Momentum-Space Wavefunction

Ask "where is the particle?" and quantum mechanics hands you $\Psi(x)$ : square it, and you have the odds for every position. But a diffraction experiment never asks where. A far-field detector, a time-of-flight spectrometer, a recoil-ion microscope — all of them ask how fast, and which way, and the honest answer to that question is a second wavefunction. The momentum-space wavefunction $\Phi(p)$ describes the very same state, with Born's rule intact: $|\Phi(p)|^2\,\mathrm{d}p$ is the probability that a momentum measurement lands in $[p, p+\mathrm{d}p]$ . Momentum Operator promised you this article. There you learned that $\hat p = -i\hbar\,\partial_x$ has plane-wave eigenfunctions, and you saw a preview of the "second home" they open up. Here the preview is cashed in full: the transform pair connecting $\Psi$ and $\Phi$ , Plancherel's guarantee that probability totals to one in either language, the famous Gaussian transform pair worked to the last constant — the calculation that saturates $\sigma_x\sigma_p = \hbar/2$ — and the operator dictionary that makes $\hat p$ trivial and $\hat x$ a derivative when you switch sides. The payoff is a habit of mind as much as a technique: one state, two complete descriptions, the way a melody is one object whether written as notes in time or as a frequency spectrum. Learn to hop between them and half of quantum mechanics gets easier — free evolution turns diagonal, uncertainty becomes bookkeeping, and the experiments that measure momentum finally have a wavefunction that speaks their language.

Key takeaways

The momentum-space wavefunction $\Phi(p)$ is the Fourier transform of $\Psi(x)$ ; the inverse transform recovers $\Psi$ exactly. Same state, two languages.
Born's rule applies unchanged: $|\Phi(p)|^2\,\mathrm{d}p$ is the probability of finding momentum in $[p, p+\mathrm{d}p]$ , just as $|\Psi(x)|^2\,\mathrm{d}x$ works for position.
Plancherel's theorem guarantees $\int|\Phi(p)|^2\,\mathrm{d}p = \int|\Psi(x)|^2\,\mathrm{d}x$ for the chosen unitary Fourier-transform convention: normalize once, and both representations agree.
A Gaussian transforms into a Gaussian with reciprocal width — the unique shape that saturates $\sigma_x\sigma_p = \hbar/2$ .
In momentum space the operators swap costumes: $\hat p$ becomes plain multiplication by $p$ , while $\hat x \to +i\hbar\,\partial/\partial p$ .
Expectation values computed in either representation agree number for number — the physics is representation-free.
Key idea: $\Phi(p)$ and $\Psi(x)$ are Fourier-transform partners — two coordinate representations of one quantum state — and $|\Phi(p)|^2$ is the probability density for momentum exactly as $|\Psi(x)|^2$ is for position.

The core insight

$\Phi(p)$ and $\Psi(x)$ are Fourier-transform partners — two coordinate representations of one Hilbert-space vector — and $|\Phi(p)|^2$ is the probability density for momentum exactly as $|\Psi(x)|^2$ is for position.

Where this lives today

Momentum distributions are measured, not just computed. In COLTRIMS reaction microscopes, an ionized atom's charged fragments are steered by weak fields onto time- and position-sensitive detectors, and the machine reconstructs each particle's full momentum vector, event by event. The accumulated momentum distribution samples $|\Phi(\mathbf p)|^2$ and can be used to study electron dynamics such as tunnelling and rescattering within an optical cycle. Electron-momentum spectroscopy can similarly probe bound electrons: in a coincidence knockout experiment, detecting both outgoing electrons allows their momenta to be reconstructed, providing information about the initial bound-state momentum distribution. Measured momentum profiles of atomic orbitals are compared directly against computed $|\Phi(p)|^2$ — a wavefunction-level test of quantum chemistry, conducted in the language this article just taught you.

The physical picture

Think of $\Psi(x)$ as a piece of music written out note by note in time, and $\Phi(p)$ as the same music written as a frequency spectrum. Neither is "the real song" — each contains the whole of it, and either can be reconstructed perfectly from the other. What the spectrum makes obvious (which pitches are present) the score hides, and vice versa. Quantum states work the same way: $\Psi(x)$ displays location at a glance, $\Phi(p)$ displays motion, and the Fourier transform is the translation between them.

