# 3.9: The Uncertainty Principle Redux - Estimating Uncertainties from Wavefunctions

Learning Objectives
• Expand on the introduction of Heisenberg's Uncertainty Principle by calculating the $$\Delta x$$ or $$\Delta p$$ directly from the wavefunction

As will be discussed in Section 4.6, the operators $$\hat{x}$$ and $$\hat{p}$$ are not compatible and there is no measurement that can precisely determine the coresponding observables ($$x$$ and $$p$$) simultaneously. Hence, there must be an uncertainty relation between them that specifies how uncertain we are about one quantity given a definite precision in the measurement of the other. Presumably, if one can be determined with infinite precision, then there will be an infinite uncertainty in the other. The uncertainty in a general quantity $$A$$ is

$\Delta A = \sqrt{\langle A^2 \rangle - \langle A \rangle ^2} \label{3.8.1}$

where $$\langle A^2 \rangle$$ and $$\langle A \rangle$$ are the expectation values of $$\hat{A^2}$$ and $$\hat{A}$$ operators for a specific wavefunction. Extending Equation \ref{3.8.1} to $$x$$ and $$p$$ results in the following uncertainties

$\Delta x = \sqrt{\langle x^2 \rangle - \langle x \rangle ^2} \label{3.8.2a}$

$\Delta p = \sqrt{\langle p^2 \rangle - \langle p \rangle ^2} \label{3.8.2b}$

These quantities can be expressed explicitly in terms of the (time-dependent) wavefunction $$\Psi (x, t)$$ using the fact that

\begin{align} \langle x \rangle &= \langle \Psi(t)\vert \hat{x}\vert\Psi(t)\rangle \label{3.8.3} \\[4pt] &=\int \Psi^{*}(x,t) x \Psi(x,t)\;dx \nonumber \end{align}

and

\begin{align} \langle x^2 \rangle &= \langle \Psi(t)\vert \hat{x}^2 \vert\Psi(t)\rangle \label{3.8.4} \\[4pt] &= \int \Psi^{*}(x,t) x^2 \Psi(x,t)\;dx \nonumber \end{align}

The middle terms in both Equations $$\ref{3.8.3}$$ and $$\ref{3.8.4}$$ are the integrals expressed in Dirac's Bra-ket notation. Similarly using the definition of the linear momentum operator:

$\hat{p}_x = - i \hbar \dfrac{\partial}{ \partial x}.$

So

\begin{align} \langle p \rangle &= \langle \Psi(t)\vert \hat{p} \vert\Psi(t)\rangle \label{3.8.5} \\&= \int \Psi^{*}(x,t) - i \hbar {\partial \over \partial x}\Psi(x,t)\,dx \nonumber \end{align}

and

\begin{align} \langle p^2 \rangle &= \langle \Psi(t)\vert \hat{p}^2\vert\Psi(t)\rangle \label{3.8.6} \\ &= \int \Psi ^{*} (x, t)\left(-\hbar^2{\partial^2 \over \partial x^2}\right)\Psi(x,t) \;dx \nonumber \end{align}

Time-dependent vs. time-independent wavefunction

The expectation values above are formulated with the total time-dependence wavefunction $$\psi(x,t)$$ that are functions of $$x$$ and $$t$$. However, it is easy to show that the same expectation value would be obtained if the time-independent wavefunction $$\psi(x)$$ that are functions of only $$x$$ are used. If $$V(x)$$ in $$\hat{H}$$ is time independent, then the wavefunctions are stationary and the expectation value are time-independent. You can easily confirm that by comparing the expectation values suing the general formula for a stationary wavefunction

$\Psi(x,t)=\psi(x)e^{-iEt / \hbar}$

and for $$\psi(x)$$.

The Heisenberg uncertainty principle can be quantitatively connected to the properties of a wavefunction, i.e., calculated via the expectation values outlined above:

$\Delta p \Delta x \ge \dfrac {\hbar}{2} \label {3.8.8}$

This essentially states that the greater certainty that a measurement of $$x$$ or $$p$$ can be made, the greater will be the uncertainty in the other. Hence, as $$Δp$$ approaches 0, $$Δx$$ must approach $$\infty$$, which is the case of the free particle (e.g, with $$V(x)=0$$) where the momentum of a particles can be determined precisely.

Example $$\PageIndex{1}$$: Uncertainty with a gaussian wavefunction

A particle is in a state described by the wavefunction

$\psi{(x)} =\left(\dfrac{2a}{π}\right)^{\frac{1}{4}} e^{−ax^2} \label{Ex1eq1}$

where $$a$$ is a constant and $$−∞≤ x ≤ ∞$$. Verify that the value of the product $$∆p∆x$$ is consistent with the predictions from the uncertainty principle (Equation \ref{3.8.8}).

