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Curious statement about operators in my QM book?

  1. Sep 29, 2005 #1
    In a discussion of the historical motivations for a move from calculus to operators, my QM book says...
    "Many mathematicians were uncomfortable with the 'metaphysical implications' of a mathematics formulated in terms of infinitesimal quantities (like dx). This disquiet was the stimulus for the development of the operator calculus."

    This is all fine and well, but I just don't see how the use of operators lets one escape the 'uncomfortable metaphysical implications' of the calculus. I mean, you're still doing the same thing to a function with the only difference (as far as I can tell) being that you use a far more concise notation to get the same job done.

    How is operator calculus so fundamentally different than plain old calculus? I just cannot see the big difference between the two.
  2. jcsd
  3. Sep 29, 2005 #2


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    I'm guessing that the author was referencing a time when people used "infinitesimals" to do calculus. What they did, in effect, was to expand the number system by adding infinitesimals. Infinitesimals are numbers so small that they are infinitely smaller than any real number.

    That stuff isn't taught any more, so maybe you haven't been exposed to it.


    Let dx be an infinitesimal (positive) number.

    Now suppose s > 4. Then s (for example, s=5) is also greater than 4+dx. And s is greater than 4+2dx. And s is greater than 4+n dx where n is any positive real number.

    Now if dx were instead considered to be a very small real number (rather than an infinitesimal), then one could always choose n to be large enough to make 4+(n dx) bigger than, for example, 5. But the theory of infinitesimals is that one can assume a set of numbers that are smaller than any real numbers. This is a consistent theory, more or less.

    Once you assume the infinitesimals, you can derive the various calculus results that you're familiar with. Look for a very very old calculus book if you want to know more.

    By the way, there's a lot of other math theory that has slowly gone away over the years. It's partly a matter of improvement, and partly a matter of changing styles. The same thing applies to physics. Students are taught the "standard" physics, but little else. There are plenty of alternative ways of getting the same results, but you have to read the literature to learn it. Some of the alternative methods are so out of style that you might have to read the literature of the 1920s or even earlier.

    Last edited: Sep 29, 2005
  4. Sep 29, 2005 #3
    Ok, but when operators were formulated, how did this new way of looking at things do away with infinitesimals?

    I guess I'm actually curious about the difference in how the two forms are defined.

    Is there a difference in their definitions?
  5. Sep 29, 2005 #4

    Tom Mattson

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    This really isn't a homework problem, and it would be a shame if this were to be buried on page 10 by tomorrow morning, so I'm going to scoot this over to Quantum Physics. Also, our best "quantum gurus" aren't known for being Homework Helpers, and it would be nice if they could see this.

    An operator (take the momentum operator for instance) is completely defined by its action its eigenstates. So if I have a pure momentum state [itex]\psi_{p'}(x)[/itex] of momentum [itex]p'[/itex], then I can define the momentum operator [itex]\hat{p}[/itex] by the following eigenvalue equation.


    In the language of differential calculus, this would read as follows.


    The operator allows one to sidestep the infinitesimals simply by not using them. Now today we don't worry about this, because we no longer think of "dy/dx" as a ratio of infinitesimal quantities. We think of it as a ratio of (real-valued) differential 1-forms. Of course they don't explicitly say that in Calculus I, but you've participated in the Differential Forms thread, so you know that. :smile:

    Yes, there is. For starters you can see that the definition of the operator, unlike that of the derivative, makes no reference to limits at all.

    Now you may be thinking, "What's the bloody difference? The two equations carry identical information", and you would be right. As the author's quote says, the difference is metaphysical, not mathematical.

    What I don't understand is why people would be bothered by the use of derivatives in QM and not in, say, Newtonian mechanics.
    Last edited: Sep 29, 2005
  6. Sep 29, 2005 #5


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    An example should suffice. What is the slope of the line given by y = cos(x) at the point x_0?

    Using infinitesimals:

    y+dy = cos(x+dx) = cos(x)cos(dx) - sin(x)sin(dx)
    = cos(x) - sin(x)dx

    So slope = dy/dx = (cos(x) - sin(x)dx - y)dx = -sin(x).

    Note that the derivation required the assumption of a "dx" that is an infinitesimal quantity.

    Modern derivation:

    = lim h -> 0 [(cos(x+h)-cos(x))/h]
    = lim h -> 0 [ (cos(x)cos(h)-sin(x)sin(h)-cos(x))/h]
    = lim h -> 0 [(cox(x)(cos(h)-1)/h] - lim h -> 0 [sin(x)sin(h)/h]
    = 0 - sin(x) = -sin(x).

    The older version, which used infinitesimals, was somewhat simpler. The newer version uses limits which are better defined, but which require a lot of theorems about limits to be proved. Perhaps you can see that the old technique was a bit easier on the student.

    One advantage of the more modern treatment is that students familiar with it will find it easier to do variational (function) problems that will show up later.

    Calculus is handy at finding maxima and minima of functions. But if you instead want a function that minimizes or maximizes some definite integral, that is called a variational problem and is approached with the "calculus of variations".

  7. Sep 29, 2005 #6


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    It is a matter of being concerned with mappings.
    Say L is an operator and F a function space
    you care that
    but you do not care about the process by which f becomes g
    High school algebra teachers love to do a similar thing with functions. They give problems like
    do the ordered pairs
    define a function (x,f(x))
    and you say no because if f is a function
    x=y -> (x,f(x))=(y,f(y))
    and pickle=pickle
    but cabage!=rock
    you do so (hopefully) without two much thought to any process by which

    For example if you wanted derivatives for polynomials on some ring you might say
    is defined by
    that is to say if a is a ring element and p a polynomial
    so you can have derivatives with out limits especialy without saying things like
    Also you could allow, say formal power series.
  8. Sep 30, 2005 #7


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    I'd be very careful about your book's veracity. Operator methods were developed both in Engineering --Oliver Heavyside's work for Laplace transforms around 1900, and in mathematics -- there are numerous examples of operator methods given in G.N.Watson's "A Treatise on the Theory of Bessel Functions, most of which were developed in the 19th century. They all were done with acceptance of the standard methods of calculus. Now make a transition to the mid 20th century to the subject of Functional Analysis -- for example, the classic Functional Analysis and Semi-Groups, by Hille and Phillips, some 750 pages dealing largely with operator methods, and again with full acceptance of calculus, included quite a bit of use of standard delta-epsilon limiting techniques from freshman calculus.

    Indeed, infitesimals are dicey, but are themselves an operator approach, a symbolic description of the standard limiting techniques. Used properly and formally, they are a fine way to go, like in Thermo and Mathematical Economics, both dense with differentials and partial derivatives

    There was(is) a branch of mathematics, developed in the early 1900's that totally eschewed limits, and things like the Axiom of Choice. If I recall correctly the main player was a very eminent Dutch mathematician named Brower. In his school, the Intuitionist School, everything had to be finite; no limits. But, this approach never went very far.

    Note: CarlB gives the standard definition of derivatives, but it is at least 150 years old, so the epithet modern is a bit misapplied. That is, calculus has been on a pretty solid mathematical basis since the early days of the 19th century.

    Reilly Atkinson
  9. Sep 30, 2005 #8


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    In the intuitionist school, every map R→R was continuous!!!

    Anyways, I figure it's worth mentioning that infinitessimals have been placed on a rigorous for fifty years or so, with the introduction of nonstandard analysis.
  10. Oct 1, 2005 #9


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    Hurkyl -- Thanks. ra
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