# Calculus of variations and integrands containing second derivatives

1. Aug 28, 2014

### ShayanJ

You know that the problem of calculus of variations is finding a y(x) for which $\int_a^b L(x,y,y') dx$ is stationary. I wanna know is it possible to solve this problem when L is a function of also y'' ?
It happens e.g. in the variational method in quantum mechanics where we say that choosing any arbitrary wave function $\psi(x)$, the energy of the ground state of the system is always smaller than $\int \psi^* H \psi dx$. But if we can find a $\psi$ for which this integral is stationary, then we have the wave function and also the energy of the ground state. But the problem is, for ordinary particle systems in non-relativistic quantum mechanics, H contains a second derivative.

2. Aug 28, 2014

### Matterwave

You can certainly do it with a function of y''. However, you will have to integrate by parts twice, so you need to make the additional constraint that the variation at the end points of y'' also vanishes.

In addition, you will end up with probably a 3rd order differential equation instead of a second order ODE, which requires usually more initial data to solve than we have. We usually just have 2 initial data points in classical physics, the initial position and momentum. A 3rd order ODE would require an additional point.

3. Aug 29, 2014

### julian

You mean variations of y' vanish at the boundary coming about from integration by parts? So that if

$\delta y = \delta y' = 0$

at the boundaries the Euler-Lagrange equations then become:

${\partial L \over \partial y} - {d \over dx} \Big( {\partial L \over \partial y_x} \Big) + {d^2 \over dx^2} \Big( {\partial L \over \partial y_{xx}} \Big) = 0.$

The Einstein-Hilbert action for general relativity is an example of an action principle that involves 2nd order derivatives of the basic variables i.e. the metric tensor. Under variations of the metric which vanish and who's normal derivatives also vanish on the boundary, the boundary of a compact region, this action is stationary if and only if the metric satisfies Einstein's equations - the Euler Lagrange equations are

${\partial L \over \partial g_{ab}} - {\partial \over \partial x^c} \Big( {\partial L \over \partial g_{ab,c}} \Big) + {\partial^2 \over \partial x^c \partial x^d} \Big( {\partial^2 L \over \partial g_{ab,cd}} \Big) = 0$

which gives

$G_{ab} = \kappa T_{ab}$

(which turn out to be second order differential equations). If the normal derivatives dont vanish you have to add a boundary term to the action principle to get out the right equations of motion in the bulk (this term is important in quantum gravity).

Last edited: Aug 29, 2014
4. Aug 29, 2014

### Matterwave

It is possible I recalled some of the discussion on higher derivative terms incorrectly. I apologize if I did. I do recall that a higher derivative term will require more initial conditions to specify the equations of motion than the regular term we are used to though.

5. Aug 29, 2014

### julian

No problem. You will have an extra term

$\int_{x_1}^{x_2} dx {\partial L \over \partial y_{xx}} {d^2 \delta y \over dx^2}$

which upon integration by parts gives

$\Big[ {d \delta y \over dx} {\partial L \over \partial y_{xx}} \Big]_{x_1}^{x_2} - \int_{x_1}^{x_2} dx {d \over dx} \Big( {\partial L \over \partial y_{xx}} \Big) {d \delta y \over dx}$
= $\Big[ {d \delta y \over dx} {\partial L \over \partial y_{xx}} \Big]_{x_1}^{x_2} - \Big[ \delta y {d \over dx} \Big( {\partial L \over \partial y_{xx}} \Big) \Big]_{x_1}^{x_2} + \int_{x_1}^{x_2} dx {d^2 \over dx^2} \Big( {\partial L \over \partial y_{xx}} \Big) \delta y (x)$

so if $\delta y = \delta y' = 0$ at the boundaries we are left with an extra term to the Euler-Lagrange equations

$+{d^2 \over dx^2} \Big( {\partial L \over \partial y_{xx}} \Big)$

In non-relativistic QM (as with any hyperbolic differential equations) it is the order of the time-derivative that dictates what initial conditions need to be specified. Shrodinger's equation is first order in time derivatives so you need only specify the state of the system a moment. It is in relativistic QMs, in particular the Klein-Gordon equation, that you encounter a differential equation of second order in time derivatives. This leads to the requirement of specifying $\phi(t_0)$ and ${\partial \over \partial t} \phi(t_0)$, which in turn leads to the well known problem of negative probability density! This, initially alarming problem, led Dirac to formulate a first order in time relativistic equation (and first order in spatial derivatives - so they are on an equal footing).

