On Sakurai page 127 he gives the formula

$$ (1)~~~~~\langle x_n,t_n|x_{n-1},t_{n-1}\rangle = \left[\frac{1}{w(\Delta t)}\right] \exp\left[\frac{im(x_n-x_{n-1})^2}{2\hbar\Delta t} \right]$$

Noting the orthonormality of Heisenberg picture position eigenkets at equal times, $$(2)~~~~~\langle x_n,t_n|x_{n-1},t_{n-1}\rangle|_{t_n=t_{n-1}} = \delta(x_n-x_{n-1})$$

we get $$(3)~~~~~~\frac{1}{w(\Delta t)}=\sqrt{\frac{m}{2\pi i\hbar\Delta t}}$$

by using

$$(4)~~~~~\int^\infty_{-\infty}d\xi \exp\left(\frac{im\xi^2}{2\hbar \Delta t}\right) = \sqrt{\frac{2\pi i \hbar \Delta t}{m}}$$

and $$(5)~~~~~~\lim_{\Delta t \rightarrow 0}\sqrt{\frac{m}{2\pi i \hbar \Delta t}}\exp\left(\frac{im\xi^2}{2\hbar \Delta t}\right)=\delta (\xi) $$

Now I just assumed we are taking the integral of (1) with respect to $\xi$ and which gives (2) for the left hand side and (4) for the right hand side. I'm missing why (5) is necessary, unless it's just a normalization check? Any help would be greatly appreciated!

  • $\begingroup$ I think what he's saying is that in the limit that $\Delta t\to 0$ what you recover is just the expression for a position eigeinstate at $\xi=0$. $\endgroup$ Aug 23, 2018 at 21:42

1 Answer 1


I am assuming the RHS of eq. 5 is saying $\delta(\xi)$.


As you said yourself, eq.2 is only true for equal times, i.e. ($t_n - t_{n-1}) = \Delta t = 0$.

$\xi$ here is $(x_n - x_{n-1})$.

So the LHS of eq.2 is $\delta(x_n - x_{n-1}) = \delta(\xi)$.

Eq.5 is simply proving the LHS of eq.2, by taking the limit $\Delta t \rightarrow 0$ of the LHS of eq.1.


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