I'm just starting of on quantum computing, specifically following the IBM Q Experience documentation [1]. In here, they are explaining the following experiment:


The expected outcomes according to the document:

  • Phase angle: $\pi/4$
  • Gates: $T$
  • Prob 0: 0.8535533
  • Prob 1: 0.1464466
  • X-length: 0.7071067

I'm trying to deduce this with math.

$T |+\rangle = \begin{bmatrix}1 & 0 \\ 0 & e^{i\pi/4}\end{bmatrix} {1\over\sqrt 2} \begin{bmatrix} 1 \\ 1 \end{bmatrix} = {1\over\sqrt 2} \begin{bmatrix}1\\e^{i\pi/4}\end{bmatrix}$

I think I now need to split this out in $|0\rangle$ and $|1\rangle$ so that I get the quantum amplitudes:

$ = {1\over\sqrt 2} \begin{bmatrix}1\\0\end{bmatrix} + {1\over\sqrt 2} e^{i\pi/4} \begin{bmatrix}0 \\1 \end{bmatrix}$

Here things are falling apart, as

$ P(0) = |{1\over\sqrt 2}|^2 = 0.5 $
$ P(1) = |{1\over\sqrt 2} e^{i\pi/4}|^2 = 0.5 $

So my question: How do I correctly calculate the probabilities and the X-length?

[1]: IBM Q: User Guide / The Weird and Wonderful World of the Qubit / Introducing Qubit Phase

  • $\begingroup$ You're Measuring in the wrong basis. you’ve done it in the z basis when you should be using the x basis. $\endgroup$ – DaftWullie Jun 28 '18 at 21:08

You are correct with your calculation that $$ T\left(\begin{array}{c} 1 \\ 1 \end{array}\right)/\sqrt{2}=\left(\begin{array}{c} 1 \\ e^{i\pi/4} \end{array}\right)/\sqrt{2}, $$ and you are correct that if you want to calculate the probability of getting a "0"$\equiv\left(\begin{array}{c} 1 \\ 0 \end{array}\right)$ measurement result, you evaluate $$ P(0)=\left|\left(\begin{array}{cc} 1 & 0 \end{array}\right)\cdot\left(\begin{array}{c} 1 \\ e^{i\pi/4} \end{array}\right)/\sqrt{2}\right|^2=\frac{1}{2}, $$ so you get both answers with probability 1/2. However, this is not what the referenced page is trying to calculate. It says

If we start with a system initially in the |+⟩ (which is done using the Hadamard), then apply multiples of the T gate and measure in the x-basis...

The X-basis is not the question of 0 or 1 that you have already calculated. Instead, it's the probability of being in $|\pm\rangle=\frac{1}{\sqrt{2}}\left(\begin{array}{c} 1 \\ \pm 1\end{array}\right)$. I think the confusion has arisen because while people often refer to 0 and 1 as being the computational basis (as I have, and you have), when you're talking about measurements where there are two possible results, you can always label the outcomes as 0 and 1, no matter what basis was used. This is what they've done.

So, $$ P(+)=\left|\frac{1}{\sqrt{2}}\left(\begin{array}{cc} 1 & 1 \end{array}\right)\cdot\left(\begin{array}{c} 1 \\ e^{i\pi/4} \end{array}\right)/\sqrt{2}\right|^2=\frac{1}{4}|1+e^{i\pi/4}|^2 $$ Expanding this gives $$ P(+)=\frac{1}{4}\left((1+\cos\frac{\pi}{4})^2+\sin^2\frac{\pi}{4}\right)=\frac{1}{4}\left((1+\frac{1}{\sqrt{2}})^2+\frac{1}{2}\right)=\frac{2+\sqrt{2}}{4} $$ If you numerically evaluate this, you'll get the required result, 0.8535533. You could repeat the calculation for $P(-)$, or just use the fact that $P(+)+P(-)=1$.

The x-length, as they call it, is $$ P(+)-P(-)=2P(+)-1=\frac{2+\sqrt{2}}{2}-1=\frac{1}{\sqrt{2}}. $$ Again, that's exactly what's claimed.

