Tensor products

• We can form tensor products of a wide variety of objects. For example:

  1.

  The tensor product of an n dimensional vector u and an m dimensional vector v is an nm dimensional vector $u \otimes v$.

  2.

  If A and B are operators on n and m dimensional vectors, respectively, then $A \otimes B$ is an operator on nm dimensional vectors.

  3.

  if $\mathit{H}_1$ and $\mathit{H}_2$ are Hilbert spaces, then $\mathit{H}_1 \otimes\mathit{H}_2$ is also a Hilbert space. If $\mathit{H}_1$ and $\mathit{H}_2$ are finite dimensional with bases $\{u_1, u_2, \ldots u_n\}$ and $\{v_1, v_2, \ldots v_m\}$ respectively, then $\mathit{H}_1 \otimes\mathit{H}_2$ has dimension nm with basis $\{u_i \otimes v_j \vert 1 \le i \le n, 1 \le j \le m\}$.

• Tensor products obey a number of nice rules. For matrices A, B, C, D, U, vectors u, v, w, and scalars a, b, c, d the following hold:
  
  $$\begin{eqnarray*}(A \otimes B) (C \otimes D) &=& AC\otimes BD\\ (A \otimes B) (... ...w)&=& u\otimes v + u\otimes w\\ au\otimes bv &=& ab(u\otimes v) \end{eqnarray*}$$
  
  

  Thus for matrices,
  

  $$\left(\begin{array}{cc}A & B\\ C & D\end{array}\right) \otime... ... U & B \otimes U\\ C \otimes U & D \otimes U\end{array}\right),$$
  
  

  which specializes for scalars to
  

  $$\left(\begin{array}{cc}a & b\\ c & d\end{array}\right) \otime... ...\left(\begin{array}{cc}a U & b U\\ c U & d U\end{array}\right).$$
  
  

• The conjugate transpose distributes over tensor products:
  
  $$(A\otimes B)^\dag = A^\dag\otimes B^\dag .$$
  
  

• The tensor product of several matrices is unitary if and only if each one of the matrices is unitary up to a constant. Let $U = A_1\otimes\dots \otimes A_n$. Then U is unitary if $A_i^\dag A_i = k_i I$ and $\prod_ik_i = 1$.
  
  $$\begin{eqnarray*}U^\dag U &=& (A_1^\dag\otimes\dots \otimes A_n^\dag )(A_1\otime... ...es A_n^\dag A_n\\ &=& k_1I\otimes\dots \otimes k_nI\\ &=& I\\ \end{eqnarray*}$$
  
  

• Note that $\langle{u \otimes v}\vert{w \otimes z}\rangle = \langle{u}\vert{w}\rangle \langle{v}\vert{z}\rangle$. This implies that $\langle{0 \otimes u}\vert{0 \otimes u}\rangle = 0$, and therefore $0 \otimes u$ must be the zero vector of the tensor product Hilbert space. This in turn implies (reminds us?) that the tensor product space is actually the equivalence classes in a quotient space. In particular, if A and B are vector spaces, F is the free abelian group on $A\times B$, and K is the subgroup of F generated by all elements of the following forms (where
  $a, a_1, a_2\in A, b, b_1, b_2\in B, \alpha$ a scalar):

  1.

  (a₁ + a₂,b) - (a₁,b) - (a₂,b)

  2.

  (a,b₁ + b₂) - (a,b₁) - (a,b₂)

  3.

  $(\alpha a,b) - (a,\alpha b)$

  then $A \otimes B$ is the quotient space F/K.