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I have a question about deriving the coordinate representation of momentum operator from the commutation relation, $[x,p]= i$.

One derivation (ref W. Greiner's Quantum Mechanics: An Introduction, 4th edi, p442) is as following: $$ \langle x|[x,p]|y \rangle = \langle x|xp-px|y \rangle = (x-y) \langle x|p|y \rangle. $$ On the other hand, $ \langle x|[x,p]|y \rangle = i \langle x|y \rangle = i \delta (x-y)$. Thus $$ (x-y) \langle x|p|y \rangle = i \delta(x-y). \tag{1} $$

We use $(x-y) \delta(x-y) = 0$. Take the derivative with respect to $x$; we have $\delta(x-y) + (x-y) \delta'(x-y) = 0$. Thus $$ (x-y) \delta'(x-y) = - \delta(x-y). \tag{2} $$

Comparing Eqs. (1) and (2), we identify $$ \langle x|p|y \rangle = -i \delta'(x-y). \tag{3} $$

In addition, we can add $\alpha \delta(x-y)$ on the right-hand side of Eq. (3), i.e. $$ \langle x|p|y \rangle = -i \delta'(x-y) + \alpha \delta(x-y), $$ and $[x,p] = i$ is still satisfied. We can also add $$ \frac{\beta}{\sqrt{|x-y|}}\delta(x-y)$$ on the RHS of Eq. (3). Here $\alpha$ and $\beta$ are two real numbers.

My question is, what is the most general expression of $\langle x|p|y \rangle$? Can we always absorb the additional term into a phase factor like Dirac's quantum mechanics book did?

Thank you very much in advance.

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Related: physics.stackexchange.com/q/53252/2451 and links therein. –  Qmechanic Jul 7 '13 at 0:09
Thanks for the link! From the link above, seems $\frac{ \beta}{ \sqrt{|x-y|}} \delta(x-y)$ is ill-defined at origin. Nevertheless even we exclude this possibility, how to find the most general expression of coordinate expression of momentum operator? Maybe it is just lack of imagination that there is something else than $\alpha \delta(x-y)$. $\alpha (x-y)^n \delta(x-y)$, $n>0$ does not count, since $\alpha (x-y)^n \delta(x-y)=0$. –  user26143 Jul 7 '13 at 0:21
@user26143 Pro formatting tip: \langle and \rangle are the bra-ket symbols, not < and >. It's more work to type, but it makes Dirac notation oh so much more pleasing to the eye :) –  Chris White Jul 7 '13 at 1:17
Thank you for your suggestion and editing work! –  user26143 Jul 7 '13 at 1:21

2 Answers 2

up vote 3 down vote accepted

We start by mentioning a couple of standard formulas

$$\tag{1} \psi(x)~=~\langle x | \psi \rangle, $$


$$\tag{2} \langle x | y \rangle ~=~\delta(x-y).$$

The canonical commutation relation (CCR) is

$$\tag{3} [\hat{x}, \hat{p}] ~=~i\hbar{\bf 1}. $$

The standard Schrödinger position representation reads

$$\tag{4}\hat{x}~=~x, \qquad \hat{p}~=~-i\hbar\frac{\partial}{\partial x}.$$

We may conjugate the standard Schrödinger position representation (4) by an unitary operator $\hat{U}=e^{-if(\hat{x})}$, where $f:\mathbb{R}\to\mathbb{R}$ is a given differentiable function. In this way we obtain an unitary equivalent position representation

$$\tag{5}\hat{x}~=~x, \qquad \hat{p} ~=~-i\hbar e^{-if(x)}\frac{\partial}{\partial x}e^{if(x)} ~=~-i\hbar\frac{\partial}{\partial x}+ \hbar f^{\prime}(x),$$

of the CCR (3). The standard Schrödinger position representation (4) corresponds to $f\equiv {\rm const}$. For a general irreducible representation of the CCR (3), see the Stone-von Neumann Theorem.

The representation (5) implies

$$\tag{6} \langle x | \hat{p} |\psi \rangle~=~(\hat {p} \psi)(x) ~=~-i\hbar e^{-if(x)}(e^{if}\psi)^{\prime}(x) ~=~-i\hbar\psi^{\prime}(x)+ \hbar f^{\prime}(x)\psi(x). $$

From (6) we conclude that the momentum matrix elements reads

$$\tag{7} \langle x | \hat{p} |y \rangle~=~-i\hbar\delta^{\prime}(x-y)+ \hbar f^{\prime}(x)\delta(x-y)$$

in the representation (5).

Finally, here and here are two other Phys.SE posts that also discuss ambiguities in $x\leftrightarrow p$ overlaps.

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I think the commutation relation is more fundamental. Eq. (1) is derived from the commutation relation. Like the canonical quantization procedure in QFT, we postulate the commnutation/anticommutation relation for the fields. –  user26143 Jul 7 '13 at 0:37
I see. Heard about the Stone–von Neumann theorem. Seems I need to go through it. Thanks a lot. –  user26143 Jul 7 '13 at 1:19
I updated the answer. Note that eq. numbers may have changed. –  Qmechanic Jul 7 '13 at 19:49

In units $\hbar=1$, from $ \langle p|p' \rangle = \delta(p-p')$ and $\langle x|p \rangle = \frac {1}{\sqrt{2 \pi}}e^{i px}$, you get :

$\langle x|\hat p|y \rangle = \int_{|p>,|p'>} \langle x|p\rangle \langle p|\hat p|p'\rangle \langle p'|y\rangle = \frac {1}{2 \pi} \int_{|p>,|p'>} e^{ipx} p' \langle p|p' \rangle e^{-ip'y}$ $= \frac {1}{2 \pi} \int dp~ p ~e^{ip(x-y)} = - i\partial_x \frac {1}{2 \pi}\int dp~ e^{ip (x - y)} = -i\delta'(x-y)$

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$\langle x | p \rangle = \frac{1}{2\pi} e ^{ipx}$ comes from $\hat{p}= - i\frac{ \partial}{\partial x}$ –  user26143 Jul 8 '13 at 0:07

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