The matrix elements of the momentum operator in position representation are:

$$\langle x | \hat{p} | x' \rangle = -i \hbar \frac{\partial \delta(x-x')}{\partial x}$$

Does this imply that $\langle x | \hat{p} | x' \rangle = 0 $ whenever $x \neq x'$?

Is the momentum operator diagonal in position representation?

I know that the momentum operator shouldn't be diagonal in position representation (otherwise solving for the eigenergies and eigenfunctions of most Hamiltonians would be trivial). I am obviously confused here. I need more than just yes/no answers to these questions. I need some explanation, or some intuition.


4 Answers 4


There is a heuristic way to look at this.

The Dirac delta function corresponds to a spike when its argument is zero. You can view it as the limit of a sequence of Gaussian functions whose areas are all one but whose width goes to zero. The derivative of a Gaussian function looks like this: enter image description here

So in the limit, the derivative of the Dirac function is something like an up spike infinitesimally to the left of the origin followed by a down spike infinitesimally to the right. So the matrix elements you're looking at aren't actually diagonal, they're infinitesimally off-diagonal.

These kinds of heuristics can be useful, but they can also be dangerous, so don't take what I say too literally.

Update: Another way to look at this is to approach the derivative Dirac delta via a discretisation. If the wavefunction is represented by a vector of equally spaced samples, the derivative can be represented by central differences. Assuming periodic boundary conditions we get a matrix like:

$\frac{1}{2}\pmatrix{ 0 & 1 & 0 & 0 & \ldots & -1 \\ -1 & 0 & 1 & 0 & \ldots & 0 \\ 0 & -1 & 0 & 1 & \ldots & 0 \\ 0 & 0 & -1 & 0 & \ldots & 0 \\ & & \vdots \\ 1 & 0 & 0 & 0 & \ldots & 0 \\ }$

We have 1's just above the diagonal and -1's just below. As the discretisation gets finer we get a matrix where they entries are more and more concentrated near the diagonal even though all of the non-zero terms are actually off-diagonal. In the limit you can again imagine something that is infinitesimally off-diagonal.


OP wrote (v1):

Does this imply that $$\tag{1}\langle x | \hat{p} | x^{\prime} \rangle = 0$$ whenever $x \neq x^{\prime}$?

For two fixed values of $x \neq x^{\prime}$, the answer is Yes $^1$. But don't try to integrate (1) over $x$ or $x^{\prime}$, i.e. treat $x$ and $x^{\prime}$ as running parameters.

Is the momentum operator diagonal in position representation?

No. If the position eigenstates $(| x \rangle)_{x\in\mathbb{R}}$ both diagonalized the position operator $\hat{x}$ and the momentum operator $\hat{p}$, this would for instance imply that they commute, which we know they don't, cf. the CCR.

The above apparent paradoxes are rooted in wrongly thinking of a distribution, say $\delta(x)$, as a function from $\mathbb{R}$ to $[0,\infty]$ (which takes the value $\infty$ at the point $x=0$). This is an insufficient picture. Distributions should either be understood as a suitable limit of ordinary functions, or defined with the help of test functions.


$^1$ This is related to that the distribution $\delta^{\prime}(x)$ only has support at $x=0$.

  • $\begingroup$ Oops; somehow I didn't see the "whenever x=x'" $\endgroup$ Commented Mar 18, 2013 at 1:24
  • 1
    $\begingroup$ I was looking for some more explanation. I mean, I already "knew" the answers to my questions. What I needed was to know why, or get some intuition. $\endgroup$
    – a06e
    Commented Mar 18, 2013 at 1:41

In addition to Qmechanic's answer, this is what happens when you integrate over $x$ with a test function, which is actually what you need to do for the expression to really become meaningful. So let's use $f(x)$ as a test function$^1$ and integrate:

$$-i \hbar \int dx \frac{\partial}{\partial x}f(x)\delta(x-x') = -i\hbar f'(x')$$

So in general this is not zero.

$^1$ Assume $f$ is differentiable.


From Dirac, for continuous eigenvalues:

1) the generalized unit operator is defined as $\delta(\xi'-\xi'')$ (analogously with the discrete case, where $ I = \delta_{ij}$).

2) A representation of a Hermitian operator $\xi$ is said to be diagonal when $\langle \xi' | \xi | \xi'' \rangle = \xi' \delta(\xi' - \xi'') $

3) Generalized matrix multiplication is defined by: $$\langle\xi' | \alpha \beta | \xi'' \rangle = \int \langle \xi' | \alpha | \xi''' \rangle d \xi''' \langle \xi''' | \beta | \xi'' \rangle $$

4) A general diagonal matrix is defined to be one that commutes with the generalized matrix of item 2) (again analogously with the discrete case.)

5) Applying this commutability criterion to the generalized matrix elements of another operator $\omega$, one finds this matrix is diagonal when:
$$(\xi' - \xi'') \langle \xi' | \omega | \xi'' \rangle = 0 $$

This relation is satisfied when: $$\langle \xi' | \omega | \xi'' \rangle = c' \delta ( \xi' - \xi'') $$ by the basic definition of the $\delta$ function, but it is not satisfied by the momentum matrix elements. Therefore the momentum is not diagonal (which you already knew).


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