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Nambu Goldstone Boson

  1. Oct 1, 2005 #1
    I'd like to ask some questions about the following example of broken symmetry and non-invariant vaccum.

    The basic argument goes as follows:

    [tex] \Cal L= \partial^\mu \phi \partial_\mu \phi \: - \: \mu^2 \phi^* \phi \: - \: \lambda (\phi^*\phi)^2

    [/tex]


    [tex]

    \frac{\partial V} {\partial \phi}=0

    [/tex]

    [tex]

    <\phi>^2=\frac{\mu^2}{2\lambda}

    [/tex]


    My first question regards the following term in the lagrangian:

    [tex]\lambda (\phi^*\phi)^2[/tex]

    does this term indicate that the field is self-interacting?
    What does that mean for a field to self interact?

    My next question regards the result.

    Now, the expectation value of the field is shown not to be equal to zero when V is minimized. I have read that

    "this implies that the vacuum is not invariant under the u(1) symmetry [tex] \phi \rightarrow e^{i\theta}\phi [/tex] therefore there must be a zero mass particle in the theory."

    I really don't understand the conclusion. Why is the vacuum not invariant as I don't see the term |0> anywhere and why does it imply that there is a zero mass particle??? I can follow the math but not what the math means.

    Any help appreciated.
     
  2. jcsd
  3. Oct 1, 2005 #2

    Haelfix

    User Avatar
    Science Advisor

    Well the zero mass particle in that theory is the nambu goldstone boson.

    Perhaps it would be best if you actually graphed the potential, its kinda hard to explain it on a message board. Now look at the value(s) that minimize the potential, notice if you naively picked a vacuum minima (called it zero say) and perturbed around this point, you would locally see a theory that does not possess the full symmetry of the entire shebang.

    Think of the mexican hat potential, what you have here.. Now, look at the saddle point, notice it possesses a U(1) symmetry. However its wildly unstable at this pt, and will want to drop into a lower vacuum state. The multiple vacua at the bottom of the potential well, here form a circle of degeneracy, and whichever one you end up with upon quantization will spontaneously break this U(1) symmetry.

    Now, as to why this implies a zero mass particle, is called Goldstone's theorem. You can think of it as a sort of pure kinetic term, since rolling around in the degenerate circle costs no energy.
     
  4. Oct 2, 2005 #3
    Thanks for your response.

    I want to ask about the Mexican hat potential. The only occurs when a specific form for V(x) is chosen (as in the example above). What is the motivation/justification for ?introducing that specific form?
     
  5. Oct 3, 2005 #4
    Ok - I know now so I don't want to waste anyone time!!

    :)
     
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