Heisenberg equations of Klein-Gordon Field in Space-Time

[martinbn]

3) A distribution (or generalized function) $f$ is a continuous functional on $\mathcal{D}(\mathbb{R}^{n})$. That is, if $\varphi_{k} \to \varphi$ as $k \to \infty$ in $\mathcal{D}(\mathbb{R}^{n})$, then $(f , \varphi_{k}) \to (f , \varphi )$ as $k \to \infty$.

In short, distribution is a continuous linear functional on the spaces of “nice” functions.

And what does $\varphi_{k} \to \varphi$ as $k \to \infty$ mean?

 

[JD_PM]

Honestly I still do not completely get it either

 

[martinbn]

Honestly I still do not completely get it either

My point was, if you have not even seen the definition how do you expect to understand or prove the continuity!?

 

[samalkhaiat]

$\mathcal{D}(\mathbb{R}^{n})$ is the vector space of operators $\varphi: \Bbb R^n \rightarrow \Bbb R^n$ having the following two properties: 1) $\varphi$ is smooth and 2) $\varphi$ has compact support.

$\mathcal{D}(\mathbb{R}^{n})$ is the space of all infinitely differentiable functions $\varphi : \mathbb{R}^{n} \to \mathbb{R}$ with compact support. The support $\mbox{supp} \ \varphi (x)$ of a function $\varphi(x)$ is the closure of the set on which $\varphi (x) \neq 0$.

Mmm I've got a little doubt here: you said $(f,\varphi)$ is a complex number, but as we're dealing with $L_f:\mathcal{D}(\mathbb{R}^{n}) \rightarrow \Bbb R$ it would make more sense to me that $(f,\varphi) \in \Bbb R$ but of course I may be missing something really evident here).

No, that is a classic misunderstanding. When we turn the set $\mathcal{D}(\mathbb{R}^{n})$ into a vector space over the field of complex numbers, the elements of $\mathcal{D}(\mathbb{R}^{n})$ become complex-valued functions of the real variables $x = (x^{1}, \cdots , x^{n})$ in the open set $\mathcal{O} \subset \mathbb{R}^{n}$. Let $\alpha$ and $\beta$ be two complex numbers and write $\varphi = \alpha \varphi_{1} + \beta \varphi_{2}$. Now, is the number $(f , \varphi )$ real or complex?

For a locally integrable function $f(x)$, the functional $F_{f} [\varphi] \equiv (f , \varphi) = \int d^{n}x \ f(x) \varphi (x)$ is linear in that, if $\varphi_{1}$ and $\varphi_{2}$ are in $\mathcal{D}$ and if $\alpha$ and $\beta$ are two (real or complex) constants, then $$F_{f}[\alpha \varphi_{1} + \beta \varphi_{2}] = \alpha \ F_{f}[\varphi_{1}] + \beta \ F_{f}[\varphi_{2}].$$

 

[samalkhaiat]

OK the idea I have in mind is showing that we get the same result either using $\partial^{k} \left( \partial^{m}f \right)$ or $\partial^{m} \left( \partial^{k} f \right)$

When $\partial^{k} \left( \partial^{m}f \right)$ we get

$$\int dx \ \partial^{k} \left( \partial^{m}f \right) \ \varphi (x) = \partial^{k} \left( \partial^{m-1}f \right) \ \varphi (x) -\int dx \partial^{k} \left( \partial^{m-1}f \right) \ \partial \varphi (x)$$

Applying integration by parts $m$ times one gets

I am afraid that was garbage. For $\varphi \in \mathcal{D}$, use the following equalities $$\partial^{r + s} \varphi = \partial^{r}(\partial^{s}\varphi) = \partial^{s}(\partial^{r}\varphi) ,$$ integrate with $f \in \mathcal{D}^{\prime}$ and multiply the equalities with $(-1)^{r + s}$: $$(-1)^{r + s} (f , \partial^{r + s} \varphi ) = (-1)^{r + s}(f ,\partial^{r}(\partial^{s}\varphi) ) = (-1)^{r + s}(f , \partial^{s}(\partial^{r}\varphi) .$$ Now, use our definition (1) once on the left-hand-side and twice in the middle and the right-hand sides. This gives you $$(\partial^{r + s}f , \varphi ) = ( \partial^{s}(\partial^{r}f) , \varphi ) = (\partial^{r}(\partial^{s}f) , \varphi ).$$ QED.

 

[samalkhaiat]

Honestly I still do not completely get it either

The functional $F_{f} [\varphi]$ is continuous in the following sense: Consider a sequence of functions $\{ \varphi_{k}(x) \}$ in $\mathcal{D} (\mathbb{R}^{n})$ with the following two properties:
1. There exists a compact set $K \subset \mathbb{R}^{n}$ such that for all $k$, $\mbox{supp}\ \varphi_{k} \subset K$.
2. $\lim_{k \to \infty} \partial^{m}\varphi_{k}(x) = 0$ uniformly for all $m = 0, 1, 2, \cdots$.
Such a sequence is said to go to $0$ in $\mathcal{D}$ and is written $\varphi_{k} \to 0, \ k \to \infty$ in $\mathcal{D}$.
We then say that the functional $F_{f}[\varphi]$ is continuous if $F_{f}[\varphi_{k}] \to 0$ for $\varphi_{k} \to 0, \ k \to \infty$ in $\mathcal{D}$.