Solution

Let's calculate the average of $$x$$:

\begin{align} \langle x\rangle &= \int ^\infty_{-\infty} \psi^{*}x \psi \,dx \nonumber \\ &= \int ^\infty_{-\infty}(2a/π)^\frac{1}{4} e^{−ax^2} x (2a/π)^\frac{1}{4} e^{−ax^2} \,dx \nonumber \\ &= \int ^\infty_{-\infty}x(2a/π)^\frac{1}{2} e^{−2ax^2}\,dx \nonumber \\ &= 0\nonumber \end{align} \nonumber

since the integrand is an odd function (an even function times an odd function is an odd function). This makes sense given that the gaussian wavefunction is symmetric around $$x=0$$.

Let's calculate the average of $$x^2$$:

\begin{align} \langle x^2\rangle &= \int ^\infty_{-\infty} \psi^{*}{x^2}\psi \,dx \nonumber \\ &= \int ^\infty_{-\infty}(2a/π)^\frac{1}{4} e^{−ax^2} (x^2) (2a/π)^\frac{1}{4} e^{−ax^2} \, dx \nonumber \\ &= \int ^\infty_{-\infty}x^2(2a/π)^\frac{1}{2} e^{−2ax^2} \, dx \nonumber \\ & =\frac{1}{4a} \nonumber \end{align} \nonumber

Let's calculate the average in $$p$$:

\begin{align} \langle p\rangle &= \int ^\infty_{-\infty} \psi^{*} p \psi \,dx \nonumber \\ &= \int ^\infty_{-\infty} \left(\dfrac{2a}{π}\right)^{\frac{1}{4}} e^{−ax^2} -i\hbar \frac{d}{dx} \left(\dfrac{2a}{π}\right)^{\frac{1}{4}} e^{−ax^2} \,dx \nonumber \\ &= \int ^\infty_{-\infty} \left(\dfrac{2a}{π}\right)^{\frac{1}{4}} e^{−ax^2} (- i \hbar ) \left(\dfrac{2a}{π}\right)^{\frac{1}{4}} e^{−ax^2} (-2ax)\,dx \nonumber \\ &=0 \nonumber \end{align} \nonumber

since the integrand is an odd function.

Let's calculate the average of $$p^2$$:

\begin{align} \langle p^2\rangle &= \int ^\infty_{-\infty} \psi^{*}{p^2} \psi dx \nonumber \\ &= -\hbar^2 \left(\dfrac{2a}{\pi}\right) ^{1/2} \int_{-\infty}^{\infty} 2a(a x^2 - 1) e^{- 2a x^2} dx \nonumber \\ &= -4\hbar^2 a^2 \left(\dfrac{2a}{\pi}\right) ^{1/2} \int_{0}^{\infty} x^2 e^{- 2a x^2} dx + 4\hbar^2 a \left(\dfrac{2a}{\pi}\right)^{1/2} \int_{0}^{\infty} e^{- 2a x^2} dx \nonumber \\ &= a\hbar^2 \nonumber \end{align} \nonumber

We use Equation \ref{3.8.1} to check on the uncertainty

\begin{align*} \Delta{x^2} &= \langle x^2 \rangle - \langle x \rangle^2 = \dfrac{1}{4a} - 0 \\[4pt] \Delta{x} &= \sqrt{\Delta{x^2}} = \dfrac{1}{2\sqrt{a}}\\[4pt] \Delta{p^2} &= \langle p^2 \rangle - \langle p \rangle^2 = a\hbar^2 - 0 \\[4pt] \Delta{p} &= \sqrt{\Delta{p^2}} = \hbar \sqrt{a} \end{align*}

Finally we have

$\Delta{p}\Delta{x} = \left(\dfrac{1}{2\sqrt{a}}\right) (\hbar \sqrt{a}) = \dfrac{\hbar}{2} \nonumber$

Not only does the Heisenburg uncertainly principle hold (Equation \ref{3.8.8}), but the equality is established for this wavefunction. This is because the Gaussian wavefunction (Equation \ref{Ex1eq1}) is special as discussed later.

Exercise $$\PageIndex{1}$$

A particle is in a state described by the ground state wavefunction of a particle in a box

$\psi = \left(\dfrac{2}{L}\right)^{1/2}\sin \left(\dfrac{\pi x}{L}\right)$

where $$L$$ is the length of the box and $$0≤ x ≤ L$$. Verify that the value of the product $$∆p∆x$$ is consistent with the predictions from the uncertainty principle (Equation \ref{3.8.8}).

The uncertainty principle is a consequence of the wave property of matter. A wave has some finite extent in space and generally is not localized at a point. Consequently there usually is significant uncertainty in the position of a quantum particle in space.