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Spontaneous Symmetry breaking,phase factor

  1. Jul 30, 2009 #1
    Book: Maggiore-A modern Introduction to QFT

    He says that if a symmetry transformation multiplies the vacuum state by a constant phase factor only, the symmetry is not broken..


    U(1) symmetry is spontaneously broken....multiplication by exp(i theta) takes the vacuum state to another state on the circle of minima..

    But isnt this multiplication by a phase factor? How can you reconcile this with the statement in the previous section?
  2. jcsd
  3. Jul 31, 2009 #2
    The first case probably refers to the vacuum state, while the second case refers to a change in the vacuum expectation value when you go to the new vacuum state. In QM, a state multiplied by a phase factor is equivalent to itself.
  4. Jul 31, 2009 #3


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    Let's take the example of a complex scalar field with a U(1) symmetry. Call the corresponding conserved charge Q. Then the unitary symmetry-transformation operator takes the form [itex]U(\alpha) = \exp(i\alpha Q)[/itex].

    Now, if the symmetry is unbroken, the vacuum is unique; call it [itex]|0\rangle[/itex]. It is an eigenstate of Q with some eigenvalue, call it [itex]Q_0[/itex]. Usually, we choose to shift Q by a constant so that [itex]Q_0=0[/itex], but it's not strictly necessary. In general, then,
    [itex]U(\alpha)|0\rangle = \exp(i\alpha Q_0)|0\rangle[/itex].

    Now, if the symmetry is broken, there is a family of vacuum states labeled by an angle; call the angle [itex]\theta[/itex], and call the corresponding vacuum state [itex]|\theta\rangle[/itex]. Now we have
    [itex]U(\alpha)|\theta\rangle = |\theta{+}\alpha\rangle[/itex].
  5. Jul 31, 2009 #4
    Another way to answer the question is to note that under infinitismal group transformation [tex]\phi_{k}(x) [/tex] changes as [tex]U^{\dagger}(\theta) \phi_{k}(x) U(\theta)=-i \theta^{a}T^{a}_{kj}\phi_{j} [/tex] where [tex] T^{a}_{kj} [/tex] is a generator of the group. So taking the case of U(1), the field under transformation is: [tex]\phi(x)-i \theta\phi(x) =\phi(x)(e^{-i \theta}) [/tex]. Therefore the vacuum expectation value of that field changes by that phase factor. If the vacuum expectation value changes by that phase factor, then [tex] U(\theta)=e^{-iQ \theta}[/tex] operated on the vacuum |0> must be a different state since looked at it from another way, [tex]<0|U^{\dagger}(\theta) \phi(x) U(\theta)|0>=<0' | \phi(x) |0'> != <0| \phi(x) |0> [/tex] where != means not equal, implying [tex]|0'>=e^{-iQ\theta}|0>[/tex] is not equal to |0>.

    What's fascinating is that no mention of SSB has been mentioned in the analysis! However, if the vacuum expectation value is zero, then |0> can equal |0'>, as they would both have the same expectation value, namely 0. But they still don't have to be the same! Obviously if the expectation value of two states is different, then the two states are different. But if the expectation value of two states is the same, then they still could be different.

    However, even if the vacuum is not unique, if the expectation value is zero in all the vacuums, then this doesn't change calculations. But technically, they could be different vacuums. At least I think.
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