2
$\begingroup$

I'm quoting what I found in a book about quantum computation:

Individual state spaces of $n$ particles combine quantum mechanically through the tensor product. If $X$ and $Y$ are vectors, then their tensor product $X\otimes Y$ is also a vector, but its dimension is $\dim(X) \times \dim(Y)$, while the vector product $X\times Y$ has dimension $\dim(X)+\dim(Y)$. For example, if $\dim(X)= \dim(Y)=10$, then the tensor product of the two vectors has dimension $100$, while the vector product has dimension $20$.

I don't understand: how can he state that the result of a vector product has dimension $\dim(X) + \dim(Y)$? What does he intend for dim?

Improve this question
$\endgroup$
2
  • $\begingroup$ Vector product is only defined in 3-D Euclidean space. $\endgroup$ – Siyuan Ren Jul 5 '12 at 12:14
  • $\begingroup$ Hello, your statement is true if you want to maintain the three basic properties of cross product (in reality also 7d maintain them). Otherwise you can have n>3 d $\endgroup$ – Peppe Jul 5 '12 at 12:22
3
$\begingroup$

OP's quote seems to originate from slide p. 45 in Dan Cristian Marinescu's keynote talk from the Computing Frontiers 2004 conference.

Individual state spaces of $n$ particles combine quantum mechanically through the tensor product. If $X$ and $Y$ are vectors, then their tensor product $X\otimes Y$ is also a vector, but its dimension is $\dim(X) \times \dim(Y)$, while the vector product $X\times Y$ has dimension $\dim(X)+\dim(Y)$. For example, if $\dim(X)= \dim(Y)=10$, then the tensor product of the two vectors has dimension $100$, while the vector product has dimension $20$.

The quote mixes the notion of a vector space and the notion of a vector living in that vector space. In particular, it confusingly speaks of a vector product, where it should have referred to a Cartesian product.

Clearly, slides are no substitute for a good textbook. Keep in mind that the speaker may have oversimplified certain points because he didn't need it later in the talk, and that he might have left out less important words on the slides, so that he could use bigger fonts.

Below we suggest a remedy marked in red.

Individual state spaces of $n$ particles combine quantum mechanically through the tensor product. If $X$ and $Y$ are vector $\color{red}{\it spaces}$, then their tensor product $X\otimes Y$ is also a vector $\color{red}{\it space}$, but its dimension is $\dim(X) \times \dim(Y)$, while the $\color{red}{\it Cartesian}$ product $X\times Y$ has dimension $\dim(X)+\dim(Y)$. For example, if $\dim(X)= \dim(Y)=10$, then the tensor product of the two vector $\color{red}{\it spaces}$ has dimension $100$, while the $\color{red}{\it Cartesian}$ product has dimension $20$.

Improve this answer
$\endgroup$
1
  • $\begingroup$ Yes you are right, the main thing which has misguided me is the fact that he intends vector spaces (now i notice the first part "individual state SPACES"), since usually we describe quantum states as vectors, because we refer to particular elements, and that is the subject in this case. However thank you for your support $\endgroup$ – Peppe Jul 5 '12 at 14:09
6
$\begingroup$

There's no wonder you're confused - the author obviously was as well.

First, the operations he's talking about are direct sum $U\oplus V$ and tensor product $U\otimes V$ of vector spaces. This has nothing to do with the vector product (an ambiguous term which most often denotes the cross product you probably know from school).

Both are two different ways to combine vector spaces into a larger one:

If $U$ has a basis $\{u_i\}_i$ and $V$ a basis $\{v_j\}_j$ then $\{u_i,v_j\}_{i,j}$ is a basis of $U\oplus V$, ie the dimensions are added. One way to construct the direct sum is via the cartesian product $U\times V$, so the basis would actually be $\{(u_i,0),(0,v_j)\}_{i,j}$.

In contrast, $\{u_i\otimes v_j\}_{i,j}$ is a basis of $U\otimes V$, ie the dimensions are multiplied. The construction of the tensor product is a bit more complicated, so I won't go into details here. What you should realize, though, is that not all vectors of the tensor product have the form $u\otimes v$; for example

$ u_1\otimes v_1 + u_2\otimes v_2 $

can't be simplified with this particular choice of basis. Physicists call such states entangled.

Improve this answer
$\endgroup$
7
  • $\begingroup$ Thank you for your answer, however I think that he is speaking about tensorial product of vectors (not vector spaces) about which I know enough to agree about what he says. My question was about the simple vector product: Do you think he made a mistake about that? $\endgroup$ – Peppe Jul 5 '12 at 11:59
  • 2
    $\begingroup$ He's not mistaken but rather made a poor choice of terminology. What he refers to as vector product is more usually known as the direct sum of vector spaces. $\endgroup$ – Emilio Pisanty Jul 5 '12 at 12:11
  • 1
    $\begingroup$ @Emilio: he also talks about the dimension of a vector, which doesn't really make sense $\endgroup$ – Christoph Jul 5 '12 at 12:41
  • $\begingroup$ @PepX: tensor products of vectors are used to denote elements of (tensor) product spaces, and it's the spaces that have a dimension; the cross product is a specific case of either Lie bracket or wedge product (in which case, however, the result isn't really a vector, but - in physicists terminology - an axial or pseudo vector); the cross product doesn't make sense here $\endgroup$ – Christoph Jul 5 '12 at 13:02
  • $\begingroup$ Yes, it's a pretty strange statement. I will try to ask further info to the authors. Thank you $\endgroup$ – Peppe Jul 5 '12 at 13:09
0
$\begingroup$

The tensor product is the natural extension of the ordinary product

$(a+b)(c+d)=ac+ad+bc+bd$.

If you have two vector $x,y$ of dimension $n$ the tensor product become

$$ x_\mu \otimes y_\nu= x_\mu y_\nu=\Theta_{\mu\nu} $$

where $\Theta_{\mu\nu}$ is a matrix of dimension $n\times n$.

The vector product of two vectors $x,y$ generate a third vector $z$ orthogonal to $x,y$. This means that if you want fully define the vector $z$ you must define $x$ and $y$ and then, in terms of degrees of freedom, you must add the degrees of freedom of $x$ and $y$.

Improve this answer
$\endgroup$
11
  • $\begingroup$ Thank you, I agree with the first part of your answer. However my real question was about the vector product. I am not able to visualize the second part effectively, so can you make a practical example with two vectors? $\endgroup$ – Peppe Jul 5 '12 at 12:07
  • $\begingroup$ This is misleading. The tensor product of vectors are still vectors, not matrices(operators). $\endgroup$ – Siyuan Ren Jul 5 '12 at 12:12
  • $\begingroup$ I disagree with your statement. Get a look here: people.rit.edu/pnveme/EMEM851n/constitutive/tensors_rect.html $\endgroup$ – Peppe Jul 5 '12 at 12:37
  • $\begingroup$ @Karsus: matrices can be used to denote elements of $U\otimes V$ as well as $U^*\otimes V\cong \mathrm{Hom}(U,V)$ $\endgroup$ – Christoph Jul 5 '12 at 13:07
  • 1
    $\begingroup$ @Karsus matrix is only a square table of numbers. Then depending on the way they tranforms we can call in a different ways. $\endgroup$ – Emanuele Luzio Jul 5 '12 at 13:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.