Heisenberg equations of Klein-Gordon Field in Space-Time

[Haorong Wu]

TL;DR Summary

1

Hi, there. I am reading An Introduction to Quantum Field Theory by Peskin and Schroeder. I am confused about some equations in section 2.4 The Klein-Gordon Field in Space-Time. It computes the Heisenberg equations of $\phi \left ( x \right )$ and $\pi \left ( x \right)$ as (in page 25)

$\begin{align} i \frac \partial {\partial t} \phi \left ( \bf {x}, t\right ) &=\left [ \phi \left ( \bf x, t \right ), \int d^3 x^{'} \{ \frac 1 2 \pi ^2 \left ( \bf { x^{'}}, t \right ) +\frac 1 2 \left ( \nabla\phi \left ( \bf { x^{'}}, t \right ) \right ) ^2 + \frac 1 2 m^2 \phi ^2 \left ( \bf { x^{'}}, t \right ) \} \right ] \nonumber \\ &=\int d^3 x^{'} \left ( i \delta ^{(3)} \left ( \bf x - \bf x^{'} \right ) \pi \left ( \bf x^{'} , t \right ) \right ) \nonumber \\ &= i \pi \left ( \bf x , t\right ) \nonumber\end{align}$;
and
$\begin{align} i \frac \partial {\partial t} \pi \left ( \bf {x}, t\right ) &=\left [ \pi \left ( \bf x, t \right ), \int d^3 x^{'} \{ \frac 1 2 \pi ^2 \left ( \bf { x^{'}}, t \right ) + \frac 1 2 \phi \left ( \bf x^{'} ,t \right ) \left ( - \nabla^2 + m^2 \right ) \phi \left ( \bf x^{'} ,t \right ) \} \right ] \nonumber \\ &=\int d^3 x^{'} \left (- i \delta ^{(3)} \left ( \bf x - \bf x^{'} \right ) \left ( - \nabla^2 + m^2 \right ) \phi \left ( \bf x^{'} ,t \right ) \right ) \nonumber \\ &= -i \left ( - \nabla^2 + m^2 \right ) \phi \left ( \bf x ,t \right ) \nonumber\end{align}$.

The first equation is a piece of cake. But I have some problems about the second equation.

First, the Hamiltonian is given by $H=\int d^3 x^{'} \{ \frac 1 2 \pi ^2 \left ( \bf { x^{'}}, t \right ) +\frac 1 2 \left ( \nabla\phi \left ( \bf { x^{'}}, t \right ) \right ) ^2 + \frac 1 2 m^2 \phi ^2 \left ( \bf { x^{'}}, t \right ) \}$. Then from $\frac 1 2 \left ( \nabla\phi \left ( \bf { x^{'}}, t \right ) \right ) ^2$ how to derive $\frac 1 2 \phi \left ( \bf { x^{'}}, t \right ) \left ( - \nabla ^2 \right ) \phi \left ( \bf { x^{'}}, t \right )$?

Second, I do not know how to derive the second line of the second equation. why does the factor $\frac 1 2$ disappear?

Also I find that the mathematics in the book is quite difficult when it comes to covariant and contravariant tensors , even though I have learned the concepts of them from special relativity. Is there any useful material I could read?

Thanks!

 

[vanhees71]

The first question is just "integration by parts". You can formally derive it using Gauß's integral theorem, writing
$$(\vec{\nabla} \phi)^2=\vec{\nabla} (\phi \vec{\nabla} \phi)-\phi \Delta \phi.$$
Now integrate over space, and the first term gets into a surface integral with the surface at infinity, where by assumption the fields vanish rapidly enough such that the surface integral vanishes.

Concerning the second question, it's clear that canonical field momenta commute and thus you have to deal only with the 2nd term. There you have to apply the general formula for commutators,
$$[\hat{A},\hat{B} \hat{C}]=\hat{A} [\hat{B},\hat{C}] + [\hat{A},\hat{B}] \hat{C}$$
and then use the canonical equal-time commutation relatins between the canonical field momenta and the field,
$$[\hat{\pi}(\vec{x},t),\hat{\phi}(\vec{x}',t)]=-\mathrm{i} \delta^{(3)}(\vec{x}-\vec{x}').$$
From the two fields you get twice the same result, which is why the factor 2 cancels.

Of course, at the end you should get the Klein-Gordon equations for the field operators,
$$(\Box+m^2) \hat{\phi}=0.$$
showing that the "canonial quantization formalism" really leads to a consistent quantum field theory with equations of motion of the field operator derived from the "quantum Hamilton equations of motion" in the Heisenberg picture.

