Energy eigeinstates written in the field operator eigenstates basis

Question

For an harmonic oscillator we can write the Hamiltonian eigenstates in the basis of the amplitude eigenstates : for example the ground state is a gaussian : $⟨x|0⟩=a.e^{-b.x^{2}}$.

I was wondering if we can do that for the energy states of a field : can we calculate $⟨a|n⟩$ with |a⟩ an eigenstate of the field operator $\phi$ (let's say a Klein-Gordon field) ?

I failed to carry out the ladder operator method, but I think we should be able to calculate (or at least have implicit equation of) any operator eigenvalue in any other operator eigenvalue basis, or is there something obvious that I missed ?

Because we always play to write state in all the basis we can : spin on position, energy on spin etc ..., but I have never seen in classes or in books energy states of a field written in the field operator basis.

EDIT : I didn't find the answer in the reference given by Peter : "Weinberg I"

Answer 1

In QM we have the coordinate representation. In this case the basis consists of eigenfunctions of the position operator $r_i$, where $i$ runs over the discrete set of degrees of freedom. Now, when we go to QFT, $i$ generalizes to the continious space-time position $x$, and $r$ generalizes to the field $\phi$: $$ r\rightarrow\phi\\ i\rightarrow x\\ r_i\rightarrow \phi(x) $$ So, the basis you are talking about contains one eigenfunction for every field configuration. However, this basis is not really physical (you measure not $\phi$, but the corresponding particles), so it is rarely used. However, if you wish to have a look at it, you can look in, e.g. Weinberg I, where he derives the functional integral from operator formalism.

Answer 2

Energy eigenstates of a field operator are those of the creation-annihilation operators for infinitely many energies. Because the field operators are simply the superposition of all the creation/annihilations of all possible energies at a position x. You'd rather not bother to work them out in the case of undetermined and changable number of particles.

So, for a field operator, you have the same exact eigenvalues of all possible energies at a position superposed by being weighted with their energies.

In order to see that, let's think we have a position eigenstate of a harmonic oscillator, namely $| x \rangle$, and we would like to write it in the energy/momentum eigenstate $| \phi_p \rangle$. Since, like any eigenbasis, $$ \int \frac{d^3 p}{\sqrt{8 \pi^3}} | \phi_p \rangle \langle \phi_p | = 1 $$ then, by multiplying it with this, we can rewrite the position eigenstate as follows: \begin{eqnarray} | x \rangle & = & \int \frac{d^3 p}{\sqrt{8 \pi^3}} | \phi_p \rangle \langle \phi_p | x \rangle \\ &=& \int \frac{d^3 p}{\sqrt{8 \pi^3}} \phi^\star_p (x) |\phi_p \rangle \end{eqnarray} where, in the second line, I have used the abbreviation, $\phi_p (x) \equiv \langle x | \phi_p \rangle$.

Now, the energy eigenstate $| \phi_p\rangle$ can be expressed as a creation operator of momentum $p$ acted on a vacuum state, i.e., $$ |\phi_p \rangle = \hat{a}^\dagger (p) \, | 0 \rangle $$ Therefore, $$ \tag{1} | x \rangle = \int \frac{d^3 p}{\sqrt{8 \pi^3}} \phi^\star_p (x) \, \hat{a}^\dagger (p) \, | 0 \rangle $$ Here, we can see that the expression before the vacuum state is the hermitian conjugate of the field operator for $\phi$, namely, $$ \tag{2} | x \rangle = \hat\Phi^\dagger (x) \, | 0 \rangle $$ what it is saying is that a position eigenstate for a particle at x is the eigenstate where a particle is created in the vacuum. This is the physical significance of a field operator (as you asked in the comments to your question).

As you can see from (1) and (2), or explicitly from the field operators, $$ \hat\Phi(x) = \int \frac{d^3 p}{\sqrt{8 \pi^3}} \phi_p (x) \, \hat{a}(p) \\ \hat\Phi^\dagger (x) = \int \frac{d^3 p}{\sqrt{8 \pi^3}} \phi^\star_p (x) \, \hat{a}^\dagger (p) $$ there are infinite range of eigenvalues for each position since the field operator is composed of infinite harmonic oscillators at infinitely many energies. So, the field operator does not reveal a specific eigenvalue but rather an infinite superposed eigenvalues at once. So, only in simple cases you would bother to compute them.