Figure

One state, two languages — squeeze one side and the other bulges

The picture to hold on to: one arrow, two coordinate grids. A vector does not change when you switch from Cartesian to polar coordinates; only its components do. $\Psi(x)$ and $\Phi(p)$ are the components of one state vector along two basis families — position spikes and plane waves. Squeeze the position components into a spike and the momentum components must spread out, because a sharp position state requires a broad range of momenta: the bandwidth theorem of Fourier analysis, with $\hbar$ setting the scale. There is also a direct experimental reading. Let a localized state fly freely for a long time and each momentum component travels a distance proportional to its momentum: fast components end up farther away, slow components nearby. After a sufficiently long flight, the position distribution can therefore provide a magnified image of the initial momentum distribution — the basic principle behind time-of-flight measurements of $|\Phi(p)|^2$ .

Figure

Free flight turns momentum into position — how experiments photograph Φ(p)

The transform pair

The plane waves are eigenfunctions of the momentum operator — their role was established in Momentum Operator — and they form a complete, delta-normalized basis. Expanding a state over that basis, with de Broglie's $p = \hbar k$ setting the scale, gives the coefficient function its own name:

\Phi(p) = \frac{1}{\sqrt{2\pi\hbar}}\int_{-\infty}^{\infty} \Psi(x)\,e^{-ipx/\hbar}\,\mathrm{d}x \qquad (1)

where: $\Phi(p)$ — the amplitude for momentum $p$ , obtained by projecting $\Psi$ onto the plane wave of that momentum; $\hbar = 1.054571817\times10^{-34}\ \mathrm{J\,s}$ ; the prefactor $1/\sqrt{2\pi\hbar}$ is chosen symmetrically so that the pair below runs in both directions without stray constants. Read it as a questionnaire: for each $p$ , the integral asks how much of that plane wave is present in $\Psi$ . The transform inverts exactly — nothing is lost in translation:

\Psi(x) = \frac{1}{\sqrt{2\pi\hbar}}\int_{-\infty}^{\infty} \Phi(p)\,e^{+ipx/\hbar}\,\mathrm{d}p \qquad (2)

where: the exponent's sign flips between (1) and (2) — a convention, but load-bearing once chosen (it fixes the sign in the operator dictionary below). The proof that (2) undoes (1) is the delta-function collapse of a double integral, worked in full in wavenumber language in Wave Packets and Their Construction; this article inherits the result. One bookkeeping bridge, because you will meet both conventions. Packet-building articles often work with the wavenumber amplitude $\phi(k)$ ; the two are one object in different units:

\phi(k) = \sqrt{\hbar}\;\Phi(\hbar k) \qquad (3)

where: the $\sqrt{\hbar}$ is the Jacobian of $p = \hbar k$ split symmetrically, so that $\int|\phi(k)|^2\,\mathrm{d}k = \int|\Phi(p)|^2\,\mathrm{d}p$ — both densities carry total probability 1 in their own variable. Forgetting this factor is the most common way to lose an $\hbar$ in a momentum-space calculation.

Born's rule in the second language

Why call $\Phi$ a wavefunction at all? Because the statistical interpretation of Born Statistical Interpretation carries over verbatim. The probability that a momentum measurement returns a value in $[p, p+\mathrm{d}p]$ is

\mathrm{d}P = |\Phi(p)|^2\,\mathrm{d}p \qquad (4)

where: $|\Phi(p)|^2$ — the momentum probability density, with SI units of $1/(\mathrm{kg\,m/s})$ , exactly parallel to $|\Psi(x)|^2$ carrying units of $1/\mathrm{m}$ . A narrow $|\Phi(p)|^2$ distribution corresponds to a well-collimated, nearly monochromatic particle beam; a broad distribution means that the state contains a wider range of possible momenta. For this to be consistent, total probability must not depend on which language you compute it in. It does not — that is Plancherel's theorem:

