# Is there a quantum analogue of Mean Value Theorem theorem?

## Background

I was thinking of Mean Value Theorem in the context of classical mechanics I have $$2$$ points $$A$$ and $$B$$ and my particle goes from $$A$$ to $$B$$ then I know the velocity of the particle at a time $$\tilde t$$ is given by $$\dot x(\tilde t)=\frac{x_a-x_b}{t_a - t_b}$$ where $$\tilde t$$ is a point between $$t_a$$ and $$t_b$$ as shown below: ## Question

My question is: is there a quantum analogy of Rolle's theorem for the sum over histories formulation of quantum mechanics? To elaborate further, we know how to define velocity via the Heisenberg picture as: $$\hat v = -i [H,x]$$ and surely one can make a statement about how Rolle's theorem would apply in the quantum realm which is each path multiplied by a phase factor? If so, what is the mathematical statement?

## Edit

Just as Rolle's theorem enabled one to make a statement about the velocity between points $$x_A$$ and $$x_B$$. Is there a statement one make about the velocity between the points $$x_A$$ and $$x_B$$.

For example consider the amplitude given by:

$$A = \langle x_A|e^{\frac{iH(t_A - t_B)}{\hbar}}| x_B \rangle$$

Splitting into $$N$$ parts:

$$A = \langle x_A|(e^{\frac{iH(\Delta t)}{\hbar}})^N| x_B \rangle$$

Let $$\hat v | v \rangle = v | v \rangle$$ where $$v$$ is the velocity operator. Then, inserting the fat identity:

$$A = \langle x_A| (e^{\frac{iH(\Delta t)}{\hbar}}) \Big( \int dv_{N-1} | v_{N-1} \rangle \langle v_{N-1} \Big| ) (e^{\frac{iH(\Delta t)}{\hbar}}) \dots(e^{\frac{iH(\Delta t)}{\hbar}}) \Big( \int dv_{1} | v_{1} \rangle \langle v_{1} \Big ) (e^{\frac{iH(\Delta t)}{\hbar}}) | x_B \rangle$$

Rather than re-derive the whole sum over histories with the velocity operator. We assume the Hamiltonian is a linear function of position and velocity. You mean that it is the sum of a function of position and a function of velocity.. Let, $$\hat H$$. Let them be $$\hat T(\hat v)$$ and $$\hat V(\hat x)$$

$$A = \langle x_A| (e^{\frac{i(T+ V)(\Delta t)}{\hbar}}) \Big( \int dv_{N-1} | v_{N-1} \rangle \langle v_{N-1} |\Big ) (e^{\frac{i(T+ V)(\Delta t)}{\hbar}}) \dots(e^{\frac{i(T+ V)(\Delta t)}{\hbar}}) \Big( \int dv_{1} | v_{1} \rangle \langle v_{1} \Big ) (e^{\frac{i(T+ V)(\Delta t)}{\hbar}}) | x_B \rangle$$

Now I conjecture it each path the $$e^{\hat T}$$ acting on $$|v \rangle$$ will get $$\frac{x_a-x_b}{t_a - t_b}$$ at least once in the limit $$N \to \infty$$.

How do I make the last line more precise?

• If you use Ehrenfest's Theorem and look at expectation values for some general state, rather than looking directly at operators, then you are dealing with ordinary real numbers and can just apply Rolle's theorem directly. – By Symmetry Oct 7 '18 at 11:26
• I think you mean to use the mean value theorem. – BioPhysicist Oct 7 '18 at 12:42
• The mean value therorem already breakd down for 2D mechanics. Why do you consider it significant? – lalala Oct 7 '18 at 13:12

## 1 Answer

1. There is no velocity in the Hamiltonian formalism. If you're using a Hamiltonian, you must use momentum, not velocity. See this question and my answer for why the Hamiltonian is a function of position and momentum, but not velocity.

2. The momentum operator is proportional to the time derivative of the position operator (in the Heisenberg picture) only on-shell, but the paths in the path integral are not on-shell, i.e. not necessarily solutions to the equations of motion, and we are using time-dependent states, i.e. the Schrödinger picture, in the derivation anyway.

Therefore, the $$\lvert v\rangle$$ in your question do not exist. You can do a similar derivation involving momentum, but it is an off-shell derivation where momentum and position are a priori unrelated. This is, in some sense, fortunate, because the paths in the path integral formalism after taking the limit are almost always not differentiable - the mathematically rigorous measure on the space of paths is a conditional Wiener measure, under which the differentiable paths have zero measure - therefore you cannot apply Rolle's theorem to a generic path anyway.

• Would it be too much to ask for an explicit example where I can clearly see the conjecture not holding but we do it with the momentum operator (as you suggested)? – More Anonymous Oct 7 '18 at 12:47
• @MoreAnonymous I'm not sure what you mean by an explicit example. Your velocity operator doesn't exist, so it (or $\mathrm{e}^{T(v)}$) can't act on any example of a path. – ACuriousMind Oct 7 '18 at 12:59
• Let's make the following changes to the question ... I restrict myself to the case of $\hat T = p^2/2m$ and I wedge $|p \rangle \langle p |$ instead .... – More Anonymous Oct 7 '18 at 13:06
• Also, we can define velocity as $\hat v =\hat p /m$ in this case ... So in the classical case one can infer the momentum as $m (x_A -x_B)/(t_A-t_B)$ – More Anonymous Oct 7 '18 at 13:09