# Find values of $x$ such that the probability density $|\Psi(x)|^2$ of a wave function is maximized

How do I find the values of $x$ for a given wave function $\Psi(x)$ such that it's probability density function $|\Psi(x)|^2$ will be maximized?

My guess is to to constrain the derivative by the conditions $\frac{d\Psi}{dx} = 0$, and then plug those values into $|\Psi(x)|^2$. Is this the right approach?

• Why would you think it wouldn't be the right approach? May 7, 2014 at 2:47
• Because you shouldn't find $d\psi/dx$, you should find $d|\psi|^2/dx$. May 7, 2014 at 3:11
• Don't forget to also check the boundaries if $x\in[a,b]$ May 7, 2014 at 7:46

To maximize any continuous function $f(x)$, (in your case $f(x) = |\Psi(x)|$)

• See where its derivative $f^\prime(x)=0$. This gives you a set of candidates of where it might be a local maximum, local minimum, or a point of inflection.
• If you are in some kind of a constrained area, like an interval $[a,b]$, or a box in 3 dimensions, then the maximum could also occur at the boundary, where the derivate does not equal zero. These boundary points need to be added to the set of candidates.

At this point, there is often a knee-jerk habit of computing the second derivative (because that is how we were taught), to see if it is a max or min. This is often a waste of time, and sometimes very difficult. Usually it is easier just to evaluate the function, at all the candidates, to find the global maximum.

At times, when the problem is very simple or symmetric, just being able to picture the shape of the function will quickly give you a clue about where the maximum might be located.

Note: If $x$ is a (3 dimensional) vector, then the partial derivate must be zero in all (three) directions.