Momentum-Space Representation

The momentum-space wavefunction is the coordinate function of an abstract state in the generalized momentum basis. For a particle on a line,

$$\phi(p)=\langle p|\psi\rangle.$$

The complex number $\phi(p)$ is the probability amplitude density associated with the momentum label $p$. The function $p\mapsto\phi(p)$ represents the same state vector as the position-space wavefunction $\psi(x)=\langle x|\psi\rangle$; the two functions are related by a unitary Fourier transform.

The basic dictionary is:

• position space makes $X$ multiplicative and $P$ differential;
• momentum space makes $P$ multiplicative and $X$ differential;
• free dynamics is diagonal in momentum;
• spatially varying potentials generally couple different momenta;
• probabilities and expectation values are representation independent when the state, operators, and measure are transformed consistently.

Momentum space is therefore both a conceptual second example of a continuous representation and a practical language adapted to translation symmetry, free motion, wave packets, and scattering.

The Generalized Momentum Basis

Let $P$ denote the momentum operator. Its formal generalized eigenkets satisfy

$$P|p\rangle=p|p\rangle.$$

For momentum on the full line, the continuum normalization convention is

$$\langle p|p'\rangle = \delta(p-p')$$

and the formal completeness relation is

$$I = \int_{-\infty}^{\infty} dp\,|p\rangle\langle p|.$$

Applying completeness to a normalizable state gives

$$|\psi\rangle = \int_{-\infty}^{\infty} dp\,\phi(p)|p\rangle.$$

These formulas are the continuous counterparts of

$$c_n=\langle e_n|\psi\rangle, \qquad |\psi\rangle=\sum_n c_n|e_n\rangle.$$

An exact $|p\rangle$ is not normally a finite-norm vector in the physical Hilbert space. It is a generalized eigenvector used under pairings and integrals. The distributional interpretation belongs to Generalized Eigenvectors; the normalization comparison belongs to Normalization.

Momentum Wavefunction and Reconstruction

The projection

$$\phi(p)=\langle p|\psi\rangle$$

extracts the momentum component of $|\psi\rangle$. Conversely, the complete function reconstructs the vector:

$$|\psi\rangle = \int dp\,\phi(p)|p\rangle.$$

For any two normalizable states,

$$\langle\chi|\psi\rangle = \int dp\,\chi(p)^*\phi(p).$$

In particular,

$$\|\psi\|^2 = \int dp\,|\phi(p)|^2.$$

Thus the momentum representation identifies the abstract one-particle Hilbert space with an appropriate $L^2$ space of momentum amplitudes. As in position space, functions that differ only on a set of momentum measure zero represent the same Hilbert-space vector.

The symbol $\phi$ is conventional rather than compulsory. Many texts use $\widetilde\psi(p)$, $\psi(p)$, or $\psi_p(p)$. A trustworthy formula declares what its argument labels and which Fourier convention is being used.

The Position–Momentum Overlap Kernel

The site fixes the one-dimensional overlap convention

$$\langle x|p\rangle = \frac{1}{\sqrt{2\pi\hbar}} e^{ipx/\hbar}.$$

Taking the adjoint gives

$$\langle p|x\rangle = \frac{1}{\sqrt{2\pi\hbar}} e^{-ipx/\hbar}.$$

These kernels are delta normalized. Formally,

$$\begin{aligned} \langle p|p'\rangle &= \int dx\, \langle p|x\rangle \langle x|p'\rangle \\ &= \frac{1}{2\pi\hbar} \int dx\, e^{i(p'-p)x/\hbar} \\ &= \delta(p-p'). \end{aligned}$$

Similarly,

$$\int dp\, \langle x|p\rangle \langle p|x'\rangle = \delta(x-x').$$

The normalization and phase of $\langle x|p\rangle$ determine the transform pair, the dimensions of $\phi(p)$, and the represented form of $X$. Mixing an overlap kernel from one convention with a transform formula from another is a common source of missing signs and factors of $2\pi$ or $\hbar$.

