quantum mechanics - Spin state of electron after measurement

I have a system of two spin 1/2 particles in a superposition of spin states in the z-direction given by:

$\psi = \frac{1}{2} |+ +\rangle + \frac{1}{2} |+ -\rangle + \frac{1}{\sqrt{2}} |- -\rangle$

where $+$ designates spin up, $-$ designates spin down and the first particle's state is the first term in each ket and the second particles' state is the second term in each ket. If I measure the spin on the first particle and get a value of $-\hbar / 2$ (corresponding to a spin down state) is the new state of the particles simply

$\psi = | - - \rangle$

meaning that the first particle is now "set" to being spin down? And if I determine the spin on the first particle to be spin up, would the subsequent state be

$\psi = \frac{1}{\sqrt{2}} |+ +\rangle + \frac{1}{\sqrt{2}} |+ - \rangle$ ?

Basically, my question is once I make a measurement of a spin of a particle, does the wavefunction stay collapsed on the spin determined? And does having a second particle affect this in any way?

Answer

Yes, you've correctly understood what happens. Let's try to be more mathematically precise to be confident that this is the case.

I'll be using tensor product notation in this answer (the Hilbert space of the system is just the tensor product of the spin-$\frac{1}{2}$ Hilbert space with itself).

The projective measurement postulate says

For an observable $O$ with spectral decomposition $O = \sum_i\lambda_iP_i$, where $P_i$ is the projector onto the eigenspace of $O$ corresponding to eigenvalue $\lambda$, the possible outcomes of the measurement are the eigenvalues of the observable, and given that outcome $\lambda_i$ occured, the state of the system immediately after measurement is \begin{align} \frac{P_i|\psi\rangle}{\sqrt{\langle\psi|P_i|\psi\rangle}} \end{align}

For a two-spin-$\frac{1}{2}$ system, the $z$-component of the spin of the first particle is represented by the following observable: \begin{align} S_z\otimes I \end{align} where $I$ is the identity on the Hilbert space of the second particle. Now, recall that $S_z$ has the following spectral decomposition \begin{align} S_z = -\frac{\hbar}{2}P_-+\frac{\hbar}{2}P_+ \end{align} where $P_-$ and $P_+$ are projectors defined as \begin{align} P_-=|-\rangle\langle-|, \qquad P_+=|+\rangle\langle+| \end{align} and it follows that the spectral decomposition of $S_z\otimes I$ is \begin{align} S_z\otimes I = -\frac{\hbar}{2}P_- \otimes I+\frac{\hbar}{2}P_+ \otimes I+ \end{align} Upon measurement of $I\otimes S_z$ on the state $|\psi\rangle$ given in the question, obtaining the result $-\hbar/2$ indicates that the state got projected as follows upon the measurement: \begin{align} |\psi\rangle \to \frac{P_-\otimes I|\psi\rangle}{\sqrt{\langle\psi |P_-\otimes I|\psi\rangle}} = \frac{\frac{1}{\sqrt{2}}|--\rangle}{\sqrt{\frac{1}{2}}} = |--\rangle \end{align} If the value $+\hbar/2$ had been obtained, then the state would have been projected has follows: \begin{align} |\psi\rangle \to \frac{P_+\otimes I|\psi\rangle}{\sqrt{\langle\psi |P_+\otimes I|\psi\rangle}} = \frac{\frac{1}{2}|++\rangle + \frac{1}{2}|+-\rangle}{\sqrt{\frac{1}{4}+\frac{1}{4}}} = \frac{1}{\sqrt{2}}|++\rangle + \frac{1}{\sqrt{2}}|+-\rangle \end{align}