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Very often in the standard textbooks on quantum mechanics, one finds that the joint Hilbert space of two systems is given by the tensor product of the individual Hilbert spaces. That is, if $H_1$ and $H_2$ are the Hilbert spaces associated with systems $S_1$ and $S_2$, then the composite Hilbert space of the entire system is given by $H_{1} \otimes H_{2}$, where $\otimes$ is the tensor product as defined here.

$\textbf{My question:}$ Under what conditions can a direct sum of two Hamiltonians, $H_1 \oplus H_2$, be used to represent the Hamiltonian in composite Hilbert space for the entire system?

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  • $\begingroup$ It looks like you are thinking of the Dirac-Nambu picture. Of course you could apply Fock Functor. $\endgroup$ – AHusain Jul 19 at 19:10
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    $\begingroup$ did you mean to write "composite *Hilbert space"? Otherwise, what's the Hamiltonian got to do with the rest of the post? $\endgroup$ – glS Jul 20 at 20:32
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@glS gave a fine answer, but given who is asking, would be remiss to not answer with some functoriality.

In section 4 of Twisted Equivariant Matter, the Dirac-Nambu space is defined. For a free fermion Hilbert space based on $(\mathcal{M},b)$ with a symmetry $(G,\phi)$ where $\phi$ encodes anti-linearity. Take $G=e$ if you want.

$$ H_{DN} \equiv \mathcal{M} \otimes \mathbb{C} $$

A choice of complex structure on $\mathcal{M}$ gives a Hermitian structure on $H_{DN}$.

In the particle-hole picture $H_{DN} = V \bigoplus \bar{V}$ is the 1-particle and 1-hole.

The amalgam of two systems is then the direct sum $H_{DN,1} \bigoplus H_{DN,2}$. This is easily visualized as one-particle or one-hole of different types.

Fock: On objects, send a vector space $V$ to $\bigoplus_{i=0}^n \Lambda^i V$. $\Lambda (V \bigoplus W) \simeq \Lambda (V) \otimes \Lambda (W)$. This is because above was talking about free fermions, so that's why alternating not $Sym$.

You can think as combine the systems as usual by tensor product, but when you apply the functor that gets you back to 1-particle/hole, you are picking out degree one summand which has a direct sum instead.

You may also like http://math.ucr.edu/home/baez/photon/tensor.htm

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What you are modelling when you use a tensor product is a space that can accommodate tuples of basis states of different kinds.

For example, say you want to describe a three-level system. You can use a three-dimensional Hilbert space $\mathcal H$ to accommodate the possible states of this system. But what if you have two three-level systems, each one of which can be in one of its three available states? In this case you have nine possible basis states ($00$, $01$, $02$, $10$, and $11$, $12$, $20$, $21$, and $22$), and thus need a nine-dimensional Hilbert space. As it happens, $\mathcal H\otimes\mathcal H$ has just the right dimensions (you could also use any other nine-dimensional Hilbert space, only it would make the notation more awkward when dealing with local operations).

On the other hand, $\mathcal H\oplus\mathcal H$ is a six-dimensional space, with basis the union of the bases of the two copies of $\mathcal H$ (see also this post over at math.SE). You could write this basis as $$\{|0\rangle,|1\rangle,|2\rangle,|0'\rangle,|1'\rangle,|2'\rangle\},$$ denoting with $|i'\rangle$ the $i$-th basis element of the second copy of $\mathcal H$.

Clearly, this does not describe a system that is obtained by combining multiple elementary systems. Rather, it can be used to describe how a high-dimensional system is "composed" of smaller-dimensional ones. Indeed, one can always think of an $n$-dimensional Hilbert space as the direct sum of $n$ copies of one-dimensional spaces.

The direct sum is, therefore, what you always do "under the hood" when you build up higher dimensional spaces from lower dimensional ones.

Similar reasoning applies to Hamiltonians or other operators. As an example, if you have a five-dimensional space $\mathcal H$, and two operators $A_1$ and $A_2$ operating on three- and two-dimensional systems, respectively, then $A_1\oplus A_2$ is a valid operator acting on states in $\mathcal H$. This represents an operation which does not correlate the first three and the last two modes (because of the block structure of $A_1\oplus A_2$).

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