- #1

ismaili

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I am confused by the concept of Fock space in interacting theories,

please tell me where is wrong in my following argument:

Consider an interacting theory of non-derivative interaction, say, a simple [tex]\phi^4[/tex] theory for a scalar field,

*At a particular instant [tex] t = 0 [/tex]*,

we can still expand the field by the free theory basis as follows:

[tex] \phi(x) = \int d^3x\frac{1}{\sqrt{2E}} \Big(a_p e^{i\mathbf{p}\cdot\mathbf{x}} + a^\dagger_p e^{-i\mathbf{p}\cdot\mathbf{x}} \Big) [/tex]

The canonical commutation relation would the same as the free theory,

because we are considering a non-derivative coupling.

So, we can get

[tex] [a,a^\dagger] = 1 [/tex]

That is, these expansion coefficients at the particular instant [tex]t=0[/tex] satisfy the algebra for creation and annihilation operators. Moreover, for later time [tex] t[/tex], we can do the time evolution of the field, so that we get

[tex] a(t) = e^{iHt} a e^{-iHt},\quad a^\dagger(t) = e^{iHt} a^\dagger e^{-iHt}[/tex]

where [tex] H [/tex] is the full Hamiltonian.

We can easily prove that [tex] [a(t) , a^\dagger(t)] [/tex], hence we have time-dependent creation and annihilation operators. But this doesn't prevent us to construct the Fock space:

[tex] a_p(t) |\Omega\rangle = 0, |1\rangle \equiv a_p^\dagger(t) |\Omega\rangle, \cdots [/tex]

But, in this way, I encountered a big problem.

We can never have the spontaneous symmetry breaking,

because the vacuum expectation value(VEV) of a field is always zero:

[tex] \langle \Omega | \phi(x)|\Omega\rangle \sim \langle \Omega| a(t) + a^\dagger(t) |\Omega\rangle = 0 [/tex]

I'm really confused by this.

I was trying to figure out the origin of the non-zero VEV,

I was wondering if the origin comes from the interaction or from the degeneracy of vacuum.

Then I made myself into this above puzzle...

Please help me, thanks so much!