2
$\begingroup$

If the state of one qubit can be described by a ray in $\mathbb{C}^2$, then the combined state of an $n$-qubit system can be described by a ray in $(\mathbb{C}^2)^{\otimes n}=\mathbb{C}^{2 n}$.

However, if $G_1$ is the Pauli group of one qubit, with the 16 elements $$G_1=\{i,-1,-i,1\}\times\{I,X,Y,Z\}\,,$$ the Pauli group on $n$ qubits is defined by $$G_n=\{i,-1,-i,1\}\times\{I,X,Y,Z\}^{\otimes n}$$ which not the tensor product of $n$ Pauli groups $G_1$ (because $G_n$ contains $4\cdot 4^n$ elements, which does not equal $16^n$). My question thus is: what kind of tensor product do we use on the space of operators on a Hilbert space $\mathbb{C}^2$, to define $G_n$ using $G_1$?

(I do understand intuitively that we should disregard global phase (and that therefore the number of operators in the $n$-qubit Pauli group is not $16^n$), and that this can be done by introducing the projective Hilbert space, but how do tensor products work on the space of operators on a projective Hilbert space?)

$\endgroup$
2
$\begingroup$

The group $(G_1)^{n}$ does act on $(\mathbb{C}^2)^{\otimes n}$ (you made a typo with $\mathbb{C}^{2n}$ vs $\mathbb{C}^{2^n}$) but the action factors through a quotient thereof. So you can write down this group initially as the set of all $(a_1 \sigma^{0,x,y,z}_1 \otimes \cdots a_n \sigma^{0,x,y,z}_n)$ where the $a_i \in \{ 1,i,-1,-1 \}$ and $\sigma^0=Id_2$. Not using the properties of $\otimes$ to shuffle the $a_i$ through yet. Each tensorand is a copy of $G_1$ and these factors commute amongst each other.

Then when we interpret the symbol $\otimes$, we see that $i \sigma^0_1 \otimes -i \sigma^0_2 \otimes 1 \sigma^0_3 \otimes \cdots \otimes 1 \sigma^0_n$ acts trivially. It is in the kernel of the map $(G_1)^{n} \to Aut((\mathbb{C}^2)^{\otimes n})$. By understanding this kernel, one sees that $(G_1)^{n}/(ker) \simeq G_n$. There are still some global phases from the $\{1,i,-1,-i\}$ factor.

Suppose one wants to work projectively from the very beginning. Then on 1-qubit we have some subgroup of $PU(2)$. In this case $PG_1 \simeq \mathbb{Z}_2^2$. Then for $n$ qubits you even lose that first global phases factor above and are only left with $(PG_1)^n \simeq PG_n$.

$\endgroup$
2
$\begingroup$

Tensor products always mod out $\mathbb C$ (or $\mathbb R$, or whatever field your vector space is defined over). Thus, there is only one phase in the tensor product. This is also true for the corresponding space of operators, and thus also for $G_1^{\otimes N}$. This is just the normal definition of the tensor product.

$\endgroup$
  • $\begingroup$ If we consider $\mathbb{R}^n$ and $\mathbb{R}^m$ as vector spaces, then $\mathbb{R}^n\otimes\mathbb{R}^m=\mathbb{R}^{nm}$, so tensor products do not always mod out $\mathbb{C}$ or $\mathbb{R}$. $\endgroup$ – Carucel Sep 10 '18 at 7:16
  • 1
    $\begingroup$ Of course they do, otherwise $(1,0)\otimes(1,0)$ and $(-1,0)\otimes(-1,0)$ would be different things! $\endgroup$ – Norbert Schuch Sep 10 '18 at 7:18
  • $\begingroup$ @NorbertSchuch: I think that Carucel understands you to mean that there are no phases at all --- that rather than modding out $(a\mathbf v \otimes b \mathbf w) - ab(\mathbf v \otimes \mathbf w)$, you're modding out $(a \mathbf v \otimes \mathbf w) - (\mathbf v \otimes \mathbf w)$ and $(\mathbf v \otimes b \mathbf w) - (\mathbf v \otimes \mathbf w)$, which is one possible reading of 'tensor products always mod out scalars'. $\endgroup$ – Niel de Beaudrap Sep 10 '18 at 13:38
  • $\begingroup$ @NieldeBeaudrap Maybe, but I'd interpret "mod out C" as "mod out C once", not "mod out C as often as you can". $\endgroup$ – Norbert Schuch Sep 10 '18 at 16:22
  • 1
    $\begingroup$ We both understand how that goes, though someone asking the same question as Carucel might not. $\endgroup$ – Niel de Beaudrap Sep 10 '18 at 16:32

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.