What is the difference between a functional and an operator? When we define an operator in physics, e.g. the momentum operator as $\hat{p} = i \frac{d}{dx}$, it is said this operator acts on the wave functions. But isn't something that takes a function as an argument also called a functional? Why do we call $\hat{p}$ momentum operator and not momentum functional?

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    $\begingroup$ Did you read wikipedia? What remained unclear? en.wikipedia.org/wiki/Operator_(mathematics) en.wikipedia.org/wiki/Functional_(mathematics) $\endgroup$
    – Sanya
    Oct 6, 2016 at 7:03
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    $\begingroup$ Maybe it's just me, but a definition from the wikipedia article like "A functional is an operator that maps a vector space to its underlying field." is not very enlightening to me. $\endgroup$
    – asmaier
    Oct 6, 2016 at 7:11
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    $\begingroup$ @asmaier, I agree with you somehow, that in such definition for momentum operator (and the like), something is missing about the wave function which this operator acts on. Perhaps, you should elaborate your question. For example in a quantization procedure: momentum -> momentum operator, ? -> wave function. $\endgroup$ Oct 6, 2016 at 7:32
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    $\begingroup$ To the person who voted to close as "unclear what you're asking": in this case the problem is with you and not with the post. I don't see how it could be clearer. $\endgroup$
    – N. Virgo
    Oct 6, 2016 at 13:17
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    $\begingroup$ Excellent question. And sadly, both "operator" and "functional" are missing from the otherwise excellent A Handbook of Mathematical Discourse, Wells. $\endgroup$
    – davidbak
    Oct 6, 2016 at 22:54

5 Answers 5


Loosely, an operator (acting on a function space) takes functions to functions (e.g., $f(x)$ to $-i f'(x)$). On the other hand, a functional takes functions to numbers (think about a certain integral, or the derivative evaluated at a certain point).

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    $\begingroup$ So a functional is a special case of an operator? $\endgroup$
    – asmaier
    Oct 6, 2016 at 7:11
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    $\begingroup$ @asmaier Yes, that is correct. $\endgroup$
    – DanielSank
    Oct 6, 2016 at 7:17
  • $\begingroup$ @asmaier In particular, the L2 function space is a Hilbert space. When viewed as functions, we apply functionals, and when viewed as vectors, we apply operators. $\endgroup$
    – A_P
    May 31, 2018 at 21:08
  1. An operator is a (not necessarily linear) map from one vector$^1$ space or module to another.

    In operator theory, it is usually implicitly assumed that operators are linear.

    In quantum mechanics, it is usually implicitly assumed that operators are linear or antilinear. (However, see Wigner's theorem!)

  2. A functional is a (not necessarily linear) map from a vector$^1$ space into a field.

    In the topics of calculus of variations and Lagrangian mechanics, the functionals are typically non-linear.

    In functional analysis, it is usually implicitly assumed that functionals are linear.


$^1$ This is Wikipedia's definition (October 2016). However, since the map is not necessarily linear, there is more generally no reason to insist on vector space structure in the first place. E.g. physicists would call the WZW action a functional, even if its domain is technically not a vector space.


Mathematically, we have lots of words that all refer to the same general idea — the precise meaning of a word is not one of universal definition, but of linguistic convention that develops in various subjects (and even then isn't always consistent).

That said, in my experience (as a mathematician), in linear algebra contexts, "functional" is nearly always reserved for scalar-valued linear functions, and "operator" is usually used for an element of some sort of algebra when one intends to work with representations of that algebra (e.g. the algebra of linear endomorphisms of the space of complex-valued functions on the reals, with its representation of acting on said space of functions).

The use of "functional" as meaning any sort of function whose domain includes functions tends to happen more in domains like formal logic or computer science.


As others have noted it is just a matter of definition. A functional is an operator, whose image is the underlying field of numbers. I'd mainly like to give examples to operators and functionals as others have explained the difference quite well.


Suppose we treat $L^2(\mathbb R)$ (i.e. the square integrable functions over $\mathbb R$) as an $\mathbb R$ vector space. Then following maps would be functionals.

The easiest (most boring) I can think of the zero functional i.e. you map every function to the number zero. $b: L^2(\mathbb R) \to \mathbb R$, $f \mapsto 0 \in \mathbb R$. Another non-trivial example would be the integral functional i.e. the map $I : L^2(\mathbb R) \to \mathbb R$, $f \mapsto \int_{\mathbb R} f(x) \, \mathrm dx$. Note that this is also a functional because we take the integral over all space, so that you get a number at the end. Another (more physical functional) is just the 'kinetic energy functional' in classical mechanics. Now it is somewhat radical to call this a functional but in fact it is one:

$$ E :\mathbb R \times \mathbb R^3 \to \mathbb R \;, \quad (m, \vec v) \mapsto \frac m2 \vec v \cdot \vec v$$

where we took $\mathbb R \times \mathbb R^3$ as a $\mathbb R$ vector space.

Operators that are not functionals

Now let's look at operators, which are not functionals. This time we take $C^1(\mathbb R) =: C^1$ (i.e. differentiable functions) as $\mathbb R$ vector space. There is again a trivial example, which maps every function to the zero function (ie. $\hat 0: \mathbb R \to \mathbb R$, $x \mapsto 0$ note that the zero function is differentiable). This operator is given as $B : C^1 \to C^1$, $f \mapsto \hat 0$. A non-trivial operator would be the translation operator as in $T_a : C^1 \to C^1$, $f(x) \mapsto f(x+a)$ for some fixed $a \in \mathbb R$. Note that we don't have any restriction as to what the image of an operator should be. So the differentiation operator $D$ is such an example. $D: C^1 \to C^0$, $f \mapsto \mathrm df/\mathrm dx $ since there are functions, which are just once differentiable (here $C^0$ means continuous functions over $\mathbb R$ taken as $\mathbb R$ vector space.) The cousin of the integral functional is the anti-derivative operator. For that let's take the continuous functions over the unit interval $I:= [0,1]$ and define the map in the following way:

$$ V : C^0(I) \to C^1(I) \;, \quad f(y) \mapsto F(x) := \left(x \mapsto \int_0^x f(y)\, \mathrm dy \right)$$

note that $F(x)$ is a function so the image of the operator $V$ is functions rather than real numbers.

I hope these examples can help you illustrate what operators and functionals are.

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    $\begingroup$ A definite integral is a functional, whereas the antiderivative is an operator. Thanks for this enlightening example. $\endgroup$
    – asmaier
    Oct 6, 2016 at 16:55
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    $\begingroup$ Exactly but don't forget that functionals are some special kind of operators. That is why I wrote above 'operators that are not functionals' ;) $\endgroup$
    – Gonenc
    Oct 6, 2016 at 17:05
  • $\begingroup$ These are great examples. $\endgroup$ Sep 14, 2022 at 18:54

Operators act between vector spaces, they take a vector (in the mathematical sense) as input and give a vector as the output. Of course, those two vector spaces don't have to be the same, in general. Momentum operator $\hat{p}$ acts on a function, which is a vector in the mathematical sense, and outputs another function.

Functionals are also operators. Again, they take an element from a vector space as input but they give a scalar as the output. By scalar, I mean an element of the underlying field (in the mathematical sense) of the original vector space. In other words, the field of coefficients of linear combinations in the vector space. $\hat{p}$ is not a functional because it doesn't give a number, it gives a function. Physicists often use sloppy language and call functionals functions of functions. It's because, in practice, they usually are, they take a function and output a number. A simple example would be the Hamiltonian, it takes $q(t)$ and $p(t)$ as inputs and outputs a number.


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