So I've become rusty in Quantum Mechanics. What is $\langle x_1 |V(\hat x)| x_2 \rangle$? Where $V$ is the potential and $|x \rangle$ is the postion eigenket? $$ \langle x_1 |\hat V(\hat x)| x_2 \rangle = ? $$

Is it with $V(x_1) \delta(x_1 -x_2)$ but since the position operator can also act on a bra is it $V(x_2) \delta(x_1 -x_2)$ or some combination of $x_1$, $x_2$ like $V(x_1,x_2) \delta(x_1 -x_2)$? If it does not make any difference can you prove so? ( I was doing a calculation where it seemed to make a difference).


Calculation where is matters

The following is a snippet of what I was doing.

Let the Hamiltonian $\hat H$ of the particle be:

\begin{equation} \hat H = \hat T + \hat V \end{equation}

where $\hat T$ is the kinetic energy and $\hat V$ is the potential energy. Now, to find the velocity:

\begin{equation} \hat v = \frac{-i}{\hbar}[\hat H , \hat x] = \frac{-i}{\hbar} [ \hat T , \hat x] = \hat T'(\hat p) \end{equation}

where $\hat T'(\hat p)$ is the velocity which is a function of momentum. Now, if we further assume $T'(p)$ is of degree $1$ and find the acceleration $\hat a$:

\begin{equation} \hat a = \frac{-i}{\hbar} [ \hat H , \hat T'(\hat p) ] =\frac{-i}{\hbar} [\hat V,\hat T'( \hat p )] = \hat a (\hat x) \end{equation}

We know that acceleration must be a function of $\hat x$ since we have already assumed $T'(p)$ is of degree $1$ and find the acceleration $\hat a$. Hence,

\begin{equation} [\hat a, \hat x] = 0 \end{equation}

Now, multiplying $| x \rangle$ on the $\hat a$ equation:

\begin{equation} \langle x ' | \hat a | x \rangle = \langle x ' | [\hat V,\hat T'( \hat p )] | x \rangle \end{equation}

Using the eigenvalue equation with eigenvalue $a$ (which represents the acceleration at a position $| x \rangle $ ):

\begin{equation} a \langle x ' | x \rangle= a \delta( x' - x) = \frac{-i}{\hbar} \langle x ' | (\hat V \hat T'( \hat p ) - \hat T'( \hat p ) \hat V) | x \rangle = \frac{-i}{\hbar} (V(x') - V(x)) \langle x' | \hat T'( \hat p ) | x \rangle \end{equation}

Dividing both:

\begin{equation} \frac{i \hbar a}{(V(x') - V(x)) } \delta( x' - x) = \langle x' | \hat T'( \hat p ) | x \rangle \end{equation}

Let us consider $\langle x' | \hat T'( \hat p ) | \psi \rangle$:

\begin{equation} \langle x' | \hat T'( \hat p ) | \psi \rangle = \int_{-\infty}^\infty \langle x' | \hat T'( \hat p ) | x \rangle \langle x| \psi \rangle dx = \int_{- \infty}^\infty \frac{i \hbar a}{(V(x') - V(x)) } \delta( x' - x) \psi(x)dx \end{equation}


\begin{equation} \langle x' | \hat v | \psi \rangle = \int_{- \infty}^\infty \frac{i \hbar a}{(V(x') - V(x)) } \delta( x' - x) \psi(x) dx \end{equation}

Due to the quantization condition of position and momentum:

\begin{equation} \frac{- i \hbar}{m }\frac{\partial }{ \partial x'} \psi (x') = \int_{- \infty}^\infty \frac{ a i \hbar}{(V(x') - V(x)) } \delta( x' - x) \psi(x) dx \end{equation}


\begin{equation} \frac{\partial }{ \partial x'} \psi (x') = \int_{- \infty}^\infty \frac{ - m a }{(V(x') - V(x)) } \delta( x' - x) \psi(x) dx \end{equation}