\int_{-\infty}^{\infty} |\Phi(p)|^2\,\mathrm{d}p \;=\; \int_{-\infty}^{\infty} |\Psi(x)|^2\,\mathrm{d}x \;=\; 1 \qquad (5)

where: the first equality is the substance — the Fourier transform preserves the norm, i.e. it is a unitary change of basis. The proof is the delta-collapse owned by Wave Packets and Their Construction in $k$ -units; equation (5) restates that result in physical momentum units via the bridge (3). Normalize once, in either language, and you are normalized forever in both — time evolution preserves the norm too, by Conservation of Normalization. Which language should you compute in? A rule of thumb worth a table:

task	better language	why
position statistics, local potentials $V(x)$	$\Psi(x)$	$V$ acts by multiplication there
momentum statistics, collimation, diffraction	$\Phi(p)$	the answer is $\vert\Phi\vert^2$ read directly
free-particle time evolution	$\Phi(p)$	each amplitude just rotates in phase — no mixing (packet story: Free Particle and Wave Packets, Wave Packet Spreading and Dispersion)
uniform force $V = -Fx$	$\Phi(p)$	$\hat x \to i\hbar\,\partial_p$ makes it first-order — momentum-representation craft, developed in the Formalism chapter

The Gaussian pair, worked to the end

This is the calculation the whole chapter has been pointing at: take the minimum-uncertainty shape, transform it honestly, and watch the reciprocity come out exact.

Derivation AThe Gaussian pair worked to the end

We are computing the momentum-space wavefunction of the normalized Gaussian $\Psi(x)=(2\pi a^2)^{-1/4}e^{-x^2/4a^2}$ , for which the position spread is $\sigma_x=a$ , and extracting the exact uncertainty product. The strategy is the classic complete-the-square move: fold the plane-wave phase into the exponent. The remaining integral is then a standard Gaussian integral. Watch two things: the complex shift of the integration variable (harmless, by contour deformation, but it deserves its one-line justification), and the way the width of the result comes out inverted — that inversion is the uncertainty principle being born.

Step 1. Set up the transform. Insert $\Psi$ into definition (1):

\Phi(p) = \frac{1}{\sqrt{2\pi\hbar}}\left(\frac{1}{2\pi a^2}\right)^{1/4}\int_{-\infty}^{\infty} e^{-x^2/4a^2}\,e^{-ipx/\hbar}\,\mathrm{d}x

Why: the momentum wavefunction is a projection of the state onto each plane wave in turn; for a Gaussian this projection can be done in closed form.

Step 2. Complete the square in the exponent:

-\frac{x^2}{4a^2} - \frac{ipx}{\hbar} = -\frac{1}{4a^2}\left(x + \frac{2ia^2 p}{\hbar}\right)^2 - \frac{a^2 p^2}{\hbar^2}

Why: this is the standard Gaussian-integral move — all the $p$ -dependence is now in the last term, outside the squared bracket.

Step 3. Do the Gaussian integral. Shifting $x \to x - 2ia^2p/\hbar$ moves the integration path off the real axis, but the integrand is analytic and dies at infinity, so the contour slides back with no change: the integral is $\sqrt{4\pi a^2} = 2a\sqrt{\pi}$ regardless of the shift. Collecting all constants:

\Phi(p) = \left(\frac{2a^2}{\pi\hbar^2}\right)^{1/4} e^{-a^2 p^2/\hbar^2}

Why: a Gaussian transforms into a Gaussian — it is the shape that is its own Fourier partner, which is precisely what lets it make both spreads small at once.

Step 4. Read off the widths and multiply. Comparing with the standard forms $e^{-x^2/2\sigma_x^2}$ and $e^{-p^2/2\sigma_p^2}$ :

\sigma_x = a, \qquad \sigma_p = \frac{\hbar}{2a}, \qquad \sigma_x\,\sigma_p = \frac{\hbar}{2}\ \text{exactly}

Why: the widths are reciprocal — squeezing the position-space Gaussian broadens the momentum-space Gaussian in inverse proportion. The general inequality $\sigma_x\sigma_p \geq \hbar/2$ is stated in Standard Deviation and Quantum Uncertainty and proved in full in the Formalism chapter's Generalized Uncertainty Principle; this Gaussian saturates the bound. *As a check: the result $\Phi(p) = (2a^2/\pi\hbar^2)^{1/4}e^{-a^2p^2/\hbar^2}$ is correctly normalised — $\int|\Phi|^2\mathrm{d}p = 1$ — and returns $\sigma_x = a$ , $\sigma_p = \hbar/2a$ , product $\hbar/2$ , confirming the saturation.*

