# Functional Derivative ${\delta q_a(t)}/{\delta q_b(t')}$

$\newcommand{\fdv}[2]{\frac{\delta #1}{\delta #2}}$ $\newcommand{\dv}[2]{\frac{\mathrm{d} #1}{\mathrm{d}#2}}$ $\newcommand{\pdv}[2]{\frac{\partial #1}{\partial #2}}$

I'm from a physics background and I've always known the definition to be related to the Euler-Lagrange Equations i.e.

$$\fdv {L(q,\dot q)} {q(t)} = \pdv{L}{q} - \dv{}{t} \pdv{L}{\dot q} \; ,$$

where $\dot q$ denotes the derivative of the function $q$ with respect to $t$.

However with this definition I cannot prove a fact that I've seen in a lecture video about Quantum Field Theory, which is

$$\fdv{q_a(t)}{q_b(t')} = \delta_{ab} \delta(t-t') \; , \tag 1$$

where $\delta(t-t')$ is the Dirac delta function/distribution and $\delta_{ab}$ is the Kroneker delta.

I don't know how I should make sense out of this equation let alone prove it because $q_a(t)$ is not a functional but a function. I'd appreciate if someone can explain what is meant by the equation (1) and give a proof thereof.

• The discrepancy lies in that actually almost nobody(!!) knows how to read these lines!!! That missconception starts already from sloppy taught mechanics classes. (Let me outline it as answer.) – C-Star-W-Star Jul 10 '15 at 15:17
• Besides it is not even a function but only a number. ;) – C-Star-W-Star Jul 10 '15 at 16:57
• You seem to have lost track of the fact that the Euler-Lagrange equation has hypotheses. Namely, along the path from a fixed point $a$ to a fixed point $b$ which minimizes the quantity $\int_0^1 L(t,q(t),q'(t)) dt$, the Euler-Lagrange equation is satisfied. But the fact that these points are held fixed is essential, in that it forces $q(0)=a$ and $q(1)=b$, and therefore forces the admissible variations to satisfy $v(0)=v(1)=0$. – Ian Jul 10 '15 at 18:34
• @Ian: Just to remark: That won't help him since there's basically no proof as these are, simply put, conventions. – C-Star-W-Star Jul 10 '15 at 18:39

First have a look at:

Integral Argument

For enough variations: $$\varphi\in\mathcal{C}^\infty:\quad\int_0^T\varphi(s)\delta\varphi(s)\mathrm{d}s=0\quad(\delta\varphi\in\mathcal{C}^\infty_0)\implies\varphi(t)\equiv0$$

That is needed for Euler-Lagrange!

Now go carefully through:

Detailed Calculation

Directional derivative: $$\frac{\delta}{\delta q}:=\frac{\partial}{\partial\delta q}:=\partial_{\delta q}$$

Consider as example: $$\mathcal{L}:\mathbb{R}^3\to\mathbb{R}:\quad\mathcal{L}(u,v;w):=\frac12mv^2-\frac{1}{w}u^2$$

For partial derivatives: $$\frac{\partial\mathcal{L}}{\partial u}=-\frac{1}{w}2u\quad\frac{\partial\mathcal{L}}{\partial v}=mv\quad\frac{\partial\mathcal{L}}{\partial w}=\frac{1}{w^2}u^2$$

Introduce evaluations: $$\eta_a:\mathcal{C}^\infty(\mathbb{R})\to\mathbb{R}:\quad\eta_a[q]:=q(a)$$ $$\vartheta_a:\mathcal{C}^\infty(\mathbb{R})\to\mathbb{R}:\quad\vartheta_a[q]:=\dot{q}(a)$$ $$\omega_a:\mathcal{C}^\infty(\mathbb{R})\to\mathbb{R}:\quad\omega_a[q]:=a$$

For directional derivative: $$\frac{\delta\eta_a}{\delta q}[q]=\lim_{\varepsilon\to0}\frac{1}{\varepsilon}\{\eta_a[q+\varepsilon\delta q]-\eta_a[q]\}=\lim_{\varepsilon\to0}\frac{1}{\varepsilon}\{q(a)+\varepsilon\delta q(a)-q(a)\}=\delta q(a)$$ $$\frac{\delta\vartheta_a}{\delta q}[q]=\lim_{\varepsilon\to0}\frac{1}{\varepsilon}\{\vartheta_a[q+\varepsilon\delta q]-\vartheta_a[q]\}\stackrel{*}{=}\lim_{\varepsilon\to0}\frac{1}{\varepsilon}\{\dot{q}(a)+\varepsilon\dot{\delta q}(a)-\dot{q}(a)\}=\dot{\delta q}(a)$$ $$\frac{\delta\omega_a}{\delta q}[q]=\lim_{\varepsilon\to0}\frac{1}{\varepsilon}\{\omega_a[q+\varepsilon\delta q]-\omega_a[q]\}=\lim_{\varepsilon\to0}\frac{1}{\varepsilon}\{a-a\}=0$$

