# The factorization of a qbit product state is not unique; how should the different factorizations be interpreted?

Consider two arbitrary qbits; their product state is as follows:

$\begin{bmatrix} a \\ b \end{bmatrix} ⊗ \begin{bmatrix} c \\ d \end{bmatrix} = \begin{bmatrix} ac \\ ad \\ bc \\ bd \end{bmatrix}$

We can factor this product state into four different tensor products of two qbits:

$\begin{bmatrix} ac \\ ad \\ bc \\ bd \end{bmatrix} = \begin{bmatrix} a \\ b \end{bmatrix} ⊗ \begin{bmatrix} c \\ d \end{bmatrix}$, $\begin{bmatrix} ac \\ ad \\ bc \\ bd \end{bmatrix} = \begin{bmatrix} -a \\ -b \end{bmatrix} ⊗ \begin{bmatrix} -c \\ -d \end{bmatrix}$,

$\begin{bmatrix} ac \\ ad \\ bc \\ bd \end{bmatrix} = \begin{bmatrix} ia \\ ib \end{bmatrix} ⊗ \begin{bmatrix} -ic \\ -id \end{bmatrix}$, $\begin{bmatrix} ac \\ ad \\ bc \\ bd \end{bmatrix} = \begin{bmatrix} -ia \\ -ib \end{bmatrix} ⊗ \begin{bmatrix} ic \\ id \end{bmatrix}$

What is the relationship between these four factorizations? Do they all represent the same state?

• the factorizations are the same up to global factors – lurscher Apr 19 '18 at 19:18
• But for the last two factorizations, there isn't a global factor. Or do you mean individual qbits still represent the same state when multiplied by any scalar? That seems a bit odd for scalars like $-1, i,$ and $-i$ which rotate you around the unit sphere. – ahelwer Apr 19 '18 at 20:09
• in each Hilbert space, vectors are defined up to a complex phase factor, both factor qbits belong to their individual Hilbert spaces – lurscher Apr 19 '18 at 20:27
• BTW, this is a good place for this type of questions: quantumcomputing.stackexchange.com – Alexander Pozdneev Apr 20 '18 at 8:04

There aren't just four different tensor products; there are an uncountably infinite number of tensor products that give the same state, because

$$\begin{bmatrix}e^{i\phi}a\\e^{i\phi}b\end{bmatrix}\otimes\begin{bmatrix}e^{-i\phi}c\\e^{-i\phi}d\end{bmatrix}=\begin{bmatrix}ac\\ad\\bc\\bd\\\end{bmatrix}$$

for any real number $\phi$. (In fact, there are even more, if you don't require the two input vectors to be normalized; multiplying one vector by some arbitrary nonzero complex number $c$ and the other by $\frac{c^*}{|c|^2}$ will also give you the same result.)

This isn't some sort of special property; it's just due to the bilinearity of the tensor product, which states that for vectors $A$ and $B$ and complex scalars $c$ and $d$,

$$(cA)\otimes (dB)=cd(A\otimes B)$$

In this case, we take $d=c^{-1}$. Bilinearity is a property that also appears in a lot of other operators, like the dot product:

$$(a\vec{v})\cdot (b\vec{w})=ab(\vec{v}\cdot\vec{w})$$

And the cross product:

$$(a\vec{v})\times(b\vec{w})=ab(\vec{v}\times\vec{w})$$

It's also trivially shared with ordinary scalar multiplication, though in that case, bilinearity is equivalent to commutativity.

So, not only is there nothing special about the four examples you provided, there is also nothing special about any particular decomposition of a vector into a tensor product of two other vectors.

• That's very interesting; how, then, do we interpret these factors? Is there a normalized factorization we choose when factoring a product state? Or even for a single qbit, are there other points on the unit sphere which are identical to that qbit? – ahelwer Apr 19 '18 at 21:27
• @ahelwer You interpret these factors in the exact same way as you would interpret them if you were taking, say, the dot product of two vectors. You are free to multiply one vector by a scalar and the other by the inverse of that scalar. If you're not satisfied with that, then think of it this way: taking the tensor product maps from two Hilbert spaces (each of which is invariant up to a global phase) to one Hilbert space (which is invariant up to a global phase). Because of the extra free parameter in the input, there is necessarily a redundancy that makes this mapping not one-to-one. – probably_someone Apr 19 '18 at 21:36
• What do you mean when you say "invariant up to a global phase"? Is that the same as saying there exists no series of operations (measurement or otherwise) which could differentiate qbit $|x\rangle$ from qbit $e^{i\phi}|x\rangle$ for any $\phi \in \mathbb{R}$? So they're equivalent for all computational purposes? – ahelwer Apr 19 '18 at 21:44
• @ahelwer Yes. There is no measurement that can distinguish those two. This is true for both input qubits, giving you two degrees of freedom in selecting functionally identical input qubits (one global phase for each). It is also true for the result, but since there is now only a single vector, we only have a single degree of freedom in selecting a functionally identical resultant qubit. This means that one degree of freedom is redundant (i.e. you're losing the information contained in the phase of one of the qubits). – probably_someone Apr 19 '18 at 21:48
• Just to be 100% clear - you say you mean precisely the opposite, but it sounds like we are saying the same thing? – ahelwer Apr 19 '18 at 21:49