QFT lagrangian problem

  • Thread starter arten
  • Start date
  • #1
3
0
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
 

Answers and Replies

  • #2
tom.stoer
Science Advisor
5,766
161
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
I think you have to find a hermitean Lagrangian from which the Schrödinger equation can be derived:

[tex]\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[/tex]
 
  • #3
tom.stoer
Science Advisor
5,766
161
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

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

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

[tex]\psi^\ast\,\partial_0\psi[/tex]

plus cc, of course.

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


[tex](\partial_i\psi^\ast)\,(\partial_i\psi)[/tex]

plus cc.

... in a manifestly hermitian way

... I'm a bit lost in the hermitian way part... What does it mean here ?
It means that

[tex]\mathcal{L}^\ast = \mathcal{L}[/tex]
 
Last edited:
  • #4
3
0
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:
  • #5
tom.stoer
Science Advisor
5,766
161
Nearly correct; the Lagrangian isn't hermitean, so you have to add the cc of the first term
 
  • #6
tom.stoer
Science Advisor
5,766
161
Now it's OK
 

Related Threads on QFT lagrangian problem

  • Last Post
Replies
1
Views
918
  • Last Post
Replies
2
Views
4K
  • Last Post
Replies
3
Views
2K
  • Last Post
Replies
0
Views
2K
  • Last Post
Replies
0
Views
3K
  • Last Post
Replies
4
Views
6K
  • Last Post
Replies
3
Views
2K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
5
Views
2K
  • Last Post
Replies
8
Views
3K
Top