# Thread: A show that question (identity matrix, trace and commutator)

1. ## A show that question (identity matrix, trace and commutator)

Hi,
I have a question that asks:
$\displaystyle [X,R] = i\hbar \textbf{I}$
Where h is a number and is the imaginary unit
However, i need to show that this cannot be true and the answer must in fact be
$\displaystyle [X,R] = i\hbar$

X and R are square matrices, complex.

Ideas:
I know to take the commutator and the trace - though a bit unsure about the trace, i know it is the sum of the diagonals but not clear how it helps here mathematically.

$\displaystyle [X, R] = XR - RX$

Given that:

$\displaystyle Tr ( A + B) = Tr(A) + Tr(B)$
and
$\displaystyle Tr(AB) = Tr(BA)$
Doing the trace above to [X,R] would be 0.

So somehow this implies that the commutator is finite within the imaginary plane. How is taking the trace useful here to help me justify that the identity matrix is not within the solution.

Thanks

2. ## Re: A show that question (identity matrix, trace and commutator)

Originally Posted by imagemania
I have a question that asks:
$\displaystyle [X,R] = i\hbar \textbf{I}$
Where h is a number and is the imaginary unit
However, i need to show that this cannot be true
The trace of $\displaystyle [X,R]$ is 0, as you say. But the trace of $\displaystyle i\hbar I$ is $\displaystyle i\hbar n$, where n is the size of the matrix. That is not 0, and therefore $\displaystyle [X,R]\ne i\hbar I.$

Originally Posted by imagemania
... and the answer must in fact be
$\displaystyle [X,R] = i\hbar$
That makes no sense at all. [X,R] is a matrix, but $\displaystyle i\hbar$ is a scalar. So they can never be equal.