I was told that in order to find the most probable location of a particle we have to differentiate the probability density of a wavefunction. I don't quite get it. If we want to find out the $x$ value for which $f(x)$ is maximum we solve $f'(x)=0$.

By similar reasoning, since the probability is given by:

$$P=\int{|\psi|^2dx}$$ $x$ value for maximum probability must be given by:

$$\frac{dP}{dx}= |\psi|^2=0$$

not: $$\frac{d|\psi|^2}{dx}=0$$

Where did I go wrong?

  • 1
    $\begingroup$ Your second equation has no $x$ dependence after integration. There's no variable with which to take the derivative. That is, $P$ is not a function of $x$. $\endgroup$
    – garyp
    Dec 4, 2020 at 17:29
  • $\begingroup$ Simply your definition of probability needs a domain of integration, if you integrate a probability density in the whole domain you will always get 1. $\endgroup$
    – ohneVal
    Dec 4, 2020 at 18:12

1 Answer 1


The probability density function is what you need to differentiate, not the integral. $$P = \int_D |\psi|^2 dx = 1$$ for any normalized state for $D$ the domain of the problem. So you have to look at the probability density, $f(x) = |\psi(x)|^2$, and find its maximum. More formally, the probability of finding the particle described by $\psi(x)$ within the infinitesimal interval $(x, x+dx)$ is given by $$\int_x^{x+dx} f(x)dx = |\psi(x)|^2 dx $$

Finding the most probable location, amounts to finding the maximum of $f(x)$ for which you can take derivative and equate it to zero, then solve for $x$. Notice that you technically require a non-zero volume (could be small, but not zero) to actually obtain a non-zero probability. Units make sense therefore, in one dimension $$\text{probability} = \frac{\text{probability}}{\text{lenth}}\cdot \text{length} = |\psi(x)|^2 \Delta x$$

P.D. Don't confuse with the expected value of the position or, the average of the position. $$\langle \psi | \hat{x} | \psi \rangle = \int x|\psi(x)|^2dx$$.

  • $\begingroup$ I was told the same thing. I don't understand why. To me it feels like differentiating density to optimise mass. $\endgroup$ Dec 4, 2020 at 17:14
  • $\begingroup$ What do you mean by optimize mass? $\endgroup$
    – ohneVal
    Dec 4, 2020 at 17:15
  • $\begingroup$ from the relation: mass=density*volume, if we wanted to optimise mass, we won't differentiate density right? $\endgroup$ Dec 4, 2020 at 17:18
  • 1
    $\begingroup$ Here, the density is NOT the mass density $\rho = \frac{m}{V}$, but the probability density. $\endgroup$
    – wyphan
    Dec 4, 2020 at 17:21
  • 2
    $\begingroup$ You need to understand better what the integral stands for. It's a DEFINITE integral BTW. $\endgroup$
    – Gert
    Dec 4, 2020 at 17:22

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