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My book on quantum mechanics introduces the notation $\mathcal O(1)$ as follows:

We represent it by the formula $\Delta x \Delta k \gtrsim \mathcal O(1)$ where $\Delta x$ and $\Delta k$ are the "widths" of the two distributions, and we imply by $\mathcal O(1)$ that this is a number that may depend on the functions that we are dealing with, but is not signifiantly smaller than 1.

This seems to differ from how the big-O notation is normally used. Is it related, or simply another function?

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This is an incorrect use of big-O notation. They want something like big-Omega:… – Qiaochu Yuan May 8 '11 at 19:39

2 Answers 2

This is the normal one, but I can see why it might appear strange. $O(1)$ just means it is bounded by some constant, but that this constant may depend on the argument in the function (some). Imagine for a moment that we could bound the function by something like $(1 + 1/n + 1/n^2)$ or $\frac{1}{1 + logn}$ - both are $O(1)$.

However, As Qiaochu has pointed out, Big-O notation refers only to upper bounds. So to say that $f > O(1)$ doesn't carry the correct meaning at all. If $f = O(g)$, then $|f| < k \cdot |g|$ for some constant k and all arguments. I will assume that your book meant that $\Delta x \Delta k$ is $\Omega (1)$, meaning that there is some constant that serves as a lower bound.

In short, this is an improper usage of Big-O.

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@Tim: Wouldn't the meaning of the 1 depend on the units you/the authors of the text are using? In natural units, it may be of the order of one ($\frac{1}{4\pi}$ if I'm not mistaken), but in the gms system it's much smaller than one. – yatima2975 May 8 '11 at 21:08
$O(1)$ does not mean "a constant." To say that $f(x) = O(1)$ is to say that $|f|$ is bounded as a function of $x$; in other words, big-O notation provides upper bounds, not lower bounds. The use of the equality symbol here is an abuse of notation; while it is helpful intuition, it should not be taken too literally. – Qiaochu Yuan May 8 '11 at 22:23
@Qiaochu Yuan: I never stated or implied equality - I meant that $O(1)$ just means bounded by some constant, which is what I had meant by the following phrase in the sentence. But I have edited my answer to say explicitly 'bounded by' at your suggestion.I also never in any way suggested a lower bound at all - where in my answer did you read that I had meant a lower bound? – mixedmath May 9 '11 at 2:31
@Qiaochu Yuan: I suddenly understand what you meant by a lower bound! It is in the question statement, which I hadn't read fully (it seems). I am now editing my answer to include this as well. – mixedmath May 9 '11 at 2:33
You could replace "If $f = O(g)$, then $|f| < k \dot g$..." by "If $f = O(g)$, then $|f| \le k |g|$..." (Both the absolute value and the $\le$ sign are necessary.) – Did May 9 '11 at 9:18

As a physicist, I'm quite positive they don't refer to the normal definition of $O(1).$ As you'll see many more times in your studies, sometimes you don't care about numerical constants as $e^2, \sqrt{2}, \pi^2/6, 1/3, \ldots$. For lots of reasons, authors might choose to summarize such a constant as $O(1)$. There's not a rigorous definition of what can be classified as $O(1)$; personally, I wouldn't classify 0.0000781232 or 33233432 as $O(1),$ but there are no solid conventions.

As an example the true Heisenberg inequality is $$\Delta x \Delta p \geq \hbar / 2,$$ or with $k := p/\hbar,$ $$\Delta x \Delta k \geq 1/2.$$ In order to make you think about the physics behind this inequality, and not just about the factor 1/2, the authors write O(1).

[Edit for clarification.] Generically, the product of the standard deviation depends on some parameters of the problem and some quantum numbers; for a wave packet, you have a wave number $k$, for discrete states you have $n \in \{0,1,2,\ldots\}$, etc. Therefore a mathematical definition of O(1) in this context might be $$\Delta x \Delta k \geq f(k,n,\ldots) \geq 0$$ for some function f, and there exists a number R and a constant C, such that for all $\sqrt{k^2 + n^2 + \ldots} > R$, $f(k,n,\ldots) < C.$ [1]

That's (likely) not what is meant; most of the time, the right hand side actually diverges as a function of these parameters. Physically: if you look at large wave numbers, or states with large quantum numbers, you'll find a large $\Delta x \Delta k.$ You use O(f) to specify what you know about the behaviour of $f$ near infinity, but in that case we only know that the global minimum of $f$ will be larger than 1/2. The factor of 1/2 is thus a 'best case scenario' that you seldomly encounter, and often you find a number that is somewhat larger, but still 'of the order of 1', or as your textbook writes, O(1).

[1] This doesn't even make a lot of sense, since x and k have different dimensions.

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The mathematical definition of $O(1)$ you suggest amounts to requiring the quantity of interest to be nonnegative. – Did May 8 '11 at 23:28
@Didier Piau: true. In fact, the $\Delta x$ and $\Delta k$ represent standard deviations $\sigma_x$ and $\sigma_k,$ so they're always nonnegative. – Gerben May 9 '11 at 0:00
You are missing the point: the definition given in the sentence For completeness: the mathematical definition of O(1) in this context might be... is empty, hence it cannot be a definition of the $O(1)$ property. – Did May 9 '11 at 0:21
@Didier: I get what you're saying and will make the edit. – Gerben May 9 '11 at 8:59

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