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I was going through one of the examples in Griffith's Quantum book and there was a few things in Example 3.3 that I didn't understand that I was hoping to get some clarification on.

For instance, we first find that the eigenfunctions of the position operator x, have to be

Starting from the eigenvalue equation

$xg_{y}(x) = yg_{y}(x)$

We get

$g_{y}(x) = A\delta(x-y)$

The first thing I didn't understand was that Griffith's states that these functions are not square integrable. I'm not quite sure what's wrong with the integral

$ \lvert A \rvert \int \delta^{2}(x-y) dx$

The integral without delta square would converge to 1, but squaring the function somehow breaks it?

The other thing I didn't understand was how to get from

$\int g_{y'}^{*}(x)g_{y}(x)dx = \lvert A \rvert \int \delta(x-y')\delta(x-y) dx = \lvert A \rvert^{2} \delta(y-y')$

How do you get from the multpilication of the two delta functions to the single delta function on the right?

Thanks for any help

Qmechanic
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user1236
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4 Answers4

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The integral without delta square would converge to 1, but squaring the function somehow breaks it?

Yes. Recall the sifting property:

$$\int_{-\infty}^{\infty}f(x)\delta(x - a)dx = \int_{-\infty}^{\infty}f(a)\delta(x - a)dx = f(a)$$

Then, it follows formally that

$$\int_{-\infty}^{\infty}\delta^2(x - a)dx = \int_{-\infty}^{\infty}\delta(x - a)\delta(x - a)dx = \int_{-\infty}^{\infty}\delta(a - a)\delta(x - a)dx = \delta(a-a) = \delta(0)$$

But $\delta(0)$ 'equals' $\infty$ so it follows that $\delta^2(x - a)$ is not square integrable.

Alfred Centauri
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The Dirac delta is defined by the equation \begin{equation} \int_a^b\mathrm{d}x \,\delta(x)f(x) = f(0) \end{equation} for $a < 0 < b$. By direct application of this definition we get \begin{equation} \int \mathrm{d}x\, \delta(x - y) \delta(x - y^\prime) = \delta(y - y^\prime) \end{equation}

If we let $a < c < 0$ we can write \begin{equation} \int_a^b\mathrm{d}x \,\delta(x)f(x) = \int_a^c\mathrm{d}x \,\delta(x)f(x) + \int_c^b\mathrm{d}x \,\delta(x)f(x) \end{equation} Now both the integral from $a$ to $b$ and the integral from $c$ to $b$ equal $f(0)$ so we can conclude that for any range not containing 0 the integral $\delta(x)f(x)$ is $0$ for all $f$ so we can say $\delta(x) = 0$ $\forall x \ne 0$. At the origin, however, we have a problem as if we take $f$ to be the constant function $1$ we find that \begin{equation} \int_{-\epsilon}^\epsilon\mathrm{d}x \,\delta(x)1 = 1 \end{equation} no matter how small we make $\epsilon$. This means the Dirac delta has a peak which encloses a finite area, despite having no width. We therefore can't give $\delta(0)$ a well defined value.

From this we can then say \begin{equation}\int_{-\infty}^\infty\mathrm{d}x \,\delta^2(x) = \delta(0) = \mathrm{undefined} \end{equation}

By Symmetry
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The $\delta$ function has the following property: $$ \int \text{d}x\; f(x)\delta(a-x)=f(a) $$ This actually answers both of your questions. First, the non-square-integrability: $$ \int \text{d}x\;\delta(x-y)\delta(x-y)=\delta(y-y)=\delta(0)=\infty $$ according to the rule above if you choose one of the $\delta$'s to be the $f$. Your second question is the same: $$ \int\text{d}x\;\delta(x-y')\delta(x-y)=\delta(y-y') $$

physicus
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  • Thanks for this. It seems obvious now but for some reason I was confused about choosing what delta to sift. I suppose it doesn't actually matter. – user1236 Oct 23 '14 at 23:20
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What happens when you integrate a function multiplied by a delta function? You get:

$$\int f(x) \delta (x-y) dx=f(y)$$

(Because the delta function is zero everywhere, except at zero where its integral gives 1.) So, when integrating over $\delta^2 (x-y)$ we get:

$$\int \delta^2 (x-y) dx=\int \delta (x-y) \delta (x-y)dx = \delta (y-y) = \infty$$

Jold
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