Let us now try $V(x) = - G \frac{m M}{x}$ (point particle potential):

\begin{equation} \frac{\partial }{ \partial x '} \psi (x') = \int_{- \infty}^\infty \frac{- a}{(-\frac{GM}{x'} + \frac{GM}{x}) } \delta( x' - x) \psi(x) dx \end{equation}


\begin{equation} GM \frac{\partial }{ \partial x '} \psi (x') = \int_{- \infty}^\infty \frac{ a x x'}{(x' - x) } \delta( x' - x) \psi(x) dx \end{equation}

Again Taylor expanding around $\psi (x)$ around $x'$:

\begin{equation} GM \frac{\partial }{ \partial x '} \psi (x') = \int_{- \infty}^\infty \frac{ a x x'}{(x' - x) } \delta( x' - x) (\psi(x') + (x-x') \partial_{x'} \psi(x') + \dots) dx \end{equation}

Plugging in the ansatz $a = - \frac{GM}{x x'}$ we get:

\begin{equation} GM \frac{\partial }{ \partial x '} \psi (x') = \int_{- \infty}^\infty - \frac{ GM}{(x' - x) } \delta( x' - x) (\psi(x') + (x-x') \partial_{x'} \psi(x') + \dots) dx \end{equation}

Note the term $\int_{- \infty}^\infty \frac{ GM}{(x' - x) } \delta( x' - x) \psi(x') $ goes to $0$ since is antisymmetric.

\begin{equation} GM \frac{\partial }{ \partial x '} \psi (x') = GM \frac{\partial }{ \partial x '} \psi (x') \end{equation}

Hence, we get a consist solution which agrees with classical calculations. Note: $a = - GM{x^{-2}}$ or $a = - GM{x'^{-2}}$ both will give wrong answers.

  • $\begingroup$ I'd be very interested to see a calculation where it makes a difference... $\endgroup$
    – Philip
    Commented Nov 15, 2020 at 8:21
  • $\begingroup$ @Philip I've included the calculation. $\endgroup$ Commented Nov 15, 2020 at 8:26
  • $\begingroup$ @All Everything is correct except the last line's claim which states note: $a = - GM{x^{-2}}$ or $a = - GM{x'^{-2}}$ both will give wrong answers.... $\endgroup$ Commented Nov 16, 2020 at 4:59

3 Answers 3


I'm not very good at the rigorous math, but the way I think about it is that "functions" like the Dirac Delta are actually distributions, meaning that they only make sense inside of an integral of the form:

$$\int_{a}^b \text{d}x\,\,f(x)\delta(x),$$

where $f(x)$ is a well-behaved (smooth, with a compact support over the interval, etc.) test function.

The quantity that you are talking about is also a distribution (it's just the product of the delta-function with the potential) and so it too must be defined in a similar way. However, it's pretty easy to show that:

$$\int_{a}^b \text{d}x_1\,\,f(x_1) V(x_1)\delta(x_1 -x_2) = \int_{a}^b\text{d}x_2\,\, f(x_2) V(x_2)\delta(x_1-x_2),$$

using a simple substitution of variables. As a result, both your answers are essentially equal, in the sense that their distributions are equal.

As to your example, there are many assumptions there that might be justified, but which I don't quite understand. However, there are some steps that I am uncomfortable with: for example, consider the step where you divide by $V(x')-V(x)$. The term on the left hand side is proportional to: $$\frac{\delta(x'-x)}{V(x')-V(x)},$$ and I suspect there are many problems with this. For starters, you will be dividing by zero when $V(x')=V(x)$, and more importantly, you cannot multiply distributions by singular functions, as the result would not be a well defined distribution.