 

[PeroK]

Concerning the second question, it's clear that canonical field momenta commute and thus you have to deal only with the 2nd term. There you have to apply the general formula for commutators,
$$[\hat{A},\hat{B} \hat{C}]=\hat{A} [\hat{B},\hat{C}] + [\hat{A},\hat{B}] \hat{C}$$
and then use the canonical equal-time commutation relatins between the canonical field momenta and the field,
$$[\hat{\pi}(\vec{x},t),\hat{\phi}(\vec{x}',t)]=-\mathrm{i} \delta^{(3)}(\vec{x}-\vec{x}').$$
From the two fields you get twice the same result, which is why the factor 2 cancels.

I get $$[\pi(x, t), -\frac{\partial^2}{\partial t^2}\phi(x',t)]$$ for the second term. I can't see how that leads to another delta function using the commutation relations. How do you deal with that differential operator?

 

[samalkhaiat]

How do you deal with that differential operator?

Differentiate the canonical commutation relation: $$\partial_{x}\big[ \varphi (x ,t) , \pi (y ,t) \big] = \big[ \partial_{x}\varphi (x,t) , \pi (y,t) \big] = i \partial_{x}\delta (x - y) ,$$ and integrate by parts:
$$-i\partial_{t}\pi (y,t) = 2 \frac{1}{2} \int dx \big[ \partial_{x}\varphi (x,t) , \pi (y,t) \big] \partial_{x}\varphi (x,t) + 2 \frac{m^{2}}{2} \int dx \ i \varphi(x,t) \delta (x-y) ,$$ $$-i\partial_{t}\pi (y,t) = i \int dx \ \partial_{x}\delta (x - y) \ \partial_{x} \varphi (x , t) + i m^{2} \varphi (y,t) ,$$ $$-i\partial_{t}\pi (y,t) = i m^{2} \varphi (y,t) - i \int dx \ \delta (x - y) \ \partial^{2}_{x} \varphi (x,t) ,$$ $$-\partial_{t}\pi (y,t) = m^{2}\varphi (y,t) - \partial^{2}_{y}\varphi (y,t) .$$

 

[Haorong Wu]

Thanks, @vanhees71 @PeroK @samalkhaiat, for explaining them so clearly.

 

[vanhees71]

I get $$[\pi(x, t), -\frac{\partial^2}{\partial t^2}\phi(x',t)]$$ for the second term. I can't see how that leads to another delta function using the commutation relations. How do you deal with that differential operator?

You could try that approach, but it's always a bit delicate to have equal-time-commutators with time derivatives involved. So it's better to use @samalkhaiat 's approach in #4.

 

[PeroK]

You could try that approach, but it's always a bit delicate to have equal-time-commutators with time derivatives involved. So it's better to use @samalkhaiat 's approach in #4.

Yes, I think I found that out!

 

[Haorong Wu]

Hi, @samalkhaiat . I am still confusing about the equation
$\partial_{x}\big[ \varphi (x ,t) , \pi (y ,t) \big] = \big[ \partial_{x}\varphi (x,t) , \pi (y,t) \big] = i \partial_{x}\delta (x - y)$.

From $\partial_{x} \left [\pi \left ( x^{'} , t\right ) , \phi \left ( x ,t \right ) \right ]$, I get
$\begin{align} \partial_{x} \left [ \phi \left ( x ,t \right ), \pi \left ( x^{'} , t\right ) \right ]&=\partial_{x} \{ \pi \left ( x^{'} , t\right ) \phi \left ( x ,t \right ) -\phi \left ( x ,t \right ) \pi \left ( x^{'} , t\right ) \} \nonumber \\ &= \partial_{x} \pi \left ( x^{'} , t\right ) \cdot \phi \left ( x ,t \right ) + \pi \left ( x^{'} , t\right ) \cdot \partial_{x} \phi \left ( x ,t \right ) -\partial_{x} \phi \left ( x ,t \right ) \cdot \pi \left ( x^{'} , t\right ) - \phi \left ( x ,t \right ) \cdot \partial_{x} \pi \left ( x^{'} , t\right ) \nonumber \end{align}$.

It seems $\partial_{x} \pi \left ( x^{'} , t\right ) \cdot \phi \left ( x ,t \right ) -\partial_{x} \phi \left ( x ,t \right ) \cdot \pi \left ( x^{'} , t\right ) =0$ from the result, but I do not know why.