Fourier Transform Pair

Insert the position identity into $\phi(p)=\langle p|\psi\rangle$:

$$\begin{aligned} \phi(p) &= \int dx\, \langle p|x\rangle \langle x|\psi\rangle \\ &= \frac{1}{\sqrt{2\pi\hbar}} \int_{-\infty}^{\infty} dx\,e^{-ipx/\hbar}\psi(x). \end{aligned}$$

The inverse relation follows by inserting the momentum identity:

$$\begin{aligned} \psi(x) &= \int dp\, \langle x|p\rangle \langle p|\psi\rangle \\ &= \frac{1}{\sqrt{2\pi\hbar}} \int_{-\infty}^{\infty} dp\,e^{ipx/\hbar}\phi(p). \end{aligned}$$

The signs and symmetric normalization factors are fixed in Fourier Transform Conventions. The transform theorem, inversion conditions, and distributional extensions are canonical in Fourier Transform.

The two wavefunctions are not independent initial data. Either complete representative determines the other, up to the same almost-everywhere and global-phase equivalences that apply to the abstract state.

Why the Transform Is Unitary

For suitable wavefunctions, direct substitution gives

$$\int dp\,\chi(p)^*\phi(p) = \int dx\,\eta(x)^*\psi(x),$$

where $\eta$ and $\chi$ represent one state in position and momentum space, and $\psi$ and $\phi$ represent the other. Setting the two states equal gives

$$\int dp\,|\phi(p)|^2 = \int dx\,|\psi(x)|^2.$$

This is the quantum-mechanical form of the Plancherel identity. It means that Fourier transformation is a unitary change of representation: it preserves inner products, norms, orthogonality, and therefore transition probabilities.

The theorem-level statement and proof are in Plancherel and Parseval Theorems. Here its physical role is the important point: normalization does not have to be reimposed after an exact representation change.

Momentum Probability Density

For a normalized pure state,

$$\int_{-\infty}^{\infty} dp\,|\phi(p)|^2=1.$$

The momentum Born rule assigns

$$\Pr(P\in\Delta) = \int_\Delta dp\,|\phi(p)|^2$$

to a measurable momentum region $\Delta$. Thus $|\phi(p)|^2$ is a probability density with respect to $dp$, not a probability at one exact value.

In one dimension,

$$[|\phi(p)|^2]=[p]^{-1}, \qquad [\phi(p)]=[p]^{-1/2}.$$

These dimensions ensure that $|\phi(p)|^2dp$ is dimensionless. They differ from the dimensions of the position wavefunction, which are $L^{-1/2}$ on the line. A unitary transform preserves norms, not the physical units of the coordinate functions.

The probability of obtaining one exact value is normally zero for an absolutely continuous distribution, even if $|\phi(p_0)|^2$ is nonzero. The full statement in terms of spectral projectors is in Born Rule for Continuous Spectra.

Exact Momentum and Wave Packets

For the generalized momentum eigenket $|p_0\rangle$, the momentum representative is

$$\langle p|p_0\rangle = \delta(p-p_0),$$

while the position representative is the plane wave

$$\langle x|p_0\rangle = \frac{1}{\sqrt{2\pi\hbar}} e^{ip_0x/\hbar}.$$

Neither representative is an ordinary $L^2$ wavefunction. The delta distribution cannot be squared to form a normalizable density, and the plane wave has constant modulus on the full line.

A physical normalizable state is instead a wave packet:

$$\psi(x) = \frac{1}{\sqrt{2\pi\hbar}} \int dp\, e^{ipx/\hbar}\phi(p),$$

with $\phi\in L^2(\mathbb R,dp)$. A packet concentrated near $p_0$ has an approximately sharp momentum but remains a superposition of continuum eigenkets. Plane-wave normalization and box regularization are developed in Plane Waves and Delta Normalization.

Momentum Versus Wave Number

Wave number $k$ and physical momentum $p$ are related by

$$p=\hbar k.$$

Changing the spectral label changes the density and the normalized generalized basis. If

$$\langle k|k'\rangle=\delta(k-k'),$$

then in one dimension

$$|k\rangle = \sqrt{\hbar}\,|p=\hbar k\rangle.$$

For

$$\widetilde\psi(k)=\langle k|\psi\rangle,$$

the amplitudes obey

$$\widetilde\psi(k) = \sqrt{\hbar}\,\phi(\hbar k).$$

Consequently,

$$|\widetilde\psi(k)|^2dk = |\phi(p)|^2dp.$$

The two functions are therefore not related by merely renaming the horizontal axis. Their Jacobian factor is required by probability conservation. In $d$ dimensions the corresponding factor is $\hbar^{d/2}$.

