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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 $\langle{x}|PX|\alpha\rangle$: $$ \begin{equation} \begin{split} \langle{x}|PX|\alpha\rangle & = \langle{x}|PX|\int_{-\infty}^{\infty}|x'\rangle\langle{x'}|\alpha\rangle dx' \\ & = \langle{x}|P|\int_{-\infty}^{\infty}x'|x'\rangle\psi_{\alpha}(x') dx' \\ & = \langle{x}|P|\int_{-\infty}^{\infty}x'\left(\int_{-\infty}^{\infty}|p'\rangle\langle{p'}|x'\rangle dp'\right)\psi_{\alpha}(x') dx' \\ & = \langle{x}|P|\int_{-\infty}^{\infty}x'\left(\int_{-\infty}^{\infty}|p'\rangle\frac{1}{\sqrt{2\pi\hbar}}e^{-\frac{ip'x'}{\hbar}} dp'\right)\psi_{\alpha}(x') dx' \\ & = \langle{x}|\int_{-\infty}^{\infty}x'\left(\int_{-\infty}^{\infty}p'|p'\rangle\frac{1}{\sqrt{2\pi\hbar}}e^{-\frac{ip'x'}{\hbar}} dp'\right)\psi_{\alpha}(x') dx' \\ & = \int_{-\infty}^{\infty}x'\left(\int_{-\infty}^{\infty}p'\langle{x}|p'\rangle\frac{1}{\sqrt{2\pi\hbar}}e^{-\frac{ip'x'}{\hbar}} dp'\right)\psi_{\alpha}(x') dx' \\ & = \int_{-\infty}^{\infty}x'\left(\int_{-\infty}^{\infty}p'\frac{1}{\sqrt{2\pi\hbar}}e^{\frac{ip'x}{\hbar}}\frac{1}{\sqrt{2\pi\hbar}}e^{-\frac{ip'x'}{\hbar}} dp'\right)\psi_{\alpha}(x') dx' \\ & = \frac{1}{{2\pi\hbar}}\int_{-\infty}^{\infty}x'\left(\int_{-\infty}^{\infty}p'e^{\frac{ip'(x-x')}{\hbar}} dp'\right)\psi_{\alpha}(x') dx' \\ \end{split} \end{equation} $$

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

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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, $\delta'(x-x')$.

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

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  • $\begingroup$ 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. $\endgroup$ – DanielSank Sep 27 '15 at 6:16
  • $\begingroup$ @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 $\endgroup$ – Discovery Sep 27 '15 at 8:56
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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.

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