Can a coherent photon state also belong to the Fock space? If yes, under what conditions? For example I read that $$\exp\bigg\{-\frac{1}{2}\sum_i|\alpha_i|^2\bigg\}\exp\bigg\{-\sum_i\alpha_ia_i^{\dagger}\bigg\}\exp\bigg\{\sum_i\alpha^*_ia_i\bigg\}|0\rangle$$ lies in the Fock space if the criterion $$\sum_i|\alpha_i|^2<\infty$$ is met. Why does this hold?

I know that the Fock space contains states whose norm is finite. If this has something to do with that, I can not see the connection. And even if something like that holds (or the above criterion is met), does the Fock space contain $n$ particle states even when $n\rightarrow\infty$? Because if we expand the exponentials, we can have such states even if the criterion is met.

Thanks a lot.

By the way, $a_i^{\dagger}$ and $a_i$ are the usual bosonic creation and annihilation operators and $\alpha_i$ and $\alpha_i^*$ are numerical coefficients.

Source: Faddeev-Kulish paper "Asymptotic conditions and infrared divergences in QED" written in 1970


1 Answer 1


Let $\mathfrak h$ be the $1$-particle Hilbert space, $|\psi_i\rangle$ an orthonormal basis and $\mathcal F$ the bosonic Fock space (edit : for indistinguishable bosons), which is given by : $$\mathcal F = \bigoplus_{n=0}^{\infty}S^n\mathfrak h$$ Here, $S^n \mathfrak h$ is the symmetrized $n$-fold tensor product of $\mathfrak h$, ie the space of $n$-particle states.

$\mathcal F$ is a Hilbert space, with the different $n$-particle subspaces being orthogonal. Therefore, there are $n$-particle states with arbitrary high values of $n$ as well as states which are superposition involving arbitrary high values of $n$.

Now, let $P_n$ be the orthogonal projection onto the $n$-particle subspace, $\underline \alpha = \{\alpha_i\}$ a collection of complex numbers and set (formally, for now) : $$|\underline \alpha\hspace{0.3mm}\rangle = \exp\bigg\{-\frac{1}{2}\sum_i|\alpha_i|^2\bigg\}\exp\bigg\{-\sum_i\alpha_ia_i^{\dagger}\bigg\}\exp\bigg\{\sum_i\alpha^*_ia_i\bigg\}|0\rangle$$

Then, we have : \begin{align} P_n |\underline \alpha\hspace{0.3mm}\rangle &= \frac{1}{n!}\exp\bigg\{-\frac{1}{2}\sum_i|\alpha_i|^2\bigg\}\bigg(-\sum_i\alpha_ia_i^{\dagger}\bigg)^n|0\rangle \end{align} In particular, we have : \begin{align} P_1|\underline \alpha\hspace{0.3mm}\rangle &= \exp\bigg\{-\frac{1}{2}\sum_i|\alpha_i|^2\bigg\}\bigg(-\sum_i\alpha_ia_i^{\dagger}\bigg)|0\rangle\\ &= -\exp\bigg\{-\frac{1}{2}\sum_i|\alpha_i|^2\bigg\}\sum_i \alpha_i|\psi_i\rangle \end{align}

This is a well-defined state of $\mathfrak h$ only if $\sum_i |\alpha_i|^2 <\infty$, ie this condition is necessary. If this condition holds, we can define : \begin{align} \sqrt{\sum_i |\alpha_i|^2} &=\sqrt{\sum_i |\alpha_i|^2} \\ b &= \frac{1}{\sqrt{\sum_i |\alpha_i|^2}}\sum_i \alpha_i^* a_i \end{align}