With the normalized $k$ basis,

$$\langle x|k\rangle = \frac{1}{\sqrt{2\pi}}e^{ikx},$$

whereas the normalized $p$ basis carries $(2\pi\hbar)^{-1/2}$. Declaring the spectral variable prevents dimensional mistakes.

Three Dimensions

For one spinless particle in three Cartesian dimensions,

$$\phi(\mathbf p) = \langle\mathbf p|\psi\rangle$$

and

$$\langle\mathbf r|\mathbf p\rangle = \frac{1}{(2\pi\hbar)^{3/2}} e^{i\mathbf p\cdot\mathbf r/\hbar}.$$

The transform pair is

$$\phi(\mathbf p) = \frac{1}{(2\pi\hbar)^{3/2}} \int d^3r\, e^{-i\mathbf p\cdot\mathbf r/\hbar} \psi(\mathbf r)$$

and

$$\psi(\mathbf r) = \frac{1}{(2\pi\hbar)^{3/2}} \int d^3p\, e^{i\mathbf p\cdot\mathbf r/\hbar} \phi(\mathbf p).$$

Normalization reads

$$\int d^3p\,|\phi(\mathbf p)|^2=1,$$

so

$$[\phi(\mathbf p)]=[p]^{-3/2}.$$

For spherical momentum coordinates,

$$d^3p = p^2\sin\theta_p\, dp\,d\theta_p\,d\varphi_p.$$

As in position space, the probability density must be interpreted together with its measure.

Internal and Many-Particle Labels

If momentum does not form a complete label, additional indices remain. A spin-$1/2$ particle has components

$$\phi_s(\mathbf p) = \langle\mathbf p,s|\psi\rangle, \qquad s\in\{\uparrow,\downarrow\}.$$

When spin is not resolved, the momentum density is

$$\rho(\mathbf p) = \sum_s|\phi_s(\mathbf p)|^2.$$

Normalization becomes

$$\sum_s \int d^3p\, |\phi_s(\mathbf p)|^2 =1.$$

For $N$ distinguishable particles, the momentum wavefunction is

$$\Phi(\mathbf p_1,\ldots,\mathbf p_N) = \langle \mathbf p_1,\ldots,\mathbf p_N |\Psi\rangle.$$

Its domain is the $3N$-dimensional momentum configuration space. Identical particles add exchange symmetry, and second quantization reorganizes the same information using modes and occupation numbers. Those many-body structures belong to Momentum-Space Representation in Many-Body Theory.

Basis Phase Conventions

The delta normalization of $|p\rangle$ does not fix its $p$-dependent phase. One may rephase the generalized basis as

$$|p\rangle' = e^{i\alpha(p)}|p\rangle.$$

The same abstract state then has components

$$\phi'(p) = e^{-i\alpha(p)}\phi(p).$$

The momentum density is unchanged, but overlap kernels and represented operators change. Under this rephasing, the standard position operator $i\hbar\,d/dp$ becomes

$$X' = i\hbar\frac{d}{dp} - \hbar\alpha'(p).$$

This is why the plane-wave overlap convention is part of the representation, not a disposable typographical choice. A basis rephasing is passive and leaves all predictions invariant when every represented object is transformed. By contrast, multiplying only the state amplitude by a $p$-dependent phase is an active change of state.

Momentum and Functions of Momentum

In its own spectral representation, momentum acts by multiplication:

$$(P\phi)(p)=p\phi(p).$$

More generally, the spectral calculus gives

$$(f(P)\phi)(p) = f(p)\phi(p).$$

Examples include

$$(P^2\phi)(p)=p^2\phi(p)$$

and the nonrelativistic kinetic energy

$$(T\phi)(p) = \frac{p^2}{2m}\phi(p).$$

Diagonalization here does not mean that every momentum-space function is an eigenstate. It means the operator acts pointwise in the spectral coordinate. An exact eigenstate would be concentrated at one $p$ as a delta distribution.

