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The propagator of a free particle in 1d is $$ K(x_b, t_b; x_a, t_a ) = \sqrt{\frac{m}{2\pi i \hbar (t_b-t_a)}} \exp \left [ \frac{i m (x_b-x_a)^2}{2 \hbar (t_b-t_a)} \quad \right ] .$$ It looks nice.

But, here we have a square root of $i$. Between the two roots, which one should be taken? Based on what rule?

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  • $\begingroup$ I'm not sure what you mean by "Which one should be taken?" $\endgroup$
    – Kyle Kanos
    Nov 20, 2014 at 21:03
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    $\begingroup$ @KyleKanos: I think the OP means that you can interpret $\sqrt{i}=\pm \frac{1+i}{\sqrt{2}}$, and that there is a sign ambiguity. $\endgroup$
    – Adam
    Nov 20, 2014 at 21:07
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    $\begingroup$ Well, it's not as if the root of a real number was unambiguous...why do you ask that question for $\sqrt{i}$ and not for $\sqrt{m}$? $\endgroup$
    – ACuriousMind
    Nov 20, 2014 at 21:12
  • $\begingroup$ Related: physics.stackexchange.com/q/90558/2451 $\endgroup$
    – Qmechanic
    Nov 20, 2014 at 21:13
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    $\begingroup$ Indeed. If it was only $m$, i would take the positive one without hesitation. But, for $i$, i am really puzzled. $\endgroup$ Nov 20, 2014 at 21:14

3 Answers 3

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That propagator is nothing but the analytic continuation of the Green function of heat equation from real positive to imaginary values of $t_b-t_a$. The cut in the complex plane to make single valued the square root has therefore to be put along the negative real axis, or however in the semiplane $x<0$. With this cut the square root is well defined.

All that means that $\sqrt{i} = e^{i\pi/4}$ is the mathematically correct choice in that formula, the one producing a Dirac delta for $t_a=t_b$.

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  • $\begingroup$ A simple way to choose the correct root is noting that if the square-root branch cut is along the negative real axis, then all square roots will have a positive real part. $\endgroup$ Nov 20, 2014 at 22:15
  • $\begingroup$ Actually here we are iterested on what happens for imaginary values analytically continued from the positive real axis, so the cut can more generally be taken in the real negative semiplane exiting from the origin, not necessarily along the negative axis. $\endgroup$ Nov 20, 2014 at 22:22
  • $\begingroup$ Yeah. But for a quick-and-ready rule of thumb, if the negative real axis works as a cut for $\sqrt{z}$ for your purposes, then $\mathrm{Re}(\sqrt{z})>0$. $\endgroup$ Nov 20, 2014 at 22:52
  • $\begingroup$ @Emilio Pisanty: I agree, the rule of thumb is put the cut along the negative real axis. $\endgroup$ Nov 21, 2014 at 8:29
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Define $\Delta t := t_b-t_a$ and $\Delta x := x_b-x_a$.

One should ensure that

$$\tag{1}{\rm Re}(i\Delta t)~>~0$$

is positive in order for the exponential factor

$$\tag{2}\exp \left [- \frac{ m}{2 \hbar}\frac{(\Delta x)^2}{ i\Delta t} \right]$$

to be exponentially damped.

Equivalently, one should perform the Feynman $i\epsilon$ prescription, i.e., substitute $ \Delta t\to\Delta t-i\epsilon$ in the propagator. This requirement (1) is to ensure that

$$\tag{3} \langle x_b ,t_b | x_a ,t_a \rangle ~\longrightarrow~\delta(\Delta x) \quad \text{for} \quad \Delta t \to 0$$

when one picks the branch of the square root that has positive real part.

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Since the propagator is defined by the relation $$ \psi(x,t) = \int K(x',x,t,t') \psi(x',t') dx'\, dt' $$ A sing ambiguity would result in a change of phase of the wavefuntion, which does not alter the predictions of quantum mechanics. So it doesn't matter which square root you take, as a matter of fix ideas i always take the root

$$ \sqrt(z) = \|z\|^{1/2} \, e^{i \arg(z) / 2} $$

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    $\begingroup$ It is physically correct, however, from the mathematical viewpoint only a phase choice produces an identity for $t=t'$ in your first identity... $\endgroup$ Nov 20, 2014 at 22:04

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