 

[samalkhaiat]

Hi, @samalkhaiat . I am still confusing about the equation
$$\partial_{x}\big[ \varphi (x ,t) , \pi (y ,t) \big] = \big[ \partial_{x}\varphi (x,t) , \pi (y,t) \big] = i \partial_{x}\delta (x - y).$$

1) $$[a(x) , b(y)] = a(x)b(y) - b(y)a(x)$$
2) $$\partial_{x}\left( a(x)b(y)\right) = \left(\partial_{x}a(x)\right) \ b(y),$$ because $b(y)$ does not depend on $x$, i.e., $\partial_{x}b(y) = 0$.
Observing (1) and (2), you get $$\partial_{x}[a(x),b(y)] = \left( \partial_{x}a(x)\right) b(y) - b(y) \left( \partial_{x}a(x)\right) = [\partial_{x}a(x) , b(y)].$$

 

[Haorong Wu]

1) $$[a(x) , b(y)] = a(x)b(y) - b(y)a(x)$$
2) $$\partial_{x}\left( a(x)b(y)\right) = \left(\partial_{x}a(x)\right) \ b(y),$$ because $b(y)$ does not depend on $x$, i.e., $\partial_{x}b(y) = 0$.
Observing (1) and (2), you get $$\partial_{x}[a(x),b(y)] = \left( \partial_{x}a(x)\right) b(y) - b(y) \left( \partial_{x}a(x)\right) = [\partial_{x}a(x) , b(y)].$$

Thanks!

 

[PeroK]

Hi, @samalkhaiat . I am still confusing about the equation
$\partial_{x}\big[ \varphi (x ,t) , \pi (y ,t) \big] = \big[ \partial_{x}\varphi (x,t) , \pi (y,t) \big] = i \partial_{x}\delta (x - y)$.

From $\partial_{x} \left [\pi \left ( x^{'} , t\right ) , \phi \left ( x ,t \right ) \right ]$, I get
$\begin{align} \partial_{x} \left [ \phi \left ( x ,t \right ), \pi \left ( x^{'} , t\right ) \right ]&=\partial_{x} \{ \pi \left ( x^{'} , t\right ) \phi \left ( x ,t \right ) -\phi \left ( x ,t \right ) \pi \left ( x^{'} , t\right ) \} \nonumber \\ &= \partial_{x} \pi \left ( x^{'} , t\right ) \cdot \phi \left ( x ,t \right ) + \pi \left ( x^{'} , t\right ) \cdot \partial_{x} \phi \left ( x ,t \right ) -\partial_{x} \phi \left ( x ,t \right ) \cdot \pi \left ( x^{'} , t\right ) - \phi \left ( x ,t \right ) \cdot \partial_{x} \pi \left ( x^{'} , t\right ) \nonumber \end{align}$.

It seems $\partial_{x} \pi \left ( x^{'} , t\right ) \cdot \phi \left ( x ,t \right ) -\partial_{x} \phi \left ( x ,t \right ) \cdot \pi \left ( x^{'} , t\right ) =0$ from the result, but I do not know why.

Let's look at the whole thing. We want to evaluate:

$i \frac \partial {\partial t} \pi \left ( \bf {x}, t\right ) =\left [ \pi \left ( \bf x, t \right ), \int d^3 x^{'} \{ \frac 1 2 \pi ^2 \left ( \bf { x^{'}}, t \right ) + \frac 1 2 \phi \left ( \bf x^{'} ,t \right ) \left ( - \nabla^2 + m^2 \right ) \phi \left ( \bf x^{'} ,t \right ) \} \right ]$.

Note that we have already used the trick of integration by parts to change the $(\nabla \phi)^2$ to $-\nabla^2 \phi$. See post #2.

