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I Conservation of charge

  1. Apr 21, 2017 #1
    What equation in QM show us that symmetry in quantum-mechanical phase implies charge conservation?
     
  2. jcsd
  3. Apr 21, 2017 #2

    PeterDonis

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  4. Apr 21, 2017 #3
    Thanks. This is more difficult to understand than I thought.
     
  5. Apr 22, 2017 #4

    vanhees71

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    A conservation law is always due to global (not local!) gauge symmetries. Since a local symmetry implies a corresponding global symmetry, you also have a conservation law (in this case of a charge-like quantity).

    Let's consider the most imple special case of the Noether theorem, which is just taylored to charge conservation. Here you have a symmetry, where the variation of the action
    $$A[\psi]=\int \mathrm{d}^4 x \mathcal{L}(\psi,\partial_{\mu} \psi)$$
    is unchanged under a socalled "external symmetry", i.e., a symmetry transformation not involving the space-time coordinates. For an infinitesimal transformation you have
    $$\delta x^{\mu}=0, \quad \delta \psi=\delta \eta \tau(x,\psi),$$
    where ##\tau(x,\psi)## is some function.

    Now the variation of the action is
    $$\delta A=\delta \eta \int \mathrm{d}^4 x \left [\frac{\partial \mathcal{L}}{\partial \psi} \tau + \frac{\partial \mathcal{L}}{\partial (\partial_{\mu} \psi)} \partial_{\mu} \tau \right] = \delta \eta \int \mathrm{d}^4 x \tau \left [\frac{\partial \mathcal{L}}{\partial \psi} \ - \partial_{\mu} \frac{\partial \mathcal{L}}{\partial (\partial_{\mu} \psi)} \right]=0.$$
    This means we have a symmetry if there exists a vector field ##j^{\mu}## such that
    $$\left [\frac{\partial \mathcal{L}}{\partial \psi} - \partial_{\mu} \frac{\partial \mathcal{L}}{\partial (\partial_{\mu} \psi)} \right] \tau=-\partial_{\mu} j^{\mu}. \qquad (*)$$
    The sign on the right-hand side is convention.

    The equations of motion are given by the Euler-Lagrange equations, i.e., the left-hand side vanishes for the solutions of the equations of motion, implying the continuity equation ##\partial_{\mu} j^{\mu}=0##, which means that the Noether charge
    $$Q=\int \mathrm{d}^3 \vec{x} j^0(t,\vec{x})=\text{const}$$
    and that this is a Lorentz scalar.

    Now we need to find the explicit expression for ##j^{\mu}## given the Lagrangian and ##\tau##. To that end we rewrite (*) a bit:
    $$\partial_{\mu} j^{\mu}=\partial_{\mu} \left [\frac{\partial \mathcal{L}}{\partial (\partial_{\mu} \psi)} \tau \right]-\tau \frac{\partial \mathcal{L}}{\partial \psi} - \frac{\partial L}{\partial (\partial_{\mu} \psi)} \partial_{\mu} \tau.$$
    So if we have a symmetry, we must have ##\Omega^{\mu}(x,\psi)## such that
    $$\partial_{\mu} \Omega^{\mu}=-\delta L/\delta \eta=-\tau \frac{\partial \mathcal{L}}{\partial \psi} - \frac{\partial L}{\partial (\partial_{\mu} \psi)} \partial_{\mu} \tau. \qquad (**)$$
    Here we have used that the symmetry transformation for the Lagrangian must not change the variation of the action for all fields, so it must be a total four-divergence of a function of only the fields and not its gradient. Then obviously
    $$j^{\mu}=\frac{\partial \mathcal{L}}{\partial (\partial_{\mu} \psi)} \tau+\Omega^{\mu}.$$
    As the most simple example let's take the free Dirac field with the Lagrangian
    $$\mathcal{L}=\overline{\psi}(\mathrm{i} \not{\partial}-m) \psi.$$
    In this case the Lagrangian is obviously invariant under a phase transformation, i.e.,
    $$\psi \rightarrow \psi'=\exp(-\mathrm{i} q \eta)\psi, \quad \overline{\psi} \rightarrow \overline{\psi}'=\exp(+\mathrm{i} q \eta) \psi,$$
    where ##q## and ##\eta## are real numbers. Now we make ##\eta \rightarrow \delta \eta## small, getting
    $$\delta \psi=-\mathrm{i} q \psi \delta \eta, \quad \delta \overline{\psi} = +\mathrm{i} q \overline{\psi} \delta \eta.$$
    Thus we have
    $$\tau=-\mathrm{i} q \psi, \quad \overline{\tau}=q \mathrm{i} \overline{\psi}.$$
    Now
    $$\frac{\partial \mathcal{L}}{\partial(\partial_{\mu} \psi )}=\overline{\psi} \mathrm{i} \gamma^{\mu}, \quad \frac{\partial \mathcal{L}}{\partial (\partial_{\mu} \overline{\psi})}=0.$$
    To determine ##\Omega## we have to use (**), which in our case reads
    $$\partial_{\mu} \Omega^{\mu}=0,$$
    i.e., we can set ##\Omega^{\mu}=0## and thus the Noether Current is
    $$j^{\mu}=q \overline{\psi} \gamma^{\mu} \psi.$$
    It is easy to check that this is indeed a conserved current for the solutions of the equations of motion, i.e., the Dirac equation
    $$(\mathrm{i} \not{\partial} +m) \psi=0.$$
    To make the symmetry local, i.e., to make the Lagrangian invariant under a transformation of the form
    $$\psi \rightarrow \psi'=\exp(-\mathrm{i} q \tau(x)) \psi, \quad \overline{\psi} \rightarrow \overline{\psi}'=\exp(+\mathrm{i} q \tau(x)) \overline{\psi},$$
    we have to introduce a gauge field ##A^{\mu}## and plug everywhere where we have a partial derivative the gauge-covariant derivative
    $$\partial_{\mu} \rightarrow \partial_{\mu} +\mathrm{i} q A_{\mu}.$$
    If we define the transformation for ##A_{\mu}## as
    $$A_{\mu}'=A_{\mu} + q \partial_{\mu} \tau$$
    we get
    $$\mathrm{D}_{\mu}' \psi'=(\partial_{\mu} + \mathrm{i} q A_{\mu}') \psi '=\exp(-\mathrm{i} q \tau)[\partial_{\mu} \psi-\mathrm{i} q \partial_{\mu} \tau \psi + \mathrm{i} q A_{\mu} \psi + \mathrm{i} q \partial_{\mu} \tau \psi]=\exp(-\mathrm{i} q \tau) [\partial_{\mu} \psi + \mathrm{i} q A_{\mu}]=\exp(-\mathrm{i} q \tau) \mathrm{D}_{\mu} \psi.$$
    Thus now indeed the Lagrangian is invariant under local gauge transformations. Of course, the current is still conserved, because the Lagrangian is still invariant under global gauge transformations.
     
