4

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?

Jiang-min Zhang
  • 4,092
  • I'm not sure what you mean by "Which one should be taken?" – Kyle Kanos Nov 20 '14 at 21:03
  • 3
    @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. – Adam Nov 20 '14 at 21:07
  • 3
    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 $\sqrt{m}$? – ACuriousMind Nov 20 '14 at 21:12
  • Related: http://physics.stackexchange.com/q/90558/2451 – Qmechanic Nov 20 '14 at 21:13
  • 1
    Indeed. If it was only $m$, i would take the positive one without hesitation. But, for $i$, i am really puzzled. – Jiang-min Zhang Nov 20 '14 at 21:14
  • @Jiang-minZhang He means that, by convention, the $\sqrt{}$ symbol means only the positive root. Even though there is a positive and negative solution to $z^2=i$, just as there is with $z^2=n$, by definition we take $\sqrt{i}$ to mean the positive root, just like $\sqrt{n}$ is the positive root. – Jold Nov 20 '14 at 21:47

3 Answers3

8

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 $t_a=t_b$.

Valter Moretti
  • 71,578
  • 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. – Emilio Pisanty Nov 20 '14 at 22:15
  • 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. – Valter Moretti Nov 20 '14 at 22:22
  • 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 $\mathrm{Re}(\sqrt{z})>0$. – Emilio Pisanty Nov 20 '14 at 22:52
  • @Emilio Pisanty: I agree, the rule of thumb is put the cut along the negative real axis. – Valter Moretti Nov 21 '14 at 08:29
2

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.

Qmechanic
  • 201,751
1

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

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

Héctor
  • 903
  • 3
    It is physically correct, however, from the mathematical viewpoint only a phase choice produces an identity for $t=t'$ in your first identity... – Valter Moretti Nov 20 '14 at 22:04