The state of a spin-$\frac12$ particle that is spin up along the axis whose direction is specified by the unit vector $n=\sin\theta\cos\phi i+\sin\theta\sin\phi j+\cos\theta k,$ is given by $$|+n \ \rangle=\cos\frac{\theta}{2}|+z \ \rangle + e^{i\phi}\sin\frac{\theta}{2}|-z \ \rangle.$$

a. Suppose that a measurement of $S_z$ is carried out on a particle in the state $|+n \ \rangle.$ What is the probably that the measurement yields $\hbar/2$, $-\hbar/2$. How about the measurements of $S_x?$

b. Determine the uncertainty $\triangle S_z$ and $\triangle S_x$ of your measurements.

I know how to the measurement of $S_z$ for parts a and b, but I am not sure how to do the measurements for $S_x$. I have:

For reference, $$|+x \ \rangle = \frac{1}{\sqrt{2}}|+ z \ \rangle+\frac{1}{\sqrt2}|-z \ \rangle$$ $$|+y \ \rangle = \frac{1}{\sqrt{2}}|+ z \ \rangle+\frac{i}{\sqrt2}|-z \ \rangle.$$

a.) Probability for $\hbar/2$: $$|\langle \ +z |+n \rangle|^2 = \cos^2\frac{\theta}{2}$$ Probability for $-\hbar/2$: $$|\langle \ -z |+n \rangle|^2 = \left|e^{i\phi}\sin\frac{\theta}{2}\right|^2=\sin^2\frac{\theta}{2}.$$

But I am not sure how to do it for $S_x?$


1 Answer 1


To find the probability of measuring $\left|+x\right\rangle$ you need to determine the expression $\left|\left\langle+x\middle|+n\right\rangle\right|^2$. You have the expression for $\left|+n\right\rangle$ and you have the expression for $\left|+x\right\rangle$. Therefore you first need to determine the bra vector $\left\langle +x\right|$ which is found by taking the complex conjugate of each coefficient in the $\left|+x\right\rangle$ expression and turning all of the ket vectors to bra vectors. Doing this, we find the following:

$$\left\langle+x\right|= {1 \over \sqrt{2}}\left\langle+z\right|+{1 \over \sqrt{2}}\left\langle-z\right|.$$

You can then use this expression along with the expression given for $\left|+n\right\rangle$ to find:

$$\left\langle+x\middle|+n\right\rangle=\left({1 \over \sqrt{2}}\left\langle+z\right|+{1 \over \sqrt{2}}\left\langle-z\right|\right)\left(\cos{\theta \over 2}\left|+z\right\rangle+e^{i\phi}\sin{\theta \over 2}\left|-z\right\rangle\right)$$ $$\Longrightarrow{1 \over \sqrt2}\cos{\theta \over 2}+{e^{i\phi} \over \sqrt2}\sin{\theta \over 2}.$$

Then, you simply find the modulus squared of this expression to obtain the probability of measuring $+{\hbar \over 2}$. The process is entirely similar for finding the probability of measuring $-{\hbar \over 2}$, the only difference being that you use the bra $\left\langle-x\right|$ rather than $\left\langle +x\right|$.

  • 1
    $\begingroup$ I've improved your equations a bit. Please see my edits to learn how to typeset equations nicer in future. In general, one should prefer \left and \right to denote starting and ending brackets/braces/parens. Then the sizes will be automatically selected. \big is only for some special cases which are quite rare. And many functions have commands to make them look like functions, e.g. \sin. If there's no such command, you can use \operatorname{func}, e.g. \operatorname{Ai}(x) will give $\operatorname{Ai}(x)$. $\endgroup$
    – Ruslan
    Jan 13, 2015 at 9:09

Your Answer

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

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