# Probability of a particle to go from one position to another

In Feynman's lectures on physics Volume:3 Ch:21. Feynman states

The amplitude that a particle goes from one place to another along a certain route when there’s a field present is the same as the amplitude that it would go along the same route when there’s no field, multiplied by the exponential of the line integral of the vector potential, times the electric charge divided by Planck’s constant $$\langle b \rvert a \rangle _{A}= \langle b \rvert a \rangle _{A=0}.exp \Biggl[\frac{iq}{\hbar}\int ^b _a A.ds \Biggl]$$

My doubt is regarding the expression $$\langle b \vert a \rangle_{A=0}$$.

I have never come across this kind of expression. As far as I know when we write $$c=\langle \Psi \vert \psi_n \rangle$$ then we mean that if we do a measurement the probability of getting the state $$\vert \psi _n \rangle$$ as the outcome if our initial state is $$\vert \Psi \rangle$$, is $$\vert c \vert ^2$$. I don't how the inner product of $$\vert a \rangle$$ with $$\vert b \rangle$$ gives the probability of the particle to go from position a to position b as inner product is a way to know how much component of $$\vert a \rangle$$ is contained in $$\vert b \rangle$$.

Also, the wavefunction is defined over the entire spatial dimension for example in one dimension wavefunction is defined from x =$$- \infty$$ to $$\infty$$ so isn't there some fault in saying wavefunction at position a since during the inner product calculation in Hilbert space we have to do integration over a certain range of $$x$$ then the expression $$\langle b \vert a \rangle$$ doesn't make any sense.

## 1 Answer

Feynman is deliberately sloppy with his notation in chapter 21 (he justifies why at the start).

But you can make sense of his $$\langle b \lvert a \rangle$$ following section 8-4.

He is considering a particle starting at position $$a$$ at time $$0$$, evolving for some time $$t$$, then having its position measured at time $$t$$. We want to calculate the probability that particle is measured to be at position $$b$$.

So the precise way to express this (assuming the Schrodinger picture which you are probably most use to) is

$$\langle b \rvert U(t,0) \lvert a \rangle$$

$$U(t,0)$$ is the unitary transformation describing the evolution of the particle wavefunction in the absence of an electrical field. This will basically be a spreading out of the wavefunction starting from a wavefunction localised at $$a$$ to $$\lvert \psi (t) \rangle = U(t,0) \lvert a \rangle$$. So at time t there will be some overlap of the wavefunction with $$\langle b \rvert$$.