    Last edited: Apr 22, 2017
  6. Apr 22, 2017 #5

    Mentz114

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    Thanks for posting this in such detail. It is not covered in any of my textbooks.

    There is a latex typo in ##\frac{\partial \mathcal{L}}{\partial(\partial_{\mu} \psi )}=\overline{\psi} \mathrm{i} \gamma^{\mu}, \quad \frac{\partial \mathcal{L}}{\frac (\partial_{\mu} \overline{\psi})}=0.##
    but it is obvious what it is meant.
     
  7. Apr 22, 2017 #6

    vanhees71

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    Thanks for the hint. I've corrected the typo.
     
  8. Apr 22, 2017 #7
    Great demonstration vanhees71. A few questions:

    - Where does ##\delta \eta## and ##\delta \alpha## come from
    - The action ##A[ \Psi]## can always be expressed by that integral for all phenomena (e.g. electromagnetism, relativity...)? And why haven't you inserted ##\sqrt{-g}##?
    - In this case, the vector field turned out to be the familiar current density and the symbol ##j## usually is the one used for it. It became the current density after you have given that example of the free Dirac field or can we regard that vector as a "generalized current"?
     
  9. Apr 22, 2017 #8

    vanhees71

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    I corrected the typo. Of course, there are only the ##\delta \eta## in the entire posting.

    All fundamental laws of physics are derivable from an action principle. Further I restricted myself to special relativity in Minkowski coordinates where ##g=-1##.

    As demonstrated for any independent one-parameter Lie-symmetry group there exists a conserved current. You can repeat everything with the Klein-Gordon field too:
    $$\mathcal{L}=(\partial_{\mu} \phi^*)(\partial^{\mu} \phi)-m^2 \phi^* \phi.$$
    The result is the Noether current
    $$j_{\mu} = \mathrm{i} q (\phi^* \partial_{\mu} \phi-\partial_{\mu} \phi^* \phi).$$
     
  10. Apr 22, 2017 #9
    Ok. But where this term comes from?
     
  11. Apr 22, 2017 #10

    vanhees71

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    It's the "infinitesimal" group parameter (in our example the phase).
     
  12. Apr 22, 2017 #11
    So, at the beggining of your post, ##\Psi## is a function of ##x, \eta##, but as you are holding ##x## constant, ##\delta \Psi = \delta \eta \ \tau (x, \Psi)##?
     
  13. Apr 23, 2017 #12

    vanhees71

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    I don't understand the confusion. Of course ##\psi## is a function of ##x## only. It denotes a general set of fields, and ##\eta## are a set of parameters of the Lie group. Take ##\psi## an ##n##-dimensional field, ##t^a## a set of generators of the Lie group. Then the Lie group acts on the fields via
    $$\psi \rightarrow \psi'=\exp(-\mathrm{i} \eta_{a} t^a) \psi.$$
    For an infinitesimal transformation you have
    $$\psi \rightarrow \psi+\delta \psi, \quad \delta \psi=\delta \eta_{a} t^{a} \psi$$
    and thus
    $$\tau^{a} =t^a \psi.$$
    For the Abelian case discussed as an example you just have a one-dimensional Lie group ##\mathrm{U}(1)## and ##t^a=1##.
     
  14. Apr 23, 2017 #13
    Oh yea, I see. Thanks.
     
  15. Apr 24, 2017 #14

    DrDu

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    I would be carefull to say it is "due to". After all, also a global gauge symmetry is a redundant symmetry, as all observables are totally symmetric under gauge transformations. Rather I would argue with Haag and Dopplicher that the gauge structure can be constructed from the observed charges.
     
  16. Apr 24, 2017 #15

    vanhees71

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    The global transformation however changes states via (anti-)unitary transformations, while a local gauge symmetry means to express things with different representatives of equivalence classes of the representation, i.e., it doesn't change states. That's also why local gauge symmetries cannot be spontaneously broken but only "Higgsed", i.e., in a Higgsed local gauge symmetry you don't get Goldstone modes but massive gauge fields. It's of course true that you can reconstruct the gauge group from the conserved charges.
     
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