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

cianfa72
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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}##.
 
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PeterDonis said:
@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.
 
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cianfa72 said:
No, I haven't.
Then you should. That will be much more helpful to you than having someone just tell you.
 
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.
 
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cianfa72 said:
an unitary vector
I'm not sure what you mean. Unitarity is a property of operators, not vectors.

cianfa72 said:
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.
 
cianfa72 said:
a unit vector with zero global phase
What would that mean? Relative to what is "zero global phase" being evaluated?
 
PeterDonis said:
I'm not sure what you mean. Unitarity is a property of operators, not vectors.
Sorry, I meant a unit vector (norm 1).

PeterDonis said:
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.
 
cianfa72 said:
Suppose there was an "hidden phase" inside ##S \in SU(2)##, then ##S = e^{j\theta} S', S' \in SU(2)##.
Why?
 
  • #10
cianfa72 said:
##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.
 
  • #11
cianfa72 said:
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)##.
 
  • #12
Ah yes, you are right :rolleyes: . 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.
 
  • #13
martinbn said:
##S## is an element of ##SU(2)##, hence it is an element of ##U(2)##.
You didn't read the next part:

PeterDonis said:
an element of ##U(2)## that is not in ##SU(2)##,
 
  • #14
cianfa72 said:
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.
 
  • #15
PeterDonis said:
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##.
 
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  • #16
PeterDonis said:
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 $$ e^{-i\frac {\pi} {4}} \begin{bmatrix} \frac 1 {\sqrt{2}} \\ \frac i {\sqrt {2}} \end{bmatrix}$$
 
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