Next, we can take the commutator inside the integral. The first term involving $\pi^2$ vanishes. Next we can deal with the $m^2$ term.
$$\int d^3x' \frac{m^2}{2} [\pi(x, t), \phi^2( x', t)]$$
We use the commutator identity: $[A,BC] = [A, B]C + B[A, C]$. In this case with $B = C = \phi$. This gives:
$$\int d^3x' \frac{m^2}{2} \big ( [\pi(x, t), \phi(x', t)]\phi(x', t) + \phi(x', t)[\pi(x, t), \phi(x', t)] \big )$$
And using $[\pi(x, t), \phi(x', t)] = -i\delta^3(x - x')$, we get:
$$\int d^3x' \frac{m^2}{2} \big (-2i \delta^3(x - x')\phi(x', t) \big ) = -im^2\phi(x, t)$$
Now we come to the final term:
$$-\frac 1 2 \int d^3x' [\pi(x, t), \phi(x', t) \nabla^2\phi(x', t)]$$
Again we use the commutator identity to give:
$$-\frac 1 2 \int d^3x' \big ( [\pi(x, t), \phi(x', t)] \nabla^2\phi(x', t) + \phi(x', t) [\pi(x, t), \nabla^2\phi(x', t)] \big )$$
The first term here is easy, again using $[\pi(x, t), \phi(x', t)] = -i\delta^3(x - x')$ gives:
$$-\frac 1 2 \int d^3x' ( -i\delta^3(x - x')) \nabla^2\phi(x', t) = i\frac 1 2 \nabla^2\phi(x, t)$$
This leaves the term involving $[\pi(x, t), \nabla^2\phi(x', t)]$ to be dealt with. Note that it might be better to write $\nabla'$ here, as differentiation is with respect to $x'$. In any case, $\pi(x, t)$ is effectively a constant here so we have:
$$\phi(x', t)[\pi(x, t), \nabla^2\phi(x', t)] = \phi(x', t)\nabla^2[\pi(x, t), \phi(x', t)] = -i\phi(x', t)\nabla^2\delta^3(x-x')$$
Now we use the trick of integration by parts again. In this case we just assume the delta function behaves like any other function in this respect. Two applications of parts are used to move the $\nabla^2$ from the second function to the first:
$$-\frac 1 2 \int d^3x'\big (-i \phi(x', t)\nabla^2\delta^3(x-x') \big ) = i \frac 1 2 \int d^3x'\nabla^2 \phi(x', t)\delta^3(x-x') = i\frac 1 2 \nabla^2\phi(x, t)$$
When we put this together we get:
$$\frac 1 2\int d^3x' [\pi(x, t), \phi(x', t)(-\nabla^2 + m^2)\phi(x', t)] = i(\nabla^2 - m^2)\phi(x, t) $$
Perhaps the experienced physicists who write these books can see all that without doing the calculations, but that is far from a one-liner. And especially without any hint about how to do the calculation it was not at all obvious.

 

[Haorong Wu]

Wow! So clear! Thanks @PeroK . Your analysis is so explicit.

 

[PeroK]

Wow! So clear! Thanks @PeroK . Your analysis is so explicit.

It was a good exercise for me!

 

[samalkhaiat]

It was a good exercise for me!

A better exercise is (using the canonical commutation relations) to calculate the equal time commutators $\big[ T^{0 \mu} (x) , \varphi (y) \big]$ and $\big[ T^{0 \mu}(x) , \pi (y) \big]$, where $$T^{0 \mu} (x) = \pi (x) \partial^{\mu}\varphi (x) - \eta^{0 \mu} \mathcal{L}(x) ,$$ is the $(0 \mu)$-components of energy-momentum tensor for a generic field theory described by $\mathcal{L}(x) \equiv \mathcal{L} \left(\varphi (x) , \partial_{\mu}\varphi (x) \right)$.

 

[JD_PM]

Now we use the trick of integration by parts again. In this case we just assume the delta function behaves like any other function in this respect. Two applications of parts are used to move the $\nabla^2$ from the second function to the first:
$$-\frac 1 2 \int d^3x'\big (-i \phi(x', t)\nabla^2\delta^3(x-x') \big ) = i \frac 1 2 \int d^3x'\nabla^2 \phi(x', t)\delta^3(x-x') = i\frac 1 2 \nabla^2\phi(x, t)$$

$$-\frac 1 2 \int d^3x'\big (-i \phi( \vec x', t)\nabla'^2\delta^3(\vec x-\vec x') \big ) =\frac{i}{2}\phi(\vec x', t) \nabla' \delta^3(\vec x-\vec x') -\frac{i}{2} \int d^3 x' \nabla' \phi(\vec x', t) \nabla' \delta^3(\vec x-\vec x')$$ $$=\frac{i}{2}\phi(\vec x', t) \nabla' \delta^3(\vec x-\vec x') - \frac{i}{2} \nabla'\phi(\vec x', t) \delta^3(\vec x-\vec x') + \frac{i}{2} \int d^3 x' \nabla'^2 \phi(\vec x', t) \delta^3(\vec x-\vec x')$$ $$= \frac{i}{2}\phi(\vec x', t) \nabla' \delta^3(\vec x-\vec x') - \frac{i}{2} \nabla'\phi(\vec x', t) \delta^3(\vec x-\vec x')+\frac{i}{2}\nabla^2 \phi(\vec x, t)$$

Naive question: why do $\frac{i}{2}\phi(\vec x', t) \nabla' \delta^3(\vec x-\vec x') - \frac{i}{2} \nabla'\phi(\vec x', t) \delta^3(\vec x-\vec x')$ cancel out?

 

[PeroK]

Assumed boundary conditions.

 

[JD_PM]

Assumed boundary conditions.

Do you mean you assumed $x = x' \Rightarrow \delta^3(\vec x - \vec x')=0$?

 

[PeroK]

Do you mean you assumed $x = x' \Rightarrow \delta^3(\vec x - \vec x')=0$?

Those terms from the integration by parts are evaluated on the boundary, where we assume they vanish.

