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

Physics Asked on January 6, 2021

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:

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

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 $$langle x’ | hat T'( hat p ) | psi rangle$$:

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

Or:

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

Due to the quantization condition of position and momentum:

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

OR:

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

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

$$begin{equation} frac{partial }{ partial x ‘} psi (x’) = int_{- infty}^infty frac{- a}{(-frac{GM}{x’} + frac{GM}{x}) } delta( x’ – x) psi(x) dx end{equation}$$

Simplifying:

$$begin{equation} GM frac{partial }{ partial x ‘} psi (x’) = int_{- infty}^infty frac{ a x x’}{(x’ – x) } delta( x’ – x) psi(x) dx end{equation}$$

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.

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 $$xneq y$$ (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 $$0=0$$, $$i=j$$.

Correct answer by user2723984 on January 6, 2021

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:

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

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.

Answered by Philip on January 6, 2021