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Dirac delta

by Rasalhague
Tags: delta, dirac
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Rasalhague
#1
Jul6-11, 11:06 AM
P: 1,402
I'm reading Daniel T. Gillespie's A QM Primer: An Elementary Introduction to the Formal Theory of QM. In the section on continuous eigenvalues, he admits to playing "fast and loose" with the laws of calculus, with respect to the Dirac delta function. I'd like to understand it better, or, if such understanding requires more advanced conceptual machinery than I have yet, at least to get some pointers. Here's what I've been able to find so far:

Formalised as a probability measure, I think the Dirac delta means

[tex]\delta_x : \Sigma \rightarrow [0,1],[/tex]

[tex]\[ \delta_x(A) = \left\{
\begin{array}{l l}
1 & \quad \mbox{if $x$ is in $A$}\\
0 & \quad \mbox{if $x$ is not in $A$}\\ \end{array} \right. \][/tex]

where [itex]\Sigma[/itex] is any sigma algebra on the configuration space of the system.

Formalised as a distribution, I think it means

[tex]\delta_x : L^2(\mathbb{R},\mathbb{C}) \rightarrow \mathbb{C},[/tex]

[tex]\delta_x(\Psi) = \Psi(x).[/tex]

A distribution is a continuous (in a certain sense) linear functional on the vector space of test functions. A test function is a smooth function with bounded support (Griffel: Applied Functional Analysis).

I wonder how these definitions relate to Gillespie's "fast and loose" Dirac delta. In an expression like

[tex](\delta_x, \Psi),[/tex]

should I reinterpret this notation, for what Gillespie calls the inner product between two state vectors, as

[tex]\delta_x(\Psi)[/tex]

(that is, the value of a dual vector on a vector) or is there a state vector in L2 dual to each Dirac delta distribution? If not, does this mean that the position operator has no actual eigenvectors, only something analogous to eigenvectors, which Gillespie informally treats as eigenvectors but which is really a set of dual vectors? How does the probability measure definition relate to the distribution definition? How does it formalise Gillespie's idea of an infinitesimal multiplied by a function of infinite value?

In the equation

[tex]f(a) \int_{-\infty}^{\infty} f(x) \delta_a(x) dx,[/tex]

should I interpret this as, with the distribution formalism, an integral of the product of [itex]f[/itex] with [itex]\delta_a \circ Id[/itex] considered as a function, where Id denotes the identity test function. And in the measure formalism, as an integral of the product of [itex]f[/itex] with [itex]\delta_a \circ g[/itex], where

[tex]g : \mathbb{R} \rightarrow \Sigma,[/tex]

[tex]g(x) = {x},[/tex]

for some appropriate sigma? It seems to me these naive guesses reintroduce exactly the problem such formalisms are supposed to solve.*

Can anyone recommend a text, online or in print, which takes up these ideas where Gillespie leaves off, beginning at a similarly elementary level.


*EDIT. Ah, perhaps the rule is, when viewing the Dirac delta as a distribution (in the sense of a continuous linear functional on test functions), take the notation

[tex]\int_{-\infty}^{\infty} \Psi(x) \delta_a(a) \; dx[/tex]

to mean, not what it would denote if delta was replaced by any other symbol (in which case, the value would always be zero), but rather take the integral notation to mean [itex]\delta_a(\Psi) \equiv \Psi(a)[/itex]. Not sure how this relates to the measure idea, but maybe some similar device is employed there.

(I wonder if there's any connection between the use of the word distribution for this kind of function, and its use in probability theory to mean a probability measure on the sigma algebra associated with an observation space.)
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mathman
#2
Jul6-11, 03:44 PM
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P: 6,031
There are at least two distinct functions involved in the discussion you have presented.

The first is usually called the characteristic function of a set, i.e. a function which = 1 for points in the set and = 0 for points outside.

The Dirac delta function is the term under the integral sign which makes the integral = value of the function at the point. It is usually written as δ(x-a), but I presume your notation is acceptable. For mathematical rigor, it is called a distribution (completely unrelated to probability distribution).

I am not familiar with the material in the middle of your note, so I can't comment on it.
Rasalhague
#3
Jul6-11, 06:03 PM
P: 1,402
Thanks you, mathman. I've been looking at

http://en.wikipedia.org/wiki/Distrib...mathematics%29

and

http://en.wikipedia.org/wiki/Dirac_measure

I'll need to get a better idea of what Lebesgue integration means to understand the latter article, but I wonder if it's correct to say that the Dirac delta distribution has no "corresponding" test function - corresponding as defined in the Basic Idea section of Wikipedia: Distribution - although it does have a corresponding probability measure, namely the Dirac measure.

