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

Step 1: Start from the fundamental canonical commutation relation:
\[ [x, p_x] = i\hbar. \]
Step 2: Use the operator identity \([AB, C] = A[B, C] + [A, C]B\) with \(A = B = x\) and \(C = p_x\):
\[ [x^2, p_x] = x[x, p_x] + [x, p_x]x. \]
Step 3: Substitute \([x, p_x] = i\hbar\):
\[ [x^2, p_x] = x(i\hbar) + (i\hbar)x = 2i\hbar x. \]
(The extra \(\pi\) written in the printed stem is a typographical artifact; the commutator itself evaluates to the following.)
\[ \boxed{[x^2, p_x] = 2i\hbar x} \]