Homework Help: QFT lagrangian problem

1. Sep 29, 2012

arten

Hello, I've started a course on QFT and I'm having some troubles trying to find the solution of this exercise:

Write the action of a non-relativistic spineless free particle in a manifestly hermitian way

The problem should be simple but I'm a bit lost in the hermitian way part... What does it mean here ?

Thanks

2. Sep 29, 2012

tom.stoer

I think you have to find a hermitean Lagrangian from which the Schrödinger equation can be derived:

$$\partial_0\frac{\partial \mathcal{L}}{\partial (\partial_0\psi^\ast)} + \partial_i\frac{\partial \mathcal{L}}{\partial (\partial_i\psi^\ast)}- \frac{\partial \mathcal{L}}{\partial \psi^\ast} = 0\;\;\Rightarrow\;\; i\hbar\partial_0\psi + \frac{\hbar^2}{2m}\partial_i^2\psi = 0$$

3. Sep 30, 2012

tom.stoer

I guess you are familiar with Lagrangians in field theory, e.g. electrodynamics (the problem here is simpler). You have to find a Lagrangian with

$$\mathcal{L} = \mathcal{L}[\psi,\psi^\ast, \partial_0\psi, \partial_0\psi^\ast, \partial_i\psi, \partial_i\psi^\ast]$$

Here ψ and ψ* are independent variables, so in principle there are two Euler-Lagrange equations, one for ψ* derived via variation w.r.t. ψ and one for ψ derived via variation w.r.t. ψ*; I wrote down the ansatz for the latter one. Of course these two equations are related via complex conjugation, so you get the Schrödinger equation and the cc Schrödinger equation.

The Schrödinger equation is of first order in the time derivative, so there can't be a square of the time derivative in the Lagrangian, you have to have something like

$$\psi^\ast\,\partial_0\psi$$

plus cc, of course.

The Schrödinger equation is of second order in the spatial derivative, so you have to have something like

$$(\partial_i\psi^\ast)\,(\partial_i\psi)$$

plus cc.

It means that

$$\mathcal{L}^\ast = \mathcal{L}$$

Last edited: Sep 30, 2012
4. Sep 30, 2012

arten

I write the lagrangian as:

L = ihψ*∂0ψ+h2/2m(∂iψ*)(∂iψ) but it's not hermitian so I rewrite as

L = ih/2[ψ*∂0ψ-ψ∂0ψ*]+h2/2m(∂iψ*)(∂iψ)

If I apply the equation to L or L* I get the schrödinger equation for ψ and ψ*

Is it correct?

Then the action is S=∫dx4L

Last edited: Sep 30, 2012
5. Sep 30, 2012

tom.stoer

Nearly correct; the Lagrangian isn't hermitean, so you have to add the cc of the first term

6. Sep 30, 2012

Now it's OK