Schrodinger's Equation validity for relativistic particles

  • Thread starter turnerre
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  • #26
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You don't :rofl:

Open any textbook on field theory and see how easy it is to derive the Hamiltonian density

sam
Except that it isn't. In order to properly do it, you need to use Dirac brackets and Dirac's treatment of constraints, since, as was said earlier, the map from Lagrangian to Hamiltonian is singular for fermions.
 
  • #27
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Too bad I missed strangerep's post, heh =)
 
  • #28
samalkhaiat
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Except that it isn't. In order to properly do it, you need to use Dirac brackets and Dirac's treatment of constraints, since, as was said earlier, the map from Lagrangian to Hamiltonian is singular for fermions.
Wow, thank you for this valuable piece of information! :smile: Do read my posts to learn more about the use of Dirac brackets in constraint systems.

regards

sam
 
  • #29
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Wow, thank you for this valuable piece of information! :smile: Do read my posts to learn more about the use of Dirac brackets in constraint systems.

regards

sam
Always welcome! (unless my sarcasm detector is malfunctioning)
 
  • #30
reilly
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Gentlemen -- You are talking about a problem that was, for all practical purposes solved many years ago.Hans has it right; although, in my opinion, his normalization is non-standard. I say this with all due respect, and follow Weinberg pp21 and on(The Quantum Theory of Fields) . Drop the Eo from his H and L, and we agree. (Hans -Could it be that the Eo depends on how the exponential wave-functions are normalized?)

The standard Dirac Lagrangian is is simply the Dirac Equation LHS(including the time deriv) sandwiched between PSIBar and PSI (Gross,Rel. QM and FT; a very nice theory book, with lots of examples and a very straightforward style.)

For the practical physicist, no surprises in the KG world since about 1930. Vector bosons, pi mesons, photons(mass=0, of course), fermions, .... all obey the KG equation. which has been used, I'm sure, thousands of times without incident, since 1930. What's to worry?
Regards,
Reilly Atkinson
 
  • #31
This is a question involving the semantic on the statement

"The SE is not Lorentz invariant"

I'm trying to prove that the SE isn't Lorentz Invariant.

I think I've showed it. I'd just like some input on my argument.

Under the change of coordinates:

[tex]x^\mu \mapsto x'^{\mu}=\Lambda^\mu_{\phantom{1}\nu} x^\nu[/tex]

Covariantly;

[tex]\partial_\mu \mapsto \partial_{'\mu}=\Lambda_\mu^{\phantom{1}\nu} \partial_{'\nu}[/tex]

And the wavefunction;

[tex]\psi(x)\mapsto \psi'(x')=S(\Lambda)\psi(x)[/tex]

where [itex]S(\Lambda)[/itex] is just some linear function.

The SE:

[tex]\{\iota \hbar\partial_t+\frac{\hbar^2}{2\mu} \partial_\nu \partial_\nu \}\psi(x)=0[/tex]

where [itex]\nu=1,2,3[/itex].

Trasforming [itex]\psi(x) \mapsto \psi'(x')[/itex] and multiplying by [itex]S^{-1}(\Lambda)[/itex], and transforming the partial derivatives, we get

[tex]\{\iota \hbar\Lambda_0^{\phantom{1}\nu} \partial_{'\nu}+\frac{\hbar^2}{2\mu} \sum^{3}_{i=1}\Lambda_i^{\phantom{1}\alpha}\Lambda_i^{\phantom{1}\beta}\partial_{'\alpha} \partial_{'\beta} \} \psi'(x')=0[/tex]

On comparing this to the "primed" SE. We find that the conditions [itex]\Lambda[/itex] must satisfy are

[tex]\Lambda_0^{\phantom{1}\nu} \partial_{'\nu}=\partial_{'0} [/tex]

and

[tex]\sum^{3}_{i=1}\Lambda_i^{\phantom{1}\alpha}\Lambda_i^{\phantom{1}\beta}\partial_{'\alpha} \partial_{'\beta} = \partial_{'\nu} \partial_{'\nu}[/tex].

Now, my question is, for me to prove that the SE isn't Lorentz invariant, do I need to show

a) that no such [itex]\Lambda \in \mbox{O(1,3)}[/itex] can exist

or

b) is it sufficient to pick some [itex]\Lambda' \in \mbox{O(1,3)}[/itex] and show that it doesn't satisfy the above condition ("counterexample").

If b) is indeed sufficient, could someone explain why, as it would seem that the SE is Lorentz invariant for certain matrices [itex]\in \mbox{O(1,3)}[/itex], and what does this mean to the statement:

"The SE equation is not Lorentz invariant"

Thank you
 
  • #32
Looking back on this question, I suppose I could cut out all the texing and just simply ask:

If I show that [tex]\exists[/tex] a [tex]\Lambda \in O(1,3)[/tex] which the Schrodinger Equation does not transform covariantly under, is this enough to say that the SE isn't Lorentz Covariant.

Because trivially, if I take the identity, it's going to work. And presumably others work. So certain transformations will work, but not all.

So my question is: is it sufficient to show a counterexample in O(1,3) (which I have) to make the "SE not Lorentz Covariant" statement.

Thank you
 

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