# What does the mean value of the wave function signify?

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.)

• What we were hoping for was that the real and imaginary parts might say something about the particle in question. – 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.
• @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. – Emilio Pisanty Apr 23 '18 at 22:05
• 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!) – Curiosity Apr 23 '18 at 22:22
• 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 $p=0$. – dodosoft Apr 23 '18 at 22:34