What is the most general expression for the coordinate representation of momentum operator?

Question

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?

Answer 1

We start by mentioning a couple of standard formulas

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

and

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

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.

Answer 2

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 :

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