Question

The commutator \([x^2, p_x]\) is equal to:

• A. \(i\hbar x\)
• B. \(2i\hbar x\)
• C. \(2i\hbar p_x\)
• D. Zero

Hint:

Use \([AB,C]=A[B,C]+[A,C]B\) with \([x,p_x]=i\hbar\), or apply \([x^n,p_x]=i\hbar n x^{n-1}\).

Answer 1

The Correct Option is B

Direct approach using the position representation: In one dimension take $p_x = -i\hbar\,\dfrac{d}{dx}$ and let the commutator act on an arbitrary test function $f(x)$.

Compute each term:
\[ x^2 p_x f = -i\hbar\, x^2 f', \]
\[ p_x (x^2 f) = -i\hbar\,\frac{d}{dx}\!\left(x^2 f\right) = -i\hbar\left(2x f + x^2 f'\right). \]
Now subtract:
\[ [x^2, p_x]f = x^2 p_x f - p_x x^2 f = -i\hbar x^2 f' - \left[-i\hbar(2xf + x^2 f')\right]. \]
The $x^2 f'$ pieces cancel, leaving
\[ [x^2, p_x]f = 2i\hbar\, x f. \]
Since $f$ is arbitrary, the operator identity is
\[ [x^2, p_x] = 2i\hbar\, x. \]
This matches the general result $[x^n, p_x] = i\hbar\, n\, x^{n-1}$ evaluated at $n = 2$.
\[ \boxed{[x^2, p_x] = 2i\hbar x} \]