Measuring the spin of a particle from a singlet state Imagine that I have a singlet state:
$|s\rangle = {1 \over \sqrt2}(| \uparrow_1\downarrow_2 \rangle - |\downarrow_1\uparrow_2\rangle)$
I want to measure the spin along the z axis of the first particle.
So I should apply the operator $S_{z1}$ to the singlet state:
$S_{z1}|s\rangle$ = ?
The possible answer is 50% $\uparrow_1$ and 50% $\downarrow_1$.
How should I write in the bra-ket notation the result of this measurement?
Thank you!
 A: There are a few ways you can formally approach this problem. 
First note that the singlet state is an element of the Hilbert space of the tensor product of two spin-1/2 particles $1$ and $2$, i.e. $| s \rangle \in \mathcal{H} = \mathcal{H}_{1} \otimes\mathcal{H}_{2}$.
To describe the system in the full space, you can form the density matrix $\rho = | s \rangle \langle s | \in D(H)$.
1st way:
Since you care only about the first system, you can take the partial trace over system $2$ yielding the reduced density matrix of system $1$:
\begin{align}
\rho_1 = \text{tr}_2(\rho) \in D(H_1).
\end{align}
This density matrix will give you the correct predictions if you choose to only do measurements on $1$. In this case, you can show that
\begin{align}
\rho_1 = \frac{1}{2}|\uparrow\rangle_1 \langle \uparrow |_1+\frac{1}{2}|\downarrow\rangle_1 \langle \downarrow |_1,
\end{align}
a mixed state.
Now the operator $S_{z1}$ has the spectral decomposition
\begin{align}
S_{z1} = +\frac{\hbar}{2} \pi_{\uparrow 1} + (- \frac{\hbar}{2}) \pi_{\downarrow 1},
\end{align}
where $\pi_{\uparrow 1}$ is the projector onto the spin-z-up axis of the 1st particle: $\pi_{\uparrow 1} = | \uparrow \rangle_1 \langle \uparrow |_1$ and similarly for the other projector.
According to the Born rule, the probability of getting $\hbar/2$ is
\begin{align}
\mathbb{P}(s_{z1}=\hbar/2) &= \text{tr}_1 (\rho_1 \pi_{\uparrow 1}) \nonumber \\
&= \sum_k \langle k|_1 \left[\frac{1}{2}|\uparrow\rangle_1 \langle \uparrow |_1+\frac{1}{2}|\downarrow\rangle_1 \langle \downarrow |_1 \right]| \uparrow \rangle_1 \langle \uparrow |_1 | k \rangle_1 \nonumber \\
& = \frac{1}{2} \sum_k \langle k|_1|\uparrow\rangle_1\langle \uparrow |_1 | k \rangle_1 \nonumber \\
& = \frac{1}{2}\langle \uparrow |_1 \sum_k | k \rangle_1\langle k|_1|\uparrow\rangle_1 \nonumber \\
& = \frac{1}{2}  \uparrow |_1|\uparrow\rangle_1 \nonumber \\
& = \frac{1}{2},
\end{align}
and similarly for $\mathbb{P}(s_{z1} = -\hbar/2)$. This is called a projective valued measurement.
2nd way: You can think of a measurement on the full system, except that you do nothing on system 2. So the relevant operator is the tensor product of the $S_{z1}$ operator with the identity operator on 2: $S_{z1} \otimes \mathbb{I}_2$, which I'll call the FULL spin operator.
The spectral decomposition is
\begin{align}
S_{z1} \otimes \mathbb{I}_2 = +\frac{\hbar}{2} \pi_{\uparrow 1} \otimes \mathbb{I}_2 + (- \frac{\hbar}{2}) \pi_{\downarrow 1}\otimes \mathbb{I}_2,
\end{align}
i.e. it has only 2 eigenvalues, but now each eigenspace is 2-dimensional.
So according to the Born rule, the probability of measuring $\hbar/2$ for the FULL spin operator is given by
\begin{align}
\mathbb{P}(s_{z1}^\text{full}=\hbar/2) = \text{tr} (\rho (\pi_{\uparrow 1} \otimes \mathbb{I}_2))
\end{align}
which you can work out to be $1/2$ again.
Note: this is the formal way of showing these calculations, following the axioms of quantum mechanics.
Expressions like $|\langle \uparrow|_1 S_{z1} |s\rangle |^2$, should be regarded as heuristic expressions that 'give' the right answer, but they are not mathematically correct because $| \uparrow \rangle_1$ is an element of $\mathcal{H}_1$, $S_{z1}$ an element of $L(\mathcal{H_1})$, but $|s \rangle$ an element of $\mathcal{H}$, so it just doesn't quite make sense.
A: $\newcommand{bra}[1]{ \langle #1 |}$
$\newcommand{ket}[1]{ | #1 \rangle}$
$\newcommand{upone}{\uparrow_1 }$
$\newcommand{downone}{\downarrow_1 }$
$\newcommand{uptwo}{\uparrow_2 }$
$\newcommand{downtwo}{\downarrow_2 }$
$\newcommand{spin}{S_{z1} }$
You are asking what $\spin \ket{s}$ is. You say you want to do it with braket notation. Well, 
$\spin \ket{s} = \spin \frac{1}{\sqrt{2}} \left( \ket{\upone \downtwo} - \ket{\downone \uptwo}\right) \\
= \frac{1}{\sqrt{2}} \left(\spin \ket{\upone \downtwo} - \spin\ket{\downone \uptwo}\right) \\
=\frac{1}{\sqrt{2}} \left(\spin \left(\ket{\upone} \otimes \ket{ \downtwo} \right) - \spin \left( \ket{\downone} \otimes \ket{ \uptwo}\right)\right)\\
=\frac{1}{\sqrt{2}} \left( \left( \spin \ket{\upone} \right) \otimes \ket{ \downtwo}  -  \left( \spin \ket{\downone} \right) \otimes \ket{ \uptwo}\right)\\
=\frac{1}{\sqrt{2}} \left( \left( \frac{\hbar}{2} \ket{\upone} \right) \otimes \ket{ \downtwo}  -  \left( -\frac{\hbar}{2} \ket{\downone} \right) \otimes \ket{ \uptwo}\right)\\
=\frac{1}{\sqrt{2}} \left( \frac{\hbar}{2} \ket{\upone} \otimes \ket{ \downtwo}  +  \frac{\hbar}{2} \ket{\downone}  \otimes \ket{ \uptwo}\right) \\
=\frac{1}{\sqrt{2}} \left( \frac{\hbar}{2} \ket{\upone \downtwo}   +  \frac{\hbar}{2} \ket{\downone \uptwo}\right)\\
=\frac{\hbar}{2\sqrt{2}} \left( \ket{\upone \downtwo}   +  \ket{\downone \uptwo}\right).$
If you can read that you are off to a good start. If you have specific questions about any notation or steps just ask. 
