The position operator x̂ and the momentum operator p̂
Position multiplies; momentum differentiates, and the i makes it Hermitian.
Every observable of a particle without spin is assembled from these two. Position is the easy one: multiply by x. Momentum is a derivative with −iħ in front, and by the end you should be able to say what each factor does, where the rule comes from, and which integral goes wrong if you drop the i.
Build the model
x̂ multiplies by x; p̂ differentiates, and the i in front of it is not decoration.
1The big idea
In wave mechanics every observable acts on ψ(x) as an operator, and for a particle without spin the two that all the others are built from are position and momentum. Position is the simple one: x̂ multiplies ψ by the real number x at each point, so it slides from one function to the other inside ∫f* x̂g dx at no cost, and it is Hermitian because x* = x. Momentum cannot be a multiplication, because ψ does not carry a momentum at each point; it carries one in how fast its phase winds. The plane wave e(ikx), which with the time factor e(−iωt) travels towards +x with de Broglie momentum ħk, comes back from d/dx multiplied by ik, so −iħ d/dx is the operation that hands back ħk. That is a motivation, not a derivation: |e(ikx)|² = 1 everywhere, the wave cannot be normalised, and the rule has to earn its place on normalisable states. It does. Integration by parts makes d/dx anti-Hermitian, moving across an integral with a minus sign, and the −i, whose conjugate is +i, supplies the second minus sign that makes p̂ Hermitian, so ∫ψ* p̂ψ dx is real for every admissible ψ. The same calculation marks where the model stops: p̂ is Hermitian only on functions that make the boundary term −iħ[f*g] vanish, x̂ only on functions for which xψ is still normalisable, and neither has a normalisable eigenfunction on the whole line.
2In one sentence
The position operator multiplies a wavefunction by x, x̂ψ = xψ, and the momentum operator differentiates it and multiplies by −iħ, p̂ψ = −iħ dψ/dx; each takes a function and returns a new function, not a number.
Example. Applied to e(ikx) with k = 1.0 × 10¹⁰ m⁻¹, p̂ returns the same function times ħk = 1.05 × 10⁻²⁴ kg m s⁻¹, which is de Broglie's h/λ for λ = 0.628 nm; x̂ returns x e(ikx), which is no constant multiple of e(ikx).
3x̂ multiplies, and that is its whole definition
The position operator takes ψ(x) and returns the function xψ(x): at each point the value of ψ is multiplied by that point's coordinate. It is linear, since x(c₁ψ₁ + c₂ψ₂) = c₁xψ₁ + c₂xψ₂, and local: x̂ψ at x = 0.3 nm depends only on ψ at 0.3 nm. Because x is real it slides freely across an overlap integral, ∫f* (x g) dx = ∫(x f)* g dx, which is all that Hermiticity asks, with no integration by parts. The one condition hides in the domain, because x̂ψ must itself be square-integrable. ψ(x) = (πb)⁻¹⁄² (1 + x²/b²)⁻¹⁄² is normalised, yet xψ tends to the constant (b/π)¹⁄² as x grows, so x̂ carries that state out of the space of normalisable functions. And xψ = x₀ψ for every x forces ψ to vanish everywhere except at x₀, so no normalisable function is an eigenfunction of x̂.
4Reading p̂ off a plane wave
Momentum cannot work the same way: ψ does not carry a momentum at each point. What it carries is a phase, and de Broglie ties momentum to how fast that phase winds. The wave e(i(kx − ωt)) keeps constant phase where kx − ωt is fixed, so for k and ω positive it travels towards +x, describing a particle with p = h/λ = ħk. Differentiating returns it multiplied by ik; multiplying by −iħ turns that into ħk, so p̂ = −iħ d/dx. An electron diffracted at λ = 0.150 nm has k = 4.19 × 10¹⁰ m⁻¹, and p̂ hands back ħk = 4.42 × 10⁻²⁴ kg m s⁻¹, exactly h/λ, a speed ħk/m = 4.85 × 10⁶ m s⁻¹ and 66.9 eV of kinetic energy. The sign is not free: with the e(−iωt) that the Schrödinger equation attaches to energy, +iħ d/dx would give the right-moving wave negative momentum. And the argument motivates rather than proves. |e(ikx)|² = 1 at every x, the normalisation integral diverges, and the plane wave is not a state; the rule is judged by what it does to normalisable ψ.