George Jones
#4
Jul6-11, 06:39 PM
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Dirac delta

Quote Quote by Rasalhague View Post
*EDIT. Ah, perhaps the rule is, when viewing the Dirac delta as a distribution (in the sense of a continuous linear functional on test functions), take the notation

[tex]\int_{-\infty}^{\infty} \Psi(x) \delta_a(a) \; dx[/tex]

to mean, not what it would denote if delta was replaced by any other symbol (in which case, the value would always be zero), but rather take the integral notation to mean [itex]\delta_a(\Psi) \equiv \Psi(a)[/itex].
Quote Quote by Rasalhague View Post
I wonder if it's correct to say that the Dirac delta distribution has no "corresponding" test function
Yes. From

http://www.physicsforums.com/showthr...48#post1346748
Quote Quote by George Jones View Post
Here's an attempt at a distributional treatment. The ideas were easier than the write-up seems to indicate. Informal reasoning usuall works well, but sometimes it leads ones astray. See this thread. Here's Roger Penrose on the question of mathematical rigour in quantum theory: "Quantum mechanics is full of irritating issues of this kind. As the state of the art stands, one either be decidedly sloppy about such mathematical niceties and even pretend that position states and momentum sates are actually states, or else spend the whole time insisting on getting the mathematics right, in which case there is a contrasting danger of getting trapped in 'rigour mortis'. ... I am not at all sure what the correct answer is for making progress in the subject!''

A distribution is a continuous linear mapping (functional) from [itex]\mathcal{T}[/itex] to [itex]\mathbb{C}[/itex], where [itex]\mathcal{T}[/itex] is the space of test (i.e., sufficiently nice) functions. Any locally integrable function [itex]g[/itex] naturally defines a distribution [itex]G[/itex]:

[tex]G\left[ f\right] =\int_{-\infty }^{\infty }g\left( x\right) f\left( x\right) dx[/tex]

for all test functions [itex]f[/itex]. Not all distributions arise in such a fashion.

Even though the Dirac delta function, defined by [itex]\delta \left[ f\right] =f\left( 0\right)[/itex], is an example of a distribution that doesn't arise in the above manner, it is convenient (and useful!) notationally to pretend that it does, i.e., that [itex]\delta[/itex] is a function such that

[tex]f\left( 0\right) =\int_{-\infty }^{\infty }\delta \left( x\right) f\left( x\right) dx.[/tex]

If [itex]g[/itex] is any function such that

[tex]\int_{-\infty }^{\infty }g\left( x\right) dx=1,[/tex]

then the family of functions [itex]g_{\varepsilon }\left( x\right) :=g\left( x/\varepsilon \right) /\varepsilon[/itex] defines a family of distributions (as above) [itex]G_{\varepsilon }[/itex], with [itex]G_{\varepsilon }\rightarrow \delta[/itex] (in the distributional or weak sense) as [itex]\varepsilon \rightarrow 0[/itex], i.e.,

[tex]f\left( 0\right) =\lim_{\varepsilon \rightarrow 0}\int_{-\infty }^{\infty }g_{\varepsilon }\left( x\right) f\left( x\right) dx.[/tex]

The definitions of differentiation of distributions and multiplication of distributions by somewhat nice functions are both motivated by functions considered as distributions. Let [itex]g[/itex] be a function that has locally integrable derivative [itex]g^{\prime }[/itex], and use [itex]g^{\prime }[/itex] to define the distribution [itex]G^{\prime }[/itex] in the usual way:

[tex]
\begin{equation*}
\begin{split}
G^{\prime }\left[ f\right] &= \int_{-\infty }^{\infty }g^{\prime }\left( x\right) f\left( x\right) dx \\
&= \left[ g\left( x\right) f\left( x\right) \right] _{-\infty }^{\infty }-\int_{-\infty }^{\infty }g\left( x\right) f^{\prime }\left( x\right) dx.
\end{split}
\end{equation*}
[/tex]

Test functions die at [itex]\pm \infty[/itex], so the first term in the last line is zero, giving [itex]G^{\prime }\left[ f\right] =-G\left[ f^{\prime }\right][/itex]. This definition works for all distributions (including the Dirac delta function), not just those that correspond to functions.