The operator dictionary

Once both languages exist, every operator must have a word in each. The momentum operator's entry is the whole point of the momentum basis: acting on the expansion (2), $\hat p = -i\hbar\,\partial_x$ differentiates only the kernel, pulling down exactly $p$ . So in momentum space $\hat p$ is bare multiplication. The surprise is the mirror entry for position:

\hat p \;\longrightarrow\; p \times, \qquad \hat x \;\longrightarrow\; +\,i\hbar\,\frac{\partial}{\partial p} \qquad (6)

where: the $+$ sign on $\hat x$ is the mirror image of $\hat p = -i\hbar\,\partial_x$ , and it is fixed by the sign convention of the transform pair (1)–(2), not by taste. The two representations are perfect duals: whichever variable you represent by multiplication, its partner becomes $\pm i\hbar$ times a derivative.

Go deeper · Derivation B · 1 minWhy position becomes a derivative in momentum space

Derivation BWhy position becomes a derivative in momentum space

We are deriving the momentum-space representation of $\hat x$ , sign included — the classic place to slip. The strategy: write the position-space matrix element of $\hat x$ , trade the factor $x$ for a derivative of the plane-wave kernel with respect to $p$ , and pull the derivative outside the integral. The one subtlety is which exponent appears: the forward transform (1) carries $e^{-ipx/\hbar}$ , and differentiating that kernel with respect to $p$ produces $-ix/\hbar$ times the kernel — so recovering a bare $x$ costs a factor $+i\hbar$ .

Step 1. Act with $\hat x$ (multiplication by $x$ ) inside the forward transform:

\big(\widehat{x\,\Psi}\big)(p) = \frac{1}{\sqrt{2\pi\hbar}}\int x\,\Psi(x)\,e^{-ipx/\hbar}\,\mathrm{d}x

Why: this is by definition the momentum-space wavefunction of the state $x\Psi(x)$ — the object we want to express through $\Phi$ alone.

Step 2. Trade $x$ for a $p$ -derivative of the kernel:

x\,e^{-ipx/\hbar} = i\hbar\,\frac{\partial}{\partial p}\,e^{-ipx/\hbar}

Why: differentiating $e^{-ipx/\hbar}$ with respect to $p$ brings down $-ix/\hbar$ ; multiplying by $i\hbar$ isolates the bare $x$ . Check the sign here every time — with the opposite kernel convention the sign flips.

Step 3. Pull the derivative outside the $x$ -integral (legitimate because the integral converges uniformly for physical states):

\big(\widehat{x\,\Psi}\big)(p) = i\hbar\,\frac{\partial}{\partial p}\left[\frac{1}{\sqrt{2\pi\hbar}}\int \Psi(x)\,e^{-ipx/\hbar}\,\mathrm{d}x\right] = i\hbar\,\frac{\partial \Phi}{\partial p}

Why: the bracket is nothing but the definition of $\Phi(p)$ — so the operator $\hat x$ has become $+i\hbar\,\partial_p$ acting on the momentum-space wavefunction. *As a check: for a Gaussian centred at $x_0$ , the expectation value $\int\Phi^*(i\hbar\,\partial_p)\Phi\,\mathrm{d}p = x_0$ exactly, while the wrong sign $-i\hbar\,\partial_p$ returns $-x_0$ — confirming that the sign is load-bearing.*

The dictionary would be useless if the two languages could disagree about a measurable number. They cannot:

Go deeper · Derivation C · 1 minSame numbers from either language

Derivation CSame numbers from either language

We are proving that expectation values are representation-independent: the momentum moments computed in position space, $\int\Psi^*(-i\hbar\,\partial_x)^n\Psi\,\mathrm{d}x$ , equal the weighted averages $\int p^n|\Phi(p)|^2\mathrm{d}p$ . The strategy: expand each $\Psi$ through the inverse transform (2) and let plane-wave orthogonality — the delta identity established in Wave Packets and Their Construction — collapse the double momentum integral. This theorem makes "one state, two languages" more than a slogan: every prediction comes out identical.

