# Direct Product Basis for Interacting Systems

1. Sep 21, 2009

### tommy01

Hi.

I found in a book on quantum optics (Vogel, Welsch - Quantum Optics) and also in a lecture script the following statements.

A System is composed of two subsystems (say A and B) which interact. So the total Hamiltionian is $$H_{AB}= H_A + H_B + H_{int}$$.
Nontheless the states of $$H_{AB}$$ are build up from direct products $$\Psi_{Ai}\otimes \Psi_{Bj}$$.

I always thought direct products are only allowed if there is no interaction.

2. Sep 21, 2009

### Bob_for_short

It is true but if there is no interaction at all, the initial state does not change with time.

An interaction, if it lasts for a limited time, makes the state evolve so the final state is different from the initial one. Think of potential scattering as an example.

Edit: Even a time dependent (evolving) state can be decomposed in eigenstates of non-interacting subsystems with variable coefficients Cj(t). Cj(∞) determines the transition amplitude to the final state |j>.

Last edited: Sep 21, 2009
3. Sep 21, 2009

### meopemuk

Your book is right. The Hilbert space of states of a compound system is build as a direct product of subsystem's spaces. This construction does not depend on interactions.

Why this is so? In my opinion, the best explanation is contained in the "quantum logic" approach to QM. In this approach, subspaces (or projections) in the Hilbert space of states are identified with experimental (or logical) "yes-no" propositions about the system. The propositions that can be formulated about the interacting systems are the same as propositions about the non-interacting system. So, their Hilbert spaces cannot be different.

4. Sep 21, 2009

### javierR

You are partially right. If there are two subsystems, the total Hilbert space is denoted by
$$H_{A}\times H_{B}$$, but elements of this space are *not* generally tensor products of $$H_{A}$$ elements with $$H_{B}$$ elements. In general, an element of the total space *can* be expanded in terms of a product of a basis of $$H_{A}$$ with a basis of $$H_{B}$$:
$$|\Psi>=\Sigma_{jk} c_{jk}\left( |a_{j}>\times |b_{k}> \right)$$, where the a's and b's are basis elements of $$H_{A}$$ and $$H_{B}$$, respectively. Only for special values of the coefficients does the sum result in $$|\Psi>=|\Psi_{A}>\times |\Psi_{B}>$$.

There are two more things to note: (1) Even without interactions, two subsystems can form a non-product, or entangled, state, such as two identical particles with spin. (2) In perturbation theory, the interaction part of a Hamiltonian is assumed to be small and we can use the states in the total Hilbert space that we would have used without the interaction term. So if the states happened to be product states $$|\Psi>=|\Psi_{A}>\times |\Psi_{B}>$$ in the non-interacting theory, we can use these as starting-point states in the perturbation theory.

5. Sep 21, 2009

### Fredrik

Staff Emeritus
First of all, it's "tensor product", not "direct product". If U, V and W are vector spaces, a bilinear function $\tau:U\times V\rightarrow W$ is said to be a tensor product if for every bilinear function $\sigma:U\times V\rightarrow X$ where X is a vector space, there's a unique linear bijection $f:X\rightarrow W$, such that $f(\sigma(u,v))=\tau(u,v)$ for all u in U and all v in V.

W is then said to be the tensor product of U and V, and we write $U\otimes V$ instead of W. We also write $u\otimes w$ instead of $\tau(u,w)$, and we define the scalar product on $U\otimes V$ by $\langle u\otimes v,u'\otimes v'\rangle_{U\otimes W}=\langle u,u'\rangle_U\langle v,v'\rangle_W$.

The definition guarantees that $U\otimes V$ is unique up to isomorphisms. To prove that a tensor product $\tau:U\times V\rightarrow W$ exists for arbitrary U and V is tricky. You do it by explicitly constructing a suitable space W and explicitly defining $\tau$. This construction involves the following steps: Find a vector space that has a basis that can be mapped bijectively onto $U\times V$. Then use it to define a vector space that consists of certain equivalence classes of vectors from the the first vector space. When U and V are Hilbert spaces, the construction involves an additional step. You define a third vector space, which consists of equivalence classes of Cauchy sequences of vectors from the second vector space.

So the construction is quite complicated, and serves no other purpose than to prove existence.

I don't know enough about quantum logic to know if it will give you the best motivation for why the tensor product is used. What I do know is that the Born rule gives you some motivation. The Born rule is the rule that says that if a system is in state $|\psi\rangle$ when we measure the operator A, the probability that we will get the result $a$ is $P(a)=|\langle a|\psi\rangle|^2$. Now consider the case where the two systems aren't interacting. The probability that the result of a simultaneous measurement of A on the first system and B on the second system will give use the results $a$ and $b$, must satisfy P(a,b)=P(a)P(b), and this is automatically satisfied when we take the Hilbert space of the combined system to be the tensor product of the Hilbert spaces of the component systems.

Suppose that the Hamiltonian can be expressed as $H=H_1\otimes I+I\otimes H_2=H_1'+H_2'$. Here the $\otimes$ symbol is defined by $X\otimes Y (|\psi\rangle\otimes|\phi\rangle)=X|\psi\rangle\otimes Y|\psi\rangle$. Now the time evolution operator can be expressed as

$$U(t)=e^{-iHt}=e^{-iH_1't}e^{-iH_2't}=e^{-iH_1t}\otimes e^{-iH_2t}=U_1(t)\otimes U_2(t)$$

which means that the systems evolve completely independently of each other. This is what we would expect if and only if the systems aren't interacting, so it looks like we have found a definition of "non-interacting". The systems are "non-interacting" if the Hamiltonian can be expressed as above.

Last edited: Sep 21, 2009
6. Sep 21, 2009

### meopemuk

The rigorous quantum logical proof that the Hilbert space of a compound system is a tensor product of the component spaces can be found in

T. Matolcsi, "Tensor product of Hilbert lattices and free orthodistributive product of
orthomodular lattices", Acta Sci. Math. (Szeged) 37 (1975), 263.

D. Aerts and I. Daubechies, "Physical justification for using the tensor product to describe two quantum systems as one joint system", Helv. Phys. Acta, 51 (1978), 661.

7. Sep 22, 2009

### tommy01

Great.

Thanks for your answers. i hope i got it (at least partially).

@Fredrik:
Thanks for your rigorous explanation. But the Hamiltonian i wrote above can't be brought in this "non-interacting" from.