Calculating Position and Momentum Expectation Values

Hello. In the previous lesson, you normalized wavefunctions so that

$${\int }_{\text{allowed domain}}|\psi (x){|}^{2}dx=1.$$

That condition makes $|\psi (x){|}^2$ a genuine probability density. We now use that density to predict the average results of repeated measurements. In this lesson, you will calculate the expectation values of position and momentum for normalized one-dimensional wavefunctions, learn why momentum requires an operator, and use symmetry to make many calculations nearly immediate.

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Expectation value: a weighted average, not a guaranteed result

Suppose many identically prepared particles are described by the same normalized wavefunction $\psi (x)$. Measure the position of one particle in each trial. Individual results can differ, but their average approaches the expectation value of position:

$$⟨x⟩={\int }_{\text{domain}}x|\psi (x){|}^{2}dx.$$

This is the continuous version of an ordinary weighted average. Each possible position $x$ is weighted by its probability density.

The notation $⟨x⟩$ is read “expectation value of $x$.” It does not mean that a single position measurement must return $⟨x⟩$. It is the mean of results over many measurements.

It is also important not to confuse the expectation value with the most probable position:

• The most probable position is where $|\psi (x){|}^2$ is largest.
• The expectation value is the probability-weighted mean position.

For a symmetric two-peaked density, for instance, the expectation value may lie at the central point even if the particle is least likely to be detected there.

For a normalized state, a compact general rule is

$$\boxed{⟨Q⟩={\int }_{\text{domain}}{\psi }^{\ast }(x)\hat{Q}\psi (x)dx}$$

where $\hat{Q}$ is the operator for observable $Q$. For position, the operator simply multiplies by $x$:

$$\hat{x}=x.$$

Therefore,

$$⟨x⟩={\int }_{\text{domain}}{\psi }^{\ast }(x)x\psi (x)dx={\int }_{\text{domain}}x|\psi (x){|}^{2}dx.$$

The second form makes the probability interpretation especially transparent.

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Reading the wavefunction, then the density

The wavefunction itself can be positive, negative, or complex. Its sign or phase is not a probability. Squaring its magnitude produces the nonnegative density that weights a position average.

In the middle pair, ${\Psi }_1$ changes sign at the center. The density $|{\Psi }_1{|}^2$ does not: both lobes contribute positive probability. Consequently, a position average must use $|\psi {|}^2$, never $\psi$ by itself.

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Position expectation value in a one-dimensional box

Return to the normalized box-state wavefunction from the previous lesson:

$${\psi }_n(x)=\{\begin{array}{ll}\sqrt{\frac{2}{L}}\sin \left(\frac{n\pi x}{L}\right),& 0\le x\le L,\\ 0,& \text{otherwise}.\end{array}$$

Its probability density is

$$|{\psi }_n(x){|}^2=\frac{2}{L}{\sin }^2\left(\frac{n\pi x}{L}\right).$$

To calculate mean position, integrate only within the box:

$$⟨x⟩=\frac{2}{L}{\int }_0^{L}x{\sin }^2\left(\frac{n\pi x}{L}\right)dx.$$

This integral evaluates to

$${\int }_0^{L}x{\sin }^2\left(\frac{n\pi x}{L}\right)dx=\frac{L^2}{4}.$$

Therefore,

$$\boxed{⟨x⟩=\frac{2}{L}\left(\frac{L^2}{4}\right)=\frac{L}{2}}$$

for every positive integer $n$.

The result follows even faster from symmetry. The density satisfies

$$|{\psi }_n(L-x){|}^2=|{\psi }_n(x){|}^2.$$

It is mirror-symmetric about the midpoint $L/2$, so the average must be the midpoint. This remains true for excited states with several probability-density lobes. More lobes do not shift the mean if the distribution stays symmetric.

A useful expectation-value checklist is:

1. Identify the domain where $\psi (x)$ is nonzero.
2. Confirm that the state is normalized.
3. Write ${\psi }^{\ast }\hat{Q}\psi$.
4. Apply $\hat{Q}$ to the wavefunction on the right.
5. Look for symmetry before doing lengthy algebra.
6. Check units. Since $x$ has units of length, $⟨x⟩$ must too.

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Momentum is represented by differentiation

Classically, momentum is a number such as $p=mv$. In position space, quantum momentum is represented by the differential operator

$$\boxed{{\hat{p}}_x=-i\hslash \frac{d}{dx}}$$

where $\hslash$ is the reduced Planck constant.

The expectation value of momentum is therefore

$$\boxed{⟨p_x⟩={\int }_{\text{domain}}{\psi }^{\ast }(x)\left(-i\hslash \frac{d}{dx}\right)\psi (x)dx}$$

The order is essential. Differentiate $\psi (x)$ first, multiply by $-i\hslash$, then multiply by ${\psi }^{\ast }(x)$, and finally integrate.