Step 1. Write the position-space matrix element with both factors expanded:

\int \Psi^*(-i\hbar\,\partial_x)^n\,\Psi\,\mathrm{d}x = \frac{1}{2\pi\hbar}\int\!\!\int\!\!\int \Phi^*(p')\,e^{-ip'x/\hbar}\,(-i\hbar\,\partial_x)^n\,\Phi(p)\,e^{+ipx/\hbar}\,\mathrm{d}p'\,\mathrm{d}p\,\mathrm{d}x

Why: each derivative acts only on the plane-wave kernel, and each application pulls down one factor of $p$ : $(-i\hbar\,\partial_x)^n e^{ipx/\hbar} = p^n e^{ipx/\hbar}$ — the eigenvalue credential of the plane waves doing its job.

Step 2. Do the $x$ -integral first. The remaining integral is the standard Fourier representation of the Dirac delta function:

\frac{1}{2\pi\hbar}\int e^{i(p-p')x/\hbar}\,\mathrm{d}x = \delta(p - p')

Why: this is the plane-wave orthogonality relation — the continuous analogue of an orthonormal basis, quoted from its chapter owner rather than re-derived.

Step 3. Collapse the double integral with the delta:

\int\!\!\int \Phi^*(p')\,p^n\,\Phi(p)\,\delta(p-p')\,\mathrm{d}p'\,\mathrm{d}p = \int p^n\,|\Phi(p)|^2\,\mathrm{d}p

Why: the delta forces $p' = p$ , and the matrix element becomes a plain average of $p^n$ over the momentum probability density — Born's rule (4) in action. *As a check: for a Gaussian boosted to momentum $p_0$ , both the position-space and momentum-space routes give $\langle p\rangle = p_0$ and $\langle p^2\rangle = p_0^2 + \hbar^2/4a^2$ exactly; the mirror calculation of $\langle x\rangle$ on a shifted Gaussian likewise agrees via $+i\hbar\,\partial_p$ .*

Limiting cases

case	result	meaning
position spike $\Psi = \delta(x-x_0)$	$\vert\Phi(p)\vert^2 = 1/2\pi\hbar$ , flat	perfect localization means total momentum ignorance — every momentum equally likely
momentum eigenstate $\Phi = \delta(p-p_0)$	$\Psi = e^{ip_0x/\hbar}/\sqrt{2\pi\hbar}$	one plane wave: definite momentum, no location; delta-normalized ideal, not a physical state (the non-normalizability verdict is Free Particle and Plane Waves')
Gaussian of width $a$	$\sigma_x\sigma_p = \hbar/2$ exactly	the unique minimum-uncertainty shape; both spreads finite, product pinned
width $a \to \infty$	$\Phi \to$ spike at $p = 0$	an ever-flatter $\Psi$ approaches a momentum eigenstate from inside Hilbert space
free flight, any state	$\vert\Phi(p,t)\vert^2 = \vert\Phi(p,0)\vert^2$	with no force, each momentum amplitude only rotates in phase; the packet-level consequences are owned by Wave Packet Spreading and Dispersion

Worked examples

Example 1 — reading a momentum-space wavefunction (confidence builder). A beam source produces electrons in a Gaussian momentum state centred at $p_0 = 5.0\times10^{-24}\ \mathrm{kg\,m/s}$ with spread $\sigma_p = 2.0\times10^{-25}\ \mathrm{kg\,m/s}$ . Find the probability that a momentum measurement lands within $\pm\sigma_p$ of $p_0$ , the mean kinetic energy, and the tightest position spread this beam could have.

Step 1. The probability is a direct Born-rule read of the Gaussian density: $P(|p - p_0| \leq \sigma_p) = \operatorname{erf}(1/\sqrt{2}) = \mathbf{0.6827}$ — the familiar 68.3% one-sigma rule, applied to momentum.