 

[JD_PM]

Those terms from the integration by parts are evaluated on the boundary, where we assume they vanish.

Oh so we simply assume they vanish on the boundary? Mmm why? Well, maybe I should read more about it before asking.

 

[PeroK]

Oh so we simply assume they vanish on the boundary? Mmm why? Well, maybe I should read more about it before asking.

It's a problem if they don't! In general a field and its derivatives must vanish at infinity.

 

[JD_PM]

I think I get it. We take the following terms to be zero

$$\left. \frac{i}{2}\phi(\vec x', t) \nabla' \delta^3(\vec x-\vec x') \right|^{\infty}_{0}, \ \left. \frac{i}{2} \nabla'\phi(\vec x', t) \delta^3(\vec x-\vec x') \right|^{\infty}_{0}$$

 

[PeroK]

The integral is over all of 3D space, so it's a boundary surface integral which vanishes in the limit as the boundary moves out to infinity.

 

[samalkhaiat]

Do you mean you assumed $x = x' \Rightarrow \delta^3(\vec x - \vec x')=0$?

Unlike ordinary function, a distribution (or generalized function) has no definite value at a single point. For example, for $\delta (x) \in \mathcal{S}^{\prime}(\mathbb{R})$, you can say that $\delta (x) = 0$ on every interval $(a,b)$ that does not contain the point $x = 0$, and that $\delta (x) \neq 0$ at $x = 0$. So, it is meaningless to say that $\delta (1) = 0$.

Obviously, you have not studied the theory of distributions. So I can only tell you that differentiation of distributions is defined by the following integration-by-parts formula: For any distribution $f$ and a “nice” function $\varphi$, we have $$\int_{\mathbb{R}^{n}} dx \ \partial^{m}f(x) \ \varphi (x) = (-1)^{|m|} \int_{\mathbb{R}^{n}} dx \ f(x) \ \partial^{m}\varphi (x) ,$$ or simply by $$\left( \partial^{m}f , \varphi \right) = (-1)^{|m|} \ \left( f , \partial^{m}\varphi \right).$$ So, for $f = \delta$, you get by $$\left( \partial^{m}\delta , \varphi \right) = (-1)^{|m|} \ \left( \delta , \partial^{m}\varphi \right) = (-1)^{|m|} \ \partial^{m}\varphi (0).$$

 

[JD_PM]

Obviously, you have not studied the theory of distributions. So I can only tell you that differentiation of distributions is defined by the following integration-by-parts formula: For any distribution $f$ and a “nice” function $\varphi$, we have $$\int_{\mathbb{R}^{n}} dx \ \partial^{m}f(x) \ \varphi (x) = (-1)^{|m|} \int_{\mathbb{R}^{n}} dx \ f(x) \ \partial^{m}\varphi (x) ,$$

Let's prove it!

Spoiler: Proof by induction (attempt)

1) Base case $m=0$

$$\int_{\mathbb{R}^{n}} dx \ \partial^{0}f(x) \ \varphi (x) = (-1)^{|0|} \int_{\mathbb{R}^{n}} dx \ f(x) \ \partial^{0}\varphi (x) \Rightarrow \int_{\mathbb{R}^{n}} dx \ f(x) \ \varphi (x) = \int_{\mathbb{R}^{n}} dx \ f(x) \ \varphi (x)$$

OK.

2) Induction hypothesis: the statement holds for $m$ (i.e. $\int_{\mathbb{R}^{n}} dx \ \partial^{m}f(x) \ \varphi (x) = (-1)^{|m|} \int_{\mathbb{R}^{n}} dx \ f(x) \ \partial^{m}\varphi (x)$ holds).

3) Induction step: show that the statement holds for $m+1$. The trick is integrating by parts (historical note: this integration technique was built up by Brook Taylor back in 1715!)

$$\int_{\mathbb{R}^{n}} dx \ \partial^{m+1}f(x) \ \varphi (x) = \partial^{m}f(x) \ \varphi (x)+(-1)\int_{\mathbb{R}^{n}} dx \ \partial^{m}f(x) \ \partial \varphi (x) $$

Applying integration by parts $m$ times one gets

$$\int_{\mathbb{R}^{n}} dx \ \partial^{m+1}f(x) \ \varphi (x) =$$ $$=\partial^{m}f(x) \ \varphi (x)-\partial^{m-1}f(x) \ \partial \varphi (x) +...+(-1)^{|m+1|}\int_{\mathbb{R}^{n}} dx \ f(x) \ \partial^{m+1} \varphi (x) $$

So we see that to finish our proof we need $\partial^{m}f(x) \ \varphi (x)-\partial^{m-1}f(x) \ \partial \varphi (x) +...=0$. This is precisely the same issue I had at #15... I guess we can assume boundary conditions to justify they do but, honestly, I still do not understand this technique sorry .

