Position Operator in Momentum Space and Uncertainty

Position Operator in Momentum Space and the Uncertainty Relation

In 1D momentum space the position and momentum operators act very differently:

Momentum operator:

$(p\psi)(p) = p\,\psi(p)$
Position operator:

$(x\psi)(p) = i\hbar \,\frac{d\psi}{dp}(p)$

That “multiply vs. differentiate” mismatch is exactly what produces uncertainty:
a state sharply localized in $p$ must vary slowly with $p$ ,
while a state sharply localized in $x$ requires $\psi(p)$ to vary rapidly with $p$ .
These requirements are incompatible in the extreme, and the quantitative statement is the
Heisenberg uncertainty relation.

1. The Commutator in Momentum Space

Let $\psi(p)$ be square-integrable and vanish at $|p|\to\infty$ . Using the operator forms above:

$[x,p]\psi = x(p\psi)-p(x\psi) = i\hbar \frac{d}{dp}\big(p\psi\big) - p\Big(i\hbar \frac{d\psi}{dp}\Big) = i\hbar\,\psi.$

So as operators,

$[x,p]=i\hbar\mathbf 1.$

2. Derivation via Cauchy–Schwarz Inequality

Define centered operators

$A := x - \langle x\rangle,\qquad B := p - \langle p\rangle,$

and the vectors

$\phi := A\psi, \qquad \chi := B\psi.$

Then

$\Delta x^2 = \lVert \phi\rVert^2 = \langle \phi|\phi\rangle,\qquad \Delta p^2 = \lVert \chi\rVert^2 = \langle \chi|\chi\rangle.$

By Cauchy–Schwarz:

Split $AB$ into commutator and anticommutator:

$AB=\tfrac12\{A,B\} + \tfrac12[A,B].$

Taking expectation values and using $[A,B]=[x,p]=i\hbar$ :

$\langle AB\rangle = \tfrac12\langle \{A,B\}\rangle + \tfrac{i\hbar}{2}.$

Hence

$|\langle AB\rangle|^2 = \Big|\tfrac12\langle \{A,B\}\rangle + \tfrac{i\hbar}{2}\Big|^2 \ge \Big|\tfrac{i\hbar}{2}\Big|^2 = \frac{\hbar^2}{4}.$

Therefore,

$\boxed{\Delta x\,\Delta p \ge \frac{\hbar}{2}}.$

This is the Heisenberg uncertainty relation.

(Equivalently, one can quote the general Robertson–Schrödinger inequality
$\Delta A\,\Delta B \ge \tfrac12 |\langle[A,B]\rangle|$ , and insert $[x,p]=i\hbar$ .)

3. Why the Derivative Form Introduces Uncertainty

Since $x$ acts as $i\hbar\,\partial/\partial p$ , making $\psi(p)$ very narrow around some $p_0$
forces $\partial\psi/\partial p$ to be large in magnitude (sharp changes near the edges),
which inflates $\Delta x$ . Conversely, making $\psi(p)$ very smooth (small derivative) spreads
it out in $p$ , increasing $\Delta p$ . The non-commutativity $[x,p]=i\hbar$ is the precise
algebraic statement of this trade-off.

4. State that Saturates the Bound

Equality holds when $(A - i\lambda B)\psi=0$ for real $\lambda$ . In $p$ -space:

$\big(i\hbar \tfrac{d}{dp} - \langle x\rangle\big)\psi(p) = i\lambda\,(p-\langle p\rangle)\psi(p).$

Solving gives a Gaussian

$\psi(p) \propto \exp\!\Big[-\frac{(p-\langle p\rangle)^2}{2\sigma_p^2} + i\,\frac{\langle x\rangle}{\hbar}\,p\Big],$

with $\Delta p=\sigma_p$ and $\Delta x=\hbar/(2\sigma_p)$ , so $\Delta x\,\Delta p=\hbar/2$ .

Summary

In momentum representation, the position operator being a derivative operator guarantees
non-commutation with the momentum operator. By applying the Cauchy–Schwarz inequality
to the commutator, we derive the uncertainty relation:

$\Delta x\,\Delta p \ge \frac{\hbar}{2}.$

This expresses the fundamental trade-off between sharpness in position and sharpness in momentum.

Deriving Uncertainty from the Position Operator in p-space