All qubit gates can be written in the form of:

$$U = \exp(i\alpha)R_n(\theta).$$

I know $R_n(\theta)$ is a rotation of $\theta$ about an arbitrary axis n in Bloch sphere, but what does $\exp(i\alpha)$ stand for? From my view, an arbitrary rotation might be enough to represent all qubit operations?

  • $\begingroup$ @ YuZi The accepted answer is incorrect. Please reconsider its status. $\endgroup$ – Emilio Pisanty May 23 at 16:05

$e^{i\alpha}$ is just an overall phase factor to capture the fact that the determinant of unitary matrices can be any complex number with norm one, while rotations always have determinant 1. This factor is usually not important though, since the overall phase of a wavefunction is not observable.

Edit: By "rotations" here I meant "rotations around the Bloch sphere", or exponentials of Pauli matrices $e^{i \mathbf{\theta} \cdot {\bf \sigma}}$ - as pointed out in the answer below this would not work if we were assuming rotations in $SO(2)$.

  • $\begingroup$ This answer is incorrect, and counter-examples are trivial to construct; see my answer for one. $\endgroup$ – Emilio Pisanty May 22 at 11:26
  • $\begingroup$ Thanks for pointing this out - I had interpreted the $R_n(\theta)$ as rotations around the Bloch sphere. I've edited the answer to clarify this. $\endgroup$ – Iris Cong May 23 at 15:01
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    $\begingroup$ The overall phase factor is not usually omitted in the case of more than one qubit (unless the $R_n(\theta)$ part represents a gate on the entire quantum system, which is unusual). $\endgroup$ – DanielSank May 23 at 15:42
  • $\begingroup$ "this would not work if we were assuming rotations in SO(2)" - the question explicitly has $U = e^{i\alpha}R_n(\theta)$, where $U$ is a $2\times 2$ complex-valued matrix. If $R_n(\theta)$ is a rotation on the Bloch sphere, then $U$ and $R_n(\theta)$ have different dimensions - they cannot possibly be in such a relationship. This answer (as of v2) is still incorrect. $\endgroup$ – Emilio Pisanty May 23 at 16:00
  • $\begingroup$ Sorry, I don't see what's incorrect in this version of the answer as I've explicitly said a Bloch sphere rotation is a matrix of the form $e^{i \theta \cdot \sigma}$. Here, $\sigma$ is the vector of 3 2-by-2 Pauli matrices and $\theta$ is a 3-component vector, so the two sides of the equation have same dimension. $\endgroup$ – Iris Cong May 23 at 17:51

The qubit gate form as you have written is incorrect; it is likely that you have misunderstood a different claim.

As a simple counter-example, consider the qubit gate $$ U = \begin{pmatrix} 1 & 0 \\ 0 & e^{i\varphi} \end{pmatrix}. $$ The claim in your question implies that it should be possible to write it as the product $U = e^{i\alpha} R(\theta)$, where $$ R(\theta) = \begin{pmatrix} \cos(\theta) & \sin(\theta) \\ -\sin(\theta) & \cos(\theta) \end{pmatrix} \in \mathrm{SO}(2) $$ is a rotation matrix. This is provably impossible; to see why, simply consider the action of $U$ on $e_1 = (1,0)^T$, which implies that you need to have $\alpha = \theta = 0$ ─ which would then leave $e_2 = (0,1)^T$ untouched, instead of giving it a phase.

From what I can tell, it is likely that you have misunderstood a different claim, namely, that any single-qubit unitary gate can be understood as a rigid rotation in the Bloch sphere representation, followed by a phase $e^{i\alpha}$ multiplying the wavefunction.

To see this in a bit more detail, consider an arbitrary state $|\psi\rangle$ of a qubit. To this, we associate two different objects:

  • The first is its density matrix, the operator $\hat \rho = |\psi\rangle \langle \psi|$.
  • The second is the Bloch-sphere vector $\mathbf n = \sum_{i=1}^3 \hat{\mathbf e}_i \, \mathrm{Tr}(\hat \rho \hat \sigma_i)$, where $\hat \sigma_i$ are the Pauli matrices.

Here, it is relatively easy to show that

  • $\mathbf n \in \mathbb R^3$ is a real-valued vector, since the Pauli matrices are hermitian,
  • $\mathbf n$ has unit norm, i.e. $\sum_{i=1}^3 \mathrm{Tr}(\hat \rho \hat \sigma_i)^2 = 1$, and
  • the absolute value of the inner product between two quantum states $|\psi\rangle$ and $|\psi'\rangle$ is in one-to-one correspondence with the dot product of their Bloch sphere vectors, $$ \mathbf n \cdot \mathbf n' = \cos(2\arccos(|\langle\psi|\psi'\rangle|)). $$

Together, these imply that any single-qubit unitary $U$ generates a rigid motion of the unit sphere in three dimensions, i.e. it can be represented in that space by a rotation matrix $R\in\mathrm{SO}(3)$. Moreover, this rotation matrix completely determines the action of the unitary on all density matrices $\hat \rho$, and the only thing that you need to do is to recover $|\psi\rangle$ from $\hat \rho$, which can be done uniquely but only up to a global phase factor of $e^{i\alpha}$.

But that does not mean that you can write $U = e^{i\alpha} R(\theta)$, which, as we saw above, is impossible.

  • $\begingroup$ I think you mis-interpret the question (which indeed is not very precise). If $R_{\vec n}(\theta) = \exp[i\theta\,\vec n\cdot\vec\sigma]$ all is good. This is certainly how I read the question right from the beginning. And I would probably term this a "rotation about $\vec n$ by $\theta$ ". $\endgroup$ – Norbert Schuch May 24 at 8:39
  • $\begingroup$ @NorbertSchuch I think you attribute far too much understanding of the OP. S/he is concerned with the meaning of a global $e^{i\alpha}$ phase; I don't think it's reasonable to assume that s/he understands in what sense $e^{i\theta\vec n\cdot\vec\sigma}$ is "a rotation about $\vec n$ by $\theta$", given that $\vec n$ does not live in the space ($\mathbb C^2$) on which the two-by-two unitary acts. $\endgroup$ – Emilio Pisanty May 24 at 10:05
  • $\begingroup$ I agree, if the OP would understand, they wouldn't ask. But if you read "I know" as "I have read somewhere", then I disagree that "The qubit gate form as you have written is incorrect". This formula as it is could be pretty much be written in a quantum information textbook (or my lecture notes). (Differently speaking, your claim that this is wrong also attributes far too much understanding of the OP as to what $R_n(\theta)$ is - and assumes that the $n$ subscript has no meaning.) $\endgroup$ – Norbert Schuch May 24 at 11:24
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    $\begingroup$ @NorbertSchuch That may be. It's pretty pointless to speculate without OP's input about what they did or did not mean, at this stage. $\endgroup$ – Emilio Pisanty May 24 at 11:31
  • $\begingroup$ Thank all you guys for your help. Sorry my description was not a clear one. What I mean by "a rotation 𝜃 about an arbitrary axis" was "rotate about axis n in the Bloch Sphere by $\theta$", so think @IrisCong 's post could be enough to answer my question. I'm not a Math or Physics student and not even learning Quantum Computing systematically so might ask stupid questions sometimes :( $\endgroup$ – Yu Zi Jun 4 at 12:16

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