Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

We normalize the wave function to $1$, but couldn't we also normalize it to $-i$ as $(-i)^2=1$?

Does this not work? Is it equivalent?

share|cite|improve this question

Normalizing $\psi$ to $1$ means that we ensure that

$$ \int|\psi|^2dx = 1 $$

normalizing it to $-i$ would presumably mean ensuring that

$$ \int|\psi|^2dx = -i $$

which is impossible because the integrand $|\psi|^2$ is positive everywhere.

share|cite|improve this answer
Oh whoops... Haha... Okay, can we just normalize to $-1$ then? I guess that doesn't matter. – user24082 Sep 19 '13 at 4:27
And do we always normalize the square? Why not just the normal wave function? – user24082 Sep 19 '13 at 4:28
@BMS I wouldn't say it's misleading because the term "normalize" arises from setting the norm of something equal to 1. But there are different kinds of norms, and the details of what it means to normalize depend entirely on which kind of norm you're using. For example the standard norm of a vector $\mathbf{v}$ is $\sqrt{v_x^2 + v_y^2 + v_z^2}$; the standard norm of a wavefunction is $\sqrt{\int\lvert\psi\rvert^2}$. If you set $\int|\psi| = 1$ instead, you wouldn't be using the norm. (Well, it'd be a norm, the L1 norm, which is rather arcane and sees essentially no use in physics.) – David Z Sep 19 '13 at 4:44
Since the integrand is positive everywhere, how can we normalize to $-1$? We normalize the square because $|\psi(x)|^2dx$ is the probability of finding the particle in the small interval $(x,x+dx)$; hence $\int |\psi|^2 dx$ is the probability of finding the probability in the interval $(-\infty, \infty)$, which better be $1$. – xuanji Sep 19 '13 at 4:44
@Tony Actually there is an interesting discussion around this. If you have real positive functions which norm to 1 in the the so-called L1 norm $\int |\psi| = 1$, you get ordinary probability theory. If you do the L2 norm $\int |\psi|^2 = 1$ that's QM and interesting. You could also ask about Lp norms $\int |\psi|^p = 1$ for any $p$... but turns out that only $p=1,2$ make any sense because of a thing called Gleason's theorem (I'm brushing past some technicalities - the moral is right). – Michael Brown Sep 19 '13 at 5:39

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.