$\newcommand{\Sym}{\text{Sym}}$ Let $H$ be a Hilbert space. Recall that the bosonic Fock space of $H$ is the completion of $\Sym(H) := \bigoplus\limits_{n \geq 0} \Sym^n(H)$ with respect to a particular inner product. Here $\Sym^n(H) := H^{\otimes n} / \Sigma_n$ is the $n$th symmetric power of $H$. My question is about the choice of inner product.

The usual way to choose the inner product -- reduce to $T(H)$:

The usual way to choose an inner product on $\Sym(H)$ is to observe that the quotient map $T(H) \to \Sym(H)$ (where $T(H) = \bigoplus_{n \geq 0} T^n(H) = \bigoplus_{n \geq 0} H^{\otimes n}$) has a natural section $\Sym(H) \to T(H)$ realizing $\Sym(H)$ as a summand of $T(H)$. Then one restricts the inner product on $T(H)$ to an inner product on $\Sym(H)$.

This strategy reduces the question to the choice of an inner product on $T(H)$. Here we choose the unique inner product on $T(H)$ such that

  1. $T^n(H) \perp T^m(H)$ for $m \neq n$; and

  2. The inner product on $T^n(H) = H^{\otimes n}$ is the usual tensor power inner product, where $\langle v_1\otimes \cdots \otimes v_n, w_1 \otimes \cdots \otimes w_n\rangle = \langle v_1,w_1\rangle \cdots \langle v_n,w_n \rangle$.

I think I can see good physical motivation for (1) and (2): (2) says that $T^n(H)$ is the state space for a system which consists of $n$ copies of the system whose state space is $H$, and (1) is equivalent to saying that there exists a self-adjoint ``number" operator which will tell us how many copies of $H$ we are dealing with.

Shortcoming of the usual approach -- why bring up $T(H)$ at all? This usual method is a nice way to get an inner product on $\Sym(H)$, except that I don't understand where $T(H)$ comes from. After all, $\Sym(H)$ is the state space for a boson whose one-particle states live in state space $H$ -- I don't see the physical relationship to $T(H)$, which is the state space for a system which has something to do with taking a bunch of copies of $H$. Thus, I don't see the physical interpretation of the embedding $\Sym(H) \to T(H)$: it just looks like a mathematical curiosity that there happens to be this embedding of state spaces. Absent a physical interpretation, I see no reason why the inner product on $T(H)$ should be relevant at all for determining the inner product on $\Sym(H)$.

Alternative approach: axiomatize the inner product on $\Sym(H)$ directly Alternatively, we could select our inner product on $\Sym(H)$ via criteria analogous to (1) and (2) above. That is, for each orthonormal basis $\mathcal E = (e_1,e_2,\dots)$ of $H$, we define the state $\mid n_1,\dots,n_i\rangle_{\mathcal E} \in \Sym^n(H)$ to be proportional to $e_{m_1}\cdots e_{m_n}$ where $n_i = \#\{j \mid m_j = i\}$ (more on the normalization in a moment). Then we stipulate that:

  1. $\Sym^n(H) \perp \Sym^m(H)$ for $m \neq n$; and

  2. For any orthonormal basis $\mathcal E$ of $H$, the states $\mid n_1,\dots,n_i,\dots \rangle_{\mathcal E}$ are pairwise orthogonal; and

  3. For any unit vector $v \in H$, the state $v^n = v\cdots v$ is a unit vector in $\Sym^n(H)$.

I believe the axioms (3,4,5) uniquely characterize a norm on $\Sym(H)$, and moreover that this norm turns out to be the same as the norm obtained via (1,2) above.

