How do you define the “natural” inner product on a tensor product space?

I'm reading Nielsen and Chuang, Quantum Computation and Quantum Information. On p. 73 it is introducing inner products and tensor products. So it says the following:

The inner product on the spaces $V$ and $W$ can be used to define a natural inner product on $V \otimes W$. Define

$$\big(\sum_i a_i \left | v_i \right\rangle \otimes \left | w_i \right\rangle, \sum_j b_j \left | v_j' \right\rangle \otimes \left | w_j' \right\rangle\big) \equiv \sum_{ij} a_i^*b_j \left\langle v_i | v_j' \right\rangle \left\langle w_i | w_j' \right\rangle.\tag{2.49}$$

This only makes sense if $V$ and $W$ have the same dimension. Say, $V$ is 2-dimensional and $W$ is 3-dimensional. What does $\left | v_3 \right\rangle$ mean? I would appreciate some clarification.

• There is no need for $V$ and $W$ to have the same dimension. – Qmechanic May 17 '18 at 2:57
• But then, what do the $i$ and $j$ indices run over? – azani May 17 '18 at 3:00
• The book is wrong. creillyucla's answer below is correct. BTW, what book is this? – NickD May 17 '18 at 3:25
• Quantum Computation and Quantum Information by Nielsen and Chuang. Page 73 of the 10th anniversary edition. – azani May 17 '18 at 3:27
• I contacted the authors with an errata request. – azani May 17 '18 at 3:37

A basis for a tensor product space are the vectors $|v_i\rangle\otimes |w_j\rangle$, where $i = 1,\ldots,N_V$ and $j = 1, \ldots, N_W$, where $N_V$ and $N_W$ are the dimensions of $V$ and $W$, respectively. Note that we include all combinations, not just ones where $i=j$. A vector $|u\rangle$ can thus be written: $$|u\rangle = \sum_i^{N_V} \sum_j^{N_W} u_{ij} |v_i\rangle\otimes |w_j\rangle$$
The natural inner product between two vectors $|a\rangle,|b\rangle$ is then: $$\langle b | a \rangle := \sum_i^{N_V} \sum_j^{N_W} \sum_{i'}^{N_V} \sum_{j'}^{N_W} b^*_{i'j'} a_{ij} \langle v_{i'} | v_i \rangle \langle w_{j'} | w_j \rangle$$ if the $\{|v_1\rangle\ldots,|v_i\rangle,\ldots,|v_{N_V}\rangle\}$ and $\{|w_1\rangle\ldots,|w_j\rangle,\ldots,|w_{N_W}\rangle\}$ bases are both orthonormal (in quantum mechanics they nearly always are) this simplifies to: $$\langle b | a \rangle := \sum_i^{N_V} \sum_j^{N_W} \sum_{i'}^{N_V} \sum_{j'}^{N_W} b^*_{i'j'} a_{ij} \delta_{ii'} \delta_{jj'} = \sum_i^{N_V} \sum_j^{N_W} b^*_{ij} a_{ij}$$
To motivate the "naturalness" of this selection, I would point out that the familiar space of functions (i.e. wavefunctions) of multiple variables (say $x,y,z$) is the tensor product of the spaces of functions of the individual variables. That can be seen simply by noting that a function $f(x,y,z)$ must specify a value for all possible combinations of coordinates $(x,y,z)$, just like a vector in the space you give is characterized by its coefficient $u_{ij}$ for all possible pairs of indices. The inner product of two functions $f(x,y,z)$ and $g(x,y,z)$ is of course: $$\int dx \int dy \int dz g^*(x,y,z) f(x,y,z)$$ which is totally analogous to the general definition given above.
Concerning the index notation: Nielsen & Chuang are considering arbitrary finite sums $$\sum_{i\in I} a_i \left| v_i \right\rangle \otimes \left| w_i \right\rangle, \qquad a_i~\in~\mathbb{C},$$ in the tensor product $V\otimes W$, where the index set $I$ is arbitrary but finite: $|I|<\infty$. There is no assumption of, say, linear independence of $$\left| v_i \right\rangle, \qquad i\in I.$$ Nor is there an assumption of linear independence of $$\left| w_i \right\rangle, \qquad i\in I.$$ In particular, it is not necessary to assume that the vector spaces $V$ and $W$ have the same dimension.
• So, just to clarify, what you're saying is that they are not summing over the basis. They're just representing the operands of the inner product as linear combination of some arbitrary vectors of the tensor product space. so the $i$ doesn't really index over $v$ and $w$ in parallel. It indexes over $v \otimes w$. Is that right? – azani May 17 '18 at 15:01