Suppose [itex]v\left( x\right) =u\left( x\right) g\left( x\right)[/itex]. Then,

[tex]V\left[ f\right] =\int_{-\infty }^{\infty }v\left( x\right) f\left( x\right) dx=\int_{-\infty }^{\infty }g\left( x\right) \left[ u\left( x\right) f\left( x\right) \right] dx=G\left[ uf\right].[/tex]

This motivates the definition of a distribution [itex]\left( uG\right) \left[ f\right] :=G\left[ uf\right][/itex] for [itex]G[/itex] an arbitrary distribution and [itex]u[/itex] a function.
Rasalhague
#5
Jul7-11, 03:12 AM
P: 1,402
Thanks, George. That's very helpful. Would I be right in thinking that the Dirac delta, although "continuous" in the distribution sense, does not belong to the continuous dual space of H = L2(R,C), since it's not continuous with respect to the norm topologies of H and C, since it's not bounded (in the bounded operator sense)? Also, I think, by the Riesz representation theorem (as stated here by Tillmann Berg, in the Remark section before Unbounded Operators), if g is the inner product function, and there exists no f such that delta = g(f,_), then delta is not in H'.

Distributions are only defined for test functions (=smooth functions with compact support); does L2(R,C) contain functions which are not test functions, and are therefore functions for which distributions, such as the Dirac delta, are undefined? And does this mean that the Dirac delta is not even an element of H*, the algebraic dual space of H = L2(R,C)?
strangerep
#6
Jul7-11, 03:25 AM
Sci Advisor
P: 1,889
Rasalhague,

You seem to want a rigorous treatment. If so, try reading up on
"rigged Hilbert space", aka "Gel'fand triples", in the context
of "generalized functions".

Wikipedia has some (rather brief) information:

http://en.wikipedia.org/wiki/Rigged_hilbert_space


Ballentine's textbook gives an introduction (though he skips some
of the rigor). There's also this useful overview paper: quant-ph/0502053

Distributions are only defined for test functions (=smooth functions with compact support);
One can also take a space of smooth functions of fast decrease at infinity (whose dual is the space of tempered distributions).

does L2(R,C) contain functions which are not test functions, and are therefore functions for which distributions, such as the Dirac delta, are undefined? And does this mean that the Dirac delta is not even an element of H*, the algebraic dual space of H = L2(R,C)?
The Hilbert space is self-dual, but the Dirac delta is not square-integrable (i.e., doesn't
have a sensible Hilbert space norm).

But it's probably best if you check out the references above first and see how many of your questions still remain afterwards.
George Jones
#7
Jul7-11, 04:54 AM
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Quote Quote by strangerep View Post
Rasalhague,

You seem to want a rigorous treatment. If so, try reading up on
"rigged Hilbert space", aka "Gel'fand triples", in the context
of "generalized functions".

Wikipedia has some (rather brief) information:

http://en.wikipedia.org/wiki/Rigged_hilbert_space


Ballentine's textbook gives an introduction (though he skips some
of the rigor). There's also this useful overview paper: quant-ph/0502053
I quite like
Quote Quote by George Jones View Post
For a rigourous overview of rigged Hilbert spaces (Gelfand triples) and Dirac notation, I recommend highly sections 11.2, 11.3, and 12.2 from Quantum Field Theory I: Basics in Mathematics and Physics (A Bridge Between Mathematicians and Physicists) and subsection 7.6.4 from Quantum Field Theory II: Quantum Electrodynamics (A Bridge Between Mathematicians and Physicists) by Eberhard Zeidler.
Fredrik
#8
Jul7-11, 05:57 AM
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P: 9,225
Quote Quote by Rasalhague View Post
Would I be right in thinking that the Dirac delta, although "continuous" in the distribution sense, does not belong to the continuous dual space of H = L2(R,C),
That's right. Distributions are however members of the continuous dual space of the set of test functions, with a more complicated topology than the ones you mentioned.
Rasalhague
#9
Jul7-11, 06:44 AM
P: 1,402
Thanks to everyone for the replies and reading suggestions!

Quote Quote by strangerep View Post
The Hilbert space is self-dual
Self-dual, I take it, meaning having natural anti-isomorphism between itself and its continuous dual; namely, if g is the inner product, [itex]f \mapsto g(f, \cdot )[/itex].


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