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I am trying to solve an exercise, but I can't seem to get it to work.

I get given this rule,

$$\langle \sigma_x \rangle^2 + \langle \sigma_y \rangle^2+\langle \sigma_z \rangle^2 =1 $$

and I am asked to verify this for $|\psi\rangle = \cos\theta|0\rangle + \sin\theta|1\rangle$.

I first expand upon the rule, by actually computing the probabilities for each base.

enter image description here

This then leads me to this: enter image description here

This leads me to this: enter image description here

I assume my mistake lies here.

When inserting $a_0 = \cos\theta$ and $a_1 = \sin\theta$ I get this: enter image description here

This is incorrect though, because the relationship at the beginning should, according to the exercise, hold for all $\theta$ and not just when $\theta = n\pi$ $\forall n \in \mathbb N$, because it asks me to verify it for that specific state ($|\psi\rangle= \cos\theta|0\rangle + \sin\theta|1\rangle$). Can anyone help me out here? Thanks, Tom

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I think that you made a mistake at

$$\frac{\vert a_0 - a_1 \vert^2}{2} = \frac{\vert \cos^2(\theta)-\sin^2(\theta)\vert^2}{2} = \frac{1}{2}(\cos^2(\theta) -\color{red}{2}\cos(\theta)\sin(\theta)+\sin^2(\theta))$$ and similarly for the other term where there seems to be a missing factor of $2$. Therefore I think that the calculations should proceed as follows, after the third image.

$$\frac{1}{4}((a_0-a_1)^2-(a_0+a_1)^2)^2 + (0)^2 + (a_0^2-a_1^2)^2=$$

$$\frac{1}{4}(a_0^2-2a_0a_1+a_1^2-(a_0^2+2a_0a_1+a_1^2))^2 + (a_0^2-a_1^2)^2=$$

$$\frac{1}{4}(-4a_0a_1)^2 + (a_0^2-a_1^2)^2 = a_0^4-2a_0^2a_1^2+a_1^4+4a_0^2a_1^2=a_0^4+2a_0^2a_1^2+a_1^4$$

$$=(a_0^2+a_1^2)^2=(\sin^2(\theta)+\cos^2(\theta))^2=1^2=1$$

So in conclusion you were correct on where you did the mistake. No worries, these calculations are big and mistakes like that are common.

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  • $\begingroup$ Thanks for the answer! I just have one problem: I think the factor of $\frac14$ at the beginning is incorrect. $\endgroup$ Mar 26, 2022 at 9:34
  • $\begingroup$ @tommasopeduzzi I don't think it is because you have ((1/sqrt(2))^2)^2 = 1/4. The first 1/sqrt(2) is the state, the first ^2 is the overlap and the second ^2 is the exp. squared. $\endgroup$
    – R.W
    Mar 26, 2022 at 13:35
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    $\begingroup$ Thanks a lot! I often overlook these kinds of details, even if I stare at my monitor for hours. $\endgroup$ Mar 26, 2022 at 17:44
  • $\begingroup$ Could you please accept the answer if possible and if you think it was the correct answer for the question. Att. R.W. $\endgroup$
    – R.W
    Mar 31, 2022 at 11:09

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