  • $\begingroup$ They do a similar derivation in Quantum Optics in Phase Space by Wolfgang P.Schleich page 38 section 2.1.2 ... I can share a screenshot (of the book) on chat if you like? $\endgroup$ Commented Nov 15, 2020 at 8:48
  • $\begingroup$ I've got the book somewhere, let me take a look. But it's possible that it's beyond my pay-grade! :) $\endgroup$
    – Philip
    Commented Nov 15, 2020 at 8:50
  • $\begingroup$ I uploaded a screenshot on the chat just incase you don't find it ;) $\endgroup$ Commented Nov 15, 2020 at 8:50
  • 1
    $\begingroup$ Ok, I must admit that they do do exactly what I say shouldn't be done. In fact, they seem to claim that $$\frac{\delta(x-x')}{x-x'}= \delta'(x-x')$$ which I really have a problem with. Now that I think about it, I feel like I have answered another question on PSE about this exact section of this book, let me see if I can find it... $\endgroup$
    – Philip
    Commented Nov 15, 2020 at 8:55

You can think of it in the sense of distributions, and then Philip's answer applies, in the sense that, as distributions

$$ f(x)\delta(x-y)=f(y)\delta(x-y)$$

but more intuitively, you can also think of $\delta(x-y)$ as having support only on $x=y$ meaning that it is equal to $0$ for all $x\neq y$ (this is not mathematically rigorous, but it works on an intuitive level), so the only point where $V(x_1)\delta(x_1-x_2)$ does not vanish is $x_1=x_2$, so you can put either of them in the potential. In fact you can think of

$$ V(x_1)\delta(x_1-x_2)=V(x_2)\delta(x_1-x_2)$$

as the continuous equivalent of

$$ \delta_{ij} V_j=\delta_{ij}V_i$$

where $\delta_{ij}=1$ if $i=j$ and $0$ otherwise and $V$ is a vector. It doesn't matter whether you take $V_i$ or $V_j$, since in all cases where the above expression is not $0=0$, $i=j$.

  • $\begingroup$ The why doesn't that apply to the section: "Calculation where is matters"? Can you include that in your answer as well? $\endgroup$ Commented Nov 15, 2020 at 9:13
  • $\begingroup$ @MoreAnonymous sorry, I'm not sure I'm following the calculation, where does the ansatz of $a$ come from? How do you get $a$ from $V$? $\endgroup$ Commented Nov 15, 2020 at 9:35
  • $\begingroup$ The ansatz came from first trying $a = G\frac{m}{x_1 ^2}$ and then $a = G\frac{m}{x_2 ^2}$ both the equations give inconsistent answers but it must agree with classical calculations ... Basically you want something of the sort $a \frac{\partial \psi}{\partial x} = \frac{V'(x)}{m} \frac{\partial \psi}{\partial x}$. $\endgroup$ Commented Nov 15, 2020 at 9:46
  • $\begingroup$ Let us experiment with the potential $V = -mgx$ (flat earth potential): $ \frac{\partial }{ \partial x '} \psi (x') = \int_{- \infty}^\infty \frac{- a}{(-gx' + gx) } \delta( x' - x) \psi(x) dx $ Taylor expanding $\psi(x)$ around $x'$ and simplifying $ \frac{\partial }{ \partial x '} \psi (x') =\frac{a}{g} \int_{- \infty}^\infty \frac{1}{( x - x') } \delta( x' - x)( \psi(x') + (x -x') \partial_{x'} \psi(x') + \dots) dx$ $\endgroup$ Commented Nov 15, 2020 at 9:46
  • $\begingroup$ Hence, cancelling the wave-function: $ a=g $ The idea is to make the quantum case also abide with classical calculations of the calculated acceleration would be. $\endgroup$ Commented Nov 15, 2020 at 9:48

To add on to the other answers, if $V(x)$ is function of a real variable, then $\hat{V}$ really means

$$ \hat{V} = \int_{-\infty}^{\infty} V(x) | x \rangle \langle x | \mathrm{d} x $$

where $| x \rangle$ is the same position eigenket as your definition $$ \hat{x} | x \rangle = x | x \rangle $$

You can verify that $$ \hat{V} |x \rangle = V(x) |x \rangle $$

And so yes, the answer to your first question is \begin{align} \langle x_1 | \hat{V} |x_2 \rangle = V(x_2) \langle x_1 |x_2 \rangle &= V(x_2) \delta(x_2 - x_1) \\ &= V(x_1) \delta(x_2 - x_1) \end{align}


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