- #1
ismaili
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Does anyone know how to construct Fock space in an interacting theory?
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 \phi^4 theory for a scalar field,
At a particular instant t = 0,
we can still expand the field by the free theory basis as follows:
\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)
The canonical commutation relation would the same as the free theory,
because we are considering a non-derivative coupling.
So, we can get
[a,a^\dagger] = 1
That is, these expansion coefficients at the particular instant t=0 satisfy the algebra for creation and annihilation operators. Moreover, for later time t, we can do the time evolution of the field, so that we get
a(t) = e^{iHt} a e^{-iHt},\quad a^\dagger(t) = e^{iHt} a^\dagger e^{-iHt}
where H is the full Hamiltonian.
We can easily prove that [a(t) , a^\dagger(t)], hence we have time-dependent creation and annihilation operators. But this doesn't prevent us to construct the Fock space:
a_p(t) |\Omega\rangle = 0, |1\rangle \equiv a_p^\dagger(t) |\Omega\rangle, \cdots
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:
\langle \Omega | \phi(x)|\Omega\rangle \sim \langle \Omega| a(t) + a^\dagger(t) |\Omega\rangle = 0
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!
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 \phi^4 theory for a scalar field,
At a particular instant t = 0,
we can still expand the field by the free theory basis as follows:
\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)
The canonical commutation relation would the same as the free theory,
because we are considering a non-derivative coupling.
So, we can get
[a,a^\dagger] = 1
That is, these expansion coefficients at the particular instant t=0 satisfy the algebra for creation and annihilation operators. Moreover, for later time t, we can do the time evolution of the field, so that we get
a(t) = e^{iHt} a e^{-iHt},\quad a^\dagger(t) = e^{iHt} a^\dagger e^{-iHt}
where H is the full Hamiltonian.
We can easily prove that [a(t) , a^\dagger(t)], hence we have time-dependent creation and annihilation operators. But this doesn't prevent us to construct the Fock space:
a_p(t) |\Omega\rangle = 0, |1\rangle \equiv a_p^\dagger(t) |\Omega\rangle, \cdots
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:
\langle \Omega | \phi(x)|\Omega\rangle \sim \langle \Omega| a(t) + a^\dagger(t) |\Omega\rangle = 0
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!
