Meant to reply to this a bit back, this is probably a stupid question, but...

The uncountable set that you would intuitively think is a basis for Hilbert space, namely the set of functions which are zero except at a single value where they are one, is in fact not even a sequence of distinct elements of Hilbert space, since all these functions are elements of , and are therefore considered to be equivalent to the zero function.

What about the semi intuitive notion of having the dirac delta distributions as a basis? ie, a basis delta(X—R) parameterized by the vector R? How does that fit into all this?

Good question! The Dirac delta distributions are a basis in a certain sense, but not in the sense that I was talking about in my previous comment (which is the sense in which mathematicians and physicists say that “the Hilbert space of quantum mechanics has a countable basis”). I realize now that I should have been more clear about what kind of basis I was talking about, which is an orthonormal basis—each element of the basis is a unit vector, and the lines spanned by distinct basis elements meet at right angles. Implicit in this formulation is the assumption that elements of the basis will be elements of Hilbert space. This is why the Dirac delta distributions are not a basis in this sense—they are not elements of Hilbert space; in fact they are not even functions but are rather generalized functions). Physicists also like to say that they are “nonrenormalizable” in the sense that “no scalar multiple of a delta function is a unit vector”—illustrating failure of the criterion of orthonormality in a more direct way.

The sense in which the Dirac delta distributions are a basis is that any element of Hilbert space can be written as a integral combination of them:

\delta_x%20\;dx)

(Both sides of this equation are considered in the distributional sense, so what this formula really means is that for any function

,
\left(\int%20g\delta_x\right)\;dx,)

which is a tautology.) This is of course a very different statement from the notion of orthonormal basis discussed above.

So what are some differences between these two notions of bases?

Orthonormal bases have the advantage that any two orthonormal bases have the same cardinality, allowing dimension to be defined consistently. By contrast, if one applies a Fourier transform to Hilbert space on [0,1], one gets Hilbert space on the integers; but the former has an uncountable basis of Dirac delta functions while the latter has a countable basis of Dirac delta functions. The Fourier transform is a unitary transformation, so intuitively that means it shouldn’t change the dimension (or other properties) of the Hilbert space. So the size of the Dirac delta basis is not a good way of talking about dimension.

Orthonormal bases take the point of view that Hilbert space is an abstract geometric object, whose properties are determined only by its elements and the distances between them as defined by the distance function I described in my previous comment. By contrast, Dirac delta bases only make sense when you go back and think of the elements of Hilbert space as functions again. Both these points of view can be useful. A big advantage of the abstract approach is that it means that unitary transformations will automatically preserve all relevant properties (e.g. Fourier transform preserving dimension as noted above).

So to summarize, both bases are useful, but the orthonormal basis is the right basis with respect with which to ask and answer the question “What is the dimension of Hilbert space?”

Nice explanation. Just to provide a reference, I would mention that the theory that extends Hilbert spaces to distribitions goes under the name of Rigged Hilbert spaces.

Aaaaarggghh! (sorry, that was just because I realized I was being stupid… specifically that I’d been thinking of the deltas as orthonormal because the integral of a delta = 1.)

Though… it occurs to me that one could construct something that acted like a “square root of a delta”, which would then make an orthonormal basis (though still not part of the hilbert space).

Though… it occurs to me that one could construct something that acted like a “square root of a delta”, which would then make an orthonormal basis (though still not part of the hilbert space).

I’m not sure what you’re trying to construct, but note that one can only multiply distributions under rather restrictive conditions. There are some even more abstract classes of distributions which permit an associative multiplication (Colombeau algebras, generalized Gevrey classes of ultradistributions, and so on) but they’re neither terribly common nor fun to work with.

Ah, nevermind then. I was thinking something like let b(x,k) = 1/sqrt(2k) when |x| < k and 0 otherwise

then define integral B(x)f(x) dx as the limit as k->0+ of integral b(x,k)f(x) dx

I was thinking that then integral (B(x))^2 f(x) dx would be like integral delta(x)f(x) dx.

