So, while the only immediate restriction on an operator evolving a quantum state in time, is that it be unitary, in quantum computation/information, it is considered somewhat common knowledge that all single-qubit operators can be "

written as the product of an element of SU(2) with an unimportant global phase factor.

(e.g. see p.618 in "Quantum Computation and Quantum Information 10th Anniversary Edition".)

It therefore seems (to me at least) to be the case, that it should be possible to show, by considering a general matrix representation of both U(2) and SU(2), that the former introduces a global phase factor, that the latter doesn't?

Specifically, taken together w. matrix multiplication one has:

$$\tag{1} U(2)\equiv\{A\in\mathbb{C}^{2\times2}:A^{\dagger}A=AA^{\dagger}=\mathbb{I}\} $$ and: $$\tag{2} SU(2)\equiv\{A\in\mathbb{C}^{2\times2}:A^{\dagger}A=AA^{\dagger}=\mathbb{I}\land\text{det}(A)=1\} $$ or equivalently that a general matrix in $U(2)$ can be written on the form: $$\tag{3} \begin{pmatrix} a & b \\ -e^{i\phi}b^{\ast} & e^{i\phi}a^{\ast} \end{pmatrix}, \quad |a|^2+|b|^2=1 $$ (resulting in $\text{det}(A)=e^{i\phi}\Leftrightarrow |\text{det}(A)|=1$), while a general matrix in $SU(2)$ can be written on the form: $$\tag{4} \begin{pmatrix} a & b \\ -b^{\ast} & a^{\ast} \end{pmatrix}, \quad |a|^2+|b|^2=1 $$ (resulting in $\text{det}(A)=1$).

In summary: Is it the case that the action of a general element from $U(2)$, on any single-qubit state, introduces a global phase, while a general element from $SU(2)$ doesn't? If this is the case, is there a more formal way to illustrate this? If this isn't the case - what is then the reason for referring to $SU(2)$ instead of $U(2)$?

  • $\begingroup$ The question is unclear. The state you "produce" at the end can be the result of either an SU(2) or an U(2) matrix, or even the identity matrix, provided you let it act on a properly chosen initial state. And as for the "way to show this" that you ask for, just let the matrix act on the vector you have, then you get the result. Probably you want to ask something else, correct? $\endgroup$ Feb 27 at 20:13
  • $\begingroup$ @JosBergervoet I tried reformulating the final part of the post - does this help clarify my question? :) $\endgroup$
    – seba2390
    Feb 27 at 21:01
  • $\begingroup$ Yes, U imprints a global phase and SU sets that global phase to 0. Of course some states may acquire phases from an SU operation (eg an eigenstate of the unitary) but a global phase adds the same phase to all states $\endgroup$ Feb 28 at 15:41

1 Answer 1


It is true that an element from U(2) can change the global phase of a state in an arbitrary way, the U(2) matrix $$\tag{1} \begin{pmatrix} e^{i\varphi} & 0 \\ 0 & e^{i\varphi} \end{pmatrix} $$ clearly changes the global phase of every vector by $\varphi$, and there is no such matrix in SU(2) except for the special case $\varphi=\pi$.

Still, if you start with a specific vector, there is always an SU(2) matrix that can change the global phase of that vector by any phase $\varphi$. For the vector $(1,0)$ that would obviously be the SU(2) matrix for rotation of $2\varphi$ around the $z$-axis: $$\tag{2} \begin{pmatrix} e^{i\varphi} & 0 \\ 0 & e^{-i\varphi} \end{pmatrix} $$ For any other given vector $\xi$ we can first rotate it to $(1,0)$ by some SU(2) matrix $U$, then rotate around the $z$-axis, and then transform it back: $$\tag{3} U^{-1} \begin{pmatrix} e^{i\varphi} & 0 \\ 0 & e^{-i\varphi} \end{pmatrix} U \ \xi = e^{i\varphi}\xi, $$ where the three matrices multiplied together will still result in an SU(2) matrix. So we can control the global phase of a vector with an SU(2) matrix if we know the vector in advance.


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