Consider the action: $$S[q]:=\int_0^T\mathcal{L}(\eta_s[q],\vartheta_s[q];s)\mathrm{d}s=:\int_0^T\hat{\mathcal{L}_s}[q]\mathrm{d}s\quad(q\in\mathcal{C}^\infty)$$

Suppose variations: $$\mathcal{C}^\infty_0:=\{\delta q\in\mathcal{C}^\infty:\delta q(0)=\delta q(T)=0\}$$

For directional derivative: $$\frac{\delta S}{\delta q}[q]=\lim_{\varepsilon\to0}\frac{1}{\varepsilon}\{S[q+\varepsilon\delta q]-S[q]\}\\ =\lim_{\varepsilon\to0}\int_0^T\frac{1}{\varepsilon}\bigg\{\mathcal{L}(\eta_s[q+\varepsilon\delta q],\vartheta_s[q+\varepsilon\delta q];s)\mathrm{d}s-\mathcal{L}(\eta_s[q],\vartheta_s[q];s)\bigg\}\mathrm{d}s\\ \stackrel{**}{=}\int_0^T\lim_{\varepsilon\to0}\frac{1}{\varepsilon}\bigg\{\mathcal{L}(\eta_s[q+\varepsilon\delta q],\vartheta_s[q+\varepsilon\delta q];s)\mathrm{d}s-\mathcal{L}(\eta_s[q],\vartheta_s[q];s)\bigg\}\mathrm{d}s\\ =\int_0^T\frac{\delta\hat{\mathcal{L}_s}}{\delta q}[q]\mathrm{d}s=\int_0^T\bigg\{\frac{\partial\mathcal{L}}{\partial u}\frac{\delta\eta_s}{\delta q}+\frac{\partial\mathcal{L}}{\partial v}\frac{\delta\vartheta_s}{\delta q}+\frac{\partial\mathcal{L}}{\partial w}\frac{\delta\omega_s}{\delta q}\bigg\}\mathrm{d}s\\ =\int_0^T\bigg\{\frac{\partial\mathcal{L}}{\partial u}\delta q(s)+\frac{\partial\mathcal{L}}{\partial v}\dot{\delta q}(s)+0\bigg\}\mathrm{d}s=\int_0^T\bigg\{\frac{\partial\mathcal{L}}{\partial u}\delta q(s)+\frac{\partial\mathcal{L}}{\partial v}\frac{\mathrm{d}}{\mathrm{d}s}\delta q(s)\bigg\}\mathrm{d}s\\ =\int_0^T\bigg\{\frac{\partial\mathcal{L}}{\partial u}\delta q(s)-\frac{\mathrm{d}}{\mathrm{d}s}\frac{\partial\mathcal{L}}{\partial v}\delta q(s)\bigg\}\mathrm{d}s=\int_0^T\bigg\{\frac{\partial\mathcal{L}}{\partial u}-\frac{\mathrm{d}}{\mathrm{d}s}\frac{\partial\mathcal{L}}{\partial v}\bigg\}\delta q(s)\mathrm{d}s$$

By the integral argument: $$\frac{\delta S}{\delta q}[q]=0\quad(\delta q\in\mathcal{C}^\infty_0)\implies\bigg\{\frac{\partial\mathcal{L}}{\partial u}-\frac{\mathrm{d}}{\mathrm{d}s}\frac{\partial\mathcal{L}}{\partial v}\bigg\}=0$$

Concluding Euler-Lagrange.

Next try it sombolically:

Symbolic Calculation

Regard action functional: $$S[q]:=\int\mathcal{L}(q(s),\dot{q}(s);s)\mathrm{d}s\quad(q\in\mathcal{C}^\infty)$$

Its derivite writes: $$\frac{\delta S}{\delta q}=\int\bigg\{\frac{\partial\mathcal{L}}{\partial q}\frac{\delta q}{\delta q}+\frac{\partial\mathcal{L}}{\partial\dot{q}}\frac{\delta\dot{q}}{\delta q}+\frac{\partial\mathcal{L}}{\partial t}\frac{\delta t}{\delta q}\bigg\}\delta q\mathrm{d}s\\ =\int\bigg\{\frac{\partial\mathcal{L}}{\partial q}\delta q+\frac{\partial\mathcal{L}}{\partial\dot{q}}\dot{\delta q}+\frac{\partial\mathcal{L}}{\partial t}0\bigg\}\delta q\mathrm{d}s=\int\bigg\{\frac{\partial\mathcal{L}}{\partial q}-\frac{\mathrm{d}}{\mathrm{d}s}\frac{\partial\mathcal{L}}{\partial\dot{q}}\bigg\}\delta q\mathrm{d}s$$

By the integral argument: $$\frac{\delta S}{\delta q}=0\quad(\delta q\in\mathcal{C}^\infty_0)\implies\frac{\delta\mathcal{L}}{\delta q}=\frac{\mathrm{d}}{\mathrm{d}s}\frac{\delta\mathcal{L}}{\delta q}$$

Concluding Euler-Lagrange.

