Isaac0427 said:
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?
This is a good definition of the Dirac delta; it has area 1,
always. However, I disagree with your normalized position eigenfunction. The Dirac delta
is the normalized position eigenfunction.
To see why this is the case, we need to talk about orthonormality. If the spectrum of an operator is
discrete (that is, there is a finite spacing between each eigenvalue) then its eigenvectors have
true orthonormality; i.e., if \left\lvert f_m\right> and \left\lvert f_n\right> are eigenvectors of some operator with a discrete spectrum, then \left<f_m\big\vert f_n\right>=\delta_{mn} However, if an operator's spectrum is
continuous, then its eigenvectors can only achieve something that Griffiths calls "Dirac orthonormality": \left<f_{x'}\big\vert f_x\right>=\delta(x-x') Considering the delta function here is zero everywhere except when x=x', this is very close to true orthonormality.
What does this have to do with the formula you have provided for the delta function? It means that if the position eigenfunctions are really Dirac deltas, we should expect to get
another Dirac delta when we take the inner product of two eigenfunctions.
The position eigenfunction should satisfy the equation xg_y(x)=yg_y(x) where y is the eigenvalue. We are looking for a function with the property that multiplying it by any x is the same as multiplying it by y. Furthermore, we know that g_y(x)=0 wherever x\not=y. However, at x=y, g_y(x) can equal any positive, nonzero value (if we chose a negative value, this would be somewhat redundant since the negative signs would just cancel in the equation above). This fits the bill for a Dirac delta function in x centered at y. Since we don't yet know that the delta function is a
normalized eigenfunction of the position operator, we'll (naively) slap a normalization constant in front of it. So g_y(x)=A\delta(x-y)
Now, the big revelation... eigenvectors of an operator with a continuous spectrum are
not square integrable. (If you have a copy of Mathematica, you can see this for yourself. Take your definition of the Dirac delta, square it, and take the integral from -\infty to \infty. If you then let a\rightarrow0, you'll see that it blows up.) However, we have another property that we need to verify: Dirac orthonormality. Our eigenvector is no good at all if it does not satisfy Dirac orthonormality. Note \left<g_{y'}\big\vert g_y\right>=|A|^2\int_{-\infty}^\infty\delta(x-y')\delta(x-y) \, dx=|A|^2\delta(y-y') To satisfy Dirac orthonormality, we let A=1, so g_y(x)=\delta(x-y) is the eigenfunction of the position operator. Unfortunately, normalization just doesn't makes sense in this context, so this is as good as we can get.
If we're to perform these same integrals with your definition of the Dirac delta (assuming A=1 for simplicity), \left<g_q'\big\vert g_q\right>=\lim_{a\rightarrow0}\lim_{a'\rightarrow0}\int_{-\infty}^\infty\frac{1}{a'\sqrt \pi}e^{-(x-q')^2/(a')^2}\frac{1}{a\sqrt \pi}e^{-(x-q)^2/a^2} \, dx=\quad\lim_{a\rightarrow0}\lim_{a'\rightarrow0}\frac{1}{aa'\sqrt{\frac{1}{a^2}+\frac{1}{(a')^2}}\sqrt{\pi}}e^{-\frac{(q-q')^2}{a^2+(a')^2}} Though this may not look familiar at first, it behaves just we expect it too. Note that \exp(-\frac{(q-q')^2}{a^2+(a')^2}) becomes arbitrarily narrow as a and a' go to zero. Furthermore, aa'\sqrt{\frac{1}{a^2}+\frac{1}{(a')^2}}\sqrt{\pi} goes to zero as a and a' go to zero, so the height of the gaussian blows up. This is precisely the behavior of a delta function. In fact, the behavior is identical to if we had gone ahead and written the resulting delta function according to your original definition: \delta(q-q')=\lim_{a \rightarrow 0}\frac{1}{a\sqrt \pi}e^{-(q-q')^2/a^2}<br />