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I'm studying Quantum Computing: A Gentle Introduction. On page 33, Section 3.1.2, after defining tensor product with 3 properties (distribution over addition on both left and right, scalar on both sides), it says all element of $V \otimes W$ have form $|v_1\rangle \otimes |w_1\rangle +|v_2\rangle \otimes |w_2\rangle +\ldots+|v_k\rangle \otimes |w_k\rangle$, where $k$ is the minimum of the dimensions of $V$ and $W$.

Assume $V$ has $k+1$ dimensions (and $W$ has $k$ dimensions), why the basis $|v_{k+1}\rangle$ is not used?

Similarly, in the classic Quantum computing and quantum information textbook, section 2.1.7, formula 2.46, it also uses a single index $i$ $$(A\otimes B)(\sum_i a_i|v_i\rangle \otimes |w_i\rangle).$$ I thought it should be $$(A\otimes B)(\sum_ {i,j} a_{i,j}|v_i\rangle \otimes |w_j\rangle).$$

Is there anything deeper to explain that a single index with a subset of bases is sufficient?

Thanks

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The reason is relatively straightforward. Consider an $m$ dimensional vector space $V$ with basis $\lbrace \vert v_1 \rangle,...,\vert v_m \rangle \rbrace$, and an $n$ dimensional vector space $W$ with basis $\lbrace \vert w_1 \rangle,...,\vert w_n \rangle \rbrace$. As your intuition suggests, we can naturally express any element $A \in V \otimes W$ in the form $$A = \sum \limits_{j=1}^{m} \sum \limits_{k=1}^n \lambda_{jk} \vert v_j \rangle \otimes \vert w_k \rangle,$$ where $\lambda_{jk}$ are scalar coefficients.

The reason we can express $A$ in $\text{min}(m,n)$ terms is that we can group the set of $m$ vectors $\lambda_{jk} \vert v_j \rangle$ into a new set of $n$ vectors $\vert a_k \rangle$ given by $$\vert a_k \rangle=\sum \limits_{j=1}^m \lambda_{jk} \vert v_j \rangle,$$ which gives an expression for $A$ with one index in $n$ terms as $$A=\sum \limits_{k=1}^n \vert a_k \rangle \otimes \vert w_k \rangle.$$ We can do the same to express $A$ in $m$ terms by $$A = \sum \limits_{j=1}^m \vert v_j \rangle \otimes \vert b_j \rangle, \;\;\;\; \vert b_j \rangle = \sum \limits_{k=1}^n \lambda_{jk} \vert w_k \rangle,$$ showing that $A$ can always be expressed in $\text{min}(m,n)$ terms.

However, to show that the $mn$ elements $\lbrace \vert v_j \rangle \otimes \vert w_k \rangle \rbrace$ form a basis in $V \otimes W$, we still need to show that these elements are linearly independent. That proof is not so easy (or concise). I would refer you to Linear Algebra via Exterior Products by Winitzki, section 1.7.3, if you want that level of rigor.

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    $\begingroup$ What do you mean by "the set of $min(m,n)$ terms form a basis in $V \otimes W$"? This space has dimension $mn$. $\endgroup$ – Danylo Y Oct 2 '19 at 12:03
  • $\begingroup$ Thank you for pointing that out. That was a mistake. Edited to fix. $\endgroup$ – Jonathan Trousdale Oct 2 '19 at 12:49
  • $\begingroup$ what is the difficulty in proving that $\{|v_j\rangle\otimes|w_k\rangle\}$ are linearly independent? They are obviously orthogonal to each other (essentially by definition of the inner product in the tensor product space), and therefore linearly independent, no? $\endgroup$ – glS Oct 4 '19 at 18:03
  • $\begingroup$ If $\{|v_j\rangle \otimes |w_k\rangle \}$ span the whole $V \otimes W$ (and they do) then they have to be linearly independent. If they are linearly dependent then they can span only subspace of dimension less than $mn$. $\endgroup$ – Danylo Y Oct 6 '19 at 20:41
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    $\begingroup$ I see, in 1.7.3 we don't know yet that dimension of tensor product is $mn$. Anyway, in the field $\mathbb{R}$ or $\mathbb{C}$ it's much easier since we can introduce the scalar product. $\endgroup$ – Danylo Y Oct 7 '19 at 6:39

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