2

I was doing a Quantum Mechanics problem and came across the quantity $(-1)^{(-{1\over2})}$ within an infinite square well wavefuntion of $\Psi(x,t)=\Psi_1+e^{i\phi}\Psi_2$ and asked to analyze its mathematical behavior when $\phi=0,\pi,\pi/2$, and $\pi/2$ in my $\Psi_2$ presented the problem. I can't find any resources in my textbook or online about this quantity, except for this post, which didn't totally address my concern. How should complex numbers be treated in such a situation? It seems almost dependent on an order of operation, but there are two possible results: $i$ and $-i$. $$(-1)^{(-{1\over2})} = {(-1)^{{(-1)^{({1\over2})}}}} = \left({1\over-1}\right)^{({1\over2})} = \left({1\cdot(-1)\over(-1)\cdot(-1)}\right)^{({1\over2})} = (-1)^{({1\over2})} = i$$ or omission of the num & denom multiplication of $(-1)$ could result in the same output as below: $$(-1)^{(-{1\over2})} = {(-1)^{{({1\over2})^{(-1)}}}} = (i)^{(-1)} = \left({1\over i}\right) = \left({1\cdot i\over i\cdot i}\right) = \left({i \over -1}\right) = \left({i\cdot(-1)\over(-1)\cdot(-1)}\right) = -i$$ As the other post mentioned, online calculators favor the second result, but I couldn't glean whether there was a reason for that bias. Between these two (or something else), is there a more correct or conventional result, especially in Quantum Mechanics? I know that square roots yield both positive and negative results, but I beleive that only applies to real numbers, and I haven't yet learned enough complex analysis to know how they're treated in complex spaces.

Any help would be appreciated!

Philip Neumeister
  • 21
  • 4
  • It does not matter. In quantum mechanics all quantities that have a physical meaning are real numbers. So for a given wavefunction $\Psi$, $-\Psi$, $i\Psi$, $-i\Psi$, and in general $e^{i\phi}\Psi$, correspond to the same probability density – Andrei Feb 06 '21 at 07:05
  • 2
    A complex power with a fractional exponent is multivalued. In other words, ambiguous. Whoever uses such notation should explain. It can be made somewhat "canonical" by describing a range for the complex logarithm, but then negative reals (as bases) usually end up sitting on a fence. – Jyrki Lahtonen Feb 06 '21 at 07:08
  • 4
    You ask "how should complex numbers be treated in such a situation," but you never really clarify what the situation is. What is this problem, and how did $(-1)^{1/2}$ arise? – anon Feb 06 '21 at 07:09
  • 1
    When you encounter a complex number with square equal to $-1$, how do you know whether it is $i$ or $-i$? – Ivan Neretin Feb 06 '21 at 07:39
  • 1
    I guess it is worth reminding you that the rule $(a^b)^c=a^{bc}$ you use is not true in general. It works, if A) both $b$ and $c$ are integers or B) $a$ is real and positive, and $b$ and $c$ are real. The latter case because in that case we can declare the power to be a positive real number to resolve branch issues like this. Look up questions tagged [tag:fake-proofs] for several examples of how things go South if you apply this rule in a situation where there is ambiguity. – Jyrki Lahtonen Feb 06 '21 at 07:49
  • @runway44 The situation was that this quantity showed up as the result of a $\phi = \pi/2$ in an $e^{(i\phi)}$ term on one of the wave functions in a linear combination of two wavefunctions $\psi_1 + e^{(i\phi)}\psi_2$ – Philip Neumeister Feb 06 '21 at 08:37
  • @JyrkiLahtonen That's good to know. I didn't know it wasn't generally true in nonreal spaces, Thanks for that. – Philip Neumeister Feb 06 '21 at 08:38
  • @Andrei It does matter for what I'm doing, since I'm supposed to be analyzing mathematical behavior at different radial intervals (see above comment), and the answer to my question will drastically change the mathematical behavior of my wavefunction. – Philip Neumeister Feb 06 '21 at 08:48
  • Edited post for context. – Philip Neumeister Feb 06 '21 at 09:02
  • Well, in the situation you describe OP, you're just using Euler's formula for $e^{i\phi}$, there is no $(-1)^{1/2}$. You shouldn't rewrite $e^{i\pi/2}$ as $(e^{i\pi})^{1/2}$. – anon Feb 06 '21 at 09:19
  • @runway44 I realized how that sounded now. It ended up as an $e^{-i\pi/2}$ term because of my $\Psi_2$ wavefunction (which then coincidentally worked out to $e^{-i\pi/2} * \Psi_2$). And yes, I used Euler's formula, because there was it was the best way to get a concise behavioral description with $\phi=\pi$. It seems it should it only not work similarly with $\pi/2$ because of the linked post in the previous comments. – Philip Neumeister Feb 06 '21 at 09:47

0 Answers0