Implementation of a unitary operator scaled by a factor

Question

Is it possible to implement a unitary operator scaled by a factor on a quantum computer?

Let's say the unitary operator is $U$:

$$U=\begin{bmatrix} u_0 & u_1 \\ u_2 & u_3 \end{bmatrix}\newcommand{\ket}[1]{|#1\rangle}$$

If I apply $fU$ with $f\in\mathbb{C}$ to a 1-qubit state $\ket{\psi}$

$$\ket{\psi} = \alpha \ket{0} + \beta\ket{1} = \begin{bmatrix}\alpha \\ \beta\end{bmatrix}$$

the result $\ket{\psi'}$ is just scaled by $f$

$$\ket{\psi'} = fU\ket{\psi} = f\begin{bmatrix}\alpha u_0 + \beta u_1 \\ \alpha u_2 + \beta u_3 \end{bmatrix}$$

Isn't $\ket{\psi'}$ equal to $U\ket{\psi}$ up to a global phase and $fU$ can, therefore, be implemented by $U$ without the factor $f$?

Answer 1

That is true. However, if you want a controlled version of the unitary, then $CfU$ is not necessarily equivalent to $CU$.

Take for example 2 qubit system starting at the $|+0\rangle$ state. Applying $C_0U_1$ with $U=X$ will result with the state $\frac{1}{\sqrt2}(|00\rangle+|11\rangle) = |\Phi^+\rangle$ while applying $C_0(fU)_1$ with $f=-1$ will result with the state $\frac{1}{\sqrt2}(|00\rangle-|11\rangle) = |\Phi^-\rangle$, which are two orthogonal bell states.

This can be demonstrated using simulation. See the following code using the classiq platform:

from classiq import *

@qfunc
def main(x: Output[QBit], y: Output[QBit]):
    prepare_state([0.5, 0.5], 0, x)
    allocate(1, y)
    control(x, lambda: X(y))
    
qmod = create_model(main)
qprog = synthesize(qmod)
show(qprog)

which results after execution (using the Execute button in the IDE that pops up, and choosing state vector simulator) on a state vector simulator with the following state:

while this code with $C_0(fU)_1$ will result with an orthogonal state:

@qfunc
def main(x: Output[QBit], y: Output[QBit]):
    prepare_state([0.5, 0.5], 0, x)
    allocate(1, y)
    control(x, lambda: (U(0, 0, 0, np.pi, y), X(y)))
    
qmod = create_model(main)
qprog = synthesize(qmod)
show(qprog)

(Disclaimer - I am a Classiq employee)