Step 2. The mean kinetic energy averages $p^2/2m$ over $|\Phi|^2$ , and picks up the spread: $\langle K\rangle = (p_0^2 + \sigma_p^2)/2m_e = \mathbf{85.78\ eV}$ . The naive $p_0^2/2m_e = 85.65\ \mathrm{eV}$ undershoots by $0.137\ \mathrm{eV}$ — the momentum spread carries real kinetic energy.

Step 3. The uncertainty bound sets the best possible localization: $\sigma_x \geq \hbar/2\sigma_p = \mathbf{0.264\ nm}$ , with equality only for the exact Gaussian pair. Any other shape with this $\sigma_p$ is more spread out in position.

Example 2 — the box state translated into momentum language (exam level). An electron sits in the ground state of an infinite square well of width $L = 1.00\ \mathrm{nm}$ (spectrum and eigenfunctions quoted from Infinite Square Well). Find $|\Phi_1(p)|^2$ , verify it against the well's known momentum spread, and compute the probability that $|p| \leq \pi\hbar/L$ .

Step 1. Transform $\psi_1(x) = \sqrt{2/L}\,\sin(\pi x/L)$ on $[0, L]$ through definition (1). The integral is elementary (two exponentials), and with $u \equiv pL/\hbar$ the density comes out as

|\Phi_1(p)|^2 = \frac{4\pi L}{\hbar}\,\frac{\cos^2(u/2)}{(\pi^2 - u^2)^2}

a single central hump. At $u=\pm\pi$ — corresponding to the classical momentum magnitude $\pi\hbar/L$ — the apparent pole is cancelled by the zero of the cosine, leaving the finite value $L/(4\pi\hbar)$ . The ground state's momentum distribution is one smooth hump centred at zero, not two bouncing spikes.

Step 2. Moments check. Numerically $\int|\Phi_1(p)|^2\,\mathrm{d}p = 1.000000$ and $\langle p^2\rangle = \pi^2\hbar^2/L^2 = 2mE_1$ , exactly as required by $E_1 = 0.376\ \mathrm{eV}$ . Since $\langle p\rangle = 0$ , $\sigma_p = \pi\hbar/L = \mathbf{3.31\times10^{-25}\ kg\,m/s}$ — the same result obtained from the position-space treatment in Momentum Operator and Infinite Square Well. Three routes, one answer.

Step 3. Born's rule in $p$ : integrating the density over the window $|p| \leq \pi\hbar/L = \sigma_p$ gives $P = \mathbf{0.698}$ — eerily close to a Gaussian's 68.3% one-sigma content, though this distribution is no Gaussian. Widening to $|p| \leq 3\pi\hbar/L$ (the first true zeros of $\Phi_1$ ) captures 99.5%: the well's ground state is, in momentum language, essentially one soft lobe.

Figure

The box ground state in momentum language — one soft hump, finite at the classical momenta

Origins · 1 minNorbert Wiener and how this was discovered

Norbert Wiener

1894–1964 · MIT, Cambridge Massachusetts

Months before Schrödinger's wave mechanics appeared, Max Born spent the winter of 1925–26 at MIT and teamed up with a 31-year-old prodigy who had entered college at eleven: Norbert Wiener. Their joint paper made the first systematic use of operators acting on function spaces to express the new quantum rules — the mathematical move this article lives on, with $\hat x$ and $\hat p$ trading costumes between representations. Wiener's deeper gift to the subject was making the Fourier side honest: his generalized harmonic analysis (1930) and the Paley–Wiener theorems turned "a function and its transform cannot both be concentrated" from folklore into theorems, tying a function's decay to its transform's smoothness — the very reciprocity that becomes $\sigma_x\sigma_p \geq \hbar/2$ here. The famously absent-minded founder of cybernetics never much cared that his hard analysis had quietly become quantum mechanics' bookkeeping; he was already three fields away.

History

1807

— Fourier's heat memoir: arbitrary functions built from sinusoids; this article's mathematics begins.

1924

— de Broglie ties momentum to wavelength,

p = h/\lambda

: momentum becomes a wave question.

1926

— Born and Wiener's operator paper; months later Born's rule makes squared amplitudes probabilities — in either representation.

1927

— Dirac and Jordan's transformation theory:

\Psi(x)

and

\Phi(p)

become two coordinate systems on one state space.

