fourier transforms and quadratic phases preserve Gaussian form.
• Continuous-variable Gaussian states are described by first moments and covariance matrices of quadrature operators, with additional quantum uncertainty constraints.
• Quadratic approximations to actions, Hamiltonians, or log-likelihoods lead to Gaussian integrals and determinants. In path integrals, Gaussian fluctuation determinants are standard only after regularization and boundary conditions are specified.

These uses are connected, but they are not identical. A Gaussian probability density, a Gaussian wavefunction, a Gaussian Wigner function, and a Gaussian functional integral carry different mathematical meanings.

Common Mistakes

• Confusing the standard deviation $\sigma$ with the variance $\sigma^2$.
• Dropping the factor of $2$ in the exponent.
• Treating the value of a continuous Gaussian density at one point as a probability.
• Forgetting that densities have units; a Gaussian in $x$ and a Gaussian in $k$ have different units and convention-dependent normalizations.
• Omitting the determinant factor in the multivariate normalization.
• Using the multivariate density formula when $\Sigma$ is singular.
• Assuming uncorrelated variables are independent outside the Gaussian family.
• Treating every bell-shaped curve as Gaussian.
• Confusing a Gaussian wavefunction with its probability density. Squaring the amplitude usually changes the width parameter.

Cross-Links

• Probability Densities
• Expectation Values
• Variance and Covariance
• Characteristic Functions
• Fourier Transform Tables for QM
• Wave Packets
• Gaussian Wave Packets
• Gaussian Variational Methods
• Gaussian States Preview
• Gamma and Beta Functions
• Asymptotic Analysis
• Path Integrals

References

• W. Feller, An Introduction to Probability Theory and Its Applications, Volume II, 2nd ed., Wiley, 1971.
• P. Billingsley, Probability and Measure, 3rd ed., Wiley, 1995.
• R. Durrett, Probability: Theory and Examples, 5th ed., Cambridge University Press, 2019.
• G. B. Folland, Fourier Analysis and Its Applications, American Mathematical Society, 1992.
• B. C. Hall, Quantum Theory for Mathematicians, Springer, 2013.
• R. Shankar, Principles of Quantum Mechanics, 2nd ed., Springer, 1994.

Exercises

1. Normalize the one-dimensional Gaussian density

$$f(x)=C\exp\!\left[-\frac{(x-\mu)^2}{2\sigma^2}\right], \qquad \sigma>0.$$

Solution

Use $z=(x-\mu)/\sigma$, so $dx=\sigma\,dz$. Then

$$1 = C\sigma \int_{-\infty}^{\infty} e^{-z^2/2}\,dz = C\sigma\sqrt{2\pi}.$$

Thus

$$C = \frac{1}{\sqrt{2\pi}\,\sigma}.$$

2. Show that the standard normal distribution has variance $1$.

Solution

For $Z\sim\mathcal N(0,1)$,

$$\operatorname{Var}(Z) = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} z^2e^{-z^2/2}\,dz.$$

Use the moment integral with $a=1/2$:

$$\int_{-\infty}^{\infty} z^2e^{-z^2/2}\,dz = \sqrt{2\pi}.$$

Therefore

$$\operatorname{Var}(Z)=1.$$

3. Let $X$ and $Y$ be independent with $X\sim\mathcal N(\mu_X,\sigma_X^2)$ and $Y\sim\mathcal N(\mu_Y,\sigma_Y^2)$. Use characteristic functions to find the distribution of $X+Y$.

Solution

The characteristic functions are

$$\chi_X(t) = \exp \left( i\mu_Xt-\frac12\sigma_X^2t^2 \right),$$

and

$$\chi_Y(t) = \exp \left( i\mu_Yt-\frac12\sigma_Y^2t^2 \right).$$

Independence gives

$$\begin{aligned} \chi_{X+Y}(t) &= \chi_X(t)\chi_Y(t)\\ &= \exp \left( i(\mu_X+\mu_Y)t - \frac12(\sigma_X^2+\sigma_Y^2)t^2 \right). \end{aligned}$$

This is the characteristic function of

$$\mathcal N \left( \mu_X+\mu_Y, \sigma_X^2+\sigma_Y^2 \right).$$

4. Prove the multivariate Gaussian integral for a symmetric positive-definite matrix $A$ by diagonalizing $A$.

Solution

Since $A$ is real symmetric, write

$$A=O^\mathsf T D O,$$

where $O$ is orthogonal and

$$D=\operatorname{diag}(\lambda_1,\ldots,\lambda_n), \qquad \lambda_j>0.$$

Let $\mathbf y=O\mathbf x$. Orthogonal transformations have unit Jacobian, so

$$\begin{aligned} \int_{\mathbb R^n} e^{-\mathbf x^\mathsf T A\mathbf x/2}\,d^n x &= \int_{\mathbb R^n} e^{-\mathbf y^\mathsf T D\mathbf y/2}\,d^n y\\ &= \prod_{j=1}^n \int_{-\infty}^{\infty} e^{-\lambda_j y_j^2/2}\,dy_j\\ &= \prod_{j=1}^n \sqrt{\frac{2\pi}{\lambda_j}}\\ &= \frac{(2\pi)^{n/2}}{\sqrt{\det A}}. \end{aligned}$$

5. Let

$$\psi(x) = N\exp\!\left[-\frac{x^2}{4s^2}\right].$$

Find $N$ for a normalized real wavefunction and identify the variance of the position density.

Solution

The probability density is

$$\lvert\psi(x)\rvert^2 = N^2\exp\!\left[-\frac{x^2}{2s^2}\right].$$

This is a centered Gaussian density with standard deviation $s$, so normalization requires

$$N^2 = \frac{1}{\sqrt{2\pi}\,s}.$$

Taking $N>0$,

$$N = (2\pi s^2)^{-1/4}.$$

The variance of the position density is

$$\operatorname{Var}(X)=s^2.$$

The amplitude has exponent denominator $4s^2$, while the probability density has exponent denominator $2s^2$.