Expectation values of suitable functions of momentum are ordinary weighted integrals:

$$\langle f(P)\rangle_\psi = \int dp\, |\phi(p)|^2 f(p).$$

For an unbounded $f(P)$, this expression also states a domain requirement: the integral must exist for the state under consideration.

Position as a Momentum Derivative

Using the standard overlap phase,

$$\langle p|X|\psi\rangle = i\hbar\frac{d\phi}{dp}.$$

Thus

$$(X\phi)(p) = i\hbar\frac{d\phi}{dp}$$

on an appropriate domain. The canonical commutator is visible directly:

$$\begin{aligned} ([X,P]\phi)(p) &= i\hbar\frac{d}{dp} \bigl(p\phi(p)\bigr) - p\,i\hbar\frac{d\phi}{dp} \\ &= i\hbar\phi(p). \end{aligned}$$

Therefore

$$[X,P]=i\hbar I$$

on vectors for which both compositions are defined.

The phrase “on an appropriate domain” is essential. A derivative formula does not by itself specify a self-adjoint operator. Regularity, endpoint behavior, and boundary conditions matter, especially when momentum space is compact or restricted. The general operator viewpoint is in Operator Representations.

General Operator Kernels

An abstract operator $A$ has momentum-space kernel

$$A(p,p') = \langle p|A|p'\rangle.$$

Its action is formally

$$(A\phi)(p) = \int dp'\, A(p,p')\phi(p').$$

Momentum itself has diagonal kernel

$$\langle p|P|p'\rangle = p\,\delta(p-p'),$$

and a function of momentum has

$$\langle p|f(P)|p'\rangle = f(p)\delta(p-p').$$

This continuum kernel notation is the integral analogue of matrix multiplication. Diagonal kernels act by multiplication; off-diagonal kernels mix amplitudes at different momentum labels.

Position-Dependent Potentials Mix Momenta

Let $V(X)$ act by multiplication with $V(x)$ in position space, and define

$$\widetilde V(q) = \frac{1}{\sqrt{2\pi\hbar}} \int dx\,e^{-iqx/\hbar}V(x).$$

Then, whenever the expressions are ordinary functions or are interpreted distributionally,

$$\bigl(V(X)\phi\bigr)(p) = \frac{1}{\sqrt{2\pi\hbar}} \int dp'\, \widetilde V(p-p')\phi(p').$$

Equivalently,

$$\langle p|V(X)|p'\rangle = \frac{1}{\sqrt{2\pi\hbar}} \widetilde V(p-p').$$

The argument $p-p'$ is the momentum transferred by the potential. A constant potential is diagonal because its transform is proportional to a delta distribution. A rapidly varying potential contains broad momentum-transfer components and can couple widely separated momenta.

For real $V(x)$,

$$\widetilde V(q)^* = \widetilde V(-q),$$

which is the kernel condition associated with Hermiticity. The full convolution calculation is developed in Momentum Representation as a Computational Tool.

Schrödinger Equation in Momentum Space

For

$$H=\frac{P^2}{2m}+V(X),$$

the momentum-space Schrödinger equation is

$$\begin{aligned} i\hbar\frac{\partial\phi(p,t)}{\partial t} &= \frac{p^2}{2m}\phi(p,t) \\ &\quad+ \frac{1}{\sqrt{2\pi\hbar}} \int dp'\, \widetilde V(p-p')\phi(p',t). \end{aligned}$$

For a free particle, $V=0$, so each momentum component evolves independently:

$$\phi(p,t) = \exp\left( -\frac{ip^2t}{2m\hbar} \right) \phi(p,0).$$

Consequently,

$$|\phi(p,t)|^2 = |\phi(p,0)|^2.$$

The free momentum distribution is time independent even though the position-space packet can translate and spread. The different momentum components acquire different phases, and their interference determines the later position profile. The canonical free-motion analysis belongs to Free Particle.

