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I'm new to the IBM Q Experience. In the Composer, I built a circuit consisting of two Hadamard Gates and an Ry:

OPENQASM 2.0;
include "qelib1.inc";
qreg q[5];
creg c[5];

h q[0];
h q[0];
ry(pi/2) q[0];
measure q[0] -> c[0];

The results returned are 50/50 probability. However, when I apply the matrices in a spreadsheet, I get a $\frac{28}{72}$ result. (See image below.) For the Hadamard matrices, I'm using one divided by the square root of two for the corresponding a, b, and c elements and minus one divided by the square root of two for the d element. For the Ry matrix elements a,b c, and d, I have $\cos\big(\frac{pi/2}{2}\big)$, $-\sin\big(\frac{pi/2}{2}\big)$, $\sin\big(\frac{pi/2}{2}\big)$ and $\cos\big(\frac{pi/2}{2}\big)$, where $\pi=180^{\text{o}}$. The output squared in the spreadsheet gives me 28/72 probability. I hate to be a pest with such a low level question, but where in the matrices am I going wrong?

For $\pi$ I used 3.14 instead of 180 degrees, and it came out 50/50.

enter image description here

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Here are the gates:

\begin{equation} H = \frac{1}{\sqrt{2}}\begin{pmatrix} 1 & 1 \\1&-1 \end{pmatrix} \qquad R_y(\theta) = \begin{pmatrix} \cos(\theta/2) & -\sin(\theta/2) \\\sin(\theta/2) & \cos(\theta/2) \end{pmatrix} \end{equation}

Matrix multiplication (note that $HH = I$):

$$U = R_y(\pi/2) H H = R_y(\pi/2) = \frac{1}{\sqrt{2}}\begin{pmatrix} 1 & -1 \\ 1 & 1 \end{pmatrix}$$

If we will apply this $U$ to the $|0\rangle$ state, we will obtain:

$$U |0\rangle = \frac{1}{\sqrt{2}}\begin{pmatrix} 1 & -1 \\ 1 & 1 \end{pmatrix} \begin{pmatrix} 1 \\ 0 \end{pmatrix} = \frac{1}{\sqrt{2}}\begin{pmatrix} 1 \\ 1 \end{pmatrix} = \frac{1}{\sqrt{2}} (|0\rangle + |1\rangle)$$

So with $50\%$ probability we will measure $|0\rangle$ (or $|1\rangle$) state.

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    $\begingroup$ Cool. Thank you. $\endgroup$ – James Arneberg May 17 '20 at 22:04
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For pi I used 3.14 instead of 180 degrees, and it came out 50/50.

Sounds like you answered your own question. Sine and cosine functions would be expecting radians by default, so using $\pi=3.14 \cdots$ is correct.

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    $\begingroup$ Thank you, sir. $\endgroup$ – James Arneberg May 17 '20 at 22:02

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