5p̂ on anything that is not a plane wave
On a general ψ, p̂ returns a function that is not a multiple of ψ, so the ratio (p̂ψ)/ψ = −iħψ′/ψ varies with x. The product rule organises the work: for ψ = e(ikx) f(x), p̂ψ = ħk e(ikx) f + e(ikx)(p̂f), the plane-wave value plus whatever p̂ does to the envelope. For the packet envelope f = A e(−x²/(4σ²)), p̂f = iħ(x/(2σ²)) f, so p̂ψ = ħ(k + ix/(2σ²))ψ: the plane-wave ħk plus an imaginary part growing linearly with x. With k = 5.0 nm⁻¹ and σ = 1.0 nm the extra part is 10% of ħk at x = σ and 20% at x = 2σ. It scales as 1/(2kσ), so ψ approaches an eigenfunction only as σ → ∞, where A = (2πσ²)⁻¹⁄⁴ goes to zero and normalisation is lost. Whether a state has a definite momentum is settled by the eigenvalue equation later in this unit; what a momentum measurement on this packet can return, and how often, is the Measurement unit's question in Unit 4.
6d/dx on its own is anti-Hermitian
Test the derivative without its prefactor. Integration by parts gives ∫f* g′ dx = [f*g] − ∫(f′)* g dx, and admissible wavefunctions on the whole line go to zero at ±∞, so the bracket vanishes and ∫f* (dg/dx) dx = −∫(df/dx)* g dx. The derivative moves to the other function and picks up a minus sign: d/dx is anti-Hermitian. Put f = g = ψ and the consequence is sharp. The number z = ∫ψ*ψ′ dx obeys z = −z*, so its real part is zero for every ψ. For the packet above, ∫ψ*ψ′ dx = ik exactly: the −x/(2σ²) piece is odd and integrates to zero, and the ik piece multiplies ∫|ψ|² dx = 1. With k = 5.0 nm⁻¹ that is 5.0i × 10⁹ m⁻¹, an imaginary number with the dimensions of a wave number. Any real multiple of d/dx inherits the defect, so it cannot represent a real observable.
7The i supplies the second minus sign
Now restore −iħ. Complex conjugation turns it into +iħ, which provides a second minus sign to cancel the first: ∫f* (−iħ g′) dx = −iħ[f*g] + iħ∫(f′)* g dx = −iħ[f*g] + ∫(−iħ f′)* g dx. With the bracket zero, ∫f* (p̂g) dx = ∫(p̂f)* g dx, and p̂ is Hermitian. For the packet, the imaginary ik becomes the real ħk = 5.27 × 10⁻²⁵ kg m s⁻¹. The same bookkeeping done at a single x, rather than under the integral, splits the integrand into ψ* p̂ψ = m j − (iħ/2) d|ψ|²/dx, with j the probability current of Unit 2. The imaginary part is a total derivative, so it integrates to the boundary term and vanishes, and ∫ψ* p̂ψ dx = m∫j dx is real. That real number is what the expectation-value lesson of this unit calls ⟨p⟩, and the general consequences of Hermiticity, real eigenvalues and orthogonal eigenfunctions, are proved later in this unit.
8The domain is part of the operator
Hermiticity was proved only for functions that kill the bracket −iħ[f*g], and p̂ also needs dψ/dx to exist and be square-integrable, so where ψ lives matters as much as the formula. On the whole line, normalisability does it. In an infinite well with walls at x = 0 and x = a, the walls force ψ(0) = ψ(a) = 0 and the bracket vanishes again; but e(ikx) never vanishes, so it is not in that domain, and the well's eigenfunctions are not eigenfunctions of p̂: p̂ sin(nπx/a) = −iħ(nπ/a) cos(nπx/a). On a ring of circumference L, where ψ(x + L) = ψ(x), the two ends are the same point and the bracket cancels, and now the plane wave is admissible and normalisable: e(i2πnx/L)/√L with integer n, momentum 2πnħ/L. For L = 2.0 nm neighbouring values are h/L = 3.31 × 10⁻²⁵ kg m s⁻¹ apart. A function that stays nonzero at a hard wall lies outside the domain, and its ∫ψ* p̂ψ dx can pick up an imaginary part: the boundary term reporting that p̂ is no longer Hermitian there.
Key equations
What each relation says, and the units it is written in.
Position operator
x̂ψ(x) = x ψ(x)
Hermitian because x is real: ∫f* (x g) dx = ∫(x f)* g dx with nothing to integrate by parts. Its only demand is that xψ stays square-integrable.
x in m. With ψ in m⁻¹⁄², x̂ψ is in m¹⁄². Multiplication point by point; no derivative anywhere.
