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

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

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