Besides I did not have to use the induction hypothesis. This has never happened to me before while using induction. I guess we need it to show $\partial^{m}f(x) \ \varphi (x)-\partial^{m-1}f(x) \ \partial \varphi (x) +...=0$?

As you can see, I stumbled upon the same boundary-condition issue

 

[JD_PM]

or simply by $$\left( \partial^{m}f , \varphi \right) = (-1)^{|m|} \ \left( f , \partial^{m}\varphi \right).$$ So, for $f = \delta$, you get by $$\left( \partial^{m}\delta , \varphi \right) = (-1)^{|m|} \ \left( \delta , \partial^{m}\varphi \right) = (-1)^{|m|} \ \partial^{m}\varphi (0).$$

I am afraid I do not understand the $( . , .)$ notation. Why is that equivalent to
$\int_{\mathbb{R}^{n}} dx \ \partial^{m}f(x) \ \varphi (x) = (-1)^{|m|} \int_{\mathbb{R}^{n}} dx \ f(x) \ \partial^{m}\varphi (x)$? This is highly likely a self-evident question; I may just need more time to get it.

 

[Haorong Wu]

I am afraid I do not understand the $( . , .)$ notation. Why is that equivalent to
$\int_{\mathbb{R}^{n}} dx \ \partial^{m}f(x) \ \varphi (x) = (-1)^{|m|} \int_{\mathbb{R}^{n}} dx \ f(x) \ \partial^{m}\varphi (x)$? This is highly likely a self-evident question; I may just need more time to get it.

It is the definition of inner product.

$\left ( f,g \right )=\int dx f^{\dagger}\left (x \right ) g\left (x \right )$ for one-dimension case.

 

[samalkhaiat]

Let's prove it!

Spoiler: Proof by induction (attempt)

1) Base case $m=0$

$$\int_{\mathbb{R}^{n}} dx \ \partial^{0}f(x) \ \varphi (x) = (-1)^{|0|} \int_{\mathbb{R}^{n}} dx \ f(x) \ \partial^{0}\varphi (x) \Rightarrow \int_{\mathbb{R}^{n}} dx \ f(x) \ \varphi (x) = \int_{\mathbb{R}^{n}} dx \ f(x) \ \varphi (x)$$

OK.

2) Induction hypothesis: the statement holds for $m$ (i.e. $\int_{\mathbb{R}^{n}} dx \ \partial^{m}f(x) \ \varphi (x) = (-1)^{|m|} \int_{\mathbb{R}^{n}} dx \ f(x) \ \partial^{m}\varphi (x)$ holds).

3) Induction step: show that the statement holds for $m+1$. The trick is integrating by parts (historical note: this integration technique was built up by Brook Taylor back in 1715!)

$$\int_{\mathbb{R}^{n}} dx \ \partial^{m+1}f(x) \ \varphi (x) = \partial^{m}f(x) \ \varphi (x)+(-1)\int_{\mathbb{R}^{n}} dx \ \partial^{m}f(x) \ \partial \varphi (x) $$

Applying integration by parts $m$ times one gets

$$\int_{\mathbb{R}^{n}} dx \ \partial^{m+1}f(x) \ \varphi (x) =$$ $$=\partial^{m}f(x) \ \varphi (x)-\partial^{m-1}f(x) \ \partial \varphi (x) +...+(-1)^{|m+1|}\int_{\mathbb{R}^{n}} dx \ f(x) \ \partial^{m+1} \varphi (x) $$

So we see that to finish our proof we need $\partial^{m}f(x) \ \varphi (x)-\partial^{m-1}f(x) \ \partial \varphi (x) +...=0$. This is precisely the same issue I had at #15... I guess we can assume boundary conditions to justify they do but, honestly, I still do not understand this technique sorry .

Besides I did not have to use the induction hypothesis. This has never happened to me before while using induction. I guess we need it to show $\partial^{m}f(x) \ \varphi (x)-\partial^{m-1}f(x) \ \partial \varphi (x) +...=0$?

As you can see, I stumbled upon the same boundary-condition issue

We don’t prove definitions:
1) I said that a (generalized) derivative $\partial^{m}f$ of the generalized function $f$ (in $\mathcal{D}^{\prime}(\mathbb{R}^{n}) \ \mbox{or} \ \mathcal{S}^{\prime}(\mathbb{R}^{n})$) is defined by the equation $$\int d^{n}x \ \partial^{m}f(x) \ \varphi (x) = (-1)^{|m|} \int d^{n}x \ f(x) \partial^{m}\varphi (x) , \ \ \ \ \ \ \ (1)$$ where $\varphi$ is a “nice” function (i.e., $\in \mathcal{D}(\mathbb{R}^{n}) \ \mbox{or} \ \mathcal{S}(\mathbb{R}^{n})$).

