# Commutator of position and momentum

I'm reading Sakurai's Quantum Mechanics. One of the problem in the book asks to use the relation $$\langle{x}|p\rangle=\frac{1}{\sqrt{2\pi\hbar}}e^{\frac{ipx}{\hbar}}$$ to evaluate $\langle{x}|[X,P]|\alpha\rangle=\langle{x}|XP|\alpha\rangle-\langle{x}|PX|\alpha\rangle$ in terms of $\psi_{\alpha}(x)=\langle{x}|\alpha\rangle$ without using the fact that in the $x$ representation, $P$ acts like $-i\hbar\frac{d}{dx}$.

I'm not sure how to proceed with this. Here is my attempt:

The eigenvalue equations for the position operator $X$ and the momentum operator $P$ are, respectively $$X|x'\rangle=x'|x'\rangle \text{ and } P|p'\rangle=p'|p'\rangle$$ So, for example, let's evaluate $\langle{x}|PX|\alpha\rangle$: $$$$\begin{split} \langle{x}|PX|\alpha\rangle & = \langle{x}|PX|\int_{-\infty}^{\infty}|x'\rangle\langle{x'}|\alpha\rangle dx' \\ & = \langle{x}|P|\int_{-\infty}^{\infty}x'|x'\rangle\psi_{\alpha}(x') dx' \\ & = \langle{x}|P|\int_{-\infty}^{\infty}x'\left(\int_{-\infty}^{\infty}|p'\rangle\langle{p'}|x'\rangle dp'\right)\psi_{\alpha}(x') dx' \\ & = \langle{x}|P|\int_{-\infty}^{\infty}x'\left(\int_{-\infty}^{\infty}|p'\rangle\frac{1}{\sqrt{2\pi\hbar}}e^{-\frac{ip'x'}{\hbar}} dp'\right)\psi_{\alpha}(x') dx' \\ & = \langle{x}|\int_{-\infty}^{\infty}x'\left(\int_{-\infty}^{\infty}p'|p'\rangle\frac{1}{\sqrt{2\pi\hbar}}e^{-\frac{ip'x'}{\hbar}} dp'\right)\psi_{\alpha}(x') dx' \\ & = \int_{-\infty}^{\infty}x'\left(\int_{-\infty}^{\infty}p'\langle{x}|p'\rangle\frac{1}{\sqrt{2\pi\hbar}}e^{-\frac{ip'x'}{\hbar}} dp'\right)\psi_{\alpha}(x') dx' \\ & = \int_{-\infty}^{\infty}x'\left(\int_{-\infty}^{\infty}p'\frac{1}{\sqrt{2\pi\hbar}}e^{\frac{ip'x}{\hbar}}\frac{1}{\sqrt{2\pi\hbar}}e^{-\frac{ip'x'}{\hbar}} dp'\right)\psi_{\alpha}(x') dx' \\ & = \frac{1}{{2\pi\hbar}}\int_{-\infty}^{\infty}x'\left(\int_{-\infty}^{\infty}p'e^{\frac{ip'(x-x')}{\hbar}} dp'\right)\psi_{\alpha}(x') dx' \\ \end{split}$$$$

but then I got stuck because the middle integral is not convergent. I sensed that I did something wrong as well.

Two main points are....

1. Generally $\langle{x}|[X,P]|\alpha\rangle \not= \langle{x}|XP|\alpha\rangle-\langle{x}|PX|\alpha\rangle$

When $[X,P]=XP-PX$ is an well-defined operator in a Hilbert space, $H=L^2([a,b])$, space of square-integrable functions in $[a,b]$, the domain of definition of $[X,P]$ is a set of functions $|\alpha\rangle$ satisfying

$|\alpha\rangle$ is in the domain of operator $X$

$|\alpha\rangle$ is in the domain of operator $P$

$P|\alpha\rangle$ is in the domain of operator $X$

$X|\alpha\rangle$ is in the domain of operator $P$

However, the domain of definition of $XP$ is a set of functions $|\alpha\rangle$ satisfying

$|\alpha\rangle$ is in the domain of operator $P$

$X|\alpha\rangle$ is in the domain of operator $P$

In the similar manner you can expect the form of the domain of $PX$.

So if you want to assert that $\langle{x}|[X,P]|\alpha\rangle = \langle{x}|XP|\alpha\rangle-\langle{x}|PX|\alpha\rangle$, you should have additional condition, $|\alpha\rangle$ is a function in the domain of $[X,P]$. Try to prove $\langle{x}|[X,P]|\alpha\rangle = \langle{x}|XP|\alpha\rangle-\langle{x}|PX|\alpha\rangle$ using $|\alpha\rangle=|p\rangle$ and Hermitianity of $X$ and $P$. You may find a contradiction.

1. Delta functional

From the last segment, $\left(\int_{-\infty}^{\infty}p'e^{\frac{ip'(x-x')}{\hbar}} dp'\right) \\$ is the form of Fourier transform of $p'$ and can be described by (Dirac) functional-derivative, $\delta'(x-x')$.

$\int_{-\infty}^{\infty}p'e^{\frac{ip'(x-x')}{\hbar}} dp'=\int_{-\infty}^{\infty}-i\hbar \frac{d}{dx}e^{\frac{ip'(x-x')}{\hbar}}dp'=-i2\pi\hbar^2 \frac{d}{dx} \delta(x-x')$.

• I kind of object to your point #1. The commutation relation guarantees that the inequality should be equality. The only real problem is that some authors and calculations use states which aren't particularly well defined. Commented Sep 27, 2015 at 6:16
• @DanielSank I agree with you, and I just want to emphasize importance of the form of state $|\alpha \rangle$. For example... a (inappropriate) paradox of equation 2.6 in p. 7 of arxiv.org/pdf/quant-ph/9907069.pdf Commented Sep 27, 2015 at 8:56

I think you'd probably need to integrate that term by parts, lowering $k = p'/\hbar$ to $1$ while raising $\exp[i~k~(x - x')]~dk$ into $[-i\hbar/(x - x')]~\exp[i~k~(x - x')].$

The result you get for the middle integral is then $$-2\pi i\hbar ~ \frac{\delta(x - x')}{x - x'}.$$ If you hold off evaluating the integral further, the other integral will be similar but with $x$ replacing the free $x'$, so you will get a value $x - x'$ which cancels the denominator.

In other words: if I just use the way you're doing this problem where you are effectively able to substitute \begin{align} \hat P \mapsto&~ \int dp~p~|p\rangle\langle p|,\\ \hat X \mapsto&~ \int dx~x~|x\rangle\langle x|, \text{ and}\\ \hat I \mapsto&~ \int dx~ |x\rangle\langle x|,\end{align} then I can just write out the term of $X P - PX = XPI - I P X$ which has a matrix element $|x\rangle\langle x'|$ by using the $x$ index for the first operator and the $x'$ index for the last, and simply have:$$[X,P] = \frac{1}{2\pi\hbar} ~ \iiint dx~dx'~dp~(x - x')~p~e^{i p (x - x')/\hbar}~|x\rangle\langle x'|.$$That's how you cancel the $(x - x')^{-1}$ to finish up the problem.