Use the momentum representation and a general function \(f\left(p_{x}\right)\) of the linear momentum to confirm that the position and momentum operators in this representation do not commute, and find the value of their commutator.

Short Answer

Expert verified
The commutator of position and momentum operators is 0

Step by step solution

01

Define Position and Momentum Operators

We will be working in the momentum representation. For quantum mechanics in one dimension, the position operator \( x \) and the momentum operator \( p \) in the momentum representation are given by \( x = i\hbar\frac{d}{dp} \) and \( p = p \). Here, \( \hbar \) is the reduced Planck constant and \( i \) is the imaginary unit.
02

Compute AB

We now apply the operators to the function consecutively in the sequence that defines the first term of the commutator, \( AB = xp \). So \( xp = xi\hbar\frac{d}{dp}f(p) = i\hbar p\frac{d}{dp}f(p) \).
03

Compute BA

Applying the operators to the function \( f(p) \) consecutively in the sequence of the second term of the commutator, \( BA = px \). So \( px = p(i\hbar\frac{d}{dp})f(p) = i\hbar p\frac{d}{dp}f(p) \).
04

Compute the Commutator

The commutator of operators \( A \) and \( B \) is given by \( [A, B] = AB - BA \), so we need to subtract the result from Step 3 from that obtained in Step 2, which gives \( [x, p] = xp - px = i\hbar p\frac{d}{dp}f(p) - i\hbar p\frac{d}{dp}f(p) = 0 \).

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