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)?


1 Answer 1


It is a perfectly well-defined expression because the tensor product is a linear space.

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. $$ \langle v_i| \otimes \langle w_j| \cdot |v_m\rangle \otimes |w_k\rangle = \langle v_i|v_m\rangle \cdot \langle w_j|w_k\rangle$$ 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.

  • $\begingroup$ 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? $\endgroup$
    – Hedra
    Aug 3, 2014 at 10:35
  • 2
    $\begingroup$ @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 $E\otimes I|\psi\rangle = \frac{1}{\sqrt{2}}E|0\rangle\otimes I|0\rangle + \frac{1}{\sqrt{2}}E|1\rangle\otimes I|1\rangle$ $\endgroup$
    – Trimok
    Aug 3, 2014 at 11:33
  • $\begingroup$ ahh, now that I see it it's obvious, but I guess I expected that given that I was confused about notation. Thanks to you both. $\endgroup$
    – Hedra
    Aug 3, 2014 at 12:00

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