Consider a potential well in 1 dimension defined by(adsbygoogle = window.adsbygoogle || []).push({});

$$

V(x)=

\begin{cases}

+\infty &\text{if}& x<0 \text{ and } x>L\\

0 &\text{if} &0\leq x\leq L

\end{cases}

$$

The probability to find the particle at any particular point [tex] x [/tex] is zero.

$$P(\{x\}) = \int_S \rho(x)\mathrm{d}x=0 ;\forall\; x \in \mathbb{R}$$.

$$S = \{x\}$$ is a null set w.r.t. to the usual integration measure and therefore for each point the assigned probability is zero.

Let's suppose that the energy level is $$E_2$$ so the wavefunction is given by $$\psi_2(x)=(\frac {2}{L})^{\frac {1}{2}} \sin(\frac {2 \pi x}{L})=\psi_{2}^*(x)$$.

Now calculate the "expectation value" of position operator $$\hat{x}$$

$$<x>_2=<E_2|\hat{x}|E_2>=\frac{2}{L}\int_0^L\psi_{2}^*(x)x\psi_2(x)dx$$

So $$<x>_2=\frac{2}{L}\int_0^L x\sin^2(\frac {2 \pi x}{L})dx=\frac {L}{2}$$

Now you say that you "will observe, on average, the particle" at $$x=\frac {L}{2}$$

Can we say that here we indeed "expect" to measure or observe this "possible" event which has "zero" probability, i.e. occurs "almost never"?

Can we say that, respecting the uncertainty principle, if you localize the particle at any particular point, that is with zero uncertainty, then its momentum tends to infinity and thus will not be in this potential well.

Do we need to redefine our common meaning for the words, expectation, possible, and probable?

Also we see that $$\psi_2(\frac {L}{2})=0 \to |\psi_2(\frac {L}{2})|^2=0$$

But what is or how we should interpret the value of $$|\psi_2(\frac {L}{2})|^2$$

Is $$x=L/2$$ any different from other points given this particular wave function?

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# Expecting the possible event of zero probability

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