Qubit Operations: Finding U Gates & Probabilities

In summary, the gates H, Rx, and Ry agree with what can be found online, but the X, Y, and Z gates differ by a factor of i due to a global phase factor. The S and T gates also differ from the usual gates due to the phase factor. The last part of the conversation has a typo and the question should be finding which gates to apply to the initial state |0> to achieve the six states given in equation 3. The transformations can be written in terms of the gates U_alpha, taking into account the complex phase factor to recover the exact states given in equation 3.
  • #1
EightBells
11
1
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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  • #2
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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1. What is a qubit and how does it differ from a classical bit?

A qubit, or quantum bit, is the basic unit of quantum information in quantum computing. It is similar to a classical bit in that it can represent either a 0 or 1, but unlike a classical bit, it can also exist in a superposition of both states at the same time. This allows for more complex and powerful calculations to be performed using qubits.

2. What are U gates and how are they used in qubit operations?

U gates, or unitary gates, are quantum logic gates that operate on qubits. They are used to manipulate the state of a qubit, such as changing its probability of being in a certain state or rotating its state in a specific direction. U gates are an essential tool in performing calculations and algorithms in quantum computing.

3. How do you find the probabilities of a qubit in a certain state?

The probabilities of a qubit being in a certain state can be found using the Born rule, which states that the probability of measuring a qubit in a particular state is equal to the square of the magnitude of the amplitude of that state. This can be calculated using the quantum state vector of the qubit.

4. What is the significance of finding U gates in qubit operations?

Finding U gates in qubit operations is significant because it allows for the manipulation and control of qubits, which is essential for performing calculations and algorithms in quantum computing. U gates also play a crucial role in quantum error correction, which is necessary for maintaining the integrity of quantum computations.

5. Can U gates be used to perform any operation on qubits?

No, U gates cannot be used to perform any operation on qubits. They are limited to specific operations, such as changing the state of a qubit or rotating its state. Other types of quantum logic gates, such as CNOT gates, are needed to perform more complex operations on qubits.

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