In a Lagrangian like the one that follows:

$$\mathcal{L} = {i} \Psi^*\dot{\Psi} - \frac{1}{2m} \nabla{\Psi} ^* \nabla \Psi.\tag{1}$$

How can I apply the Euler-Lagrangian equation, shown in $(2)$, to obtain an equation like $(3)$. I don't understand how I can deal with all the different fields?

$$ \frac{d}{dt}\left( \frac{\partial \mathcal{L}}{\partial \dot{\Psi}}\right)= \frac{\partial \mathcal{L}}{\partial \Psi} \tag{2}$$

$$i \frac{\partial \Psi}{\partial t} = - \frac{\nabla^2}{2m} \Psi \tag{3}$$

I have seen in some notes that I must treat $\Psi$ and $\Psi^*$ as different fields and must first calculate the different derivatives, shown in $(4)$:

$$\frac{\partial \mathcal{L}}{\partial \dot{\Psi^*}}= i\Psi \hspace{5mm}; \hspace{5mm} \frac{\partial \mathcal{L}}{\partial \nabla{\Psi^*}}= - \frac{1}{2m} \nabla \Psi \hspace{5mm} ; \hspace{5mm} \frac{\partial \mathcal{L}}{\partial {\Psi^*}}= i\dot{\Psi} \tag{4}$$

But how I can apply $(2)$ to these to obtain $(3)$? And do I not need the $\frac{\partial \mathcal{L}}{\partial \nabla\Psi^*}$ ?

How can I use $(2)$ if I do not have a $\Psi$ variable or a $\dot\Psi^*$ to complete the equation?

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    $\begingroup$ What is $\dot{\Psi}$? $\endgroup$ Jan 14, 2022 at 1:13
  • $\begingroup$ Hint: integrate by parts first. $\endgroup$ Jan 14, 2022 at 7:07
  • $\begingroup$ @ArtemAlexandrov , $\dot{\Psi}$ is the time derivative of the field, $\Psi$. $\endgroup$ Jan 14, 2022 at 9:22
  • $\begingroup$ For anyone else in doubt, another good source I found was: people.phys.ethz.ch/~babis/Teaching/QFT1/qft1.pdf $\endgroup$ Jan 14, 2022 at 15:05

2 Answers 2


I'll try to give a slightly more precise view which might be confusing at first but will pay off later.