Momentum operator
p̂ψ(x) = −iħ dψ/dx
A derivative, not a multiplication: p̂ψ depends on how ψ changes, not on its value, and for a real ψ it is purely imaginary at every x.
ħ = 1.055 × 10⁻³⁴ J s. dψ/dx is in m⁻³⁄², so p̂ψ is in kg m s⁻¹ × m⁻¹⁄²: a momentum times ψ's unit.
Reading p̂ off a plane wave
p̂ e(ikx) = ħk e(ikx)
Fixes the −iħ, with the sign set by the direction of travel under e(−iωt). e(ikx) cannot be normalised, so this motivates the rule rather than proving it.
k in m⁻¹; ħk in kg m s⁻¹, equal to h/λ. The wave e(−ikx) returns −ħk.
d/dx moves across with a minus sign
∫ f* (dg/dx) dx = [f* g] − ∫ (df/dx)* g dx
With the bracket gone, d/dx is anti-Hermitian, so ∫ψ* (dψ/dx) dx equals minus its own conjugate and is purely imaginary for every ψ.
[f* g] is f*g at the upper limit minus at the lower; zero for admissible f and g on the whole line, which vanish at ±∞
p̂ and its boundary term
∫ f* (p̂g) dx − ∫ (p̂f)* g dx = −iħ [f* g]
The conjugate of −i is +i, the second minus sign d/dx lacked. Whatever makes the bracket vanish defines the functions on which p̂ is Hermitian.
f*g in m⁻¹, so −iħ[f* g] is in kg m s⁻¹, like p̂. Zero at ±∞, at walls where ψ = 0, or on a ring.
The momentum integrand, split
ψ* p̂ψ = m j(x) − (iħ/2) d|ψ|²/dx
The imaginary part is a total derivative and integrates to the boundary term, so ∫ψ* p̂ψ dx = m∫j dx is real, and zero for any real ψ.
j = (ħ/m) Im(ψ* dψ/dx), the probability current in s⁻¹; m in kg, so each term is in kg s⁻¹
Interactive lab
Change one variable. Predict first, then read the result.
Controls
Try this. At θ = 90° the lower solid curve is a positive bump and the dashed one is odd, so the integral is the real ħk; at θ = 0°, where Ô = ħ d/dx, they trade places and it is purely imaginary. Back at 90°, raise σ: the upper dashed curve settles onto the solid one as 1/(2kσ) falls.
Live results
- RE ∫ψ*Ôψ dx ÷ ħ = k sin θ
- 5.00 nm⁻¹
- IM ∫ψ*Ôψ dx ÷ ħ = k cos θ
- 0.00 nm⁻¹
- IM(p̂ψ/ψ)/ħk AT x = σ
- 0.100
- NORMALISATION A = (2πσ²)⁻¹⁄⁴
- 0.632 nm⁻¹⁄²
Catch the common trap
The mistake this lesson is built to prevent.
“Momentum is real, so p̂ψ must be a real function; a real wavefunction such as the well's ground state √(2/a) sin(πx/a) would then have imaginary momentum, so the −i must be a slip.” Both halves confuse a function with a number. p̂ψ is a new wavefunction, not a value of momentum, and Hermiticity is a statement about integrals, ∫f* (p̂g) dx = ∫(p̂f)* g dx, not about pointwise values. For real ψ, p̂ψ = −iħψ′ is purely imaginary at every x, and that is exactly what makes the integral real: ∫ψ p̂ψ dx = −iħ∫ψψ′ dx = −iħ[ψ²/2] between the ends, which is zero once ψ vanishes there. A real wavefunction has zero net momentum, in step with the zero probability current it carries. Remove the −i and this particular integral is still zero, but for any travelling packet ∫ψ* (ħψ′) dx comes out imaginary.
Test it: what changes?
- Im(p̂ψ/ψ)/ħk at x = σ falls as 1/(2kσ)
- ψ nears, but never reaches, a p̂ eigenfunction
- Normalisation A = (2πσ²)⁻¹⁄⁴ falls
- ∫ψ* p̂ψ dx stays ħk
- Re ∫ψ*Ôψ dx follows ħk sin θ
- Im ∫ψ*Ôψ dx follows ħk cos θ
- Real only at θ = 90°, where Ô = p̂
- At θ = 0°, ħ d/dx gives purely imaginary iħk
Practice & worked examples
Three tiers. Try each before opening the solution.