Finally try to shorten it by:

Shortened Calculation

Regard action functional: $$S[q]:=\int\mathcal{L}(q(s),\dot{q}(s);s)\mathrm{d}s\quad(q\in\mathcal{C}^\infty)$$

For singular variations: $$\delta q_a\notin\mathcal{C}^\infty_0:\quad\delta q_a(t):=\delta(t-a)$$

Agree on the convention: $$\frac{\delta S}{\delta q(a)}:=\frac{\delta S}{\delta q_a}\quad\frac{\partial\mathcal{L}}{\partial q(t)}:=\frac{\partial\mathcal{L}}{\partial q}(t)\quad\frac{\partial\mathcal{L}}{\partial\dot{q}(t)}:=\frac{\partial\mathcal{L}}{\partial\dot{q}}(t)\quad\frac{\delta q(t)}{\delta q(a)}:=\frac{\delta q}{\delta q_a}(t)$$

By the previous calculation: $$0=\frac{\delta S}{\delta q(t)}=\int\bigg\{\frac{\partial\mathcal{L}}{\partial q(s)}-\frac{\mathrm{d}}{\mathrm{d}s}\frac{\partial\mathcal{L}}{\partial\dot{q}(s)}\bigg\}\frac{\delta q(s)}{\delta q(t)}\mathrm{d}s\\ =\int\bigg\{\frac{\partial\mathcal{L}}{\partial q(s)}-\frac{\mathrm{d}}{\mathrm{d}s}\frac{\partial\mathcal{L}}{\partial\dot{q}(s)}\bigg\}\delta(s-t)\mathrm{d}s=\frac{\partial\mathcal{L}}{\partial q(t)}-\frac{\mathrm{d}}{\mathrm{d}t}\frac{\partial\mathcal{L}}{\partial\dot{q}(t)}$$

That killed the integral argument!

...now you're ready for QFT. ;)

*Linearity of derivative!

**Regularity of Lagrangian!

• Trust me: Try to follow the second part carefully!! – C-Star-W-Star Jul 10 '15 at 18:32
• I followed you answer to almost to the very end, which by the way I should say an excellent answer. I just didn't get the part with singular variations. Why do I choose my function $q_a$ to be the dirac delta function? I'm talking about the last orange box: $\delta q_a\notin\mathcal{C}^\infty_0:\quad q_a(t):=\delta(t-a)$ – Gonenc Jul 10 '15 at 20:20
• Because you want to overleap the integral argument and end up immediately at Euler-Lagrange. That way it is not rigid but it does its job. In case one requires rigorosity one can still do the derivation properly. – C-Star-W-Star Jul 10 '15 at 20:25
• Shouldn't it be $\delta q_a$ instead of $q_a$? – Gonenc Jul 10 '15 at 20:30
• Oh right, thanks! (Corrected.) – C-Star-W-Star Jul 10 '15 at 20:31

Fix $t=t_0$, so you can pointwise identify $q_a(t_0)$ with a functional $$\mathcal F[u]=\int_{\mathbb R}\mathrm d t' u(t')\delta(t_0-t')$$ as $q_a(t_0)\equiv\mathcal F[q_a]$. Then you have the "Euler-Lagrange equation": $$\frac{\delta \mathcal F[q_a]}{\delta q_b(t')}=\frac{\partial}{\partial q_b(t')}q_a(t')\delta(t_0-t')$$ the same as for a Lagrangian (the $d/dt\dots$ part is of course missing). Now finally, using $\partial q_a(t')/\partial q_b(t')=\delta_{ab}$, the result follows.

• The Dirac measure has no derivative and therefore the derivation of the Euler-Lagrange equation is meaningless then. Rather one exploits the fundamental theorem of calculus of variations. – C-Star-W-Star Jul 10 '15 at 18:41
• It has weak derivatives – krvolok Jul 10 '15 at 18:43
• Yes, but the result is the same. I wanted to appeal to what OP knows – krvolok Jul 10 '15 at 18:43
• Yep but that is simply overkill: The whole thing works on the basis of ordinary calculus. – C-Star-W-Star Jul 10 '15 at 18:43
• And besides I assume the OP is not familiar with distribution theory. – C-Star-W-Star Jul 10 '15 at 18:45