Let us discuss this definition of continuity of linear functionals on the space $\mathcal{D}$. Continuity is a topological property. The space $\mathcal{D}$ is a linear vector space. It is made into a topological vector space by defining the neighbourhood of the “point” $\varphi (x) = 0$ by a sequence of semi-norms. One can show (and that one is definitely not you) that the two conditions required above in the definition of $\varphi_{k} \to 0$ in $\mathcal{D}$ follow from the conditions used to define the neighbourhood of $\varphi(x) = 0$. The definition of continuity of linear functional on $\mathcal{D}$ can then be based on the weak or strong topologies of space $\mathcal{D}^{\prime}$. It so happens that the definitions of continuity based on these topologies are equivalent to the above definition of $F_{f}[\varphi_{k}] \to 0$ if and only if $\varphi_{k} \to 0$ in $\mathcal{D}$. Furthermore, since $\mathcal{D}$ is a linear space, we can define $\varphi_{k} \to \varphi , \ k \to \infty$ in $\mathcal{D}$ if $(\varphi_{k} - \varphi ) \to 0, \ k \to \infty$ in $\mathcal{D}$, and because $F_{f}[\varphi]$ is linear, we can also say that $F_{f} [\varphi]$ is continuous on $\mathcal{D}$ if when $\varphi_{k} \to \varphi$, we have $F_{f} [\varphi_{k}] \to F_{f} [\varphi]$.

Now, if you have digest the above, proving that $f \mapsto \partial^{m}f$ is continuous from $\mathcal{D}^{\prime}$ into $\mathcal{D}^{\prime}$ is “easy”. First, recall our definition $$(\partial^{b}f , \varphi ) = (-1)^{|m|} (f , \partial^{m}\varphi ) . \ \ \ \ \ \ \ \ \ \ \ \ \ \ (1)$$ Since $\varphi \mapsto (-1)^{|m|}\partial^{m}\varphi$ is linear and continuous, the functional $\partial^{m}f$ defined by the right-hand-side of (1) is a generalized function in $\mathcal{D}^{\prime}(\mathbb{R}^{n})$, i.e., the derivative of a distribution is also a distribution. Now, suppose $f_{k} \to 0, \ k \to \infty$ in $\mathcal{D}^{\prime}(\mathbb{R}^{n})$. Then for all $\varphi \in \mathcal{D}(\mathbb{R}^{n})$ we have $$(\partial^{m}f_{k} , \varphi ) = (-1)^{|m|} (f_{k} , \partial^{m}\varphi ) \to 0, \ k \to \infty ,$$ meaning that $\partial^{m}f_{k} \to 0, \ k \to \infty$ in $\mathcal{D}^{\prime}(\mathbb{R}^{n})$. QED

 

[Alfred1999]

3) Given a generalized function $f \in \mathcal{D}^{\prime}(\mathbb{R}^{n})$ and a good function $a \in C^{\infty}(\mathbb{R}^{n})$, how do you know that the Leibniz rule holds for the differentiation of the product $af$? You don’t. But, if you use the definition (1), you can show that $$\partial^{m}\left( af \right) = \sum_{k \leq m} \begin{pmatrix} k \\ m \end{pmatrix} \ \partial^{k}a \ \partial^{m - k}f .$$

To properly answer this, first let's set up the proper definitions so everything checks out latter.
Let $V$ be a vector space over the scalar field $F$ which can be either $\mathbb{R}$ or $\mathbb{C}$. Given two elements of $V$ such as $u, v \in V$ be elements of $V$, then we define the inner product as map

$$(.,.): V \times V \to F$$
$$ u, v \to (u, v) \in F$$

such that it satisfies the following properties:

1. Conjugate symmetry:​

$$(u, v) = (u, v)^{*}$$

where the notation means that given $z \in \mathbb{C};$ then $z^{*}$ is just the complex-conjugate of $z$.​

2. Linearity in the first argument:​

$$\alpha, \beta \in F; \quad u, v, b \in V; \quad (\alpha u + \beta v, b) = \alpha(u, b) + \beta(v, b)$$

3. Positive definite:​

$$(x, x) \ge 0 \quad \forall x \in V \setminus \{0\}$$

Now, let $\mathcal{D}(\mathbb{R}^n)$ be the vector space over the field $\mathbb{R}$ (if it was over the field $\mathbb{C}$, then we'd say $\mathcal{D}(\mathbb{C}^n)$) such that its elements are "nice" functions $\varphi :\mathbb{R}^n \to \mathbb{R}$. We say that a function is "nice" if:

