quantum Mech question

[rich_loco]

QUBIT QUESTION:

I have measurement 1, probability of success as a function of x as:

(1 + sin^2 x)/2

and have measurement 2, probability of success as a function of x as:

cos^2(pi/4 - x/2)

im then given the following approximation for Theta --> 0

sin Theta = 0, sin^2(pi/4 - Theta) = 1/2 - Theta
QUestion is: use these approximations to estimate the probability of success for measurement 1 and 2 above as x-->0 AS A FUNCTION OF X.

 

[HallsofIvy]

You know that \(\displaystyle cos^2(\theta)= 1- sin^2(\theta)\), do you not?

But your question does not really make sense. The "limit" as x goes to 0 would not be a function of x! Perhaps you mean "find a linear approximation for x close to 0"?

 

[DrPhil]

QUBIT QUESTION:

I have measurement 1, probability of success as a function of x as:

(1 + sin^2 x)/2

and have measurement 2, probability of success as a function of x as:

cos^2(pi/4 - x/2)

im then given the following approximation for Theta --> 0

sin Theta = 0, sin^2(pi/4 - Theta) = 1/2 - Theta
QUestion is: use these approximations to estimate the probability of success for measurement 1 and 2 above as x-->0 AS A FUNCTION OF X.

The approximation of the sine at small Theta should always be
\(\displaystyle \sin \theta \approx \theta \)
thus
\(\displaystyle P_1(x) \approx \dfrac{1+x^2}{2}\)

If you neglect any terms of higher order than \(\displaystyle x\), this reduces to 1/2.

For the other, the approximation they give corresponds to \(\displaystyle \sin \theta \approx \theta \) and \(\displaystyle \cos \theta \approx 1 \) for small \(\displaystyle \theta\). Just substitute \(\displaystyle x/2\) for \(\displaystyle \theta\). [Of course you already know, as HallsofIvy reminded you, that \(\displaystyle \cos^2\theta = 1 - \sin^2\theta\).]