5
$\begingroup$

I've been experimenting with quantum circuits and can't quite fathom how the difference between states comes together.

Speaking in terms of simulations using qiskit, the following code yelds the same results:

circuit = QuantumCircuit(1)
circuit.h(0)
state = Statevector.from_instruction(circuit)
display(plot_bloch_multivector(state, title="H", reverse_bits=False))

circuit = QuantumCircuit(1)
circuit.ry(0.5*np.pi,0)
state = Statevector.from_instruction(circuit)
display(plot_bloch_multivector(state, title="Y", reverse_bits=False))

enter image description here

According to this page, the H-Gate is equivalent to the following circuit:

enter image description here

The state vector remains the same, which makes sense as it's only rotating around the x-axis.

Even negating the initial qubit state and make it a $|1\rangle$ does not bring any difference to the table.

So, I went deeper and looked at the maths behind it. Applying the H gate to $|0\rangle$ results in:

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

Now, using the RY-Gate, we can construct a matrix using $\frac{\pi}{2}$. This is where my understanding of the mathematical application stops tho, and I can't quite figure out how to do the rest of the calculation. This is as far as I've come, but I can't quite "translate" the result into a comparable qubit state:

$$ \begin{pmatrix} \cos(\frac{\pi}{4}) & -\sin(\frac{\pi}{4})\\ \sin(\frac{\pi}{4}) & \cos(\frac{\pi}{4}) \end{pmatrix}|0\rangle =\ \begin{pmatrix} 0.707 & -0.707\\ 0.707 & 0.707\end{pmatrix} \begin{pmatrix} 1 \\ 0 \end{pmatrix} =\ \begin{pmatrix} 0.707 \\ 0.707 \end{pmatrix} $$

Reason for this question is that I am trying different circuits to classify IRIS for comparison, and I am seeing much better accuracy when using my basic Y-Rotation based circuit in comparison to qiskits ZZFeature and RealAmplitudes circuit.

Improve this question
$\endgroup$
2
  • $\begingroup$ Interesting that $\ket{0}$ works in the preview window but not in the post itself? $\endgroup$
    – Ricardo
    Oct 24, 2021 at 11:20
  • 1
    $\begingroup$ The issue with ket happens to me sometimes. What is even worse, \ket works usually fine with other SE sites, but not the one where you need them the most, here in QC SE! you can still write them as | and langle or range $\endgroup$
    – user206904
    Oct 24, 2021 at 13:52

2 Answers 2

Reset to default
12
$\begingroup$

While Hadamard gate is defined as $$ H= \frac{1}{\sqrt{2}} \begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix}, $$ $y$-rotation by $\pi/2$ leads to gate $$ Ry(\pi/2)= \frac{1}{\sqrt{2}} \begin{pmatrix} 1 & -1 \\ 1 & 1 \end{pmatrix}. $$ So, there is a difference in position of -1 in the second column. Application of the $X$ gate returns the -1 in $Ry(\pi/2)$ to right place to obtain Hadamard gate.

You can easily see that $H|0\rangle = \frac{1}{\sqrt{2}}(|0\rangle + |1\rangle)$ and $Ry(\pi/2)|0\rangle = \frac{1}{\sqrt{2}}(|0\rangle + |1\rangle)$. Application of $X$ after $Ry$ leads to the same quantum state.

However, while $H|1\rangle = \frac{1}{\sqrt{2}}(|0\rangle - |1\rangle)$, $Ry(\pi/2)|1\rangle = \frac{1}{\sqrt{2}}(-|0\rangle + |1\rangle)$. Application of $X$ after $Ry$ switches states $|0\rangle$ and $|1\rangle$ and leads to $XRy(\pi/2)|1\rangle = \frac{1}{\sqrt{2}}(|0\rangle - |1\rangle)$.

This all means that $XRy(\pi/2)=H$.

Improve this answer
$\endgroup$
7
  • 1
    $\begingroup$ You cannot neglect the $X$ even if you choose to ignore the global phase. Consider that $\sqrt2 H\begin{bmatrix}a\\b\end{bmatrix} = \begin{bmatrix}a+b\\a-b\end{bmatrix}$ and $\sqrt2 R_y(\pi/2)\begin{bmatrix}a\\b\end{bmatrix} = \begin{bmatrix}a-b\\a+b\end{bmatrix}$ which in general differ by more than just the global phase. The reason that it looked like the difference is in global phase only is that $|+\rangle$ and $|-\rangle$ just happen to be the eigenstates of $X$. So the last paragraph could use an update. The rest LGTM. $\endgroup$
    – Adam Zalcman
    Oct 24, 2021 at 18:49
  • $\begingroup$ Could you elaborate on how you went from $RY(\frac{\pi}{2})$ to that matrix? Seems as if I am missing a key step there $\endgroup$
    – Ricardo
    Oct 24, 2021 at 18:51
  • 1
    $\begingroup$ @AdamZalcman: Thanks Adam, you are absolutely right. We can neglect the global phase only in case it is involving in all basis states, i.e. the whole matrix is multiplied by some, generally complex, number. $\endgroup$
    – Martin Vesely
    Oct 24, 2021 at 21:01
  • 1
    $\begingroup$ @Redimo: calculated sinus and cosinus values, then factored out the $\frac{1}{\sqrt{2}}$ and finally multiplied by $X$ from left side. This proves the equivalence of Hadamard gate and $XRy(\pi/2)$. $\endgroup$
    – Martin Vesely
    Oct 25, 2021 at 9:29
  • 1
    $\begingroup$ Oh wow, that went right over my head. Thanks a lot! $\endgroup$
    – Ricardo
    Oct 25, 2021 at 9:42
1
$\begingroup$

Just to emphasise the point on the accepted answer...

The two gates, when applied to a specific state, give the same output. That does not mean they are the same, because the action on another state may be different (and I don't just mean a global phase).

For example, if you take the state $|+\rangle=(|0\rangle+|1\rangle)/\sqrt{2}$ then $$ H|+\rangle=|0\rangle, $$ but $$ R_Y(\pi/2)|+\rangle=|1\rangle. $$ The two outcomes are orthogonal. So, in this case, you can definitely tell which of the two you have just by measuring the output.

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.