1933

— Paley and Wiener publish the decay–smoothness duality; the uncertainty principle's Fourier skeleton is theorem-grade.

Going further

References. Griffiths, Introduction to Quantum Mechanics, 3rd ed., §3.4 (momentum space) and §1.5 · Shankar, Principles of Quantum Mechanics, 2nd ed., Ch. 4 & 7 · Sakurai & Napolitano, Modern Quantum Mechanics, 2nd ed., §1.7 · Cohen-Tannoudji, Diu & Laloë, Quantum Mechanics, Vol. I, Ch. II & Complement E_II · Townsend, A Modern Approach to Quantum Mechanics, 2nd ed., Ch. 6 The inverse-transform identity rests on $\frac{1}{2\pi\hbar}\int e^{ip(x-x')/\hbar}\mathrm{d}p=\delta(x-x')$ ; the Gaussian transform follows by completing the square; $\sigma_x\sigma_p=\hbar/2$ for the Gaussian and $\geq\hbar/2$ in general. $\hbar=1.054571817\times10^{-34}\,\mathrm{J\,s}$ (CODATA). The box-state momentum density satisfies normalization $1.000000$ and $\langle p^2\rangle = 2mE_1$ numerically.

Builds on: The Wavefunction · Born Statistical Interpretation · Probability Density and Normalization · de Broglie Matter Waves

Leads to: Time-Dependent Schrödinger Equation · Stationary States and Separation of Variables · Wave–Particle Duality

Is Φ(p) just Ψ(x) with p substituted for x?

No — and this is the most common error. $\Phi(p)$ comes from the integral transform (1): its value at one momentum depends on $\Psi$ at all positions. For the box ground state, $\psi_1$ is a sine but $\Phi_1$ is a smooth hump. If you catch yourself writing $\Psi(p)$ by letter-swapping, stop and transform.

Why does the prefactor have ħ in it — my math course wrote the Fourier transform with just 2π?

Because the conjugate variable here is momentum $p = \hbar k$ , not wavenumber. Changing variables produces a Jacobian $\hbar$ , split symmetrically as $\sqrt{\hbar}$ each way — the bridge $\phi(k) = \sqrt{\hbar}\,\Phi(\hbar k)$ . Both conventions carry the same physics; the $\hbar$ -form makes $|\Phi(p)|^2$ a probability density per unit momentum, what a spectrometer calibrated in kg·m/s reads.

Which is more fundamental, Ψ(x) or Φ(p)?

Neither. The state is a basis-free vector; $\Psi$ and $\Phi$ are its components along two bases (position spikes, plane waves). Every measurable number comes out identical either way — exactly what the third derivation proves. Convenience decides: local potentials favour $x$ ; free flight and momentum questions favour $p$ .

Can I get the average position from the momentum density?

Not from the density alone. $\langle x\rangle = \int\Phi^*(i\hbar\,\partial_p)\Phi\,\mathrm{d}p$ , and the derivative acts on the phase of $\Phi$ as much as its modulus. Keep only $|\Phi(p)|^2$ and you have destroyed all position information — the warning box in equation form. Even both densities together do not determine the state.

Does Φ(p) change with time?

It obeys the same Schrödinger equation, rewritten in momentum language. For a free particle the result is spectacular: the Hamiltonian is pure multiplication by $p^2/2m$ , so each amplitude just rotates — $\Phi(p,t) = \Phi(p,0)\,e^{-ip^2t/2m\hbar}$ — and $|\Phi(p,t)|^2$ never changes. The packet-level consequences are developed in Wave Packet Spreading and Dispersion.

The plane waves are not normalizable. How can it be legal to expand over them?

Because the result is normalizable even though the basis elements are not — just as a position spike is not a state, yet integrating over spikes reconstructs any state. Delta-normalization bookkeeping, $\langle p|p'\rangle = \delta(p-p')$ , makes this systematic; the honest mathematical home for such improper bases is built in the Formalism chapter.

Momentum-Space Wavefunction

Where this lives today

The physical picture

The transform pair

Born's rule in the second language

The Gaussian pair, worked to the end

The operator dictionary

Limiting cases

Worked examples

Going further

Before this

Where this leads