Spatial Translations Become Phases

The unitary translation by $a$ is

$$T(a) = e^{-iaP/\hbar}.$$

Because $P$ is multiplicative in momentum space,

$$(T(a)\phi)(p) = e^{-iap/\hbar}\phi(p).$$

In position space the same active transformation is

$$(T(a)\psi)(x) = \psi(x-a).$$

A spatial translation therefore leaves the momentum density unchanged:

$$|e^{-iap/\hbar}\phi(p)|^2 = |\phi(p)|^2.$$

It changes relative phases across momentum components, which encode the packet’s position. Conversely, the momentum-displacement unitary

$$B(q)=e^{iqX/\hbar}$$

acts as

$$(B(q)\phi)(p)=\phi(p-q).$$

The generator and conservation-law interpretation is canonical in Translations and Momentum.

Translation Symmetry and Momentum Conservation

If the Hamiltonian is invariant under all spatial translations, then

$$[H,P]=0.$$

Momentum is conserved, and the Hamiltonian can be decomposed into momentum sectors. For one free particle this decomposition is completely diagonal:

$$H(p,p') = \frac{p^2}{2m} \delta(p-p').$$

For systems with internal degrees of freedom, $H$ may be a finite matrix within each momentum sector. For interacting many-particle systems, translation invariance conserves total momentum even when individual particle momenta are exchanged.

On a lattice, continuous momentum is replaced by crystal momentum in the Brillouin zone, and momenta differing by a reciprocal-lattice vector are equivalent labels. Those are symmetry and many-body extensions, not changes to the basic representational principle.

Discrete Momentum in a Periodic Box

Momentum need not have a continuous spectrum. On a periodic interval of length $L$, normalized modes are

$$\langle x|n\rangle = \frac{1}{\sqrt L} e^{ip_nx/\hbar}, \qquad p_n=\frac{2\pi\hbar n}{L},$$

with $n\in\mathbb Z$. The momentum representation is then the discrete sequence

$$c_n=\langle n|\psi\rangle$$

with

$$\sum_{n\in\mathbb Z}|c_n|^2=1.$$

The spacing is

$$\Delta p=\frac{2\pi\hbar}{L}.$$

In a large periodic box, continuum and discrete amplitudes are related schematically by

$$c_n \simeq \sqrt{\Delta p}\,\phi(p_n),$$

so that

$$\sum_n|c_n|^2 \longrightarrow \int dp\,|\phi(p)|^2.$$

Boundary conditions determine which momentum operator and spectrum are available. One should not import the periodic-box spectrum into a hard-wall problem without checking the operator domain.

Example: Gaussian State

Consider the normalized position wavefunction

$$\psi(x) = \frac{1}{(2\pi\sigma_x^2)^{1/4}} \exp\left[ -\frac{(x-x_0)^2}{4\sigma_x^2} +\frac{ip_0x}{\hbar} \right].$$

Its momentum representative is

$$\begin{aligned} \phi(p) &= \left( \frac{2\sigma_x^2}{\pi\hbar^2} \right)^{1/4} \\ &\quad\times \exp\left[ -\frac{\sigma_x^2(p-p_0)^2}{\hbar^2} -\frac{i(p-p_0)x_0}{\hbar} \right]. \end{aligned}$$

The momentum density is centered at $p_0$:

$$|\phi(p)|^2 = \sqrt{\frac{2\sigma_x^2}{\pi\hbar^2}} \exp\left[ -\frac{2\sigma_x^2(p-p_0)^2}{\hbar^2} \right].$$

Its standard deviation is

$$\Delta p = \frac{\hbar}{2\sigma_x}.$$

The position shift $x_0$ appears only as a linear phase in momentum space, while the momentum shift $p_0$ appears as the center of the momentum density. This is the translation dictionary in a concrete normalizable state.

The product $\Delta x\,\Delta p=\hbar/2$ is the Gaussian equality case of the uncertainty relation. Its canonical derivation is in Position–Momentum Uncertainty.