For the real box state,

$${\psi }_n(x)=\sqrt{\frac{2}{L}}\sin \left(\frac{n\pi x}{L}\right),$$

the derivative is

$$\frac{d{\psi }_n}{dx}=\sqrt{\frac{2}{L}}\frac{n\pi }{L}\cos \left(\frac{n\pi x}{L}\right).$$

Thus,

$$⟨p_x⟩=-i\hslash \frac{2n\pi }{L^2}{\int }_0^L\sin \left(\frac{n\pi x}{L}\right)\cos \left(\frac{n\pi x}{L}\right)dx.$$

Using the identity

$$\sin u\cos u=\frac{1}{2}\sin (2u),$$

the integral over the complete box is zero. Hence,

$$\boxed{⟨p_x⟩=0}$$

for every stationary sine state in the infinite box.

This does not say the particle has zero momentum in every measurement, or that it is motionless. It says that the average momentum is zero. Measurements can yield positive or negative momentum values whose average cancels. A later lesson on the particle in a box will make the associated nonzero energy explicit.

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Symmetry shortcuts and what they mean physically

Symmetry is not a trick; it expresses a physical balance in the state.

Centered distributions and mean position

If a probability density is even about the origin,

$$|\psi (-x){|}^2=|\psi (x){|}^2,$$

then $x|\psi (x){|}^2$ is odd. Over a symmetric domain, such as from $-a$ to $a$,

$$⟨x⟩={\int }_{-a}^{a}x|\psi (x){|}^{2}dx=0.$$

Probability on the left and right contributes equally but with opposite signed positions.

Real wavefunctions and mean momentum

For a real wavefunction that vanishes at the boundaries of its domain, or decays to zero at infinity,

$$⟨p_x⟩=-i\hslash \int \psi (x)\frac{d\psi }{dx}dx.$$

Because

$$\psi \frac{d\psi }{dx}=\frac{1}{2}\frac{d}{dx}\left({\psi }^2\right),$$

we obtain

$$⟨p_x⟩=-\frac{i\hslash }{2}{\left[{\psi }^2(x)\right]}_{\text{boundaries}}=0.$$

This is why the real sine box states have zero mean momentum. Their spatial variation produces momentum uncertainty, but no preferred direction.

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Phase can change momentum without changing position density

The probability density does not contain all the physical information in a wavefunction. In particular, a position-dependent complex phase can change momentum behavior.

Consider the normalized state

$${\psi }_{n,k}(x)=\{\begin{array}{ll}\sqrt{\frac{2}{L}}\sin \left(\frac{n\pi x}{L}\right)e^{ikx},& 0\le x\le L,\\ 0,& \text{otherwise},\end{array}$$

where $k$ is real. Since

$$|e^{ikx}{|}^2=1,$$

the probability density is unchanged:

$$|{\psi }_{n,k}(x){|}^2=\frac{2}{L}{\sin }^2\left(\frac{n\pi x}{L}\right).$$

So its mean position remains

$$⟨x⟩=\frac{L}{2}.$$

However, differentiation detects the phase. Let

$$\alpha =\frac{n\pi }{L}.$$

Then

$$\frac{d{\psi }_{n,k}}{dx}=\sqrt{\frac{2}{L}}{e}^{ikx}\left[\alpha \cos (\alpha x)+ik\sin (\alpha x)\right].$$

Applying the momentum operator gives

$${\hat{p}}_x{\psi }_{n,k}=\sqrt{\frac{2}{L}}{e}^{ikx}\left[-i\hslash \alpha \cos (\alpha x)+\hslash k\sin (\alpha x)\right].$$

The cosine term averages to zero, exactly as in the real box state. The sine term contributes $\hslash k$ times the normalization integral, which equals one. Therefore,

$$\boxed{⟨p_x⟩=\hslash k}$$

This comparison is central:

State feature	${\psi }_n(x)$	${\psi }_{n,k}(x)$
Probability density	Same	Same
Mean position	$L/2$	$L/2$
Mean momentum	$0$	$\hslash k$

A spatially varying phase can encode directed momentum even though the position probability density remains identical. This is one reason quantum mechanics uses the full complex wavefunction, not merely $|\psi {|}^2$.

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Key takeaways

For a normalized wavefunction, the expectation value of position is

$$\boxed{⟨x⟩=\int x|\psi (x){|}^{2}dx}$$

over the domain where the state exists. It predicts the average of many position measurements, not necessarily the most probable individual result.

Momentum in position space requires the operator

$$\boxed{{\hat{p}}_x=-i\hslash \frac{d}{dx}}$$

and therefore

$$\boxed{⟨p_x⟩=\int {\psi }^{\ast }(x)\left(-i\hslash \frac{d}{dx}\right)\psi (x)dx}$$

with the derivative acting on the right-hand $\psi$.

For the real stationary states of a one-dimensional box,

$$⟨x⟩=\frac{L}{2},⟨p_x⟩=0.$$

Symmetry often establishes these results before any integral is evaluated. Finally, position probability density determines position statistics, but spatial phase is crucial for momentum: two states can have the same $|\psi {|}^2$ and different $⟨p_x⟩$.

Next, you will build on this operator-based framework to identify eigenfunctions and eigenvalues of basic quantum-mechanical operators.