- #1

Jano L.

Gold Member

- 1,333

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it is usually said that the wavefunction of the localized particle at position [tex]x_0[/tex] is

[tex]

\psi_{x_0} = \delta (x-x_0).

[/tex]

Is there some good reason for this? I mean, this function cannot be normalized, which means that [tex]|\psi_{x_0}|^2[/tex] cannot be interpreted as a probability density. There is also problem that we can not use it to calculate the expectation value of position operator:

[tex]

\langle \psi |x|\psi\rangle = \int \delta(x-x_0)x\delta(x-x_0) = ?

[/tex]

What if we tried to solve this problem by square-rooting the delta function? We would have

[tex]

\langle \psi |x|\psi\rangle = \int \sqrt{\delta(x-x_0)}x\sqrt{\delta(x-x_0)} = x_0

[/tex]

What do you think? Why do we use delta function, which does not follow the rules for calculating expectation values?