spidey said:
while studying lagrangian i got this doubt..if the lagrangian is invariant in time,space,rotation, then we have corresponding conservation laws..
In spontaneous symmetry breaking, the lagrangian is not invariant in ground state and the symmetry breaks spontaneously.so, the conserved quantity is now not conserved in ground state. so, are the conservation laws break in ground state?
You should have posted this in QM or particles forums.
Any way, you are confusing between the two levels in which a symmetry operate. These are:
1) the dynamical level; the invariance of the action integral. This leads to a conserved (Noether) charge:
\delta \int d^{4}x \ \mathcal{L} = 0 \Rightarrow \partial_{a}J^{a}= 0 \Rightarrow \frac{d}{dt}Q = 0
2) the level of the representation of field operators and their action in the Hilbert space. At this level we ask whether the symmetry can be represented by unitary transformations generated by the above conserved charge:
U = \exp \left( i \alpha Q \right)
Here, two things can happen:
i) If U exists, i.e., Q is well-defined Hermitian operator ( say, a^{\dagger}a ), then
Q|0 \rangle \ = \ a^{\dagger}a |0 \rangle = 0
and the symmetry is said to be manifest.
ii) If U does not exist; Q is ill-defind operator, then
Q|0 \rangle \neq 0
and the symmetry is said to be spontaneously broken (hidden). This does not mean that Q depends on the time. The action is invariant even when the ground state is not.
Spontaneous breakdown of symmetry is the lack of degenercy in particle spectra in a SYMMETRIC theory. Or in the language of QFT, it is the breakdown of symmetry at the representation level, not at the dynamical level ( the action is invariant and Q is conserved).
To see that Q is conserved even when the symmetry is hidden, let us consider the theory
\mathcal{L} =(1/2) \partial_{a} \phi \partial^{a} \phi
which is invariant under field translation
\bar{\phi} = \phi + \chi
The corresponding conserved charge is given by
Q = \int d^{3}x \ \partial_{0} \phi (x)
This invariance, however, can not be unitarily represented in Hilbert space; for if we write
\bar{\phi} = \phi + \chi = U(\chi) \phi U^{-1}(\chi )
with
U = \exp \left( i \chi Q \right), \ \ Q = Q^{\dagger} = a^{\dagger}a
i.e., if we assume that the ground state is invariant U |0 \rangle = |0 \rangle, then
\langle 0 |\bar{\phi} |0 \rangle = \langle 0|\phi |0 \rangle + \chi = \langle 0|\phi |0 \rangle
Or, since \langle \phi \rangle = 0,
\langle \bar{\phi} \rangle = \chi = 0
but this can not be right because \chi is arbitrary. Thus the symmetry can not be represented by unitary operator (constructed from the conserved Q) on Hilbert space, i.e., the symmetry of the action is not a symmetry of the ground state.
regards
sam
