Science Forums 2

Ben · 07-07-2009

Quote:

Originally Posted by Jay-qu

Beginning with the Schrodinger equation for a particle in free space
$i\partial _t \Psi = -\frac{\hbar^2}{2m}\nabla^2\Psi$

Missing an hbar on the LHS? Not important. But I am slightly troubled by the absence of a potential term in your Hamiltonian. Is this to do with the qualification "in free space"?

Quote:

But why stop at just global phase changes? What if we want the wavefunction to be invariant under an arbitrary transformation $\Psi -> \Psi e^{ie\theta(\vec{x},t)}$

Yeah, I have something similar here, but without $e$ in the exponent. Is this strictly required? And why have you vectorized the spatial component x? (if that's what it is)

Quote:

When transforming the wavefunction by this arbitrary phase change the Schrodinger equation is changed.

OK, so I got, as $\Psi \mapsto \Psi e^{i \theta(x,t)} = \Psi'$ then, after a bit of manipulation, using God's units, and ditching the $e$ in the exponent

$i\partial_t \Psi -\partial_t\theta(x,t) = \frac{1}{2m}[\nabla+ i\nabla\theta(x,t)]^2 \Psi$ where I have also "cancelled" the exponential factor. This is clearly nothing to do with Erwin.

Quote:

we can essentially summarise the changes as
$\vec{A} -> \vec{A}' = \vec{A}+\nabla \theta$
$\phi -> \phi ' = \phi-\partial_t \theta$

Um. I got a mass term in there - $\phi' = \phi -2m \partial_t \theta$ Do I need to worry about that?

Also, I think I may have an argument that says that your `vector potential' $A$ is actually a connection 1-form on the $U(1)$ bundle. Leave it with me a while.

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