# What is $\langle x_1 |\hat V(\hat x)| x_2 \rangle$?

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).

Thanks

## Calculation where is matters

The following is a snippet of what I was doing.

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

$$$$\hat H = \hat T + \hat V$$$$

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

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

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 $$\hat a$$:

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

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,

$$$$[\hat a, \hat x] = 0$$$$

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

$$$$\langle x ' | \hat a | x \rangle = \langle x ' | [\hat V,\hat T'( \hat p )] | x \rangle$$$$

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

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

Dividing both:

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

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

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

Or:

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

Due to the quantization condition of position and momentum:

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

OR:

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

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

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

Simplifying:

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

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

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

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

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

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

$$$$GM \frac{\partial }{ \partial x '} \psi (x') = GM \frac{\partial }{ \partial x '} \psi (x')$$$$

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.

• I'd be very interested to see a calculation where it makes a difference... Nov 15, 2020 at 8:21
• @Philip I've included the calculation. Nov 15, 2020 at 8:26
• @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.... Nov 16, 2020 at 4:59

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.

• 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? Nov 15, 2020 at 8:48
• I've got the book somewhere, let me take a look. But it's possible that it's beyond my pay-grade! :) Nov 15, 2020 at 8:50
• I uploaded a screenshot on the chat just incase you don't find it ;) Nov 15, 2020 at 8:50
• 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... 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 $$0=0$$, $$i=j$$.

• The why doesn't that apply to the section: "Calculation where is matters"? Can you include that in your answer as well? Nov 15, 2020 at 9:13
• @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 $V$? Nov 15, 2020 at 9:35
• 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 $a \frac{\partial \psi}{\partial x} = \frac{V'(x)}{m} \frac{\partial \psi}{\partial x}$. Nov 15, 2020 at 9:46
• 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 $\frac{\partial }{ \partial x '} \psi (x') =\frac{a}{g} \int_{- \infty}^\infty \frac{1}{( x - x') } \delta( x' - x)( \psi(x') + (x -x') \partial_{x'} \psi(x') + \dots) dx$ Nov 15, 2020 at 9:46
• 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. 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}