You are minimizing the action $S[\psi,\psi^*]$ where the square brackets indicate that $S$ is a functional: an object which takes a function as argument and spits out a number. Here $\psi$ and $\psi^*$ are still considered completely independent functions. In the following equations I will drop the dependence on $\psi^*$ since otherwise it would be too long but you can just as easily put them back in. The action then looks like $$S[\psi]=\int\mathrm dx\,\mathcal L(\psi,\dot\psi,\nabla\psi,\nabla\dot\psi)\tag{1}$$ where $x$ is four vector so it includes time as well. Here I emphasize that $\mathcal L$ is just a regular function: it takes in 4 arguments and spits out a number. I could also fill in $a,b,c,d$ instead of $\psi,\dot\psi,\nabla\psi,\nabla\dot\psi$ in the Lagrangian (I will use $\psi$ instead of $\psi^*$ for brevity). $$\mathcal L(a,b,c,d)=\frac i 2ab-\frac 1{2m}dc\tag{2}$$ Now to minimize the action we will have to calculate the functional derivative. The functional derivative, $\frac{\delta S}{\delta\psi}( x)$, is formally defined as $$\int\mathrm d x\frac{\delta S}{\delta \psi}( x)\eta(x)=\lim_{\epsilon\rightarrow\infty}\frac{S[\psi+\epsilon\eta]-S[\psi]}{\epsilon}\tag{3}$$ where $\eta( x)$ is an arbitrary test function that satisfies the right boundary conditions, in this case the condition that $\eta$ goes to zero as $ x\rightarrow\infty$. $\eta$ is often written as $\delta\psi$. We can now calculate $S[\psi+\epsilon\eta]$ by Taylor expanding the Lagrangian: \begin{align} S[\psi+\epsilon\eta]=&\int\mathrm d x\,\mathcal L(\psi+\epsilon\eta,\dot\psi+\epsilon\dot\eta,\nabla\psi+\epsilon\nabla\eta,\nabla\dot\psi+\epsilon\nabla\dot\eta)\\ &=\int\mathrm dx\left[\mathcal L(\psi,\dot\psi,\nabla\psi,\nabla\dot\psi)+\\ \epsilon\,\eta(x)\frac{\partial}{\partial\psi}\mathcal L(\psi,\dot\psi,\nabla\psi,\nabla\dot\psi)+\\ \epsilon\,\dot\eta(x)\frac{\partial}{\partial\dot\psi}\mathcal L(\psi,\dot\psi,\nabla\psi,\nabla\dot\psi)+\\ \epsilon\nabla\eta(x)\frac{\partial}{\partial\nabla\psi}\mathcal L(\psi,\dot\psi,\nabla\psi,\nabla\dot\psi)+\\ \epsilon\nabla\dot\eta(x)\frac{\partial}{\partial\nabla\dot\psi}\mathcal L(\psi,\dot\psi,\nabla\psi,\nabla\dot\psi) \right]\tag{4} \end{align} This almost has the form of the LHS of (2) but $\eta$ still has derivatives acting on it. We can fix this by integrating by parts to move the derivative from $\eta$ to the other factor at the cost of introducing a minus sign. Plugging this into (3) gives \begin{align}\int\mathrm d x\frac{\delta S}{\delta \psi}( x)\eta(\mathbf x)&=\int\mathrm dx\left[\frac{\partial\mathcal L}{\partial\psi}-\frac{\partial}{\partial t}\frac{\partial\mathcal L}{\partial\dot\psi}-\nabla\cdot\left(\frac{\partial\mathcal L}{\partial\nabla\psi}\right)\pm\frac{\partial}{\partial t}\nabla\cdot\left(\frac{\partial\mathcal L}{\partial\nabla\dot\psi}\right)\right]\eta(x) \end{align} We recognize the term between brackets as the functional derivative and setting it to zero will give us the EL equations. I'm not so sure if the $\pm$ should be plus or minus but I think it's $+$. If we had included $\psi^*$ as a separate field from the start we would have got a similar equation but with $\psi^*$ instead of $\psi$.

I believe there's an error in your Lagrangian and that $\nabla\dot\psi$ should be $\nabla\psi$. The factor $\frac i 2\psi^*\dot\psi$ should either be $i\psi^*\dot\psi$ or $\frac i 2(\psi^*\dot\psi-\psi\dot\psi^*)$. See this question or this question. If you now calculate the EL you get the desired formula.

  • $\begingroup$ If $\Psi^*$ had been included would that mean that: $S[\psi \psi^*]=\int\mathrm dx\,\mathcal L(\psi,\dot\psi,\nabla\psi,\nabla\dot\psi) + \int dx \mathcal{L} \mathcal L(\psi^*,\dot\psi^*,\nabla\psi^*,\nabla\dot\psi^*)$ or does it instead mean that: $S[\psi \psi^*]=\int\mathrm dx\,\mathcal L(\psi,\dot\psi,\nabla\psi,\nabla\dot\psi , \psi^*,\dot\psi^*,\nabla\psi^*,\nabla\dot\psi^*)$ ? $\endgroup$ Jan 14, 2022 at 15:07
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    $\begingroup$ @user7077252 The latter. Note that based on my comment in the last paragraph you don't actually need the term $\nabla\dot\psi$ or $\nabla\dot\psi^*$. I only realized this after the fact so that's why I included it. $\endgroup$ Jan 14, 2022 at 15:10
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    $\begingroup$ Thank you so much for your help. I think I get it now. $\endgroup$ Jan 14, 2022 at 15:38

I'm not 100% confident in my answer as I am new to this topic, but I hope I am of help.

Edit: Your lagrangian is a slightly strange, as pointed out by AccidentalTaylorExpansion, so you may want to double check that it is correct. The below should hold for the lagrangian you have given however.

