# What is the meaning when the expected value is equal to zero for a free particle?

Let there be a free particle with mass $$m$$.
At time $$t=0$$ it can be described as following wave packet: $$\psi(x) = \frac{1}{\sqrt{2a}} \Theta(a-|x|)$$ where $$\Theta$$ is the Heaviside step function and $$a > 0$$

The expected value is given by: $$\langle x \rangle = \int x |\psi(x,t)|^2dx$$

The function is equal $$1$$ from $$-a$$ to $$a$$ and everywhere else its equal $$0$$.
Therefore, the particle can only be found between $$-a$$ and $$a$$.

$$\Rightarrow \langle x \rangle = \int_{-a}^{a} x |\psi(x)|^2dx = \frac{1}{2a} \int_{-a}^{a} x | \Theta(a-|x|)|^2dx = \frac{1}{2a} \int_{-a}^{a}xdx = 0$$

However, now I am beginning to doubt my thought process, since I'm not sure what the expected value represents for the particle if its equal to zero.
Any help is appreciated.

If $$\langle x\rangle=0$$, then if you take a bunch of position measurements of the particle in this state, the average of those measurements will be zero. This makes sense, since the wavefunction is evenly distributed around zero.