Proof involving tensor product I am trying to prove when the following holds:
$$|a\rangle |b\rangle \langle c|\langle d| = |a\rangle \langle c| \otimes |b\rangle \langle d|$$
where $\otimes$ stands for tensor product and the $a,b,c,d$ are $N$ dimensional state vectors.
I can see that for $N=2$ it holds true for $a,b,c,d\ \in\ \lbrace |0\rangle , |1\rangle \rbrace$.  Are there any other values?
 A: The identity is true in any dimension.
To see this, notice that both the left and right hand sides of the equation you wrote down are linear operators on the tensor product of vector spaces, so to show that they are equal, it suffices to show that they agree on a basis.  Since the basis for a tensor product is the set of all tensor products of basis elements, it suffices to show that the two expressions agree on every vector of the form $|\psi\rangle|\phi\rangle$.
To do this, note that when we apply the left hand side to such a vector, we obtain
\begin{align}
  |a\rangle|b\rangle\langle c|\langle d| (|\psi\rangle|\phi\rangle)
&= \langle c|\psi\rangle\langle d|\phi\rangle|a\rangle|b\rangle ,
\end{align}
while when we apply the right hand side, we get
\begin{align}
  |a\rangle\langle c|\otimes |b\rangle\langle d|(|\psi\rangle|\phi\rangle)
&= \Big(|a\rangle\langle c|(|\psi\rangle)\Big)\otimes\Big(|b\rangle\langle d|(|\phi\rangle)\Big) \\
&= \langle c|\psi\rangle\langle d|\phi\rangle|a\rangle\otimes|b\rangle  \\
&= \langle c|\psi\rangle\langle d|\phi\rangle |a\rangle|b\rangle
\end{align}
so the expressions agree.