Mixed States in Momentum Space

A general state is represented by a density operator $\rho$. Its momentum kernel is

$$\rho(p,p') = \langle p|\rho|p'\rangle.$$

The trace condition becomes

$$\int dp\,\rho(p,p)=1,$$

and the momentum density is

$$w(p)=\rho(p,p).$$

For a pure state,

$$\rho_\psi(p,p') = \phi(p)\phi(p')^*.$$

For an ensemble,

$$\rho(p,p') = \sum_j q_j \phi_j(p)\phi_j(p')^*.$$

The diagonal determines momentum-measurement probabilities. Off-diagonal entries encode coherence between different momenta and matter for observables that are not diagonal in momentum, including position. A single momentum wavefunction does not represent a general mixed state.

When Momentum Space Helps

Momentum representation is especially effective when:

• the Hamiltonian is dominated by functions of momentum;
• the system has translation symmetry;
• free propagation is central;
• a wave packet is prepared by its momentum distribution;
• scattering is organized by incoming, outgoing, and transferred momentum;
• convolution in position space becomes multiplication in momentum space;
• derivatives in position space become algebraic powers of momentum;
• total momentum sectors simplify a many-body problem.

Position representation is often more effective for localized potentials, spatial boundaries, and direct geometry. Neither representation is more physical in general. The useful choice is the one that makes the state, operator, symmetry, and boundary conditions easiest to express together.

Numerical Momentum Representations

A discrete Fourier transform does not directly produce the continuum function $\phi(p)$ without scale factors. A spatial grid with spacing $\Delta x$ and length $L=N\Delta x$ determines a reciprocal grid with spacing approximately

$$\Delta p = \frac{2\pi\hbar}{L}$$

and a finite Nyquist range set by $\Delta x$. The exact ordering of positive and negative frequencies depends on the transform library.

Numerical checks should include:

• discrete norm preservation with the chosen quadrature weights;
• the factors of $\Delta x$, $\Delta p$, $2\pi$, and $\hbar$;
• consistent frequency ordering or an explicit shift operation;
• enough position range to suppress wraparound;
• enough spatial resolution to avoid aliasing high momenta;
• boundary conditions compatible with the discrete transform.

The fast Fourier transform is an algorithm for a discrete transform. It does not remove the need to map discrete coefficients to dimensionful continuum amplitudes.

Representation-Invariant Checks

A correct position-to-momentum transformation preserves:

$$\begin{gathered} \langle\chi|\psi\rangle, \qquad \|\psi\|, \qquad \Pr(\text{physical event}),\\ \langle\psi|A|\psi\rangle, \qquad \operatorname{spec}(A). \end{gathered}$$

Useful practical checks are:

1. verify the declared Fourier convention;
2. check that the exponent is dimensionless;
3. check the physical units of the transformed amplitude;
4. verify norm preservation;
5. transform the operator as well as the state;
6. distinguish an ordinary $L^2$ state from a delta-normalized eigenfunction;
7. include Jacobians when relabelling $p$ by $k$, energy, or angular variables;
8. verify domain and boundary assumptions for derivative operators.

Common Mistakes

• Treating $\phi(p)$ as a different physical state from $\psi(x)$.
• Calling $|\phi(p)|^2$ a probability rather than a density with respect to $dp$.
• Treating an exact momentum eigenket or delta distribution as a normalizable wavefunction.
• Confusing momentum $p$ with wave number $k$ and omitting the Jacobian factor.
• Dropping $\hbar$ from $e^{ipx/\hbar}$ while still integrating over $p$.
• Combining forward and inverse transforms from different sign or normalization conventions.
• Assuming that every momentum spectrum is continuous, regardless of boundary conditions.
• Forgetting that a local potential becomes a momentum-transfer kernel rather than multiplication by $V(p)$.
• Using $X=i\hbar\,d/dp$ without checking its domain or the phase convention of the momentum basis.
• Interpreting a momentum-dependent phase as physically irrelevant merely because $|\phi(p)|^2$ is unchanged.
• Reading a raw FFT output as a continuum momentum wavefunction without grid weights and units.