Now that I think about it more carefully, especially in light of your comment, perhaps that was naive and that wouldn’t actually work. (Yeah, I can see now my reasoning wasn’t actually valid there. Whoops.)

Meant to reply to this a bit back, this is probably a stupid question, but...

What about the semi intuitive notion of having the dirac delta distributions as a basis? ie, a basis delta(X—R) parameterized by the vector R? How does that fit into all this?

Good question! The Dirac delta distributions are a basis in a certain sense, but not in the sense that I was talking about in my previous comment (which is the sense in which mathematicians and physicists say that “the Hilbert space of quantum mechanics has a countable basis”). I realize now that I should have been more clear about what kind of basis I was talking about, which is an orthonormal basis—each element of the basis is a unit vector, and the lines spanned by distinct basis elements meet at right angles. Implicit in this formulation is the assumption that elements of the basis will be elements of Hilbert space. This is why the Dirac delta distributions are not a basis in this sense—they are not elements of Hilbert space; in fact they are not even functions but are rather generalized functions). Physicists also like to say that they are “nonrenormalizable” in the sense that “no scalar multiple of a delta function is a unit vector”—illustrating failure of the criterion of orthonormality in a more direct way.

The sense in which the Dirac delta distributions are a basis is that any element of Hilbert space can be written as a integral combination of them:

\delta_x%20\;dx)(Both sides of this equation are considered in the distributional sense, so what this formula really means is that for any function

, \left(\int%20g\delta_x\right)\;dx,)which is a tautology.) This is of course a very different statement from the notion of orthonormal basis discussed above.

So what are some differences between these two notions of bases?

Orthonormal bases have the advantage that any two orthonormal bases have the same cardinality, allowing dimension to be defined consistently. By contrast, if one applies a Fourier transform to Hilbert space on [0,1], one gets Hilbert space on the integers; but the former has an uncountable basis of Dirac delta functions while the latter has a countable basis of Dirac delta functions. The Fourier transform is a unitary transformation, so intuitively that means it shouldn’t change the dimension (or other properties) of the Hilbert space. So the size of the Dirac delta basis is not a good way of talking about dimension.

Orthonormal bases take the point of view that Hilbert space is an abstract geometric object, whose properties are determined only by its elements and the distances between them as defined by the distance function I described in my previous comment. By contrast, Dirac delta bases only make sense when you go back and think of the elements of Hilbert space as functions again. Both these points of view can be useful. A big advantage of the abstract approach is that it means that unitary transformations will automatically preserve all relevant properties (e.g. Fourier transform preserving dimension as noted above).

So to summarize, both bases are useful, but the orthonormal basis is the right basis with respect with which to ask and answer the question “What is the dimension of Hilbert space?”

Nice explanation. Just to provide a reference, I would mention that the theory that extends Hilbert spaces to distribitions goes under the name of Rigged Hilbert spaces.

Aaaaarggghh! (sorry, that was just because I realized I was being stupid… specifically that I’d been thinking of the deltas as orthonormal because the integral of a delta = 1.)

Though… it occurs to me that one could construct something that acted like a “square root of a delta”, which would then make an orthonormal basis (though still not part of the hilbert space).

(EDIT: hrm… maybe not)

Anyways, thank you.

I’m not sure what you’re trying to construct, but note that one can only multiply distributions under rather restrictive conditions. There are some even more abstract classes of distributions which permit an associative multiplication (Colombeau algebras, generalized Gevrey classes of ultradistributions, and so on) but they’re neither terribly common nor fun to work with.

Ah, nevermind then. I was thinking something like let b(x,k) = 1/sqrt(2k) when |x| < k and 0 otherwise

then define integral B(x)

f(x) dx as the limit as k->0+ of integral b(x,k)f(x) dxI was thinking that then integral (B(x))^2

f(x) dx would be like integral delta(x)f(x) dx.Now that I think about it more carefully, especially in light of your comment, perhaps that was naive and that wouldn’t actually work. (Yeah, I can see now my reasoning wasn’t actually valid there. Whoops.)

Ah well. thank you for correcting me then. :)