  • $\begingroup$ Your math appears to work out, and thereby answer my question, so thank you for that. Still, I have some concerns/unclear parts. According to inst.eecs.berkeley.edu/~cs191/fa14/lectures/lecture89.pdf: $|\phi\rangle = A_0 \times |0 \rangle + A_1 \times |1 \rangle$; $P(0)=|A_o|^2$ instead of $P(0)=||0 \rangle \times A_o|^2$; Without the $|0 \rangle$, $P(+) = |(\alpha + \beta)/sqrt(2))|^2$ (as discussed in the link) works out. My remaining question: how could you add $(\begin{bmatrix} 1 & 0 \end{bmatrix}$, aka $|0 \rangle$ into the $P(0)=|A_0|^2$? $\endgroup$ – Thomas Hubregtsen Jun 30 '18 at 11:57
  • 1
    $\begingroup$ This seems a bit jumbled/muddled. Can I suggest expanding into a separate question, then i’ve got a better chance of understanding what you're asking, and giving a decent answer? $\endgroup$ – DaftWullie Jun 30 '18 at 12:00
  • $\begingroup$ You use $P(0)=|A_o \times |0 \rangle |^2$. Should it not be $P(0)=|A_o |^2$ based on inst.eecs.berkeley.edu/~cs191/fa14/lectures/lecture89.pdf $\endgroup$ – Thomas Hubregtsen Jun 30 '18 at 12:03
  • $\begingroup$ I believe you turned the use of $|+\rangle$ and $T|+\rangle$ around, taking the complex conjugate transpose of $|+\rangle$ instead of $T|+\rangle$, which eventually did not change the outcome thanks to taking the absolute value? $\endgroup$ – Thomas Hubregtsen Jul 22 '18 at 16:30
  • $\begingroup$ I believe I've done it correctly for the question as specified. The state produced was $T|+\rangle$, so the probability of being found in the + state is $|\langle +|T|+\rangle|^2$, which is exactly what I've calculated. $\endgroup$ – DaftWullie Jul 23 '18 at 7:20

The other answers were (almost?*) correct, and pointed me in the right direction for computing any probability for a measurement(especially the notion that I was measuring in the wrong basis), but I missed the definition.

Assuming you are in state $|\psi\rangle$, and you want to know the probability of measuring outcome $|\phi\rangle$:
$P=|\langle\psi|\phi\rangle|^2$ (https://ocw.tudelft.nl/course-lectures/0-3-1-measuring-qubits-standard-basis/ @1:48)
where $\langle\psi|$ is the complex conjugate transpose of the $|\psi\rangle$

More, concrete, if we are in $T |+\rangle$ ($\equiv{1\over\sqrt 2} \begin{bmatrix}1\\e^{i\pi/4}\end{bmatrix}$), and want to know the probability of measuring $|+\rangle$ ($\equiv{1\over\sqrt 2}\begin{bmatrix} 1 \\ 1 \end{bmatrix})$, we calculate:
$P=|{1\over\sqrt 2} \begin{bmatrix}1 & e^{-i\pi/4}\end{bmatrix}{1\over\sqrt 2}\begin{bmatrix} 1 \\ 1 \end{bmatrix}|^2$
$P={1\over4}|1+e^{-i\pi/4}|^2 \approx 0.853553$ (or use Euler's formula for the exact outcome)

P.S. I hope I took the right complex conjugate.
* I also have the $\psi$ and $\phi$ the other way around compared to DaftWullie his/her answer.

  • $\begingroup$ I would recommend putting this in a separate question, or editing your main question with this additional piece. Since I think people will probably miss that you had a followup and assume that @DaftWullie took care of it! But you are correct, and the order of $|\psi\rangle$ and $|\phi\rangle$ doesn't matter since that only changes the conjugation, which doesn't matter when you are taking the norm. $\endgroup$ – Dripto Debroy Jul 23 '18 at 3:13
  • $\begingroup$ Well, this is a matter of practicality. I do the calculation by saying "I have state $|\psi\rangle$, so the amplitude for being in state $|\phi\rangle$ is $\langle\phi|\psi\rangle$, and hence the probability is $|\langle\phi|\psi\rangle|^2$ but you don't have to think about it like that. The point is simply that $|\langle\phi|\psi\rangle|^2=\langle\phi|\psi\rangle\langle\psi|\phi\rangle=\langle\psi|\phi\rangle\langle\phi|\psi\rangle$, so once you take the mod-square, ordering is irrelevant. You're taking a marginally less conventional ordering from that lecture, but it makes no difference. $\endgroup$ – DaftWullie Jul 23 '18 at 7:26

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