5.3 Quantum theory, formally (continued)

Recalling Chapter 4, where we said that we were missing a key part of the formalism of quantum theory, we can now fill in this hole. Our mathematical formalism of choice behind the quantum theory of composite systems is based on the tensor product of Hilbert spaces.

5.3.1 Tensor products

Let the states of some system \mathcal{A} be described by vectors in an n-dimensional Hilbert space \mathcal{H}_{\mathcal{A}}, and the states of some system \mathcal{B} by vectors in an m-dimensional Hilbert space \mathcal{H}_{\mathcal{B}}. The combined system of \mathcal{A} and \mathcal{B} is then described by vectors in the (nm)-dimensional tensor product space \mathcal{H}_{\mathcal{A}}\otimes\mathcal{H}_{\mathcal{B}}. Given bases \{|a_1\rangle,\ldots,|a_n\rangle\} of \mathcal{H}_{\mathcal{A}} and \{|b_1\rangle,\ldots,|b_m\rangle\} of \mathcal{H}_{\mathcal{B}}, we form a basis of the tensor product by taking the ordered pairs |a_i\rangle\otimes|b_j\rangle, for i=1,\ldots,n and j=1,\ldots,m. The tensor product space \mathcal{H}_{\mathcal{A}}\otimes\mathcal{H}_{\mathcal{B}} then consists of all linear combination of such tensor product basis vectors:70 |\psi\rangle = \sum_{ij} c_{ij}|a_i\rangle\otimes|b_j\rangle. \tag{5.3.1}

The tensor product operation \otimes is distributive: \begin{gathered} |a\rangle \otimes \left( \beta_1|b_1\rangle + \beta_2|b_2\rangle \right) = \beta_1|a\rangle\otimes|b_1\rangle + \beta_2|a\rangle\otimes|b_2\rangle \\\left( \alpha_1|a_1\rangle + \alpha_2|a_2\rangle \right) \otimes |b\rangle = \alpha_1|a_1\rangle\otimes|b\rangle + \alpha_2|a_2\rangle\otimes|b\rangle. \end{gathered} The tensor product of Hilbert spaces is again a Hilbert space: the inner products on \mathcal{H}_{\mathcal{A}} and \mathcal{H}_{\mathcal{B}} give a natural inner product on \mathcal{H}_{\mathcal{A}}\otimes\mathcal{H}_{\mathcal{B}}, defined for any two product vectors by \left( \langle a'|\otimes\langle b'| \right) \left( |a\rangle\otimes|b\rangle \right) = \langle a'|a\rangle\langle b'|b\rangle and extended by linearity to sums of tensor products of vectors, and, by associativity71, to any number of subsystems. Note that the bra corresponding to the tensor product state |a\rangle\otimes|b\rangle is written as (|a\rangle\otimes|b\rangle)^\dagger = \langle a|\otimes\langle b|, where the order of the factors on either side of \otimes does not change when the dagger operation is applied.

Some joint states of \mathcal{A} and \mathcal{B} can be expressed as a single tensor product, say |\psi\rangle=|a\rangle\otimes|b\rangle (often written as |a\rangle|b\rangle, or |a, b\rangle, or even |ab\rangle), meaning that the subsystem \mathcal{A} is in state |a\rangle, and the subsystem \mathcal{B} in state |b\rangle. If we expand |a\rangle=\sum_i\alpha_i|a_i\rangle and |b\rangle=\sum_i\beta_j|b_j\rangle, then |\psi\rangle=\sum_{ij}\alpha_i\beta_j|a_i\rangle\otimes|b_j\rangle and we see that, for all such states, the coefficients c_{ij} in Equation (5.3.1) are of a rather special form: c_{ij} = \alpha_i\beta_j. We call such states separable (or just product states). States that are not separable are said to be entangled.

A useful fact about tensor products is that \lambda a\otimes b = a\otimes\lambda b (where a and b are vectors, and \lambda is a scalar). This means that we don’t need to worry about brackets, and can write something like \lambda(a\otimes b).

We will also need the concept of the tensor product of two operators. If A is an operator on \mathcal{H}_{\mathcal{A}} and B an operator on \mathcal{H}_{\mathcal{B}}, then the tensor product operator A\otimes B is an operator on \mathcal{H}_{\mathcal{A}}\otimes\mathcal{H}_{\mathcal{B}} defined by its action on product vectors via (A\otimes B)(|a\rangle\otimes|b\rangle) = (A|a\rangle)\otimes (B|b\rangle) and with its action on all other vectors determined by linearity: A\otimes B \left( \sum_{ij} c_{ij}|a_i\rangle\otimes|b_j\rangle \right) = \sum_{ij}c_{ij} A|a_i\rangle\otimes B|b_j\rangle.


  1. If the bases \{|a_i\rangle\} and \{|b_j\rangle\} are orthonormal then so too is the tensor product basis \{|a_i\rangle\otimes|b_j\rangle\}.↩︎

  2. ({\mathcal{H}}_a \otimes {\mathcal{H}}_b)\otimes {\mathcal{H}}_c = {\mathcal{H}}_a \otimes ({\mathcal{H}}_b\otimes {\mathcal{H}}_c).↩︎