# The Tensor Product

In quantum mechanics if we want to describe a composite system consisting of two subsystem we take the tensor product of the spaces describing each subsystem. We say that the space $$U_n\otimes U_m$$ is the tensor product of the spaces $$U_n$$ and $$U_m$$ if the mapping $$\otimes$$ satisfies the following conditions
1. $$\ket{u}\otimes(\ket{v}+\ket{w})=\ket{u}\otimes\ket{v}+\ket{u}\otimes\ket{w}$$.
$$c(\ket{u}\otimes\ket{v})=(c\ket{u})\otimes\ket{v}=\ket{u}\otimes(c\ket{v})$$ (bilinearity).
2. if $$\{\ket{u_1},...,\ket{u_n}\}$$ and $$\{\ket{v_1},...,\ket{v_m}\}$$ are bases in their respective spaces then $$\{\ket{u_i}\otimes\ket{v_j}|i=1,...,m, j=1,...,n\}$$ is a basis in their tensor
3. product.
4. $$\sprod{u\otimes v}{u'\otimes v'}=\sprod{u}{u'}\sprod{v}{v'}$$ (scalar product definition).
The second condition defines the orthogonal basis of uncorrelated vectors in the tensor product. Since there is $$n\cdot m$$ of basis vectors in this orthonormal basis the dimension of the product space is also $$n\cdot m$$. One way to think about the tensor product is that "on top" of each vector in one of the factors there is a copy of the other factor space. It is easy to show that the dual vector of a tensor product of two vectors is the tensor product of their respective dual vectors: $$\bra{u\otimes v}=\bra{u}\otimes \bra{v}.$$ Here we again se the convenience of the Dirac notation. Usually the $$\otimes$$ symbol is ommited in tensor product expressions. Also, all lines corresponding to bras and kets from the same space are joined and the third condition is written in Dirac notation as: $$\sprod{u\otimes v}{u'\otimes v'}=\bra{u}\bra{v}\ket{u'}\ket{v'}=\bra{u}\ket{u'}\bra{v}\ket{v'}=\sprod{u}{u'}\sprod{v}{v'}.$$ It usefull to show some details on how the tensor product is performed in a particular representations (i.e. in $$\cn^n$$). Let us consider the square of a unitary space $$U_n\otimes U_n$$. The coordinates of the tensor product of two vectors, $$\ket{u}$$ and $$\ket{v}$$, expressed in the basis of uncorrelated vectors $$\ket{e_i}\ket{e_j}$$ are calculated using formula 2 from the previous section $$\bra{e_i}\bra{e_j}\ket{u}\ket{v}=\bra{e_i}\bra{e_j}\sum_{k=1}^n u^k\ket{e_k}\sum_{l=1}^n v^l\ket{e_l}=\sprod{e_i}{e_k}\sprod{e_j}{e_l}u^kv^l=u^iv^j$$ which when written in the representation of the uncorrellated basis gives $$\left( \begin{array}{c} u^1\\ u^2\\ \vdots \\ u^n \end{array} \right)\otimes\left( \begin{array}{c} v^1\\ v^2\\ \vdots \\ v^n \end{array} \right)= \left( \begin{array}{c} u^1 \cdot \left( \begin{array}{c} v^1\\ v^2\\ \vdots \\ v^N \end{array} \right)\\ u^2 \cdot \left( \begin{array}{c} v^1\\ v^2\\ \vdots \\ v^N \end{array} \right)\\ \vdots \\ u^N \cdot \left( \begin{array}{c} v^1\\ v^2\\ \vdots \\ v^N \end{array} \right) \end{array} \right)= \left( \begin{array}{c} u^1v^1\\ u^1v^2\\ \vdots \\ u^1v^n \\ u^2v^1\\ u^2v^2\\ \vdots \\ u^2v^n \\ \vdots \\ u^nv^1\\ u^nv^2\\ \vdots \\ u^nv^n \\ \end{array} \right).$$

## Entangled Vectors

Let us examine the following vector from the product $$\cn^2\otimes\cn^2$$, i.e. the tensor product of two spin 1/2 systems $$\ket{\phi}=\frac{1}{\sqrt{2}}[\ket{+}\ket{-}+\ket{-}\ket{+}]. \label{entangled}$$ The most general form of an uncorrelated (a simple tensor product) vector that belongs to $$\cn^2\otimes\cn^2$$ is \begin{eqnarray} \ket{\psi}&=&[c_{1+}\ket{+}+c_{1-}\ket{-}]\otimes[c_{2+}\ket{+}+c_{2-}\ket{-}]\\ &=&c_{1+}c_{2+}\ket{+}\ket{+}+c_{1+}c_{2-}\ket{+}\ket{-}+c_{1-}c_{2+}\ket{-}\ket{+}+c_{1-}c_{2-}\ket{-}\ket{-}. \label{general} \end{eqnarray} Setting equal \ref{entangled} to \ref{general} we get the following equations for the coefficient in the expandion of the factors in \ref{general} $$c_{1+}c_{2+}=0 \;\;\;\;\;\; c_{1+}c_{2-}=\frac{1}{\sqrt{2}}\;\;\;\;\;\; c_{1-}c_{2+}=\frac{1}{\sqrt{2}}\;\;\;\;\;\; c_{1-}c_{2-}=0. \label{contradiction}$$ The middle two equations of \ref{contradiction} imply that all coefficients have to be nonzero while the first an last equations imply that at least two of them have to be zero. This contradiction proves that \ref{entangled} is not a tensor product vector (uncorellated vector) but a so called entangled vector. Entangled vectors inherently belong to the tensor product in the sense that they correlate the two factors regardless of a basis choice, as shown by our attempt in \ref{general}.