# While finding out the most probable location from wavefunction, why don't we set the derivative of probability to zero?

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?

• 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 $x$. Dec 4, 2020 at 17:29
• 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. Dec 4, 2020 at 18:12

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. $$\langle \psi | \hat{x} | \psi \rangle = \int x|\psi(x)|^2dx$$.

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