In quantum physics, the relation

$$ \int_{-\infty}^{\infty} (\psi[x,t]^*)(\psi[x,t]) dx=1 \tag{1} $$

is paramount. What would the consequence be of defining the normalization condition as

$$ \int_{-\infty}^{\infty} \sqrt{(\psi[x,t]^*)(\psi[x,t])}dx=1 \tag{2} $$

Of course, it goes without saying the mathematics will now be more complicated due to the square root.

However, (2) is simply the complex norm and thus I feel it is closer to my natural intuition on how complex probabilities ought to connect to real probabilities. So is there at least a trade-off, or perhaps an equivalence?

  • $\begingroup$ If you want something like the quantum version of a probability distribution then you probably want the Wigner function. Might be going towards the same place you are. en.wikipedia.org/wiki/Wigner_quasiprobability_distribution $\endgroup$ – Dast Jun 23 at 13:51
  • 2
    $\begingroup$ What are "complex probabilities"? $\endgroup$ – probably_someone Jun 23 at 14:07
  • 1
    $\begingroup$ Related: physics.stackexchange.com/q/73329/2451 , physics.stackexchange.com/q/116595/2451 and links therein. $\endgroup$ – Qmechanic Jun 23 at 14:08
  • 5
    $\begingroup$ "(2) is simply the complex norm" -- not really. The correct extension to the norm of a complex vector in $\mathbb C^n$ is (1). The second equation is the norm in the $L^1$ sense. This is a well-studied concept in functional analysis, along with the rest of the $L^p$ spaces, but only the $L^2$ norm (i.e. your eq. (1)) arises from an inner product. If your intuition finds (2) more natural, it's because of lack of exposure to the mathematics involved there. $\endgroup$ – Emilio Pisanty Jun 23 at 14:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.