Thursday, January 8, 2015

Lagrangian of Schrodinger field


The usual Schrodinger Lagrangian is $$ \tag 1 i(\psi^{*}\partial_{t}\psi ) + \frac{1}{2m} \psi^{*}(\nabla^2)\psi, $$ which gives the correct equations of motion, with conjugate momentum for $\psi^{*}$ vanishing. This Lagrangian density is not real but differs from a real Lagrangian density $$ \tag 2 \frac{i}{2}(\psi^{*}\partial_{t}\psi -\psi \partial_{t}\psi^{*} ) + \frac{1}{2m} \psi^{*}(\nabla^2)\psi $$ by a total derivative.



My trouble is that these two lagrangian densities lead to different conjugate momenta and hence when setting equal time commutation relations, I am getting different results (a factor of 2 is causing the problem). I can rescale the fields but then my Hamiltonian also changes. Then applying Heisenberg equation of motion, I don't get the operator Schrodinger equation.


Is it possible to work with the real Lagrangian density and somehow get the correct commutation relations? I would have expected two Lagrangians differing by total derivative terms to give identical commutation relations (since canonical transformations preserve them). But perhaps I am making some very simple error. Unless all conjugate momenta are equivalent for two Lagrangians differing by total derivatives, how does one choose the correct one?


I guess the same thing happens for other first order systems like Dirac Lagrangian also.




No comments:

Post a Comment

classical mechanics - Moment of a force about a given axis (Torque) - Scalar or vectorial?

I am studying Statics and saw that: The moment of a force about a given axis (or Torque) is defined by the equation: $M_X = (\vec r \times \...