The lagrangian you have is a 'lagrangian density'. If the lagrangian is by L and the lagrangian density denoted by $\mathcal L$, then:

$$ \int L \ dt = \int \mathcal L \ dt dx^3 $$

The Euler Lagrange equation associated with extremising the action for the lagrangian density $\mathcal L(\psi, \dot \psi, \partial_{\mu} \psi, \partial_{\mu} \dot \psi)$ looks a bit different (Einstein summation in implied):

$$ \frac{\partial}{\partial t} \left ( \frac{\partial \mathcal L}{ \dot \psi } \right) + \frac{\partial}{\partial x^\mu} \left ( \frac{\partial \mathcal L}{ \partial (\partial_{\mu} \psi) } \right) + \frac{\partial^2}{\partial x^\mu \partial t} \left ( \frac{\partial \mathcal L}{ \partial (\partial_{\mu} \dot \psi) } \right) = \frac{\partial \mathcal L}{\partial \psi}$$

Where $\mu$ runs from 1 to 3 for the three spatial dimensions.

For complex fields, you treat the complex and complex conjugate as two separate fields. This is because there are two degrees of freedom (see: Complex Scalar Field - Euler Lagrange equation).

Meaning that you also get the equation: $$ \frac{\partial}{\partial t} \left ( \frac{\partial \mathcal L}{ \partial \dot \psi^* } \right) + \frac{\partial}{\partial x^\mu} \left ( \frac{\partial \mathcal L}{ \partial (\partial_{\mu} \psi^*) } \right) + \frac{\partial^2}{\partial x^\mu \partial t} \left ( \frac{\partial \mathcal L}{ \partial (\partial_{\mu} \dot \psi^*) } \right)= \frac{\partial \mathcal L}{\partial \psi^*}$$

Thus: $$ \frac{\partial \mathcal L}{\partial \dot \psi^* } = 0 $$ $$ \frac{\partial \mathcal L}{ \partial_{\mu} \psi^* } = -\frac{i}{2m} \partial^{\mu} \dot \psi $$ $$ \frac{\partial \mathcal L}{\partial \psi^*} = \frac{1}{2} \dot \psi$$ $$ \frac{\partial \mathcal L}{ \partial (\partial_{\mu} \dot \psi^*) } = 0 $$

So: $$ \dot \psi = -\frac{i}{m} \partial_{\mu} \partial^{\mu} \dot \psi $$

Additionally, from the EL equation with the real $\psi$:

Thus: $$ \frac{\partial \mathcal L}{\partial \dot \psi } = \frac{i}{2} \psi^* $$ $$ \frac{\partial \mathcal L}{ \partial_{\mu} \psi } = 0 $$ $$ \frac{\partial \mathcal L}{\partial \psi} = 0 $$ $$ \frac{\partial \mathcal L}{ \partial (\partial_{\mu} \dot \psi) } = - \frac{1}{2m} \partial_{\mu} \psi^*$$

So: $$ \frac{i}{2} \dot \psi^* = \frac{1}{2m} \partial_{\mu} \partial^{\mu} \dot \psi^*$$

  • $\begingroup$ Nothing seems wrong in your answer, but your answer seems to be different from the eq (3) my lecturer got in his notes. Thank you so much for your help, I will try to see if there was some mathematical error that can be fixed in your answer. $\endgroup$ Jan 14, 2022 at 12:01
  • $\begingroup$ @user7077252 What page is it in the notes? $\endgroup$
    – Dedados
    Jan 14, 2022 at 12:03
  • $\begingroup$ this comes from an exercise booklet in the following link , but similar things can be seen in the Cambridge QFT notes in page 9 . Thank you once again for all the help, I have been stuck doing this for ages. $\endgroup$ Jan 14, 2022 at 12:05
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    $\begingroup$ @user7077252 I appended a term to the EL equation which AccidentalTaylorExpansion correctly included in his answer (it is not normally needed though). Something is probably wrong with the lagrangian though! $\endgroup$
    – Dedados
    Jan 14, 2022 at 14:25
  • $\begingroup$ I did notice a mistake in my Lagrangian and have now corrected it, my sincere apologies for that. Thank you so much for your help, it all makes more sense now I think. $\endgroup$ Jan 14, 2022 at 14:27

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