In GR it happens to turn out to be a second order in derivatives problem, higher order derivatives dont appear. The calculation using the Euler-Lagrange equations is horrendous, apparently. There is what is called a first order formulism where the metric and connection are taken to be independent variables and the connection appears only with first order derivatives making it much easier. The Euler-Lagrange equations for the connection give the connection as the metric connection. And the Euler-Lagrange equations for the (densitized) contravariant metric trivially give Einstein's equations.

Last edited: Aug 29, 2014
6. Aug 30, 2014

### ShayanJ

OK, thanks.
But now I want to find a $\psi$ for which $\int \psi^*H\psi dx$ is stationary.
For one particle in potential $V(x)$, $H=-\frac{\hbar^2}{2m}\frac{d^2}{dx^2}+V(x)=-a \partial^2+V$, so we have $L=-a\psi^*\partial^2\psi+\psi^* V \psi$ and using the Euler-Lagrange-(julian) equation:

$V\psi^*-a\psi^*=0 \Rightarrow \psi^*=0 \Rightarrow \psi=0$!!!

I think its not that easy to say $\delta y'=0$!

7. Aug 30, 2014

### julian

It isn't as simple as setting the variation equal to zero...there is the constraint that $\int |\psi (x)|^2 dx = 1$ which means you have to introduce Lagrangian multipliers...

8. Aug 30, 2014

### julian

You then have ${\delta \over \delta \psi^*} \int \psi^* H \psi dx = \lambda \psi$ where $\lambda$ is a Lagrange multiplier. Which is basically

$H \psi = E \psi$

Last edited: Aug 30, 2014
9. Aug 30, 2014

### julian

Hang on a moment...shouldn't you get a term:

${d^2 \over dx^2} \Big( {\partial \over \partial \psi_{xx}} (-a \psi^* \psi_{xx}) \Big) = -a {d^2 \over dx^2} \psi^*$?

And oh I should have been careful about the difference between initial conditions and boundary conditions on the wavefunction at spatial locations. It is the latter that is important here.

10. Aug 31, 2014

### ShayanJ

Yeah...that's right. I don't know why I did that mistake. Actually when I did the calculations, I wondered where is the $E \psi$ but I posted that when I was busy a little and so the statement of my problem changed to that wrong thing!!! But that problem is solved now. Thanks

And about $\delta y'=0$. I still have some doubt about it. Here we're actually excluding variations for which $\delta y' \neq 0$. It means we know the value of wavefunction's derivative at the boundaries. We also assume $\delta y=0$ which means we know the value of wavefunction itself at the boundaries. So we're assuming that we have four boundary conditions where we need only two!!!
Its strange that its working. What's the meaning of this?

11. Sep 9, 2014

### julian

A vaguely remember from undergrad that you solved Schrodinger's equation in separate regions and then had to match up solutions by both equating wavefunction at the boundary as well sometimes matching the derivative of the wavefunction at the boundary.

There is a more subtle issue, which as far as I know has only been brought up in what is called background-independent scattering amplitudes of loop quantum gravity, where we wish QM's equations to hold in a finite region but where we specify boundary data as coming from the experimental set-up. These can be motivated by: "finite closed boundaries represent the way real experiments are set up more directly than constant-time surfaces. A realistic experiment is confined to a finite region of spacetime." Anyway, this is called the "general boundary" approach. This seems exclusively targeted at QFT. Would be interesting to see how this can work for non-relativistic QM.