2) How do you know that the result of differentiation does not depend on the order of differentiation? You don’t. However, if you use the definition (1), you can easily show that (for a generalized $f \in \mathcal{D}^{\prime}(\mathbb{R}^{n})$) $$\partial^{k + m}f = \partial^{k} \left( \partial^{m}f \right) = \partial^{m} \left( \partial^{k} f \right) .$$

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

4) And the most important thing is the following: Since $\varphi \mapsto (-1)^{|m|} \partial^{m}\varphi$ is linear and continuous (from $\mathcal{D}(\mathbb{R}^{n}) \ \mbox{into} \ \mathcal{D}(\mathbb{R}^{n})$), then (1) can be used to show that the operation of differentiation $f \mapsto \partial^{m}f$ is linear and continuous (from $\mathcal{D}^{\prime}(\mathbb{R}^{n}) \ \mbox{into} \ \mathcal{D}^{\prime}(\mathbb{R}^{n})$).

 

[samalkhaiat]

I am afraid I do not understand the $( . , .)$ notation. Why is that equivalent to
$\int_{\mathbb{R}^{n}} dx \ \partial^{m}f(x) \ \varphi (x) = (-1)^{|m|} \int_{\mathbb{R}^{n}} dx \ f(x) \ \partial^{m}\varphi (x)$? This is highly likely a self-evident question; I may just need more time to get it.

1) A distribution (or generalized function) $f$ is a functional on the space of “nice” functions $\mathcal{D}(\mathbb{R}^{n})$. That is, for each $\varphi \in \mathcal{D}(\mathbb{R}^{n})$ there corresponds a (complex) number $(f , \varphi)$ (some textbooks use the notation $f[\varphi]$) defined by $$(f , \varphi) \equiv \int d^{n}x \ f(x) \varphi(x) .$$

2) A distribution (or generalized function) $f$ is a linear functional on $\mathcal{D}(\mathbb{R}^{n})$. That is, if $\varphi$ and $\psi$ are in $\mathcal{D}(\mathbb{R}^{n})$ and $\alpha$ and $\beta$ are complex numbers, then $$(f , \alpha \varphi + \beta \psi ) = \alpha \ (f , \varphi ) + \beta \ (f , \psi ) .$$

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.

 

[JD_PM]

Alright, I think I got the basic idea thanks! This is what I learned

$\mathcal{D}(\mathbb{R}^{n})$ 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.

A distribution (or generalized function) $L_f$ is a continuous linear functional on the space of “nice” functions $\mathcal{D}(\mathbb{R}^{n})$ (i.e. $L_f:\mathcal{D}(\mathbb{R}^{n}) \rightarrow \Bbb R:\varphi \mapsto (f,\varphi)$, where $\varphi \in \mathcal{D}(\mathbb{R}^{n}), \ (f,\varphi) \in \mathbb{C}$. 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).

Spoiler: Proving $L_f$ is linear

Given $\varphi_1, \varphi_2 \in \mathcal{D}(\mathbb{R}^{n})$ and $\alpha, \beta \in \mathbb{C}$

$$L_f ( \alpha \varphi_1 + \beta \varphi_2) = (f, \alpha \varphi_1 + \beta \varphi_2) = \alpha (f, \varphi_1)+\beta (f, \varphi_2) = \alpha L_f(f, \varphi_1) + \beta L_f(f, \varphi_2)$$

QED.

(Probably) the most famous example of distribution (or generalized function) is the Dirac-delta $\delta$, defined as

$$(\delta, \varphi):=\int_{\Bbb R^n} \delta (0+x) \varphi(x) dx^n=\varphi(0)$$

2) How do you know that the result of differentiation does not depend on the order of differentiation? You don’t. However, if you use the definition (1), you can easily show that (for a generalized $f \in \mathcal{D}^{\prime}(\mathbb{R}^{n})$) $$\partial^{k + m}f = \partial^{k} \left( \partial^{m}f \right) = \partial^{m} \left( \partial^{k} f \right) .$$

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

4) And the most important thing is the following: Since $\varphi \mapsto (-1)^{|m|} \partial^{m}\varphi$ is linear and continuous (from $\mathcal{D}(\mathbb{R}^{n}) \ \mbox{into} \ \mathcal{D}(\mathbb{R}^{n})$), then (1) can be used to show that the operation of differentiation $f \mapsto \partial^{m}f$ is linear and continuous (from $\mathcal{D}^{\prime}(\mathbb{R}^{n}) \ \mbox{into} \ \mathcal{D}^{\prime}(\mathbb{R}^{n})$).