Shortcoming of the alternative approach -- relative "weight" of $\Sym^n(H)$ vs. $\Sym^m(H)$ That's all well and good, and I think I can see physical justifications for (3) and (4) above -- they are equivalent to saying there exists a self-adjoint "number" operator which tells us how many particles we have in each state of $H$. But I'm hung up on justifying (5). For instance, rather than (5) we could have chosen the alternative

  1. For each orthonormal $v_1,\dots,v_n \in H$, the state $v_1\cdots v_n \in \Sym^n(H)$ has norm 1.

Putting together (3,4,6) instead of (3,4,5) would result in a different inner product on $\Sym(H)$. I believe this inner product is $\sqrt{n!}$ times the original one. If we're interested in just one of the summands $\Sym^n(H)$, then this is just an overal scale factor and makes no difference. But when we want to start comparing states with different numbers of particles, it does seem to make a difference. I can't think of a physical argument which tells me that (5) is more reasonable than (6), except by appealing to the embedding into $T(H)$, which I also don't understand from a physical perspective.

I suppose this also makes me question the choice of relative "weights" of $T^n(H)$ versus $T^m(H)$.


  1. What is a good physical motivation for the usual choice of inner product on the bosonic Fock space $\Sym(H)$?

  2. For that matter, why choose the inner product we choose on $T(H)$, rather than some alternative inner product which scales each summand $T^n(H)$ differently?

  3. Is there even a physical way to distinguish between the different possible choices?

  • $\begingroup$ Regarding 2., 3.: amplitudes are ratios, so finding a physical way to distinguish is tricky. The class of norms you consider are equivalent so presumably you can find unitary maps that map them all to one another. But really you'd want to show that all "reasonable" inner products satisfy some universal property, and then argue that that property is physically relevant. $\endgroup$
    – user21299
    Apr 16, 2020 at 1:49

1 Answer 1


Stop worrying about axiomatizations of the inner product on the various spaces here. The natural way is to just let the procedure of building the Fock space determine the product and take all operations in the category of Hilbert spaces to begin with. That is, we don't build the Fock space by somehow first constructing a vector space and then endowing the result with an inner product - we're dealing with Hilbert spaces every step of the way and so you should just be taking the tensor product of Hilbert spaces instead of worrying about choosing some unnatural inner product on the tensor product of the underlying vector spaces, and likewise you should just be taking the direct sum of Hilbert spaces.

Note also that if you do not take all the tensor products as products of Hilbert spaces, you have to complete the result at the end w.r.t. to the norm induced by the inner product otherwise the space is not complete.

Physically, we're clearly starting from a one-particle space $H$ and building the many-particle space out of it. So we're building a two-particle space $H\otimes H$, a three-particle space $H\otimes H\otimes H$, and so on. This is just combining subsystems (one particle + one particle + ...) and so we take the ordinary tensor product that we accept generally to yield the combined state space of subsystems. Then we realize there's an additional constraint on our particle - it's a boson - so we take the symmetric quotient. Note that this really is a constraint - in non-relativistic QM without the spin-statistic theorem there is no reason to restrict to bosons or fermions a priori, so it's very natural to first construct the most general state space and then impose specific constraints. And again, the quotient of Hilbert spaces carries a natural inner product and there isn't really any reason we should think about choosing an unnatural one.

Lastly, we look at the collection of $\mathrm{Sym}^n H$ and realize that they are superselection sectors for non-interacting particles, so we just add them side by side with the direct sum of Hilbert spaces.

  • $\begingroup$ Thanks! So you're saying that in non-relativistic QM, we really do think of collections of bosons as multiple particles which just happen to satisfy the "boson constraint" in their wavefunctions -- it's only in QFT that we have access to a way of thinking about bosons as sui generis objects. I suppose that gives me another reason to learn some QFT! The point that $Sym^n H$ inherits an inner product directly from being a quotient of $T^n H$ is a good one too, I think -- even though most treatments use its realization as a subspace, it's surely equivalent. $\endgroup$
    – tcamps
    Apr 12, 2020 at 13:58
  • $\begingroup$ I suppose the perspective of seeing $Sym^n H$ as a quotient (rather than suspace) of $T^n H$ also gives an alternate perspective on bosons even in QM (as opposed to QFT) -- rather than viewing bosons as "constrained" particles, we can view them as particles whose wavefunctions are a bit redundant, and are to be identified along the $\Sigma_n$ action. $\endgroup$
    – tcamps
    Apr 12, 2020 at 14:02

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