# Transformation of position operator

Consider a dilation of space $$x\mapsto ax$$ for some non-vanishing number $$a$$. Let $$Q$$ be the position operator defined by $$(Q\psi)(x)=x\psi(x)$$ on function $$\psi$$ of space. Suppose $$\psi$$ transforms like a scalar under these dilations, i.e. $$\psi\mapsto\psi'$$ at $$\psi'(x)=\psi(x/a)$$. Then how does $$Q\psi$$ transform? My intuition is that $$Q\psi\mapsto Q\psi'$$, where, to be explicit, $$Q\psi'(x)=x\psi'(x)=x\psi(x/a)$$. Is this correct? In a solution to a homework problem I require that is transforms into $$ax\psi(x/a)$$, which make no sense to me.

Suppose that instead of keeping track of one single function $$x\psi(x)$$ you were actually keeping track of a "tuple" (ordered pair) of functions $$(f(x) ,g(x))$$ where $$f(x)=x$$ and $$g(x)=\psi(x)$$.

You already have stated that $$f(x)=x$$ has to transform into $$ax = af(x)\;,$$ so apparently $$f(x)$$ is not a scalar function. (It's kind transforming more like a vector-valued function might transform under a rotation)

You have also already stated that $$g(x)=\psi(x)$$ has to transform into $$\psi(x/a)=g(x/a)\;,$$ so, $$g$$ is a scalar, since it is just identically equal to $$\psi$$, which is a scalar.

This means the tuple $$(f(x),g(x))$$ transforms into $$(af(x),g(x/a))$$

If I feel like it, I can always just take my tuple and multiply the two components together. Nothing stops me from doing this and the simple act of multiplication doesn't "know" about any transformation. So, I can define: $$h(x) = MULTIPLY(f(x),g(x)) = f(x)*g(x) = x\psi(x)\;.$$ By the above-specified rules this transforms into $$MULTIPLY(af(x),g(x/a)= af(x)*g(x/a) = ax\psi(x/a)\;.$$

Renaming your $$(Q\psi)(x)$$ to $$h(x)$$ shows explicitly by the above-argument that $$(Q\psi)(x)=x\psi(x)$$ transforms into $$ax\psi(x/a)$$.

As a mechanical rule, it looks like you can just remember that whenever there is an explicit factor of $$x$$ you can just replace it with $$ax$$ and whenever there is an explicitly scalar function $$\psi(x)$$ you can just replace it with $$\psi(x/a)$$ to effect the transformation. For example $$x^2\psi(x) \to a^2x^2\psi(x/a)$$