4
$\begingroup$

I'm confused about how I can measure $\langle \hat{X}\rangle$ and $\langle \hat{Y}\rangle$ using counts. Here's my code for X:

x-basis:

# Measurement in x-basis. 
quanc_x = QuantumCircuit(1)
quanc_x.u(1,2,3,0) # prepare some random state
quanc_x.h(0)
quanc_x.measure_all()
quanc_x.draw(output='mpl')
# number of repetitions
N = 10000

backend = Aer.get_backend( 'qasm_simulator' )
job = execute( quanc_x, backend, shots=N )
result = job.result()

measurement_result = result.get_counts( quanc_x )
print( measurement_result )
plot_histogram( measurement_result )
cos_phi_est = ( measurement_result['0'] - measurement_result['1'] ) / N  #<--Question
print( "cos(phi) estimated: ", cos_phi_est )

My question of this code is marked above. I'm not pretty sure if that looks correct. For Pauli X, we have $$ \langle \hat{X}\rangle=\langle\psi|0\rangle\langle1|\psi\rangle+\langle\psi|1\rangle\langle0|\psi\rangle $$ Can I simplify that further? Should that correspond to my code with the question mark? How can I apply that to $\langle \hat{Y}\rangle$? Thanks for the help!

$\endgroup$
4
$\begingroup$

That looks right to me.

Since, $HZH = X$ then we have that $\langle \psi | X | \psi \rangle = \langle \psi | HZH | \psi \rangle = \langle \psi H | Z | H\psi \rangle $.

In your code, you generate $|\psi \rangle$ with a $U_3(\theta, \phi, \lambda) $ gate applied to $|0\rangle$. Then you applied the Hadamard gate ($H$) before measuring which is what needed to measure in the $X$ basis as discussed above.


For $\langle Y \rangle$ you should note that $(SH)Z(HS^\dagger) = Y $

$$\langle \psi |Y| \psi \rangle = \langle \psi | (SH)Z (HS^\dagger) | \psi \rangle = \langle \psi SH | Z | H S^\dagger \psi \rangle $$

Thus, here you want to apply $S^\dagger$ follow by the Hadamard gate $H$ before measurement.

enter image description here

$\endgroup$
4
  • 1
    $\begingroup$ Thanks for the answer! If I want to find the expectation value from counts, is my code also correct for $X$? (( measurement_result['0'] - measurement_result['1'] ) / N)Should that be the same as $Y$? $\endgroup$
    – ZR-
    Jul 30 at 23:25
  • 2
    $\begingroup$ Yes. The counts remain the same as you wrote it. $\endgroup$
    – KAJ226
    Jul 30 at 23:28
  • $\begingroup$ Thanks, can I understand the measured expectation value as the difference of some probabilities? Why it is not result[1]- result[0]? $\endgroup$
    – ZR-
    Jul 30 at 23:31
  • 1
    $\begingroup$ You already rotate to the computational basis once you apply the rotation $H$ or $H S^\dagger $ to your state $|\psi \rangle$. In the $Z$ basis, $|0\rangle$ has eigenvalue of $+1$ and $|1\rangle$ has eigenvalue of $-1$ since $Z|0\rangle = 1|0\rangle$ and $Z|1\rangle = -1 |1\rangle$. $\endgroup$
    – KAJ226
    Jul 30 at 23:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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