- #1
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Hi,
Consider this definition of the Dirac delta:
$$\delta (x-q)=\lim_{a \rightarrow 0}\frac{1}{a\sqrt \pi}e^{-(x-q)^2/a^2}$$
First, this would make a normalized position eigenfunction
$$\psi (x)=\lim_{a \rightarrow 0}\frac{1}{\sqrt{a\sqrt \pi}}e^{-x^2/2a^2}$$
right?
If that is so, why do people say that the Dirac delta is a normalized position eigenfunction? The integral of the delta function squared clearly is not equal to one.
I have another question involving the Fourier transform, but it is dependent on the answer to this one, so I will ask it in this thread or a new thread after I receive an answer.
Consider this definition of the Dirac delta:
$$\delta (x-q)=\lim_{a \rightarrow 0}\frac{1}{a\sqrt \pi}e^{-(x-q)^2/a^2}$$
First, this would make a normalized position eigenfunction
$$\psi (x)=\lim_{a \rightarrow 0}\frac{1}{\sqrt{a\sqrt \pi}}e^{-x^2/2a^2}$$
right?
If that is so, why do people say that the Dirac delta is a normalized position eigenfunction? The integral of the delta function squared clearly is not equal to one.
I have another question involving the Fourier transform, but it is dependent on the answer to this one, so I will ask it in this thread or a new thread after I receive an answer.