Free particle probability to go from $a$ to $b$

Question

Feynman and Hibbs write that the probability for a particle to go from $a$ to $b$ is \begin{equation*} P(b,a)=|K(b,a)|^2 \end{equation*}

The kernel for a free particle is given as \begin{equation*} K(b,a)=\left(\frac{m}{2\pi i\hbar(t_b-t_a)}\right)^{1/2} \exp\left(\frac{im(x_b-x_a)^2}{2\hbar(t_b-t_a)}\right) .\tag{3.3} \end{equation*}

Hence for a free particle the probability is \begin{equation*} P(b,a)=|K(b,a)|^2=\frac{m}{2\pi\hbar(t_b-t_a)} \end{equation*}

Clearly $P(b,a)$ is not dimensionless, but instead has units of inverse area.

Please explain how to interpret $P(b,a)$ since it is not dimensionless.

Answer 1

I don't have a copy of Feynman & Hibbs but I suspect that what you have there is a probability density. To get a plain-vanilla probability, you would have to integrate the probability density over a range of $a$ and $b$. So the probability for the particle to go from $a$ between $a_1$ & $a_2$ to a value of $b$ between $b_1$ and $b_2$ is $$ \int_{a_1}^{a_2} \int_{b_1}^{b_2} P(a,b) \, da \, db. $$ Since $a$ and $b$ have dimensions of length, the result of this integral will be dimensionless, as required.