# Different statevectors with different quantity of complex numbers for same single-qubit states in Qiskit

I have several questions, but before ask, I want to write some theoretical. As we know, we can represent any single-qubit quantum state by the next representation:

$$|\psi\rangle=c_0|0\rangle+c_1|1\rangle,$$ using two numbers, $$c_0$$ and $$c_1$$, called probability amlitudes.

We can put these probability amplitudes to vector-column:

$$\begin{bmatrix} c_0 \\ c_1 \\ \end{bmatrix}$$

In other side, we can represent any single-qubit quantum state next:

$$\theta$$ and $$\phi$$ represents next angles in Bloch sphere:

As we can see, enough only one complex number to represent single-qubit quantum state, using $$|0\rangle$$ and $$|1\rangle$$ base. In this case $$c_0 = \cos{\frac{\theta}{2}}$$ and $$c_1 = e^{i\phi}\sin{\frac{\theta}{2}}$$. By watching Bloch Sphere and using probability amplitudes we can write 6 the most usual quantum states:

$$|0\rangle = \cos{\frac{0}{2}}|0\rangle + e^{i0}\sin{\frac{0}{2}}|1\rangle = 1|0\rangle+0|1\rangle=\begin{bmatrix} 1 \\ 0 \\ \end{bmatrix}$$

$$|1\rangle = \cos{\frac{\pi}{2}}|0\rangle + e^{i0}\sin{\frac{\pi}{2}}|1\rangle = 0|0\rangle+1|1\rangle=\begin{bmatrix} 0 \\ 1 \\ \end{bmatrix}$$

$$|+\rangle = \cos{\frac{\pi}{4}}|0\rangle + e^{i0}\sin{\frac{\pi}{4}}|1\rangle = \frac{1}{\sqrt{2}}|0\rangle+\frac{1}{\sqrt{2}}|1\rangle=\begin{bmatrix} \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} \\ \end{bmatrix}$$

$$|-\rangle = \cos{\frac{\pi}{4}}|0\rangle + e^{i\pi}\sin{\frac{\pi}{4}}|1\rangle = \frac{1}{\sqrt{2}}|0\rangle+(-1)*\frac{1}{\sqrt{2}}|1\rangle=\frac{1}{\sqrt{2}}|0\rangle-\frac{1}{\sqrt{2}}|1\rangle=\begin{bmatrix} \frac{1}{\sqrt{2}} \\ -\frac{1}{\sqrt{2}} \\ \end{bmatrix}$$

$$|+i\rangle = \cos{\frac{\pi}{4}}|0\rangle + e^{i\frac{\pi}{2}}\sin{\frac{\pi}{4}}|1\rangle = \frac{1}{\sqrt{2}}|0\rangle+i*\frac{1}{\sqrt{2}}|1\rangle=\begin{bmatrix} \frac{1}{\sqrt{2}} \\ \frac{i}{\sqrt{2}} \\ \end{bmatrix}$$

$$|-i\rangle = \cos{\frac{\pi}{4}}|0\rangle + e^{i\frac{3\pi}{2}}\sin{\frac{\pi}{4}}|1\rangle = \frac{1}{\sqrt{2}}|0\rangle+(-i)*\frac{1}{\sqrt{2}}|1\rangle=\frac{1}{\sqrt{2}}|0\rangle-i*\frac{1}{\sqrt{2}}|1\rangle=\begin{bmatrix} \frac{1}{\sqrt{2}} \\ -\frac{i}{\sqrt{2}} \\ \end{bmatrix}$$

Let me now use Qiskit. We can use U3 universal matrix to create any single-qubit quantum state. For example, if I use

qc = QuantumCircuit(1)
qc.u3(pi/2,pi/2,0,0)


we will have $$|+i\rangle$$ state with next statevector:

(1)

However, if we do next code:

qc = QuantumCircuit(1)
qc.h(0)
qc.rz(pi/2,0)


we will have next statevector:

(2)

As I understand, here is used global phase. But how can I connect these two statevectors with each other? And why it's difference between each other? Honestly, I had experience only with single-qubit representation, when we use only one complex number, and $$i$$ can be only in second coefficient of Jones vector. But when I see two complex numbers - I am little bit confused.

Additionally, if we start from $$|+\rangle$$ state and rotate state around Z for $$\frac{\pi}{4}$$ angle, we can do it by using "rc" gate:

qc = QuantumCircuit(1)
qc.h(0)
qc.rz(pi/4,0)


and we will have next result:

Or we will use next gate:

qc = QuantumCircuit(1)
qc.h(0)
qc.u1(pi/4,0)


and we will have next result:

Again, in first case we will have two complecs numbers, in second case only one, but states are same.

Could you, please, explain these differences?

Thank you!

As you say, the difference is in the global phase. Let me explain using the first of your examples, $$\left[\begin{array}{cc} \frac{1-i}{2} & \frac{1+i}{2} \end{array}\right].$$ Mathematically, this is the same as $$\left[\begin{array}{cc} \frac{e^{-i\pi/4}}{\sqrt{2}} & \frac{e^{i\pi/4}}{\sqrt{2}} \end{array}\right]=\left[\begin{array}{cc} e^{-i\pi/4}\times\frac{1}{\sqrt{2}} & e^{-i\pi/4}\frac{e^{i\pi/2}}{\sqrt{2}} \end{array}\right]=e^{-i\pi/4}\times\left[\begin{array}{cc} \frac{1}{\sqrt{2}} & \frac{e^{i\pi/2}}{\sqrt{2}} \end{array}\right]=e^{-i\pi/4}\times\left[\begin{array}{cc} \frac{1}{\sqrt{2}} & \frac{i}{\sqrt{2}} \end{array}\right].$$ So, if you neglect the global phase factor (any factor multiplying the whole state, with modulus 1), you see that $$\left[\begin{array}{cc} \frac{1-i}{2} & \frac{1+i}{2} \end{array}\right].$$ and $$\left[\begin{array}{cc} \frac{1}{\sqrt{2}} & \frac{i}{\sqrt{2}} \end{array}\right]$$ are the same thing.