- #1
Jano L.
Gold Member
- 1,333
- 75
Hi everybody,
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?
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?