# How to translate a 4-qubit Grover's algorithm circuit into a state Matrix?

Grover's algorithm circuit may be implemented as follows:

(from here)

It is shown very elegantly by @MartinVesely (How to interpret a 4 qubit quantum circuit as a matrix?) how to translate a 4 qubit circuit into its complete matrix representation.

Building on the previous question, I would like translate the full Grover's ciruit into a single state matrix. I attempted as follows:

Op 1: $$H \otimes H \otimes H \otimes H$$

Op 2: $$X \otimes X \otimes X \otimes X$$

Op 3: ?

Op 4: $$CNOT \otimes I \otimes I$$

Op 5: ?

...

Does anyone know how to correctly translate operation 3 and operation 5 in particular into its state matrix, please?

• Could you clarify which two qubit gates you are using? I see you are using CX/CNOT. Is the other two qubit gate a controlled Rz or a controlled phase rotation? Jan 3 at 21:17
• @CallumMacpherson the diagram is not mine either. Link to the source is given above. I am not clear myself what gate is Operation 3 using? And how to "skip lines" using some formula similar to what's mentioned in quantumcomputing.stackexchange.com/questions/9614/…? Jan 3 at 21:20
• Ah okay thanks, looks like its a U1 gate. Can give an answer now. Jan 3 at 21:22

What you have looks like a controlled-U1 gate, which for two qubits has the form:

$$\begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & e^{i \theta} \end{bmatrix}$$ Notice that the cu1 gate doesn't distinguish between the control qubit and the target qubit. Both give the same result.

The matrix for a cu1 embedded into multiple qubits is a more complicated diagonal matrix, with several of the diagonal entries having the values $$e^{i \theta}$$

You can use

qc = QuantumCircuit(4)
qc.cu1(.79, 1, 3)

array_to_latex(Operator(qc).data, max_size=100)


to see one example.

• thank you. If this controlled-U1 gate "skips over" other lines while connecting its 2 lines, how should the 16x16 matrix be constructed for say operation 3? I am trying to do all the steps using matrix algebra. Jan 4 at 0:23
• It's probably worth noting that the blog article in the question gives a full listing for the code for the circuit (and, observe that the angle is $\pi/4$ rather than .79). So you could just use the above method to directly compute the unitary for the whole circuit, as well as identify what the individual parts are. Jan 4 at 7:30

You can figure out 3) and 5) using the formula given by @MartinVesely in the page you linked to.

As was discussed in the post or the matrix of a CU gate with a control and a target separated by $$k$$ qubits is as follows.

$$$$CU_k = \begin{pmatrix} I_\frac{N}{2}& O_\frac{N}{2} \\ O_\frac{N}{2} &I_\frac{N}{4} \otimes U \end{pmatrix}$$$$

Here $$N=2^{k+2}$$, $$O$$ is the all-zero matrix and $$I$$ is the identity. The gates in your diagram seem to be Controlled U1 gates where U1 is represented by the following unitary.

$$$$U1(\lambda) = \begin{pmatrix} 1& 0 \\ 0 &e^{i \lambda} \end{pmatrix}$$$$

Now we see there are two qubits between the control and the target qubit of the CU1 gate. Therefore $$k=2$$ and $$N=16$$). Substituting in $$N=16$$ and $$U=U1$$ we get the following... $$$$CU1_{k=2} = \begin{pmatrix} I_8& O_8 \\ O_8 &I_4 \otimes U1 \end{pmatrix}$$$$.

If you have the patience you can write down the full $$16 \times 16$$ matrix from this by expanding the four entries. Maybe you could factorise this matrix into a tensor product of $$2 \times 2$$ matrices. I'd have to think about it more as I'm not sure there is a nice form.

Edit: Similarly to Frank I calculated the unitary (I did the first CU1 gate in the circuit). Wouldn't want to do this by hand.

I did this symbolically with pytket. Note the factor of $$\pi$$ due to different conventions. $$(\lambda = \pi \theta)$$

from pytket import Circuit, OpType
from pytket.utils import circuit_to_symbolic_unitary
from sympy import Symbol

theta = Symbol("theta") #theta = 1 / 4 in your example
circ = Circuit(4).add_gate(OpType.CU1, [theta], [0, 3])
circuit_to_symbolic_unitary(circ)

• thank you. yes, the same formula should work for any gate! Jan 4 at 0:37