How do we create an arbitrary vector of the following form in Qiskit?

Say, we want to initialize a qubit with a vector, $$ \vert \psi \rangle = \frac{1+i}{\sqrt{3}}\vert 0 \rangle - \frac{i}{\sqrt{3}}\vert 1 \rangle $$

If I put it in the form $$ \vert \psi \rangle = \frac{1+i}{\sqrt{3}}\vert 0 \rangle - \frac{i}{\sqrt{3}}\vert 1 \rangle = cos(\frac{\theta}{2})|0> + e^{i\phi}sin(\frac{\theta}{2})|1> $$ where
$0 < \theta < \pi$ and $0 < \phi <2\pi$ and then calculate $\theta$ and $\phi$

So, $$ cos(\frac{\theta}{2}) = \frac{1+i}{\sqrt{3}}\\ e^{i\phi}sin(\frac{\theta}{2}) = - \frac{i}{\sqrt{3}} $$ Therefore, $$ \theta = 2 * \arccos{\frac{1+i}{\sqrt{3}}} \\ \phi = i * ln(\frac{- \frac{i}{\sqrt{3}}}{sin(\frac{\theta}{2})}) $$

Now, I am using the following code..

#We create the quantum state manually first
arb_quantum_state = ((1+1.j)/math.sqrt(3))*ket_0 - (1.j/math.sqrt(3))*ket_1

theta = 2*cmath.acos((1+1.j)/cmath.sqrt(3))
print('theta : ',theta)
sinValue = cmath.sin(theta/2)
phase = -1*(1.j/cmath.sqrt(3))/sinValue
phi = cmath.log(phase)/1.j
print('phi : ',phi)

# Use these theta and phi to create the circuit
circ = QuantumCircuit(1,1)
#Verify why complex values are not allowed

results = execute(circ, backend=Aer.get_backend('statevector_simulator')).result()
quantum_state = results.get_statevector(circ, decimals=3)
print (quantum_state)

The above code creates the gate alright, but the execute function is returning the following error,

TypeError: can't convert complex to float

However, if I use just the real values of theta and phi, then the execute function returns a state vector, which is different than the one it should be.


1 Answer 1


The problem is that you're trying to equate $\cos(\theta/2)$ (a real number) with $(1+i)/\sqrt{3}$ (a complex number). The way around this is you need to take into account a global phase $\gamma$ such that $$ e^{i\gamma}|\psi\rangle=\cos\frac{\theta}{2}|0\rangle+e^{i\phi}\sin\frac{\theta}{2}|1\rangle. $$ To do this, it helps to express your initial state as complex exponentials:

As a first step your state can be reduced to $$ |\psi\rangle=\sqrt{\frac{2}{3}}\left(\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}i\right)|0\rangle+\frac{1}{\sqrt{3}}\left(0 - i*1\right)|1\rangle $$ Then to, $$ |\psi\rangle=\sqrt{\frac{2}{3}}e^{i\pi/4}|0\rangle+\frac{1}{\sqrt{3}}e^{i\frac{3}{2}\pi}|1\rangle $$ so that we can rewrite it as $$ |\psi\rangle=e^{i\pi/4}\left(\sqrt{\frac{2}{3}}|0\rangle+\frac{1}{\sqrt{3}}e^{i\frac{5}{4}\pi}|1\rangle\right). $$ Now you can easily see that $$ \cos\frac{\theta}{2}=\sqrt{\frac{2}{3}},\qquad \phi=\frac{5\pi}{4},\qquad \gamma=\frac{\pi}{4}. $$

  • $\begingroup$ Many thanks DaftWullie $\endgroup$ Commented Apr 21, 2020 at 3:58

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