Quantum mechanics. Comutators

4. Two parts

(a) (10 points) Prove the following commutator identity:

$[AB, C] = A[B, C] + [A, C]B$.

(b) (10 points) Show that

$[x^n,P] = i*\hbar*x^{n-1}$.

Answers

 a) $[AB,C] = ABC -CAB$

$A[B,C] +[A,C]B = A(BC) – A(CB) + (AC)B-(CA)B =$

$= ABC -A C B +A C B -CAB = ABC -CAB$

Hence $[AB,C] = A[B,C] +[A,C]B$

b) We know that

$[x,P] = i*\hbar$

$[x^n,P] = [x*x^{n-1}, P] = x[x^{n-1}, P] +[x,P]x^{n-1} = x[x^{n-1},P] +i*\hbar*x^{n-1}=$

$=x[x*x^{n-2},P] +i\hbar*x^{n-1} =$

$=x^2[x^{n-2},P] +[x,P]*x^{n-1} +i\hbar*x^{n-1} =$

$= x^2[x^{n-2},P] +2*i*\hbar*x^{n-1} =$

….

$=x^m[x^{n-m},P] +m*i*\hbar*x^{n-1} =$

we take $m = n-1$

$=x^{n-1}[x,P] +(n-1)*i*\hbar*x^{n-1} = n*i*\hbar*x^{n-1}$