We had proved the canonical commutation relation \([\hat{x}, \hat{p}]=i \hbar\) in generality in the previous chapter. Of course, then, any matrix realization of the position and momentum operators must satisfy the commutation relation. In this chapter, we had constructed the momentum operator in the basis of energy eigenstates of the infinite square well, and we can verify the canonical commutation relation if we also have the position operator in the same basis.

(a) Using a similar procedure as that for the momentum operator in Sec. 5.1.3, determine the matrix elements of the position operator \((\hat{x})_{m n}\) in the basis of energy eigenstates of the infinite square well.

(b) The matrix form of the canonical commutation relation is \([\hat{x}, \hat{p}]=i \hbar \mathbb{I}\), where II is the appropriate identity matrix in the corresponding basis. Thus, the commutator has no non-zero elements off the diagonal. Simplify the expression for the off-diagonal elements of the commutator as much as possible. Using mathematical software, approximately evaluate the sum that you find for different values of row \(m\) and column \(n\). Does the sum seem to vanish?

(c) Now, simplify the expression for the diagonal elements of the commutator, \(([\hat{x}, \hat{p}])_{n n}\), as much as possible. Again, using mathematical software, sum the first few (about 20) terms for a few values of n. Do you find that every diagonal entry of the commutator is indeed \(i \hbar\) ?

(d) Take the trace of the commutator as a matrix, \(\operatorname{tr}[\hat{x}, \hat{p}]\); what do you find? How do you reconcile this with the result of part (c)?