Euler-Lagrange Equations for Schördinger Eq.

  • Thread starter Thread starter gulsen
  • Start date Start date
  • Tags Tags
    Euler-lagrange
gulsen
Messages
215
Reaction score
0
Euler-Lagrange equations for the Lagrangian density \mathcal{L} = V\psi \psi^* + \frac{\hbar^2}{2m}\frac{\partial \psi}{\partial x}\frac{\partial \psi^*}{\partial x} + \frac{1}{2}\left(i\hbar \frac{\partial \psi^*}{\partial t} \psi- i\hbar \frac{\partial \psi}{\partial t} \psi^*\right) gives (complex conjuagate of) Schördinger equation, when it's considered as the minumum of \int \int \mathcal{L} dx dt. Is this of any use?

edit: corrected Lagrangian density. This LD results in \left(-\frac{\hbar^2}{2m}\frac{\partial^2 \psi}{\partial x^2} + V\psi - i\hbar \frac{\partial \psi}{\partial t}\right) + \left(-\frac{\hbar^2}{2m}\frac{\partial^2 \psi^*}{\partial x^2} + V\psi^* + i\hbar \frac{\partial \psi^*}{\partial t}\right) = 0
 
Last edited:
Physics news on Phys.org
I have never heard that complex conjugate of SE is used for something physically meaningful.But who knows...
 
I was trying to ask whether such Lagrangian approach is useful.
 
Last edited:
Same negative answer .
 
I was not talking about complex conjugate of S.E. I've tried to clarify what the LD means by editing the first post. This should be a mathematically valid LD. I was just wondering if this Lagrangian approach can be preferred over Hamiltonian approach in some cases. (Though I've never studied the subject, I've seen that QFT prefer Lagrangians for some reason.)
 
gulsen said:
I was not talking about complex conjugate of S.E. I've tried to clarify what the LD means by editing the first post. This should be a mathematically valid LD. I was just wondering if this Lagrangian approach can be preferred over Hamiltonian approach in some cases. (Though I've never studied the subject, I've seen that QFT prefer Lagrangians for some reason.)

It *is* useful, if you do quantum mechanics using the path integral formulation. In introductory quantum mechanics, one usually learns only the Hamiltonian approach. But it's possible to do non-relativistic QM using path integrals, in which case the Lagrangian is the fundamental quantity.

in QFT, one may either use the Hamiltonian (then called "canonical quantization") approach or the path integral approach. But the PI approach is more powerful for many things (and makes the invariance of the results more transparent) which si why there is much more emphasis on the Lagrangian in QFT.
 
Ah, thanks. That answered my question.

nrqed said:
But it's possible to do non-relativistic QM using path integrals, in which case the Lagrangian is the fundamental quantity.

I'd like to learn that. Do you know any resource/book that has this approach?
 
The Hamiltonian and Lagrangian approaches to QFT are equivalent by a theorem proved by Marc Henneaux and co.

Feynman and Hibbs wrote a book about path integral in QM.

Daniel.
 
Last edited:
dextercioby said:
Feynman and Hibbes wrote a book about path integral in QM.

Daniel.

Out of print, and insanely difficult to learn from.
 
  • #10
gulsen said:
Ah, thanks. That answered my question.



I'd like to learn that. Do you know any resource/book that has this approach?

I would highly recommend "Principles of Quantum Mechanics" by Ramamurti Shankar. It's a bit pricey ($68) but is an *excellent* book on quantum mechanics and is worth every penny. out of almost 700 pages there is only about 100 pages on the path integral formulation but the whole book is extremely good. And he gives a review of the lagrangian/hamiltonian formulations of classical mechanics as well. On amazon.com you can actually read excerpts, so ask for "path integral" quotes and read a few pages to get a feel. If you have access to a college/university library they surely have a copy.

Patrick
 
  • #11
Strangely enough, Goldsteins book on classical mechanics contains a lagrangian formulation of the Schrodinger field. As was said it can be very useful, especially if you dislike the use of creation and annihilation operators.
 
  • #12
I seem to remember reading somewhere that starting with a Lagrangian makes it difficult to formulate a relativistically covariant theory unless the Lagrangian is quadratic in the velocities, whereas starting from the Hamiltonian makes it easier.
 
  • #13
The Dirac lagrangian is NOT "quadratic in the <<velocities>>", yet it provides a relativistic covariant theory. And the same with the Rarita-Schwinger lagrangian.

Daniel.
 
Back
Top