Reading Nielsen and Chuang, I am under the impression that a linear operator on the tensor product can be written in two ways:

\begin{equation} (\left|a\right> \otimes \left|b\right>)(\left<c\right| \otimes \left<d\right|) = \left|a\right>\left<c\right| \otimes \left|b\right>\left<d\right| \end{equation}

As if I take a vector $\left|e\right> \otimes \left|f\right>$ from a tensor product and apply the operator, I get in the first case:

\begin{align} (\left|a\right> \otimes \left|b\right>)(\left<c\right| \otimes \left<d\right|) (\left|e\right> \otimes \left|f\right>) =& (\left|a\right> \otimes \left|b\right>) \left<c|e\right> \left<d|f\right>\\ =& \left<c|e\right> \left<d|f\right> \left|a\right> \otimes \left|b\right> \end{align}

And, in the second case,

\begin{align} (\left|a\right>\left<c\right| \otimes \left|b\right>\left<d\right|) (\left|e\right> \otimes \left|f\right>) =& \left|a\right>\left<c|e\right> \otimes \left|b\right>\left<d|f\right> \\ =& \left<c|e\right> \left<d|f\right> \left|a\right> \otimes \left|b\right> \end{align}

In the first case, I consider an operator on the tensor product, and in the second case, a tensor product of operators. Are the two equivalent ? All state vectors considered are arbitrary.


2 Answers 2


Indeed, they are equivalent. The relation you start with is a special case of the following identity: \begin{equation} (A \otimes B)(C \otimes D) = (AC) \otimes (BD) \end{equation} where $A,B,C,D$ are all operators in some Hilbert space with a tensor product structure $\mathcal{H} = \mathcal{H}_1\otimes\mathcal{H}_2$, and $A,C$ are operators that belong to $\mathcal{H}_1$, and $B,D$ to $\mathcal{H}_2$. Operators that act on different Hilbert spaces commute, i.e. $[M_1 \otimes I, I \otimes M_2]=0$, because anything you do to objects in $\mathcal{H}_1$ cannot affect $\mathcal{H}_2$ and vice versa. This is why you are allowed to "pass" $C$ to the left, through $B$, and combine the operators in "like" Hilbert spaces.

In the example you provided, the operators take the form of projectors onto 1-dimensional subspaces with these Hilbert spaces, i.e. kets and bras. Kets are column vectors which operate on row vectors, and bras are row vectors which operate on column vectors. In both cases, the output space is the complex numbers.


Yes they are equivalent, as you correctly computed. You are essentially just observing that $|a\rangle\!\langle c|d\rangle$ can be understood either as an inner product ($\langle c|d\rangle$) multiplied scalarly with a vector ($|a\rangle$), or as an outer product ($|a\rangle\!\langle c|$) applied to a vector ($|d\rangle$).

A more formal way to observe that this is not completely trivial might be using a more standard algebraic notation. Let $V$ be a Hilbert space, and $u,v,w\in V$ arbitrary vectors. Denote with $v^*\in V^*$ elements of the dual of $V$ (i.e. $v^*:V\to\mathbb C$ is the linear functional such that $v^*(v)=1$ and $v^*(v_\perp)=0$ if $\langle v,v_\perp\rangle=0$). Denote with $uv^*$ the outer product between $u$ and $v$. You can understand this as the linear operator $(uv^*): V\ni x\mapsto u \langle v,x\rangle \in V$.

Then the observation is that (I'll be pedantic with the parentheses to clarify the order of operations): $$(uv^*)(w) = u \cdot (v^*(w)).$$ Note that in the RHS $v^*(w)\in\mathbb C$ and thus $\cdot$ denotes a scalar product in $V$, whereas in the LHS we have the operator $(uv^*)$ acting on its input $w$.


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