Scope and Canonical Neighbors

This page owns the conceptual and operational momentum-space dictionary. It does not duplicate several neighboring developments:

• Fourier Transform Conventions fixes the site’s transform pair;
• Momentum Representation as a Computational Tool develops transform calculations and convolution examples;
• Plane Waves and Delta Normalization develops generalized free-particle modes and box regularization;
• Position–Momentum Uncertainty owns the uncertainty derivation and interpretation;
• Translations and Momentum owns the generator and conservation-law derivation;
• Operator Representations owns the general basis-dependent operator formalism.

References

• P. A. M. Dirac, The Principles of Quantum Mechanics, 4th ed., Oxford University Press, 1958, Chapters II and III.
• J. J. Sakurai and J. Napolitano, Modern Quantum Mechanics, 3rd ed., Cambridge University Press, 2021, Chapters 1 and 2.
• R. Shankar, Principles of Quantum Mechanics, 2nd ed., Springer, 1994, Chapters 1, 4, and 5.
• L. E. Ballentine, Quantum Mechanics: A Modern Development, 2nd ed., World Scientific, 2014, Chapters 2 and 3.
• B. C. Hall, Quantum Theory for Mathematicians, Springer, 2013, Chapters 3, 7, and 10.
• G. B. Folland, Fourier Analysis and Its Applications, American Mathematical Society, 1992, Chapters 1 and 2.

Exercises

1. Derive the transform pair

Starting from $\phi(p)=\langle p|\psi\rangle$, derive the forward transform by inserting the position identity. Then derive the inverse transform by inserting the momentum identity into $\psi(x)=\langle x|\psi\rangle$.

Solution

For the forward direction,

$$\begin{aligned} \phi(p) &= \langle p|\psi\rangle \\ &= \int dx\, \langle p|x\rangle \langle x|\psi\rangle \\ &= \frac{1}{\sqrt{2\pi\hbar}} \int dx\, e^{-ipx/\hbar}\psi(x). \end{aligned}$$

For the inverse direction,

$$\begin{aligned} \psi(x) &= \langle x|\psi\rangle \\ &= \int dp\, \langle x|p\rangle \langle p|\psi\rangle \\ &= \frac{1}{\sqrt{2\pi\hbar}} \int dp\, e^{ipx/\hbar}\phi(p). \end{aligned}$$

Both formulas use the same normalized overlap kernel.

2. Momentum and wave-number amplitudes

Let $p=\hbar k$. If $\phi(p)$ is normalized with respect to $dp$, find the normalized amplitude $\widetilde\psi(k)$ with respect to $dk$. Verify the dimensions and probability equality.

Solution

Probability invariance requires

$$|\widetilde\psi(k)|^2dk = |\phi(p)|^2dp.$$

Since $dp=\hbar\,dk$,

$$\widetilde\psi(k) = \sqrt\hbar\,\phi(\hbar k)$$

up to a convention-dependent phase. Therefore

$$\int dk\,|\widetilde\psi(k)|^2 = \int dp\,|\phi(p)|^2 =1.$$

In one dimension, $[\phi]=[p]^{-1/2}$ and $[\widetilde\psi]=[k]^{-1/2}=L^{1/2}$, as required by their respective measures.

3. The canonical commutator

Using

$$(P\phi)(p)=p\phi(p), \qquad (X\phi)(p)=i\hbar\phi'(p),$$

verify $[X,P]\phi=i\hbar\phi$ on a common invariant domain.

Solution

Compute

$$\begin{aligned} (XP\phi)(p) &= i\hbar\frac{d}{dp} \bigl(p\phi(p)\bigr) \\ &= i\hbar\phi(p) + i\hbar p\phi'(p), \end{aligned}$$

whereas

$$(PX\phi)(p) = i\hbar p\phi'(p).$$

Subtracting gives

$$([X,P]\phi)(p) = i\hbar\phi(p).$$

The domain qualification ensures that both compositions and their difference are defined.

4. Translation in the two representations

Let $T(a)=e^{-iaP/\hbar}$. Show that its momentum-space action is multiplication by a phase and recover $(T(a)\psi)(x)=\psi(x-a)$ by inverse Fourier transform.

