Spin probability of a particle state

[Zack K]

Homework Statement

Suppose that a measurement of $S_x$ is carried out on a particle in the state $|\textbf{+n}\rangle$. What is the probability that the measurement yields (i) $\frac{\hbar}{2}$ (ii) $\frac{-\hbar}{2}$?

Relevant Equations

$|\textbf{+n}\rangle = cos\frac{\theta}{2}|+\textbf{z}\rangle + e^{i\phi}sin\frac{\theta}{2}|-\textbf{z}\rangle$

Starting with finding the probability of getting one of the states will make finding the other trivial, as the sum of their probabilities would be 1.

Some confusion came because I never represented the states $|\pm \textbf{z}\rangle$ as a superposition of other states, but I guess you would not need to.

My attempt:$$\langle+\textbf{x}|+\textbf{n}\rangle=cos\frac{\theta}{2}\langle+\textbf{x}|+\textbf{z}\rangle + e^{i\phi}sin\frac{\theta}{2}\langle+\textbf{x}|-\textbf{z}\rangle$$

Since $|\textbf{x}\rangle = \frac{1}{\sqrt{2}}|+\textbf{z}\rangle + \frac{1}{\sqrt{2}}|-\textbf{z}\rangle$ has only real parts, the bra vector coefficients look the same so then $\langle+\textbf{x}|\pm\textbf{z}\rangle =\frac{1}{\sqrt{2}}$. Therefore we get

$$\langle+\textbf{x}|+\textbf{n}\rangle=\frac{1}{\sqrt{2}}cos\frac{\theta}{2} + \frac{1}{\sqrt{2}}e^{i\phi}sin\frac{\theta}{2}$$

The probability is this expression squared, which expands into

$$\frac{1}{2}\left(cos^2\frac{\theta}{2}+sin(2\theta)cos(\phi)+sin^2\frac{\theta}{2}cos(2\theta)+i(sin^2\frac{\theta}{2}sin(2\phi)+sin(2\theta)sin(\phi))\right)$$

The answer is given as $\frac{1}{2}(1+sin(\theta)cos(\phi))$

I have plugged in some values of theta and phi and my answer and that of the textbook evaluated to the same answer, though I am still skeptical on my answer. I cannot see how I could reduce this, I have tried using wolframalpha to no luck. Though I am more interested in knowing if my steps were correct.

 

[PeroK]

Homework Statement:: Suppose that a measurement of $S_x$ is carried out on a particle in the state $|\textbf{+n}\rangle$. What is the probability that the measurement yields (i) $\frac{\hbar}{2}$ (ii) $\frac{-\hbar}{2}$?
Relevant Equations:: $|\textbf{+n}\rangle = cos\frac{\theta}{2}|+\textbf{z}\rangle + e^{i\phi}sin\frac{\theta}{2}|-\textbf{z}\rangle$

Starting with finding the probability of getting one of the states will make finding the other trivial, as the sum of their probabilities would be 1.

Some confusion came because I never represented the states $|\pm \textbf{z}\rangle$ as a superposition of other states, but I guess you would not need to.

My attempt:$$\langle+\textbf{x}|+\textbf{n}\rangle=cos\frac{\theta}{2}\langle+\textbf{x}|+\textbf{z}\rangle + e^{i\phi}sin\frac{\theta}{2}\langle+\textbf{x}|-\textbf{z}\rangle$$

Since $|\textbf{x}\rangle = \frac{1}{\sqrt{2}}|+\textbf{z}\rangle + \frac{1}{\sqrt{2}}|-\textbf{z}\rangle$ has only real parts, the bra vector coefficients look the same so then $\langle+\textbf{x}|\pm\textbf{z}\rangle =\frac{1}{\sqrt{2}}$. Therefore we get

$$\langle+\textbf{x}|+\textbf{n}\rangle=\frac{1}{\sqrt{2}}cos\frac{\theta}{2} + \frac{1}{\sqrt{2}}e^{i\phi}sin\frac{\theta}{2}$$

The modulus squared of a complex number $w$ is given by $|w|^2 = ww^*$, where $w^*$ is the complex conjugate. The resulting expression is real and cannot have a factor of $i$ in it.

What is the complex conjugate of
$$\frac{1}{\sqrt{2}}cos\frac{\theta}{2} + \frac{1}{\sqrt{2}}e^{i\phi}sin\frac{\theta}{2}$$