Studying it

 

[JD_PM]

2) How do you know that the result of differentiation does not depend on the order of differentiation? You don’t. However, if you use the definition (1), you can easily show that (for a generalized $f \in \mathcal{D}^{\prime}(\mathbb{R}^{n})$) $$\partial^{k + m}f = \partial^{k} \left( \partial^{m}f \right) = \partial^{m} \left( \partial^{k} f \right) .$$

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

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

Where I've assumed boundary conditions. Integrating $k$ times the RHS integral and assuming boundary conditions again, we get

$$\int dx \ \partial^{k} \left( \partial^{m}f \right) \ \varphi (x) = (-1)^{|m+k|} \int dx f\ \partial^{k}(\partial^{m} \varphi (x)) \tag{2}$$

By analogous work on $\partial^{m} \left( \partial^{k} f \right)$ we get

$$\int dx \ \partial^{m} \left( \partial^{k}f \right) \ \varphi (x) = (-1)^{|m+k|} \int dx f\ \partial^{m}(\partial^{k} \varphi (x)) \tag{3}$$

By comparing $(2)$ and $(3)$ we see that

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

Thus $$\partial^{k + m}f = \partial^{k} \left( \partial^{m}f \right) = \partial^{m} \left( \partial^{k} f \right) .$$ QED.

 

[JD_PM]

4) And the most important thing is the following: Since $\varphi \mapsto (-1)^{|m|} \partial^{m}\varphi$ is linear and continuous (from $\mathcal{D}(\mathbb{R}^{n}) \ \mbox{into} \ \mathcal{D}(\mathbb{R}^{n})$), then (1) can be used to show that the operation of differentiation $f \mapsto \partial^{m}f$ is linear and continuous (from $\mathcal{D}^{\prime}(\mathbb{R}^{n}) \ \mbox{into} \ \mathcal{D}^{\prime}(\mathbb{R}^{n})$).

I think I know how to show linearity; given the differential operator $D^{\prime}:\mathcal{D}^{\prime}(\mathbb{R}^{n}) \rightarrow \mathcal{D}^{\prime}(\mathbb{R}^{n}):f \mapsto \partial^{m}f$, we let $\varphi_1, \varphi_2 \in \mathcal{D}^{\prime}(\mathbb{R}^{n})$, $\alpha, \beta \in \mathbb{C}$ and get

$$D^{\prime}f =\partial^m f$$

Multiplying the above equation both sides by $(\alpha \varphi_1+ \beta \varphi_2)$ (on the right) allows us to use the well-known definition to show linearity

$$D^{\prime}f (\alpha \varphi_1+ \beta \varphi_2) =\partial^m f (\alpha \varphi_1+ \beta \varphi_2)=(-1)^m f\partial^m (\alpha \varphi_1+ \beta \varphi_2)$$ $$=\alpha (-1)^m f\partial^m ( \varphi_1)+\beta (-1)^m f\partial^m ( \varphi_1)=\alpha f(\varphi_1)+\beta f(\varphi_2)$$

Where I've used the fact that $D:\mathcal{D}(\mathbb{R}^{n}) \rightarrow \mathcal{D}(\mathbb{R}^{n}):\varphi \mapsto (-1)^{|m|}\partial^{m} \varphi$ is linear.

Argh I'm having trouble to show continuity though.

 

[martinbn]

Argh I'm having trouble to show continuity though.

Are you studying some particular book/notes or just post in this thread? I may have missed it, but what continuous means for distributions wasn't said in this thread.

 

[JD_PM]

Hi martinbn

Are you studying some particular book/notes or just post in this thread? I may have missed it, but what continuous means for distributions wasn't said in this thread.

I was trying to show what samalkhaiat wrote here

4) And the most important thing is the following: Since $\varphi \mapsto (-1)^{|m|} \partial^{m}\varphi$ is linear and continuous (from $\mathcal{D}(\mathbb{R}^{n}) \ \mbox{into} \ \mathcal{D}(\mathbb{R}^{n})$), then (1) can be used to show that the operation of differentiation $f \mapsto \partial^{m}f$ is linear and continuous (from $\mathcal{D}^{\prime}(\mathbb{R}^{n}) \ \mbox{into} \ \mathcal{D}^{\prime}(\mathbb{R}^{n})$).

I showed linearity but got stuck in showing continuity.

 

[martinbn]

Hi martinbn
I was trying to show what samalkhaiat wrote here
I showed linearity but got stuck in showing continuity.

What does it mean to be continuous?

 

[JD_PM]

What does it mean to be continuous?

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 $k \to \infty$.