EasyExample 1
Two electron beams are described by ψ₁ = A e(ikx) and ψ₂ = A e(−ikx), with k = 4.0 × 10⁹ m⁻¹. Apply p̂ to each, find the constant it returns, and check it against de Broglie and against the classical kinetic energy. Then apply p̂ to ψ₃ = A cos(kx) and decide whether ψ₃ comes back as a multiple of itself.
View solution →
- dψ₁/dx = ikA e(ikx), so p̂ψ₁ = −iħ(ik)ψ₁ = ħkψ₁, with ħk = (1.055 × 10⁻³⁴ J s)(4.0 × 10⁹ m⁻¹) = 4.22 × 10⁻²⁵ kg m s⁻¹.
- For ψ₂ the derivative brings down −ik, so p̂ψ₂ = −ħkψ₂: the same size, −4.22 × 10⁻²⁵ kg m s⁻¹, for the wave travelling towards −x.
- Units: J s × m⁻¹ = kg m² s⁻¹ × m⁻¹ = kg m s⁻¹, a momentum. De Broglie: λ = 2π/k = 1.571 nm and h/λ = 6.626 × 10⁻³⁴ J s ÷ 1.571 × 10⁻⁹ m = 4.22 × 10⁻²⁵ kg m s⁻¹, the same number.
- Classical correspondence: v = ħk/m = 4.22 × 10⁻²⁵ kg m s⁻¹ ÷ 9.109 × 10⁻³¹ kg = 4.63 × 10⁵ m s⁻¹, and ½mv² = 9.77 × 10⁻²⁰ J = 0.610 eV, a slow electron with v/c ≈ 0.0015, so the non-relativistic p = mv is safe.
- For ψ₃: dψ₃/dx = −kA sin(kx), so p̂ψ₃ = iħkA sin(kx). A sine is not a multiple of a cosine, so ψ₃ is not returned as a multiple of itself; writing cos(kx) = ½(e(ikx) + e(−ikx)) shows it holds the +ħk and −ħk waves in equal parts.
Answer. p̂ψ₁ = (4.22 × 10⁻²⁵ kg m s⁻¹)ψ₁ and p̂ψ₂ = (−4.22 × 10⁻²⁵ kg m s⁻¹)ψ₂, each h/λ in size for λ = 1.57 nm; p̂ψ₃ = iħkA sin(kx), which is not a multiple of ψ₃.
MediumExample 2
On the interval 0 ≤ x ≤ a with a = 0.50 nm, take three normalised trial functions: ψ₁ = √(3/a³) x, ψ₂ = √(30/a⁵) x(a − x), and ψ₃ = ψ₂ e(iqx) with q = 2π/a. Evaluate ∫ψ* p̂ψ dx over the interval for each, and use the boundary term to explain why one of the three results is not real.
View solution →
- Normalisation first: ∫3x²/a³ dx from 0 to a is 1, and ∫x²(a − x)² dx = a⁵/30, so ψ₂ is normalised, and ψ₃ has the same |ψ|² as ψ₂. For a real ψ, ψ p̂ψ = −iħψψ′ = −(iħ/2) d(ψ²)/dx, so the integral is −(iħ/2)[ψ²] between 0 and a: only the end values matter.
- ψ₁(0) = 0 but ψ₁(a)² = 3/a = 6.0 × 10⁹ m⁻¹, so ∫ψ₁ p̂ψ₁ dx = −(iħ/2)(6.0 × 10⁹ m⁻¹) = −3.16i × 10⁻²⁵ kg m s⁻¹. Its partner ∫(p̂ψ₁)* ψ₁ dx = +3.16i × 10⁻²⁵ kg m s⁻¹, and the difference, −6.33i × 10⁻²⁵ kg m s⁻¹, is exactly −iħ[ψ₁²] between 0 and a: the Hermiticity identity with a surviving boundary term.
- ψ₂ vanishes at both ends, so the bracket is zero and ∫ψ₂ p̂ψ₂ dx = 0: real, as Hermiticity requires, and zero, as for any real function in the domain.
- For ψ₃, dψ₃/dx = (ψ₂′ + iqψ₂) e(iqx), so ψ₃* p̂ψ₃ = −iħψ₂ψ₂′ + ħqψ₂². The first term integrates to zero exactly as for ψ₂, and ∫ψ₂² dx = 1 leaves ∫ψ₃* p̂ψ₃ dx = ħq = h/a = 1.33 × 10⁻²⁴ kg m s⁻¹, real. Multiplying by e(iqx) added ħq and nothing else.