Solution

Because $P$ acts by multiplication,

$$(T(a)\phi)(p) = e^{-iap/\hbar}\phi(p).$$

Transforming back,

$$\begin{aligned} (T(a)\psi)(x) &= \frac{1}{\sqrt{2\pi\hbar}} \int dp\, e^{ipx/\hbar} e^{-iap/\hbar} \phi(p) \\ &= \frac{1}{\sqrt{2\pi\hbar}} \int dp\, e^{ip(x-a)/\hbar} \phi(p) \\ &= \psi(x-a). \end{aligned}$$

The phase changes the packet’s position while leaving its momentum density unchanged.

5. Free time evolution

Solve the free momentum-space Schrödinger equation for arbitrary initial $\phi(p,0)$. Show that every moment of the momentum distribution that exists is time independent.

Solution

For each $p$,

$$i\hbar\frac{\partial\phi(p,t)}{\partial t} = \frac{p^2}{2m}\phi(p,t),$$

so

$$\phi(p,t) = e^{-ip^2t/(2m\hbar)} \phi(p,0).$$

The phase has unit modulus, hence

$$|\phi(p,t)|^2 = |\phi(p,0)|^2.$$

Therefore, whenever the integral exists,

$$\langle P^n\rangle_t = \int dp\,p^n|\phi(p,t)|^2 = \langle P^n\rangle_0.$$

6. Momentum transfer from a cosine potential

Let

$$V(x)=V_0\cos(q_0x/\hbar).$$

Show directly, without manipulating products of delta distributions, that the potential term couples $\phi(p)$ only to amplitudes displaced by $\pm q_0$.

Solution

Write

$$V(x) = \frac{V_0}{2} \left( e^{iq_0x/\hbar} +e^{-iq_0x/\hbar} \right).$$

Multiplication by $e^{iq_0x/\hbar}$ in position space shifts the momentum amplitude by $q_0$, while multiplication by the opposite phase shifts it by $-q_0$. Therefore

$$\bigl(V(X)\phi\bigr)(p) = \frac{V_0}{2} \left[ \phi(p-q_0) + \phi(p+q_0) \right].$$

The periodic potential transfers momentum only in the two amounts $\pm q_0$.

7. Gaussian transform

Fourier transform

$$\psi(x) = \frac{1}{(2\pi\sigma_x^2)^{1/4}} \exp\left[ -\frac{(x-x_0)^2}{4\sigma_x^2} +\frac{ip_0x}{\hbar} \right]$$

and identify the momentum mean, variance, and phase associated with $x_0$.

Solution

Completing the square in the Gaussian integral gives

$$\begin{aligned} \phi(p) &= \left( \frac{2\sigma_x^2}{\pi\hbar^2} \right)^{1/4} \\ &\quad\times \exp\left[ -\frac{\sigma_x^2(p-p_0)^2}{\hbar^2} -\frac{i(p-p_0)x_0}{\hbar} \right]. \end{aligned}$$

Hence

$$|\phi(p)|^2 = \sqrt{\frac{2\sigma_x^2}{\pi\hbar^2}} \exp\left[ -\frac{2\sigma_x^2(p-p_0)^2}{\hbar^2} \right].$$

The mean is $p_0$ and the variance is

$$(\Delta p)^2 = \frac{\hbar^2}{4\sigma_x^2}.$$

The position center $x_0$ appears as the linear phase $e^{-i(p-p_0)x_0/\hbar}$ and does not alter the momentum density.

8. Pure-state momentum kernel

For a normalized $\phi(p)$, show that

$$\rho_\psi(p,p') = \phi(p)\phi(p')^*$$

has unit trace and is idempotent under kernel composition.

Solution

The trace is

$$\operatorname{Tr}\rho_\psi = \int dp\,\rho_\psi(p,p) = \int dp\,|\phi(p)|^2 =1.$$

For the square,

$$\begin{aligned} (\rho_\psi^2)(p,p') &= \int dq\, \rho_\psi(p,q) \rho_\psi(q,p') \\ &= \phi(p)\phi(p')^* \int dq\,|\phi(q)|^2 \\ &= \rho_\psi(p,p'). \end{aligned}$$

The kernel is therefore the momentum representation of the rank-one projector $|\psi\rangle\langle\psi|$.