Algebra involving operator of x and p

Question

For operators x and p in QM with $p= {h\over i}{d \over dx} $, how can I find the combination of operator such as $$(xp-px)^2 $$ or $$(x+p)(x-p) $$ Can I just expand them by using normal algebra such as $$(ab-ba)^2=a^2b^2+b^2a^2-abba-baab $$ $$ (a+b)(a-b)=a^2-b^2 $$

Answer 1

I recommend using the commutator $[x,p]=i\hbar.$ This can be shown by $$ (px)f = p(xf) = -i\hbar\frac{d}{dx}(xf) = -i\hbar(f+xf') = -i\hbar f + x(-i\hbar\frac{d}{dx}f) \\ = -i\hbar f + x(pf) = -i\hbar f + (xp)f, $$ so that $$ [x,p]f = (xp-px)f = (xp)f - (px)f = i\hbar f $$ for every smooth function $f$ which means that $[x,p]=i\hbar.$

Then, $$ (xp-px)^2 = [x,p]^2 = (i\hbar)^2 = -\hbar^2 $$ and $$ (x+p)(x-p) = x^2+px-xp-p^2 = (x^2-p^2)-(xp-px) = (x^2-p^2)-i\hbar. $$

Answer 2

You kind of have the right idea with your examples, but there are mistakes. Remember however that operator multiplication is actually function composition, so $AB$ should really be viewed as the operator defined by $f\mapsto A(B(f))$. To make sense of operators, they should be thought of as acting on functions. Because of this connection to function composition, you cannot conclude that $(a+b)(a-b) = a^2 - b^2$ in general. It should actually be $a^2 - ab + ba - b^2$, meaning that this is only equal to $a^2 - b^2$ if $a$ and $b$ commute (i.e. $ab = ba$ which is not in general true for operators). So for instance,

\begin{align} (xp - px)f(x) &= \bigg(x\bigg(-i\frac{d}{dx}\bigg) - \bigg(-i\frac{d}{dx}\bigg)x\bigg)f(x) \\ &= -i x\frac{df}{dx} + i \frac{d}{dx}(xf) \\ &= -ixf'(x) + ixf'(x) + i f(x) \\ &= i f(x) \end{align}

Thus $xp-px$ is really the same as multiplication by $i$. Can you take care of the examples in your post?

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There's a connection to polynomials in two variables that can be made if I recall correctly. You can think of this as $\mathbb{C}[x,y]/\langle xy-yx = i\rangle$. The notation might be slightly wrong but you think about the ring of polynomials in two variables quotiented by the ideal $xy-yx = i$.