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Is this an example of a gauge theory? How?

  1. Sep 5, 2014 #1

    ShayanJ

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    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:
    [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
     
  2. jcsd
  3. Sep 5, 2014 #2

    atyy

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  4. Sep 6, 2014 #3

    ShayanJ

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    Yeah, that made it better:
    [itex]

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

    [/itex]

    But does this mean that I should replace [itex] \partial [/itex] by [itex] D=\partial+A-i\partial \chi [/itex] in the Lagrangian?
    But that doesn't seem right!
    My reason for doing this, is getting the interaction term from gauge invariance but this doesn't seem to give that result!
     
  5. Sep 6, 2014 #4

    atyy

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    Actually, I haven't understood what you are doing. But I googled "Maxwell-Schroedinger Lagrangian" and this looks like it may be helpful http://arxiv.org/abs/math-ph/0505059v4 (Eq 2.6). My guess is that you should transform ψ and A together, so ψ' = exp(iλ(x))ψ(x) and A' = A + ∇λ(x), where I did not get signs, constants, imaginary things etc correct. It looks like the gauge transformations with the correct details are given in Eq 7.5.
     
    Last edited: Sep 6, 2014
  6. Sep 6, 2014 #5

    ShayanJ

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    I thought this thing that I was doing, is a well-known procedure. Because I, somehow, was following the approach explained here.
    The calculations in the page suggest, that the initial Lagrangian density has no interaction term. Then requiring it to have a local symmetry, changes the derivative to a covariant derivative which introduces a new term to the Lagrangian density that causes the interaction between the particle and the gauge field.
    But other sources, like the one you mentioned, are assuming the existence of potentials(gauge fields) and just find out their symmetry transformations.
     
  7. Sep 6, 2014 #6

    atyy

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    Yes, I understood you were trying to use the "gauge principle" to get the interaction term. I just couldn't follow the detailed steps in your posts, although I understood what you were trying to do. I was just saying that because I wanted to post a link you might find helpful, even though I'm too slow to do the calculation.

    In the old days, the "gauge principle" was magic. Nowadays we understand that it is like the equivalence principle. So both the gauge principle and the equivalence principle are "minimal coupling principles". The idea is that actually there are many other gauge invariant terms one could add, but these are the "simplest", and if they work, hooray!
     
  8. Sep 7, 2014 #7

    samalkhaiat

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    The Schrodinger equation for the field [itex]\psi[/itex] can be derived by the following equivalent ways
    1) from the Lagrangian density
    [tex]\mathcal{ L } = \frac{ \hbar }{ 2 i } \left( \psi^{ * } \partial_{ t } \psi - \psi \partial_{ t } \psi^{ * } \right) + \frac{ \hbar^{ 2 } }{ 2 m } \nabla \psi^{ * } \cdot \nabla \psi + V \psi^{ * } \psi ,[/tex]
    using the Euler-Lagrange equation
    [tex]\frac{ \partial \mathcal{ L } }{ \partial \psi^{ * } } = \partial_{ t } \left( \frac{ \partial \mathcal{ L } }{ \partial ( \partial_{ t } \psi^{ * } ) } \right) + \nabla \cdot \left( \frac{ \partial \mathcal{ L } }{ \partial ( \nabla \psi^{ * } ) } \right) ,[/tex]
    2) from the Lagrangian
    [tex]L = \int d^{ 3 } x \mathcal{ L } ,[/tex]
    using the (tricky) functional Euler-Lagrange equation
    [tex]\frac{ \delta L }{ \delta \psi^{ * } } = \partial_{ t } \left( \frac{ \delta L }{ \delta ( \partial_{ t } \psi^{ * } ) } \right) ,[/tex]
    or 3) from the action integral
    [tex]S = \int d t \ L = \int d t d^{ 3 } x \ \mathcal{ L } ,[/tex]
    using the principle of least action
    [tex]\frac{ \delta S }{ \delta \psi^{ * } } = 0 .[/tex]

    Sam
     
  9. Sep 7, 2014 #8

    samalkhaiat

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    This is explained for a general field theory and arbitrary gauge group, in the attached PDF at the end of this post

    https://www.physicsforums.com/showpost.php?p=4813808&postcount=26

    You can apply the same method on the free Schrodinger Lagrangian density.

    Sam
     
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