Action of an Hermitian unitary operator member of SU(2)

[cianfa72]

TL;DR

About the action of an Hermitian unitary operator in $SU(2)$ on the state space of QM system.

QM uses separable Hilbert spaces as model to represent quantum system's states. Take for instance a 1/2-spin particle: its quantum pure state is represented by a ray in the abstract Hilbert space $\mathcal H_2$ of dimension 2.

Take an observable represented by an Hermitian unitary operator $S \in SU(2)$. The particle quantum pure state $\ket{\psi}$ is represented by a point on the Bloch sphere $$\ket{\psi} = \cos {(\theta /2)}\ket{\uparrow} + e^{j\phi}\sin {(\theta /2)} \ket{\downarrow}$$
Now consider the action of $S$ on it, i.e. $S\ket{\psi}$. Is this a unit vector with zero global phase ? I mean a vector in the form above without any further explicit multiplication by some global phase $e^{j\alpha}$.

 

[PeterDonis]

@cianfa72 have you tried computing the answer for various members of SU(2)?

 

[cianfa72]

@cianfa72 have you tried computing the answer for various members of SU(2)?

No, I haven't. Maybe the fact that in any orthonormal basis the representative of the operator $S \in SU(2)$ , say $\hat {S}$, has det $\hat {S} = +1$ is relevant for that.

$SU(2)$ elements don't add a global phase to the the state they act on.

 

[PeterDonis]

No, I haven't.

Then you should. That will be much more helpful to you than having someone just tell you.

 

[cianfa72]

As far as I can tell, any element $T \in U(2)$ is given as $T = e^{j\phi} S$ for some $\phi$ and $S \in SU(2)$. For any $\ket{\psi}$ on the Bloch sphere, $\ket{{\psi}^{'}} =T\ket{\psi}$ is an unitary vector, hence $\ket{{\psi}^{'}} =e^{j\phi} S\ket{\psi}$. Therefore factorizing out $e ^{j\phi}$ we get the result, namely $S\ket{\psi}$ is always in that form (no global phase) without any further need to explicitly multiply it by some global phase.

 

[PeterDonis]

an unitary vector

I'm not sure what you mean. Unitarity is a property of operators, not vectors.

factorizing out $e ^{j\phi}$ we get the result, namely $S\ket{\psi}$ is always in that form (no global phase)

No, that doesn't follow. Factorizing out one phase doesn't mean there can't be another one hidden inside $S$ itself.

 

[PeterDonis]

a unit vector with zero global phase

What would that mean? Relative to what is "zero global phase" being evaluated?

 

[cianfa72]

I'm not sure what you mean. Unitarity is a property of operators, not vectors.

Sorry, I meant a unit vector (norm 1).

No, that doesn't follow. Factorizing out one phase doesn't mean there can't be another one hidden inside $S$ itself.

Suppose there was an "hidden phase" inside $S \in SU(2)$, then $S = e^{j\theta} S', S' \in SU(2)$. Hence $S$ would be in the form of an element of $U(2)$ that is not.

 

[PeterDonis]

Suppose there was an "hidden phase" inside $S \in SU(2)$, then $S = e^{j\theta} S', S' \in SU(2)$.

Why?

 

[PeterDonis]

$S = e^{j\theta} S', S' \in SU(2)$. Hence $S$ would be in the form of an element of $U(2)$

Why? All you know at this point is that, for any element $T$ of $U(2)$, $T = e^{i \theta} S$, where $S$ is an element of $SU(2)$. But you haven't shown that any operator of the form $e^{i \theta} S$, where $S$ is an element of $SU(2)$, must be an element of $U(2)$ that is not in $SU(2)$, which is what you're assuming here.

 

[martinbn]

Sorry, I meant a unit vector (norm 1).


Suppose there was an "hidden phase" inside $S \in SU(2)$, then $S = e^{j\theta} S', S' \in SU(2)$. Hence $S$ would be in the form of an element of $U(2)$ that is not.

$S$ is an element of $SU(2)$, hence it is an element of $U(2)$.

 

[cianfa72]

Ah yes, you are right . So, which is the difference in applying an element of SU(2) vs an element of U(2) to the state $$\ket{\psi} = \cos {(\theta /2)}\ket{\uparrow} + e^{j\phi}\sin {(\theta /2)} \ket{\downarrow}$$
Both are unitary operators hence the norm 1 of the transformed vector doesn't change.

 

[PeterDonis]

$S$ is an element of $SU(2)$, hence it is an element of $U(2)$.

You didn't read the next part:

an element of $U(2)$ that is not in $SU(2)$,

 

[PeterDonis]

which is the difference in applying an element of SU(2) vs an element of U(2) to the state $$\ket{\psi} = \cos {(\theta /2)}\ket{\uparrow} + e^{j\phi}\sin {(\theta /2)} \ket{\downarrow}$$

The best way for you to answer this question is to write down the math yourself and see what it says. You've already chosen a representation for qubits (Bloch sphere points) in the OP of this thread. Write down a general member of $SU(2)$ and a general member of $U(2)$ in the same representation, apply each of them to the representation of the qubit, and see what you get.

 

[cianfa72]

You've already chosen a representation for qubits (Bloch sphere points) in the OP of this thread. Write down a general member of $SU(2)$ and a general member of $U(2)$ in the same representation, apply each of them to the representation of the qubit, and see what you get.

Here my argument: a point on the Bloch sphere $$\ket{\psi} = \cos {(\theta /2)}\ket{\uparrow} + e^{j\phi}\sin {(\theta /2)} \ket{\downarrow}$$ represents a full ray in $\mathbb C^2$, i.e. all vectors $\lambda \ket{\psi}, \lambda \neq 0$ are represented by the same point on it. Restricting to only normalized/unit vectors in $\mathbb C ^2$, the above holds up to a global phase $\lambda = e^{j\alpha}, \alpha \in [0, 2\pi)$.

We said that any element $T \in U(2)$ can be always written in the form $T = e^{j\beta}S, S \in SU(2)$.

Suppose to apply an element of $SU(2)$ to unit vectors in $\mathbb C^2$. From the point of view of their representatives on the Bloch sphere, that action corresponds to rotate the sphere itself.

As before, the action of an element $T$ basically adds a global phase to the action of the $SU(2)$ element entering its representation $T=e^{j\beta}S$. Therefore, since the Bloch sphere representation is up a global phase, it acts the same as the element $S$.

 

[cianfa72]

The best way for you to answer this question is to write down the math yourself and see what it says. You've already chosen a representation for qubits (Bloch sphere points) in the OP of this thread.

Eventually I found a counterexample to my question in the OP. Take the following qubit state given in the abstract $\mathcal H_2$ standard basis $\{ \ket{\uparrow}, \ket{\downarrow} \}$ as $$\begin{bmatrix} \frac {1} {\sqrt{2}} \\ \frac {1} {\sqrt{2}} \end{bmatrix}$$
Now applying to it the $SU(2)$ matrix $$e^{-i\frac {\pi} {4}} \begin {bmatrix} 1 & 0 \\ 0 & i \end{bmatrix}$$ we get $$ e^{-i\frac {\pi} {4}} \begin{bmatrix} \frac 1 {\sqrt{2}} \\ \frac i {\sqrt {2}} \end{bmatrix}$$