and check that $[b,b^\dagger] = 1$ : $b$ is an annihilation operator and : $$|\underline \alpha\hspace{0.3mm}\rangle = \exp\left(-\frac{1}{2}|\underline\alpha|^2\right)\exp(-|\underline\alpha|b^\dagger)\exp(|\underline\alpha|b)|0\rangle $$ We can compute the norm of this state by projecting it onto every $n$-particle subspace : \begin{align} \langle \hspace{.3mm} \underline\alpha|\underline \alpha\hspace{0.3mm}\rangle &= e^{-|\underline\alpha|^2} \frac{1}{n!^2}\sum_{n=0}^\infty\langle 0 |e^{-|\underline\alpha|b}P_ne^{-|\underline\alpha|b^\dagger}|0\rangle \\ &= e^{-|\underline\alpha|^2} \sum_{n=0}^\infty\frac{|\underline\alpha|^{2n}}{n!^2}\langle 0 |b^n (b^\dagger)^n|0\rangle \\ &=e^{-|\underline\alpha|^2} \sum_{n=0}^\infty\frac{|\underline\alpha|^{2n}}{n!} \\ &= 1 \end{align}

Edit : The Fock space is generated by states with a definite and finite number of particles. But, given an orthonormal basis of such states, you can form linear combination with $L^2$ coefficients to form other states in the Fock space. In general, a (normalized) state $|\Psi\rangle\in\mathcal F$ has : $$1 = \langle \Psi|\Psi\rangle = \sum_{n=0}^{+\infty} \langle\Psi|P_n|\Psi\rangle$$ This state has a probability $p_n = \langle\Psi|P_n|\Psi\rangle$ to contain $n$-particles. This probability distribution does not need to have finite support and it is even possible for it to have an infinite expectation value : $\sum_{n=0}^{+\infty} np_n = +\infty$. However, it must be a probability distribution, ie it must sum up to $1$.

Edit 2 : As mentionned in the comments, given an orthonormal basis of $\mathfrak h$, we can build an orthonormal basis of the bosonic Fock space using occupation numbers, with vectors of the forms $|\{n_i\}\rangle$ where the $n_i$ are non-negative integers, all but a finite number of whom are zero.

If we consider the subset of those vectors with $\sum_i n_i = n$, we get an orthonormal basis for the $n$-particle subspace. By concatening those, we get a basis for the full Fock space.

An $n$-particle subspace (with $n$ fixed) can be finite dimensional if $\dim\mathfrak h<\infty$, but the bosonic Fock space is always infinite dimensional.

  • $\begingroup$ Hi @SolubleFish and thanks for the answe. It seems that you are implying that the criterion of the sum of the square of the numerical coefficients needs to be finite such that we are able to normalize our states. That is something I understand now. However, by employing the argument involving the orthogonal projection operator and since the aforesaid operator can have arbitrarily large dimensionality, doesn't this imply that it is an infinite dimensional operator and hence the Fock space allows for particle states with infinite number of particles?? $\endgroup$
    – schris38
    May 4, 2022 at 8:22
  • $\begingroup$ I am asking whether or not a state in Fock space can contain an infinite number of particles because in the paper I am reading, the authors claim that the Fock space is formed by states that are nonvanishing for sequences for which $\sum_in_i<\infty$, where $n_i$ are the integers indicating the occupancy of the state with label $i$ $\endgroup$
    – schris38
    May 4, 2022 at 8:27
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    $\begingroup$ The statement is true, but when they say that the Fock state "is formed by those states", they mean that those state form an orthonormal basis, not that any state is of this form. This does not prohibit states with an infinite expected number of particles $\langle \hat N\rangle_\Psi = \infty$ (see the edit in my answer) $\endgroup$ May 4, 2022 at 9:12
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    $\begingroup$ Well yes, though I am not sure where the confusion lies. Even if the one-particle subspace $\mathfrak h$ has finite dimension $d$, then the $n$-particle subspace has dimension ${d+n- 1 \choose n}$ and the Fock space is infinite dimensional. $\endgroup$ May 4, 2022 at 12:10
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    $\begingroup$ The Fock space is always infinite dimensional, I have not claimed the contrary. I tried to add an edit to explain things further. The particle number eigenstates have finite particle number (by definition). This does not prohibit us from building other states, which are no longer eigenstates, with an infinite expected particle number. $\endgroup$ May 4, 2022 at 13:22

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