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Philcorp

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Phil :!)

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- Thread starter Philcorp
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Philcorp

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Phil :!)

- #2

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"Time evolution of quantum systems" is an awfully imprecise statement. By virtue of the first two postulates of quantum mechanics, a quantum system is described through physical states and physical observables. So the time evolution is ascribed either to quantum physical states (Schrödinger picture), or to quantum (usually Hamiltonian) observables (Heisenberg picture), or to both states & observables (interaction/Dirac-Tomonaga-Schwinger picture).

So get a serious QM book and make sure postulates + formulations, descriptions & representations are crystal clear.

Daniel.

- #3

Philcorp

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/boggled

Phil

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Assuming the classical analysis of the KG field is clear, which picture of the 3 in QM would you choose?

Daniel.

- #5

reilly

Science Advisor

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The answer to your question lies with the development of covariant perturbation theory. And, we are talking the basis of field theory, which can be found in any number of references -- there is no quick way to get to covariant dynamics. Weinberg's, Gross's texts on field theory, Schweber's QED and the Men Who Made It, Feynman's original papers, ...

Yes, done properly, the appropriate exponential of the interaction Hamiltonian is covariant, and forms the basis for developing the S-matrix.

Regards,

Reilly Atkinson

- #6

vanesch

Staff Emeritus

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reilly said:Yes, done properly, the appropriate exponential of the interaction Hamiltonian is covariant, and forms the basis for developing the S-matrix.

In the limit of T-> infinity (the S-matrix). It's harder to talk about *finite* time evolution in QFT, although there may be techniques now that I don't know of.

cheers,

Patrick.

- #7

Juan R.

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Philcorp said:

Phil :!)

I see nobody offers you a simple and easy answer.

Of course, there is only a evolution of quantum systems. That is the Schödinger like picture.

The evolution (

Weinberg said:First some good neew: quantum field theory is based on the same quantum mechanics that was invented by Schrödinguer, Heisemberg, Pauli, Born, and others in 1925-26, and has been used ever since in atomic, molecular, nuclear, and condensed mater physics.

There is not relativistic quantum mechanics. In fact, it cannot exist.

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Philcorp

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Phil :!)

- #9

James Jackson

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Juan R. said:There is not relativistic quantum mechanics. In fact, it cannot exist.

Errm, what? As a (simple) example, the Dirac equation is the result of making Schrodinger's Equation Lorentz invarient.

- #10

Philcorp

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James Jackson said:Errm, what? As a (simple) example, the Dirac equation is the result of making Schrodinger's Equation Lorentz invarient.

This was also my understanding of the matter, however the Dirac equation is only for spin-1/2 paticles, correct? Is there a more general equation which is Lorentz invariant?

Phil

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James Jackson said:Errm, what? As a (simple) example, the Dirac equation is the result of making Schrodinger's Equation Lorentz invarient.

Nope. For a free particle, the Lorentz covariant form of the Schrödinger equation is the Klein-Gordon equation. For heaven's sake, don't you know that Schrödinger initially obtained the relativistic equation and just then considered the nonrelativistic case? Read the first chapter in Weinberg's book, first volume and the references therein.

Daniel.

- #12

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Philcorp said:This was also my understanding of the matter, however the Dirac equation is only for spin-1/2 paticles, correct?

Yes. Massive, parity invariant , electrically charged 1/2 spin particles.

Philcorp said:Is there a more general equation which is Lorentz invariant?

Nope, there's no general case. The equations of motion are simply the Lagrange field equation for each classical field corresponding to all finite dimensional nonunitary reps of the restricted Lorentz group.

Daniel.

- #13

Philcorp

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Phil :!)

- #14

George Jones

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dextercioby said:The equations of motion are simply the Lagrange field equation for each classical field corresponding to all finite dimensional nonunitary reps of the restricted Lorentz group.

Another way of looking at the field equations for non-interacting fields is that they naturally characterize the infinite-dimensional unitary irreducible representations of the (universal cover of the) Poincare group.

Regards,

George

- #15

Juan R.

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Philcorp said:

Phil :!)

It is the nonrelativistic Schrodinger equation that is not Lorentz invariant.

- #16

Juan R.

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James Jackson said:Errm, what? As a (simple) example, the Dirac equation is the result of making Schrodinger's Equation Lorentz invarient.Juan R. said:There is not relativistic quantum mechanics. In fact, it cannot exist.

No. It is not so

[tex]

i \hbar \frac{\partial \Psi}{ \partial t} = H \Psi

[/tex]

Note: Dirac equation is not the result of making Schrodinger's Equation Lorentz invarient: that is the well-known relativistic Schrodinger equation.

Dirac equation was derived from the searching of a first order (in time) representation of Klein-Gordon equation more spin corrections for hidrogen atom.

Last edited:

- #17

Juan R.

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Philcorp said:

Phil :!)

Use the canonical formalism. Obtain the corresponding hamiltonian from the lagrangian and use it directly in the S-matrix.

The path integral can offer wrong answers.

- #18

Juan R.

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dextercioby said:Nope. For a free particle, the Lorentz covariant form of the Schrödinger equation is the Klein-Gordon equation. For heaven's sake, don't you know that Schrödinger initially obtained the relativistic equation and just then considered the nonrelativistic case? Read the first chapter in Weinberg's book, first volume and the references therein.

Daniel.

The Lorentzian version of the Schrödinger equation is the relativistic Schrödinger equation, which is not the Klein-Gordon equation. In fact, the relativistic Schrodinger equation has no problems with negative probabilities for example and is very used in semirelativistic quantum chemistry .

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