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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.

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1 Answer 1

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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.

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