# Matrix representation of operators

1. Jan 2, 2016

### Happiness

Let the operators $\hat{A}$ and $\hat{B}$ be $-i\hbar\frac{\partial}{\partial x}$ and $x$ respectively.

Representing these linear operators by matrices, and a wave function $\Psi(x)$ by a column vector u, by the associativity of matrix multiplication, we have

$\hat{A}(\hat{B}$u$)$=$(\hat{A}\hat{B})$u.

By the definitions of $\hat{A}$ and $\hat{B}$, we have

LHS $= -i\hbar\frac{\partial}{\partial x}\big(x\ \Psi(x)\big) = -i\hbar\big(x\frac{\partial\Psi(x)}{\partial x} + \Psi(x)\big)$

RHS $= -i\hbar\Psi(x)$

LHS $\neq$ RHS, a contradiction. Where is the mistake?

The extract from a textbook below only talks about the matrix representation of a Hermitian operator. But is it true that all linear operators can be represented by matrices, not just those that are Hermitian?

2. Jan 2, 2016

### andrewkirk

The mistake is in the RHS calc. It arises from the abuse of notation of writing $\hat B$ as $x$. THe correct representation of $\hat B$ is $\Psi(x)\mapsto x\Psi(x)$. That is, it is the operator that, given a function of $x$, returns a new function that is the original function, multiplied by x. So we get:
$\hat A\hat B=-i\hbar\partial_x \Big(\Psi(x)\mapsto x\Psi(x)\Big)= \Psi(x)\mapsto -i\hbar \Big(\Psi(x)\partial_x( x)+x\partial_x\Psi(x)\Big)$ which will match the LHS when applied to a specific ket.
Yes. The matrix element given in formula 5.156 is just as meaningful for non-Hermitian operators. It will just be (in most cases) a non-Hermitian matrix.