- Checks: every result is ħ divided by a length, in kg m s⁻¹. As a grows, the ψ₁ result −3iħ/(2a) and ħq both fall as 1/a. ψ₁ is the odd one out because it does not vanish at the wall; outside the domain where the boundary term dies, p̂ is not Hermitian and the integral acquires an imaginary part.
Answer. ∫ψ₁* p̂ψ₁ dx = −3.16i × 10⁻²⁵ kg m s⁻¹, not real because ψ₁(a) ≠ 0; ∫ψ₂* p̂ψ₂ dx = 0; ∫ψ₃* p̂ψ₃ dx = ħq = 1.33 × 10⁻²⁴ kg m s⁻¹.
HardExample 3
Take the real, normalised functions f(x) = N₁ e(−x²/(2b²)) and g(x) = N₂ x e(−x²/(2b²)) on the whole line, with b = 0.10 nm, N₁ = (b√π)⁻¹⁄² and N₂ = (√π b³/2)⁻¹⁄². Build the 2 × 2 table of integrals ∫u* (Âv) dx, with u and v each chosen from f and g, for  = d/dx, for  = p̂ and for  = x̂, and say which tables have the symmetry Hermiticity requires.
View solution →
- Tools: f′ = −(x/b²) f and g′ = N₂(1 − x²/b²) e(−x²/(2b²)). The Gaussian integrals are ∫e(−x²/b²) dx = b√π and ∫x² e(−x²/b²) dx = √π b³/2, which confirm that N₁ and N₂ normalise f and g, and give N₁N₂ = √2/(√π b²).
- d/dx: the diagonal entries ∫f f′ dx and ∫g g′ dx are [u²/2] from −∞ to +∞, both zero. Off the diagonal, ∫f g′ dx = N₁N₂ ∫(1 − x²/b²) e(−x²/b²) dx = N₁N₂(b√π − b√π/2) = 1/(√2 b), while ∫g f′ dx = −N₁N₂ ∫(x²/b²) e(−x²/b²) dx = −1/(√2 b). With b = 0.10 nm these are ±7.07 × 10⁹ m⁻¹: an antisymmetric table, the anti-Hermitian signature.
- p̂: multiply every entry by −iħ. The diagonal stays zero, ∫f p̂g dx = −iħ/(√2 b) = −7.46i × 10⁻²⁵ kg m s⁻¹ and ∫g p̂f dx = +7.46i × 10⁻²⁵ kg m s⁻¹. Each entry is the complex conjugate of its mirror image, ∫u* (p̂v) dx = (∫v* (p̂u) dx)*, which is what Hermiticity requires.
- x̂: nothing is differentiated. x f² and x g² are odd, so the diagonal is zero, and ∫f x g dx = ∫g x f dx = N₁N₂ √π b³/2 = b/√2 = 7.07 × 10⁻¹¹ m. A real symmetric table: Hermitian with no i needed.
- Checks: the d/dx entries are in m⁻¹, the p̂ entries in kg m s⁻¹ and the x̂ entries in m. Widen the functions and the x̂ entry grows as b while the p̂ entry falls as 1/b, as it must when broad functions change slowly; the product of the two magnitudes is ħ/2 = 5.27 × 10⁻³⁵ J s for every b.
Answer. d/dx: zero diagonal, off-diagonal +7.07 × 10⁹ and −7.07 × 10⁹ m⁻¹ (antisymmetric). p̂: zero diagonal, −7.46i and +7.46i × 10⁻²⁵ kg m s⁻¹ (conjugate-symmetric, Hermitian). x̂: zero diagonal, 7.07 × 10⁻¹¹ m both ways (real symmetric).
Quick check
Choose an answer, then check it. Explain to yourself why the others fail.
Key takeaway
p̂ = −iħ d/dx is Hermitian only where boundary terms vanish.
Use x̂ψ = xψ and p̂ψ = −iħ dψ/dx; each returns a function, not a number. Integration by parts makes d/dx anti-Hermitian, and the −i supplies the second sign flip, so ∫ψ* p̂ψ dx is real, provided −iħ[f*g] vanishes: at ±∞, at walls where ψ = 0, or round a ring.
Next: Kinetic energy, potential energy and the Hamiltonian →