Let $\Psi(x,t)$ be the wave function of a particle, for instance satisfying the 1-d Schrodinger equation $$ i \hbar \, \partial_t \Psi = - \frac{\hbar^2}{2m} \partial_x^2 \Psi + V \Psi $$ What does the ``mean value'' of the wave function, by which I mean, $$ \int_{-\infty}^\infty \Psi(x,t)\, dx $$ represent physically, if anything? My study group couldn't figure this out, does anyone know of a physical interpretation? (This is not a Homework question, just a question someone brought up.)

  • $\begingroup$ What we were hoping for was that the real and imaginary parts might say something about the particle in question. $\endgroup$ – Curiosity Apr 23 '18 at 22:06

Since one can obtain the wave function in momentum space with a Fourier transform as $$\tilde{Ψ}(t,p) = \frac{1}{\sqrt{2π\hbar}}∫_{-∞}^∞ \mathrm{e}^{-ipx/\hbar} \, Ψ(t,x) \; \mathrm{d}x \;,$$ it follows that the "mean value" you are referring to is $\sqrt{2π\hbar}\,\tilde{Ψ}(t,0)$, i.e. it is proportional to the value of the wave function in momentum space, evaluated at zero momentum.

  • $\begingroup$ thanks, but this is a mathematical explanation, not a physical one. Is there a physical interpretation of the "wave function in momentum space evaluated at zero momentum"? $\endgroup$ – Curiosity Apr 23 '18 at 21:53
  • $\begingroup$ Its square modulus is related to the probability of finding the quantum particle at rest. Although technically it is a probability density: it doesn't really make sense to talk about the probability of finding the particle at exactly zero momentum. In order to obtain a probability, you should integrate over some interval in the momentum variable. $\endgroup$ – dodosoft Apr 23 '18 at 22:05
  • $\begingroup$ @Curiosity Sure there is - release the particle from any trapping potential and let it propagate freely under the kinetic hamiltonian $H = p^2/2m$; then the probability amplitude of it appearing at the origin is proportional to $\tilde \Psi (t,0)$. This is exactly analogous to how the Poisson-Arago spot, in the middle of the shadow of a circular object, is given by the $k=0$ component of the near-field's Fourier transform. $\endgroup$ – Emilio Pisanty Apr 23 '18 at 22:05
  • $\begingroup$ When thinking of "probability density" or "probability amplitute" I think of a non-neg. real-valued function, usually the abs. valued squared like $|\Psi(x,t)|^2$. However, I do not see how the "wave function in momen. space at zero momen." (a complex-valued function) is proportional to a density (say the abs. valued squared of a function). (Forgive my constant questions, I really want to understand this!) $\endgroup$ – Curiosity Apr 23 '18 at 22:22
  • $\begingroup$ Note that I started my comment specifying "its square modulus". In fact, you're right, it is not the wavefunction itself, but its square modulus $|\tilde{Ψ}(t,p)|^2$ which gives the probability density in momentum space. Your integrand, after rescaling and squaring, gives the value of this probability density at the specific value of $p=0$. $\endgroup$ – dodosoft Apr 23 '18 at 22:34

This is called normalizing the wave function and it gives the probability of finding the particle between x and +dx

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    $\begingroup$ No, it is not. He is not integrating the square modulus of the wave function, just the wave function itself. $\endgroup$ – dodosoft Apr 23 '18 at 21:49

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