- #31
cathalcummins
- 43
- 0
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:
x^\mu \mapsto x'^{\mu}=\Lambda^\mu_{\phantom{1}\nu} x^\nu
Covariantly;
\partial_\mu \mapsto \partial_{'\mu}=\Lambda_\mu^{\phantom{1}\nu} \partial_{'\nu}
And the wavefunction;
\psi(x)\mapsto \psi'(x')=S(\Lambda)\psi(x)
where S(\Lambda) is just some linear function.
The SE:
\{\iota \hbar\partial_t+\frac{\hbar^2}{2\mu} \partial_\nu \partial_\nu \}\psi(x)=0
where \nu=1,2,3.
Trasforming \psi(x) \mapsto \psi'(x') and multiplying by S^{-1}(\Lambda), and transforming the partial derivatives, we get
\{\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
On comparing this to the "primed" SE. We find that the conditions \Lambda must satisfy are
\Lambda_0^{\phantom{1}\nu} \partial_{'\nu}=\partial_{'0}
and
\sum^{3}_{i=1}\Lambda_i^{\phantom{1}\alpha}\Lambda_i^{\phantom{1}\beta}\partial_{'\alpha} \partial_{'\beta} = \partial_{'\nu} \partial_{'\nu}.
Now, my question is, for me to prove that the SE isn't Lorentz invariant, do I need to show
a) that no such \Lambda \in \mbox{O(1,3)} can exist
or
b) is it sufficient to pick some \Lambda' \in \mbox{O(1,3)} 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 \in \mbox{O(1,3)}, and what does this mean to the statement:
"The SE equation is not Lorentz invariant"
Thank you
"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:
x^\mu \mapsto x'^{\mu}=\Lambda^\mu_{\phantom{1}\nu} x^\nu
Covariantly;
\partial_\mu \mapsto \partial_{'\mu}=\Lambda_\mu^{\phantom{1}\nu} \partial_{'\nu}
And the wavefunction;
\psi(x)\mapsto \psi'(x')=S(\Lambda)\psi(x)
where S(\Lambda) is just some linear function.
The SE:
\{\iota \hbar\partial_t+\frac{\hbar^2}{2\mu} \partial_\nu \partial_\nu \}\psi(x)=0
where \nu=1,2,3.
Trasforming \psi(x) \mapsto \psi'(x') and multiplying by S^{-1}(\Lambda), and transforming the partial derivatives, we get
\{\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
On comparing this to the "primed" SE. We find that the conditions \Lambda must satisfy are
\Lambda_0^{\phantom{1}\nu} \partial_{'\nu}=\partial_{'0}
and
\sum^{3}_{i=1}\Lambda_i^{\phantom{1}\alpha}\Lambda_i^{\phantom{1}\beta}\partial_{'\alpha} \partial_{'\beta} = \partial_{'\nu} \partial_{'\nu}.
Now, my question is, for me to prove that the SE isn't Lorentz invariant, do I need to show
a) that no such \Lambda \in \mbox{O(1,3)} can exist
or
b) is it sufficient to pick some \Lambda' \in \mbox{O(1,3)} 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 \in \mbox{O(1,3)}, and what does this mean to the statement:
"The SE equation is not Lorentz invariant"
Thank you