# Inner products containing the tensor product of two operators

The book Nielsen & Chuang "Quantum Computation and Quantum Information" presents the concept of tensor products as follows.

Suppose we have the vectors $|v\rangle$ and $|w\rangle$ which exist in vector spaces $V$ and $W$ respectively. We also define the linear operators $A$ and $B$ which exists in same respective vector spaces. Then we can define the tensor product of these vectors and operators which behaves as follows

$$\left(A\otimes B\right)\left(|v\rangle\otimes|w\rangle\right) = A|v\rangle\otimes B|w\rangle \tag{1}$$

I can accept this as definition, but my question arises from an exercise where it asks to evaluate $$\langle\psi\,|E\otimes I|\,\psi\rangle\tag{2}$$ where $E$ is a positive operator and $|\psi\rangle$ is any of the four Bell states. However the book does not describe the behaviour of the expression $$(A\otimes B)(|v\rangle)\tag{3}$$ My assumption is that this is not a valid expression since $|v\rangle$ does not exist in the vector space $V\otimes W$ on which the operator is defined. Am I correct in thinking this? How does one expand the inner product in equation (2)?

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The vectors $|v\rangle\otimes |w\rangle$ form a basis of the whole tensor-product vector space, so any vector (including Bell's state) in this space may be written as linear combinations of such basis vectors. $$|\psi \rangle = \sum_{ij} c_{ij} |v_j\rangle\otimes |w_j\rangle$$ Because the operators are linear and we know how to act on each term, the result of the action of the operator $L$ is $$L |\psi \rangle = \sum_{ij} c_{ij} L |v_j\rangle\otimes |w_j\rangle$$ where your formulae already say how to evaluate the individual terms e.g. for $L = E\otimes I$.
The natural inner product of two vectors on the tensor product space is given by the simple product of the factors. Choose a basis like above, and write the inner product of two basis vectors as products in the most straightforward way. $$\langle v_i| \otimes \langle w_j| \cdot |w_k\rangle \otimes |v_m\rangle = \langle v_i|v_m\rangle \cdot \langle w_j|w_k\rangle$$ This again defines the inner product for any two vectors, by linearity. Decompose each of the two general vectors in the tensor product space that enter the inner product as a linear combination of the simple $vw$ basis vectors above, apply the distributive law to calculate the inner product of each term, and sum the terms with the same coefficients.
Thank you very much for your response. So in the case of the Bell state $|\psi\rangle =\frac{|00\rangle + |11\rangle}{\sqrt{2}}$, it is equivalent to expressing it as $|\psi\rangle = \frac{1}{\sqrt{2}}|0\rangle\otimes|0\rangle + \frac{1}{\sqrt{2}}|1\rangle\otimes|1\rangle$. Which then implies $\langle\psi|E\otimes I|\psi\rangle = \frac{1}{2}\langle 0|\,E\,|1\rangle\langle 0|\,I\,|1\rangle$ Is this correct? – Hedra Aug 3 '14 at 10:35
@Hedra : Yes to your first question. It is exactly the same thing, the first notation is more compact. No to your second question, you have $4$ terms, note that $E\otimes I|\psi\rangle = \frac{1}{\sqrt{2}}E|0\rangle\otimes I|0\rangle + \frac{1}{\sqrt{2}}E|1\rangle\otimes I|1\rangle$ – Trimok Aug 3 '14 at 11:33