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From the qiskit documentation (here), a general form of a single qubit unitary is defined as $$ U(\theta, \phi, \lambda) = \begin{pmatrix} \cos\left(\frac{\theta}{2}\right) & -e^{i\lambda} \sin\left(\frac{\theta}{2}\right) \\ e^{i\phi} \sin\left(\frac{\theta}{2}\right) & e^{i(\lambda + \phi)} \cos\left(\frac{\theta}{2}\right) \end{pmatrix}. $$ Where $0≤𝜃≤𝜋, 0≤𝜙<2𝜋, \text{and} \ 0≤𝜆<2𝜋$. However, when I tried to put some arguments out of the range, the gate still operates. For example, if I set $\theta = -1,\phi=8,\lambda=7$,

simulator = Aer.get_backend('statevector_simulator')
quancir = QuantumCircuit(1)  
quancir.u3(-1,8,7,0)
results = execute(quancir, simulator).result()
resvec = results.get_statevector(quancir)
bloch_sphere([conv(resvec)])

I can still visualize how the $U_3$ gate operates on the initial state $|0\rangle$, and plot the final vector: enter image description here

I'm wondering if my arguments aren't in the expected range, like in this case, what really happened to the $U_3$ gate? Am I still getting the vector I want, or do I need to convert the arguments myself to make sure the output vector is correct? Thanks:)

Update: I tried to take the mod of those parameters but it looks like the output vector is different (points toward the opposite direction):

quancir = QuantumCircuit(1)  
T = float(-1%pi)
P = float(8%(2*pi))
L = float(7%(2*pi))
quancir.u3(T,P,L,0)
results = execute(quancir, simulator).result()
resvec = results.get_statevector(quancir)
bloch_sphere([conv(resvec)])

enter image description here

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    $\begingroup$ These are periodic functions. So for instance, $\sin(3) = \sin(3 + 2\pi)$. $\endgroup$
    – KAJ226
    Mar 25 at 21:36
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    $\begingroup$ quancir = QuantumCircuit(1) quancir.u3(-1+2*pi,8 - 2*pi ,7 - 2*pi,0) resvec =quancir.statevector() bloch_sphere([resvec]) Try this... $\endgroup$
    – KAJ226
    Mar 25 at 22:10
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    $\begingroup$ It is the same as $-1\%(2\pi)$. Note in the function, the factor is $\theta/2$. $\endgroup$
    – KAJ226
    Mar 25 at 22:23
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    $\begingroup$ It should be. If you plug in different parameters then you can see it. $\endgroup$
    – KAJ226
    Mar 25 at 22:30
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    $\begingroup$ @KAJ226 Thank you so much:) $\endgroup$
    – ZR-
    Mar 25 at 22:40
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In this case all input parameters will be mod $4\pi$, $2\pi$, and $2\pi$ for $\theta$, $\phi$, and $\lambda$ respectively. You will obtain the same vector that you would have received if you took the mod of these parameters yourself.

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  • $\begingroup$ Thanks for the answer! I tried that but it turns out that the vector is different (I just updated my question). $\endgroup$
    – ZR-
    Mar 25 at 21:37
  • $\begingroup$ You're right, I have corrected my answer. $\theta$ is $4\pi$ periodic due to its factor of $1/2$ as seen in its matrix representation. $\endgroup$
    – user47787
    Mar 28 at 15:57

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