I have a question about deriving the coordinate representation of momentum operator from the canonical 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?

  • $\begingroup$ Related: physics.stackexchange.com/q/53252/2451 and links therein. $\endgroup$
    – Qmechanic
    Jul 7, 2013 at 0:09
  • $\begingroup$ 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 $\alpha (x-y)^n \delta(x-y)=0$. $\endgroup$
    – user26143
    Jul 7, 2013 at 0:21

2 Answers 2


We start by mentioning a couple of standard formulas

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


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

The canonical commutation relation (CCR) is

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

The standard Schrödinger position representation reads

$$\begin{align} \hat{x}~=~&x, \cr \hat{p}~=~&-i\hbar\frac{\partial}{\partial x}.\end{align}\tag{4}$$

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

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

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

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

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

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

in the representation (5).

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


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 :

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

  • 1
    $\begingroup$ $\langle x | p \rangle = \frac{1}{2\pi} e ^{ipx}$ comes from $\hat{p}= - i\frac{ \partial}{\partial x}$ $\endgroup$
    – user26143
    Jul 8, 2013 at 0:07

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