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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"

agemO
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  • actually i would like to add to the question. what is the physical significance of the field operator for particles such as electrons? for the photon, for eg., it is the 4-vector potential from which electric and magnetic fields can be calculated. how do we interpret the electron field? What is the physical meaning of measuring this field? In quantum physics every operator corresponds to an observable. Which observable does this correspond to? – guru Jun 28 '13 at 21:23
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    @guru: Please re-ask this question in its own post. (Its a good question but, very few people will see it here.) – user1504 Jul 11 '13 at 15:09
  • Related. Note, in your first sentence, the first "eigenvalues" should be "eigenstates". Relevant. – Cosmas Zachos Apr 15 '18 at 00:27
  • Of course the energy eigenvalues of field operator ladder states are infinite, as they should be. – Cosmas Zachos Apr 15 '18 at 00:29
  • Below I answered according to the previous version of the question but I think it is sufficient. If not, please comment there. – Oktay Doğangün Apr 15 '18 at 07:28
  • My bad it was indeed eigenstates and not eigenvalues. @Oktay I don't see how to go from your answer to computing for example the probability that a klein gordon vacuum is in a given field configuration (like a gaussian or whatever given f(x) function) – agemO Apr 15 '18 at 17:29
  • Thanks for the feedback @agemO I will edit my answer. Nevertheless, as I stated there in the answer, the field operator is a combination of all possible creation/annihilation operators. So, if you can calculate the eigenstate of a creation operator in good ol' single particle QM, then you can also calculate the field operator by weighing them all for each momentum and integrate them (cf. eq. 1 in my answer). I will try to edit my answer according to this version of the question when I am less lazy :) – Oktay Doğangün Apr 15 '18 at 18:09
  • Btw unless I'm missing something using eq.1 will lead to writing $|x>$ as a function of the energy eigenstate i.e. the inverse of the original problem which is writing energy eigenstates as functions of field eigenstates. – agemO Apr 16 '18 at 09:10

2 Answers2

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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.

Peter Kravchuk
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  • "so it is rarely used." Yes that exactly what my classmates told me "we don't care its useless", I agree but in this case the harmonic oscillator amplitude is also useless. "So, the basis you are talking about contains one eigenfunction for every field configuration." Exactly what I meant, I'll look in the reference you gave thanks ! – agemO May 10 '13 at 23:12
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    As I said I didn't find the answer in Weinberg, I also tried other books – agemO May 08 '15 at 16:40
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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.

Oktay Doğangün
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