# I Products of operators : products of matrices

1. Apr 26, 2017

### dyn

Hi.
If I have an operator in matrix form eg. < i | x | j > then the matrix of the operator x2 is given by the square of the former matrix. This seems like common sense but how would i prove this using Dirac notation ?
Thanks

2. Apr 26, 2017

When you square the matrices, i.e. multiply one matrix by the same matrix, you wind up doing a summation over $k$ in the following way (to get the ij element of the x matrix multiplied by the x matrix) : $<i|x|k><k|x|j>$. in the Dirac notation $|k><k|$ summed over all states is the identity operator and can be removed.

3. Apr 27, 2017

### dyn

Thanks for that

4. Apr 28, 2017

### vanhees71

The full calculation is
$$\langle i|\hat{x}^2|j \rangle=\sum_{k} \langle i|\hat{x} k \rangle \langle k|\hat{x} j \rangle,$$
i.e., in matrix notation
$$(x^2)_{ij}=\sum_k x_{ik} x_{kj}.$$
That's like matrix multiplication in finite-dimensional vector spaces, only that the matrix here is "infinite dimensional".

5. Apr 28, 2017