Let define the operators $A = \frac{1}{\sqrt{2}}(x+\partial_x)$ and $B = \frac{1}{\sqrt{2}}(x-\partial_x)$. I am suppossed to check the identity $AB-BA=1$ but I cannot proof it. Is the identity correct?
My attempt: $$AB = \frac{x}{\sqrt{2}}\left(\frac{1}{\sqrt{2}}\left(x-\partial_x\right)\right) + \frac{\partial_x}{\sqrt{2}}\left(\frac{1}{\sqrt{2}}\left(x-\partial_x\right)\right) = \frac{x^2}{2}-\frac{x}{2}\partial_x+\frac{1}{2}-\frac{1}{2}\partial_x^2,$$ $$BA = \frac{x}{\sqrt{2}}\left(\frac{1}{\sqrt{2}}\left(x+\partial_x\right)\right) - \frac{\partial_x}{\sqrt{2}}\left(\frac{1}{\sqrt{2}}\left(x+\partial_x\right)\right) = \frac{x^2}{2}+\frac{x}{2}\partial_x-\frac{1}{2}-\frac{1}{2}\partial_x^2.$$ So, $$AB-BA = -x\partial_x+1.$$
Is this correct?
The identity is correct. The problem is $\partial_x\cdot x$ is not equal to $1$. Think about it this way:
$$(\partial_x\cdot x)(f) =\partial_x(xf)=f+x\partial_x f$$
So the operator $\partial_x\cdot x=1+x\partial_x$. If you do this for both $AB$ and $BA$, you will get the required identity.