1. It's smooth
2. It has compact support

And finally, given $f, g \in \mathcal{D}(\mathbb{R}^n)$, we take the our inner product to be:
$$(f, g) := \displaystyle\int d^n x \bigg( f(x) g(x) \bigg)$$

With all this in place, we can finally define a "distribution". If $V = \mathcal{D}(\mathbb{R}^n)$ and $V^{*}$ is its dual space, then $T_f \in V^{*}$ is a linear form such that if $f, \phi \in V$ are nice functions, then $\quad T_f[\phi] = (f, \phi) \in \mathbb{R}$. We call $T_f$ a "distribution", but usually we abuse the notation by calling $f$ itself the distribution. Because of that, it is desirable to define the derivative of a distribution such that $\partial (T_f) = T_{\partial f}$. Using integration by parts and the fact that the nice functions have compact support:

$$(\partial f, \phi) = \displaystyle\int d^n x \bigg( \partial f(x) \phi(x) \bigg) = \cancel{\bigg[ f(x) \phi(x) \bigg]_{-\infty}^{\infty}} - \displaystyle\int d^n x \bigg( f(x) \partial \phi(x) \bigg) = -(f, \partial \phi)$$

$$\implies T_{\partial f}[\phi] = -T_{f}[\partial \phi]$$

From now on we'll abuse the notation too and refer to the distribution $T_f$ just as $f$.

Now, before proving the statement, I would like to point out that there's an error in the binomial number that you've written there, the $m$ and $k$ are swapped. I think the correct formula is:

$$\partial^{m}\left( af \right) = \sum_{k \leq m} \begin{pmatrix} m \\ k \end{pmatrix} \ \partial^{k}a \ \partial^{m - k}f = \sum_{k = 0}^m \begin{pmatrix} m \\ k \end{pmatrix} \ \partial^{k}a \ \partial^{m - k}f$$

based on what I see on wikipedia here.

Anyway, let's begin with the proof:
We'll first prove that the distribution $f$ satisfies the Leibniz rule as usual $\partial(af) = (\partial a) f + a(\partial f)$ . Then the proof for the given formula for a function $a$ and a functional $f$ is the same as it would be for two functions.

Let $f(x), \varphi(x) \in V$ be nice functions and $f$ be a distribution. Also let $\varphi(x) = a(x) \phi(x)$ where $a, \phi \in V$. We'll understand that when we write $f(x)$ we refer to the function and when we write $f$ or $f[.]$ we refer to the distribution.

$$f[\varphi] = (f(x), \phi(x)) = \displaystyle\int d^n x \bigg( f(x) \varphi(x) \bigg) = \displaystyle\int d^n x \bigg( a(x) f(x) \phi(x) \bigg) = $$

$$ = \displaystyle\int d^n x \bigg( [a(x) f(x)] \phi(x) \bigg) = (af)[\phi] \equiv g[\phi]$$

where we have defined the distribution $g$ as $af$ and the function $g(x) := a(x)f(x)$. Now

$$(\partial f, \varphi) = \displaystyle\int d^n x \bigg( \partial f(x) \varphi(x) \bigg) = -\displaystyle\int d^n x \bigg( f(x) \partial\{ a(x) \phi(x) \} \bigg) = $$

$$= -\displaystyle\int d^n x \bigg( \partial a(x) f(x) \phi(x) \bigg) -\displaystyle\int d^n x \bigg( g(x) \partial \phi(x) \bigg) = -(\partial a(x) f(x), \phi(x)) -(g(x), \partial \phi(x))$$

$$(\partial f, \varphi) = \displaystyle\int d^n x \bigg( \partial f(x) \varphi(x) \bigg) = \displaystyle\int d^n x \bigg( [a(x) \partial f(x)] \phi(x) \bigg) = (a(x)\partial f(x), \phi(x))$$

$$(a\partial f)[\phi] = -(\partial af)[\phi] - g[\partial \phi] = -(\partial af)[\phi] + (\partial g)[\phi] \implies$$

$$\implies \bigg( \partial(af) \bigg) = \bigg( (\partial a) f \bigg) + \bigg(a(\partial f) \bigg)$$

and so we've proven that distributions obey the Leibniz rule.

Q.E.D.​

And so, because distributions behave the same way under differentiation as functions do, you can copy-paste the proof for the formula

$$\partial^{m}\left( af \right) = \sum_{k \leq m} \begin{pmatrix} m \\ k \end{pmatrix} \ \partial^{k}a \ \partial^{m - k}f = \sum_{k = 0}^m \begin{pmatrix} m \\ k \end{pmatrix} \ \partial^{k}a \ \partial^{m - k}f$$

for $a$ and $f$ being functions (which is trivial) into the one where they are distributions; thus proving the formula for distributions as well.
You can check that proof here.