# Fock space in interaction theory

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...

strangerep
Does anyone know how to construct Fock space in an interacting theory?
Not in general. It's a major outstanding problem in QFT.

[...] we can still expand the field by the free theory basis [...]
In general, that's not true for inf-dim Hilbert spaces.
textbook for a fairly readable account of what can go wrong
in the infinite-dimensional case.

The thing about nonzero VEVs is kinda related to the notion
of unitarily inequivalent Hilbert spaces (meaning you've got
different vacua states that don't live in the same Hilbert
space).

(If you search back through this forum using some of the
buzzwords I've mentioned above, you'll probably find

Thank you very much for your discussion!
Not in general. It's a major outstanding problem in QFT.
So, in an interacting QFT, usually, we define the one-particle state by
$$a^\dagger_{\mathbf{p}} |0\rangle$$
where the creation operator and vacuum state are free theory
creation operator and free theory vacuum state,
since we are doing perturbation theory?
Or..., actually in an interacting theory, we don't have a clear "particle picture"?

In general, that's not true for inf-dim Hilbert spaces.
textbook for a fairly readable account of what can go wrong
in the infinite-dimensional case.

The thing about nonzero VEVs is kinda related to the notion
of unitarily inequivalent Hilbert spaces (meaning you've got
different vacua states that don't live in the same Hilbert
space).

(If you search back through this forum using some of the
buzzwords I've mentioned above, you'll probably find
Thanks. I will look for the theorem and book you mentioned.

Wow, I didn't know that interaction picture actually does not exist in interacting theory!
So, in the textbook, such as Peskin, the authors just try to quickly set up a framework which enables us to calculate some interesting quantities, which is coincidentally correct?!

BTW, I found my argument is wrong at the step you pointed out.
I thought we can still expand the field like the free theory in a particular instant of time,
which is wrong.
By the way, Peskin's text also says that we can expand a field as in the free theory case in a particular instant, this also deepened my belief that what I did is correct.

Actually, if we fourier transform the field
$$\int d^4p \tilde{\phi}(p) e^{ip\cdot x} = \phi(x)$$
and substitute it into the interacting equation of motion,
we would find that we couldn't expand it like the free theory case.
The nonlinear EoM prevents the simply expansion of plane waves.

------

As for the non-zero VEV, I haven't considered its origin before.
They just assumed some nonzero VEV and started to expand around that vacuum to get SSB.

BTW, one more question, I'm a little bit confused by the jargon "vacuum" in SSB.
Usually, in a QFT, vacuum means the ground state $$|\Omega\rangle$$ which is a "state." But in SSB, we often called the minimum of the effective potential of theory vacuum, what's the relation between the vacuum state of Hilbert space, and the vacuum which is the points of the potential??

strangerep
Or..., actually in an interacting theory, we don't have a clear "particle
picture"?
Typically, we only have a (physical) "particle picture" at asymptotic times,
assumed to coincide with the space of free particles and one assumes
"asymptotic completeness". But, within orthodox QFT, Haag's theorem
reveals weaknesses in this assumption. (Also, where massless fields
are involved - such as in QED - one may find that the interaction
doesn't vanish at infinity, and special techniques are needed to
figure out a suitable space of asymptotic particle-like states.)

And fyi, there's at least one simpler model (the Lee model) where
an exact 1-particle solution has been found.

So, in the textbook, such as Peskin, the authors just try to quickly set up a
framework which enables us to calculate some interesting quantities, which is
coincidentally correct?!
[...]
By the way, Peskin's text also says that we can expand a field as in the free
theory case in a particular instant, this also deepened my belief that
what I did is correct.
Much of renormalization theory is about working around these difficulties.

BTW, look at P&S section 4.6 (bottom of p108 and top of p109). They
mention about how "interactions affect not only the scattering of distinct
particles but also the form of the single-particle states themselves".
But then, (being an introductory text), they continue on assuming that
all is well.

I'm a little bit confused by the jargon "vacuum" in SSB.

Sometimes it's helpful to think in terms of how the lowest-energy
state is not annihilated by all the electroweak generators,
meaning it's not invariant under all those transformations.
(I.e., contrast this with how the vacuum _is_ annihilated by the
Poincare generators.)

You might also try asking more about SSB over in the HEP forum.