Proving that an operator squared equals its matrix squared

[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

 

[Charles Link]

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.

 

[dyn]

Thanks for that

 

[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".

 

[Charles Link]

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".

It should be noted that there are many other operators for which it is simply a matrix multiplication in a finite dimensional space.

 

[vanhees71]

Yes, e.g., for spins.