In the presence of a magnetic field with vector potential [itex] \vec A [/itex] and an electric field, the Schrodinger equation for a charged particle with charge q and mass m becomes:(adsbygoogle = window.adsbygoogle || []).push({});

[itex]

\frac{1}{2m} (\frac{\hbar}{i} \vec \nabla-q\vec A)^2 \psi =(E-q \phi)\psi

[/itex]

Another fact is that, Schrodinger equation can be derived by finding a [itex] \psi [/itex] for which [itex] \int_{x_1}^{x_2} \psi^* (-\frac{\hbar^2}{2m}\frac{d^2}{dx^2}+V(x))\psi dx [/itex] is stationary (we also need [itex] \int_{x_1}^{x_2} |\psi|^2 dx=1 [/itex] but that's irrelevant here). So, if we take [itex] a=\frac{\hbar^2}{2m} [/itex] and [itex] \partial=\frac{d}{dx} [/itex], we can say that Schrodinger equation has the Lagrangian density [itex] L=-a\psi^* \partial^2 \psi+\psi^* V \psi [/itex]. But this Lagrangian density is not invariant under [itex] \psi \rightarrow e^{i \chi(x)} \psi [/itex] so I try to replace [itex] \partial [/itex] by something that makes up for the extra terms that are brought by the transformation.

My first question is that, is this an example of a gauge theory?

My second question is this:

When I assume [itex] D=\partial+A(x) [/itex] and try to see whether this makes the Lagrangian density above invariant under the mentioned transformation, I only get [itex] \chi(x)=const [/itex] which means the wavefunction has only a global symmetry, not a local one and so we don't need a gauge field at all which we know is not the case.

Here's one method of doing it:

[itex]

(\partial+A)(e^{i\chi}\psi)=e^{i\chi}(\partial+A) \psi \Rightarrow i \psi \partial \chi+\partial \psi+A \psi=\partial \psi+A \psi \Rightarrow \partial \chi=0 \Rightarrow \chi=const

[/itex]

I also tried comparing the original Lagrangian density with the transformed part to see what should be zero and that gave two differential equations which gave the same result.

What's wrong here?

Thanks

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# Is this an example of a gauge theory? How?

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