Qubit Operations: Finding U Gates & Probabilities

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The discussion centers on the calculation of quantum gate operations and their corresponding matrices, specifically U gates like U_H, U_X, U_Y, and others. There is a discrepancy noted between calculated values for the X, Y, and Z gates, which include an extra factor of i compared to standard references, possibly due to a global phase factor. The S and T gate solutions also differ from expected values, raising questions about convention and phase factors in quantum mechanics. Additionally, the probability amplitude calculations for various states are discussed, with a suggestion that there may be a typo in the problem regarding the states to be achieved from the initial state |0>. The conversation emphasizes the importance of understanding complex phases in quantum state definitions.
EightBells
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Homework Statement
a) A single qubit operation on qubit states can be defined as the unitary
transformation
U_α = exp(i*θ_α*n_α*σ/2) = cos(θ_α/2)σ_0 + i*n_α*σ *sin(θ_α/2), (1)
corresponding to the rotation around unit vector n_α on angle θ_α, σ = {σ_x, σ_y, σ_z} are Pauli
matrices, and σ_0 is the identity matrix. Write down operators U_α in the basis of |0 > and
|1> for the standard single qubit gates
α = {H, X, Y, Z, S, T, Rx, Ry}. (2)
Also, define corresponding n_α and θ_α for each of these gates. Note that these gates may
differ by a global phase factor compared to the operators presented elsewhere.
b) In the basis of |0> and |1> states, write down probability amplitudes in the vector
form for the following six states:
|0>, |1>, |±x> =(|0> ± |1>)/√2, |±y> =(|0> ± i|1>)/√2. (3)
If the qubit is initialized in state |0>, how do you obtain all states in Eq. (2) using a
sequence of one or more standard gates U_α or their inverse from Eq.(2)?
Relevant Equations
the unitary transformation, given in equation 1 of the homework statement
Part a:
GateHXYZSTR_xR_y
Thetapipipipipi/2pi/4pi/2pi/2
n_alpha(1/sqrt(2))*(1,0,1)(1,0,0)(0,1,0)(0,0,1)(0,0,1)(0,0,1)(1,0,0)(0,1,0)

Using the info from the table and equation 1, I find:

U_H=(i/sqrt(2))*[1,1;1,-1]
U_X=i*[0,1;1,0]
U_Y=i*[0,-i;i,0]
U_Z=i*[1,0;0,-1]
U_S=[exp(i*pi/4),0;0,exp(-i*pi/4)]
U_T=[exp(i*pi/8,0;0,exp(-i*pi/8)]
U_Rx=(1/sqrt(2))*[1,i;i,1]
U_Ry=(1/sqrt(2))*[1,1;-1,1]

My problem is, the H, Rx, and Ry gates agree with what I can find online (i.e. on IBM's Quantum Experience page). The X, Y, and Z gates I calculated differ by a factor of i from what I find online. The S and T gate solutions I find online are S=[1,0;0,i] and T=[1,0;0,exp(i*pi/4)]. I figured the S and T gate solutions could be different due to the "global phase factor" mentioned in the homework question, but I'm not sure, and I'm not sure where this factor comes from. I'm also wondering why I'm finding an extra i in my X, Y, and Z gates, or is it that U_X, U_Y, and U_Z differ from the actual X, Y, and Z gates?

Part b:

I know the probability of a particle in state |b> to be found in state |a> is P=|<a|b>|^2, so is what I'm looking for, the probability amplitude, just <a|b>, where <a| are the states |0>, |1>, |±x>, |±y> from eqn. 3, and |b> are the basis states |0> and |1>? If so, I find:

<0|0>=1 <0|1>= 0
<1|0>=0 <1|1>= 1
<+x|0>=1/sqrt(2) <+x|1>=1/sqrt(2)
<-x|0>=1/sqrt(2) <-x|1>=-1/sqrt(2)
<+y|0>=1/sqrt(2) <+y|1>=-i/sqrt(2)
<-y|0>=1/sqrt(2) <-y|1>=i/sqrt(2)

For the last part of b, I am assuming there is a typo and I'm supposed to find which gates to apply to initial state |0> to achieve the six states given in equation 3 (though the problem says the states given in equation 2 and this doesn't seem possible since equation 2 gives the quantum gates, but correct me if I'm wrong please). Therefore:

Z|0>=|0> [1,0;0,-1]*[1;0]=[1;0]
X|0>=|1> [0,1;1,0]*[1;0]=[0;1]
(Ry)^(-1)|0>=|+x> (1/sqrt(2))*[1,-1;1,1]*[1;0]=(1/sqrt(2))*[1;1]
Ry|0>=|-x> (1/sqrt(2))*[1,1;-1,1]*[1;0]=(1/sqrt(2))*[1;-1]
Rx|0>=|+y> (1/sqrt(2))*[1,i;i,1]*[1;0]=(1/sqrt(2))*[1;i]
(Rx)^(-1)|0>=|-y> (1/sqrt(2))*[1,-i;-i,1]*[1;0]=(1/sqrt(2))*[1;-i]
 

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EightBells said:
My problem is, the H, Rx, and Ry gates agree with what I can find online (i.e. on IBM's Quantum Experience page). The X, Y, and Z gates I calculated differ by a factor of i from what I find online. The S and T gate solutions I find online are S=[1,0;0,i] and T=[1,0;0,exp(i*pi/4)]. I figured the S and T gate solutions could be different due to the "global phase factor" mentioned in the homework question, but I'm not sure, and I'm not sure where this factor comes from. I'm also wondering why I'm finding an extra i in my X, Y, and Z gates, or is it that U_X, U_Y, and U_Z differ from the actual X, Y, and Z gates?
The additional I is indeed simply a complex phase factor, corresponding to a multiplication by ##e^{i \phi}## with ##\phi = \pi/2##.

It means indeed that the ##U_\alpha## gates differ from the usual gates. But this is also simply a matter of convention, as a quantum state is defined up to an arbitrary global complex phase.

EightBells said:
For the last part of b, I am assuming there is a typo and I'm supposed to find which gates to apply to initial state |0> to achieve the six states given in equation 3 (though the problem says the states given in equation 2 and this doesn't seem possible since equation 2 gives the quantum gates, but correct me if I'm wrong please).
I think it is indeed a typo.

EightBells said:
Therefore:

Z|0>=|0> [1,0;0,-1]*[1;0]=[1;0]
X|0>=|1> [0,1;1,0]*[1;0]=[0;1]
(Ry)^(-1)|0>=|+x> (1/sqrt(2))*[1,-1;1,1]*[1;0]=(1/sqrt(2))*[1;1]
Ry|0>=|-x> (1/sqrt(2))*[1,1;-1,1]*[1;0]=(1/sqrt(2))*[1;-1]
Rx|0>=|+y> (1/sqrt(2))*[1,i;i,1]*[1;0]=(1/sqrt(2))*[1;i]
(Rx)^(-1)|0>=|-y> (1/sqrt(2))*[1,-i;-i,1]*[1;0]=(1/sqrt(2))*[1;-i]
You need to write the transformations in terms of ##U_\alpha##, which you haven't done. And here you have to pay attention to the complex phase in order to recover exactly the states given in eq. (3).
 
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